by Walter Lewin
So I project these four beautiful equations on different screens on all the walls of the lecture hall. “Look at them,” I say. “Inhale them. Let them penetrate your brains. Only once in your life will you see all four of Maxwell’s equations for the first time in a way that you can appreciate them, complete and beautiful and talking to each other. This will never happen again. You will never be the same. You have lost your virginity.” In honor of this momentous day in the lives of the students, as a way of celebrating the intellectual summit they’ve reached, I bring in six hundred daffodils, one for each student.
Students write me many years afterward, long after they’ve forgotten the details of Maxwell’s equations, that they remember the day of the daffodils, the day I marked their new way of seeing with flowers. To me this is teaching at the highest level. It’s so much more important to me for students to remember the beauty of what they have seen than whether they can reproduce what you’ve written on the blackboard. What counts is not what you cover, but what you uncover!
My goal is to make them love physics and to make them look at the world in a different way, and that is for life! You broaden their horizon, which allows them to ask questions they have never asked before. The point is to unlock the world of physics in such a way that it connects to the genuine interest students have in the world. That’s why I always try to show my students the forests, rather than take them up and down every single tree. That is also what I have tried to do in this book for you. I hope you have enjoyed the journey.
ACKNOWLEDGMENTS
Without the intelligence, foresight, business sense, and moral support of our exceptional literary agent, Wendy Strothman, For the Love of Physics would have remained little more than wishful thinking. She brought the two of us together, found the right home for this book at Free Press, read numerous draft chapters with an editorial eye honed by her years as a publisher, gave the book its title, and helped keep us focused on the end product. We are also the happy and fortunate recipients of her staunch friendship, which buoyed us throughout the project.
It would be hard to overstate the contributions of our editor, Emily Loose, at Free Press, whose vision for this book proved infectious and whose extraordinarily close attention to prose narrative provided an education for both of us. Despite the enormous pressure in the publishing industry to cut corners on behalf of the bottom line, Emily insisted on really editing this book, pushing us always to greater clarity, smoother transitions, and tighter focus. Her skill and intensity have made this a far better book. We are grateful as well to Amy Ryan for her deft copyediting of the manuscript.
Walter Lewin:
Every day I receive wonderful, often very moving email from dozens of people all over the world who watch my lectures on the web. These lectures were made possible due to the vision of Richard (Dick) Larson. In 1998 when he was the director of the Center for Advanced Educational Services and a professor in the Department of Electrical Engineering at MIT, he proposed that my rather unconventional lectures be videotaped and made accessible to students outside MIT. He received substantial funding for this from the Lord Foundation of Massachusetts and from Atlantic Philanthropies. Dick’s initiative was the precursor of e-learning! When MIT’s OpenCourseWare opened its doors in 2001, my lectures reached all corners of the world and are now viewed by more than a million people each year.
During the past two years, even during the seventy days that I was in the hospital (and almost died), this book was always on my mind. At home I talked about it incessantly with my wife, Susan Kaufman. It kept me awake many nights. Susan patiently endured all this and managed to keep my spirits up. She also trained her astute editorial eye on a number of chapters and improved them markedly.
I am very grateful to my cousin Emmie Arbel-Kallus and my sister, Bea Bloksma-Lewin, for sharing with me some of their very painful recollections of events during World War II. I realize how difficult this must have been for both of them, as it was for me. I thank Nancy Stieber, my close friend for thirty years, both for always correcting my English and for her invaluable comments and suggestions. I also want to thank my friend and colleague George Clark, without whom I would never have become a professor at MIT. George let me read the original American Science and Engineering proposal submitted to the Air Force Cambridge Research Laboratories that led to the birth of X-ray astronomy.
