Feynman Physics Lectures V2 Ch19 1962-12-03 Principle of Least Action

Feynman Physics Lectures V2 Ch19 1962-12-03 Principle of Least Action

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Unformatted text preview: 19 The Principle of Least Action A special lecture—almost verbatim* “When I was in high school, my physics teacher——whose name was Mr. Bader ——called me down one day after physics class and said, ‘You look bored; I want to tell you something interesting.’ Then he told me something which I found ab- solutely fascinating, and have, since then, always found fascinating. Every time the subject comes up, I work on it. In fact, when I began to prepare this lecture I found myself making more analyses on the thing. Instead of worrying about the lecture, I got involved in a new problem. The subject is this—the principle of least action. “Mr. Bader told