I am grateful to Scott Hughes, Enectali Figueroa-Feliciano, Nathan Smith, Alex Filippenko, Owen Gingerich, Andrew Hamilton, Mark Whittle, Bob Jaffe, Ed van den Heuvel, Paul Murdin, George Woodrow, Jeff McClintock, John Belcher, Max Tegmark, Richard Lieu, Fred Rasio, the late John Huchra, Jeff Hoffman, Watti Taylor, Vicky Kaspi, Fred Baganoff, Ron Remillard, Dan Kleppner, Bob Kirshner, Paul Gorenstein, Amir Rizk, Chris Davlantes, Christine Sherratt, Markos Hankin, Bil Sanford, and Andrew Neely for helping me, when help was needed.
Finally I can’t thank Warren Goldstein enough for his patience with me and for his flexibility; at times he must have felt overwhelmed (and perhaps frustrated) with too much physics in too little time.
Warren Goldstein:
I would like to thank the following people for their willingness to talk with me about Walter Lewin: Laura Bloksma, Bea Bloksma-Lewin, Pauline Broberg-Lewin, Susan Kaufman, Ellen Kramer, Wies de Heer, Emanuel (Chuck) Lewin, David Pooley, Nancy Stieber, Peter Struycken. Even if they are not quoted in For the Love of Physics, each one added substantially to my understanding of Walter Lewin. Edward Gray, Jacob Harney, Laurence Marschall, James McDonald, and Bob Celmer did their best to keep Walter and me from making mistakes in their fields of expertise; as much as we’d prefer to put the onus on them, we take full responsibility for any remaining errors. I also want to thank William J. Leo, a 2011 graduate of the University of Hartford, for his assistance at a critical moment. Three of the smartest writers I know—Marc Gunther, George Kannar, and Lennard Davis—all gave me invaluable advice early in the project. In different ways Dean Joseph Voelker and Assistant Provost Fred Sweitzer of the University of Hartford made it possible for me to find the time to finish this book. I am deeply grateful to my wife, Donna Schaper—minister and organizer extraordinaire, and author of thirty books at last count—for understanding and celebrating my immersion in a foreign world. Our grandson, Caleb Benjamin Luria, came into the world October 18, 2010; it has been a delight to watch him undertake his own series of remarkable experiments in the physics of everyday life. Finally, I want here to express my deep gratitude to Walter Lewin, who taught me more physics in the last few years than either of us would have thought possible and rekindled a passion in me that had lain dormant far too long.
APPENDIX 1
Mammal Femurs
It’s reasonable to assume that the mass of a mammal is proportional to its volume. Let’s take a puppy and compare it with a full-grown dog that is four times bigger. I am assuming that all linear dimensions of the bigger dog are four times larger than that of the puppy—its height, its length, the length and the thickness of its legs, the width of its head, everything. If that is the case, then the volume (and thus the mass) of the bigger dog is about sixty-four times that of the puppy.
One way to see this is by taking a cube with sides a, b, and c. The volume of this cube is a × b × c. When you make all sides four times larger, the volume becomes 4a × 4b × 4c, which is 64abc. If we express this a bit more mathematically, we can say that the volume (thus the mass) of the mammal is proportional to its length to the third power. If the bigger dog is four times larger than the puppy, then its volume should be about 4 cubed (43) times larger, which is 64. So, if we call the length of the femur “l,” then by comparing mammals of different size, their mass should be roughly proportional to l cubed (l3).
Okay, that’s mass. Now, the strength of the mammal’s femur supporting all that weight has to be proportional to its thickness, right? Thicker bones can support more weight—that’s intuitive. If we translate that idea to mathematics, the strength of the femur should be proportional to t
he area of the cross section of the bone. That cross section is roughly a circle, and we know that the area of a circle is πr2, where r is the radius of the circle. Thus, the area is proportional to d2 if d is the diameter of the circle.
Let’s call the thickness of the femur “d” (for diameter). Then, following Galileo’s idea, the mass of the mammal would be proportional to d2
(so that the bones can carry the weight of the mammal), but it is also proportional to l3 (that is always the case, independent of Galileo’s idea). Thus, if Galileo’s idea is correct, d2 should be proportional to l3, which is the same as stating that d is proportional to l3/2.
If I compare two mammals and one is five times bigger than the other (thus the length l of its femur is about five times larger than that of the smaller mammal), I may expect that the thickness, d, of its femur is about 53/2 = 11 times greater than the thickness of the smaller animal’s femur. In lectures I showed that the length l of the femur of an elephant was about 100 times larger than the length of the femur of a mouse; we may therefore expect, if Galileo’s idea is correct, that the thickness, d, of the elephant’s femur is about 1003/2 = 1,000 times thicker than that of the mouse.
Thus at some point, for very heavy mammals, the thickness of the bones would have to be the same as their lengths—or even greater—which would make for some pretty impractical mammals, and that would then be the reason why there is a maximum limit on the size of mammals.
APPENDIX 2
Newton’s Laws at Work
Newton’s law of universal gravitation can be written as
Here, Fgrav is the force of gravitational attraction between an object of mass m1 and one of mass m2, and r is the distance between them. G is called the gravitational constant.
Newton’s laws made it possible to calculate, at least in principle, the mass of the Sun and some planets.
Let’s see how this works. I’ll start with the Sun. Suppose m1 is the mass of the Sun, and that m2 is the mass of a planet (any planet). I will assume that the planetary orbit is a circle of radius r and let the orbital period of the planet be T (T is 365.25 days for the Earth, 88 days for Mercury, and almost twelve years for Jupiter).
If the orbit is circular or nearly so (which is the case for five of the six planets known in the seventeenth century), the speed of a planet in orbit is constant, but the direction of its velocity is always changing. However, whenever the direction of the velocity of any object changes, even if there is no change in speed, there must be an acceleration, and thus, according to Newton’s second law, there must be a force to provide that acceleration.
It’s called the centripetal force (Fc), and it is always exactly in the direction from the moving planet toward the Sun. Of course, since Newton was Newton, he knew exactly how to calculate this force (I derive the equation in my lectures). The magnitude of this force is
Here v is the speed of the planet in orbit. But this speed is the circumference of the orbit, 2πr, divided by the time, T, it takes to make one revolution around the Sun. Thus we can also write:
Where does this force come from? What on earth (no pun implied) is the origin of this force? Newton realized that it must be the gravitational attraction by the Sun. Thus the two forces in the above equations are one and the same force; they are equal to each other:
If we massage this a bit further by rearranging the variables (this is your chance to brush up on your high school algebra), we find that the mass of the Sun is
Notice that the mass of the planet (m2) is no longer present in equation 5; it does not enter into the picture; all we need is the planet’s mean distance to the Sun and its orbital period (T). Doesn’t that surprise you? After all, m2 shows up in equation 1 and also in equation 2. But the fact that it is present in both equations is the very reason that m2 is eliminated by setting Fgrav equal to Fc. That’s the beauty of this method, and we owe all this to Sir Isaac!
Equation 5 indicates that is the same for all planets. Even though they all have very different distances to the Sun and very different orbital periods, is the same for all. The German astronomer and mathematician Johannes Kepler had already discovered this amazing result in 1619, long before Newton. But why this ratio—between the cube of the radius and square of the orbital period—was constant was not understood at all. It was the genius Newton who showed sixty-eight years later that it is the natural consequence of his laws.
In summary, equation 5 tells us that if we know the distance from any planet to the Sun (r), the orbital period of the planet (T), and G, we can calculate the mass of the Sun (m1).
Orbital periods were known to a high degree of accuracy long before the seventeenth century. The distances between the Sun and the planets were also known to a high degree of accuracy long before the seventeenth century but only on a relative scale. In other words, astronomers knew that Venus’s mean distance to the Sun was 72.4 percent of Earth’s and that Jupiter’s mean distance was 5.200 times larger than Earth’s. However, the absolute values of these distances were an entirely different story. In the sixteenth century, in the day of the great Danish astronomer Tycho Brahe, astronomers believed that the distance from the Earth to the Sun was twenty times smaller than what it actually is (close to 150 million kilometers, about 93 million miles). In the early seventeenth century Kepler came up with a more accurate distance to the Sun, but still seven times smaller than what it is.
Since equation 5 indicates that the mass of the Sun is proportional to the distance (to a planet) cubed, if the distance r is too low by a factor of seven, then the mass of the Sun will be too low by a factor of 73, which is 343—not very useful at all.
A breakthrough came in 1672 when the Italian scientist Giovanni Cassini measured the distance from the Earth to the Sun to an accuracy of about 7 percent (impressive for those days), which meant that the uncertainty in r3 was only about 22 percent. The uncertainty in G was probably at least 30 percent. So my guess is that by the end of the seventeenth century the mass of the Sun may have been known to an accuracy no better than 50 percent.
Since the relative distances from the Sun to the planets were known to a high degree of accuracy, knowing the absolute distance from the Sun to the Earth to 7 percent accuracy meant that the absolute distances to the Sun of the other five known planets could also be calculated to that same 7 percent accuracy by the end of the seventeenth century.
The above method to calculate the mass of the Sun can also be used to measure the mass of Jupiter, Saturn, and the Earth. All three planets had known moons in orbit; in 1610 Galileo Galilei discovered four moons of Jupiter, now known as the Galilean moons. If m1 is the mass of Jupiter, and m2 the mass of one of its moons, then we can calculate the mass of Jupiter, using equation 5, in the same way that we can calculate the mass of the Sun, except that now r is the distance between Jupiter and its moon, and T is the orbital period of that moon around Jupiter. The four Galilean moons (Jupiter has sixty-three moons!) have orbital periods of 1.77 days, 3.55 days, 7.15 days, and 16.69 days.
Accuracies in distances and in G have greatly improved over time. By the nineteenth century G was known to about 1 percent accuracy. It is now known to an accuracy of about 0.01 percent.
Let me show you a numerical example. Using equation 5, let’s calculate together the mass of the Earth (m1) by using the orbit of our Moon (with mass m2). To use equation 5 properly, the distance, r, should be in meters, and T should be in seconds. If we then use 6.673 × 10–11 for G, we get the mass in kilograms.
The mean distance to the Moon (r) is 3.8440 × 108 meters (about 239,000 miles); its orbital period (T) is 2.3606 × 106 seconds (27.32 days). If we plug these numbers into equation 5, we find that the mass of the Earth is 6.030 × 1024 kilograms. The best current value of Earth’s mass is close to 5.974 × 1024 kilograms, which is only 1 percent lower than what I calculated! Why the difference? One reason is that the equation we used assumed that the Moon’s orbit is circular, when in fact it is elongated, what we call elliptical. As a
result, the smallest distance to the Moon is about 224,000 miles; the largest is about 252,000 miles. Of course, Newton’s laws can also easily deal with elliptical orbits, but the math may blow your mind. Perhaps it already has!
There is another reason why our result for the mass of the Earth is a little off. We assumed that the Moon circles around the Earth and that the center of that circle is the center of the Earth. Thus in equations 1 and 3, we assumed that r is the distance between the Earth and the Moon. That is correct in equation 1; however, as I discuss in more detail in chapter 13, the Moon and the Earth actually each orbit the center of mass of the Moon-Earth system, and that is about a thousand miles below the Earth’s surface. Thus r, in equation 3, is a little less than r in equation 1.
Since we live on Earth, there are other ways of calculating the mass of our home planet. One is by measuring the gravitational acceleration near the surface. When dropped, any object of mass m (m can have any value) will be accelerated with an acceleration, g, close to 9.82 meters per second per second.* Earth’s average radius is close to 6.371 × 106 meters (about 3,960 miles).
Now let’s revisit Newton’s equation 1. Since F = ma (Newton’s second law), then
Here, r is the radius of the Earth. With G = 6.673 × 10–11, g = 9.82 meters per second per second, and r = 6.371 × 106 meters, we can calculate mearth in kilograms (you try it!). If we simplify equation 6 somewhat, we get
I find that mearth is 5.973 × 1024 kilograms (impressive, right?).
Notice that the mass, m, of the object we dropped does not show up in equation 7! That should not surprise you, as the mass of the Earth could not possibly depend on the mass of the object that you drop.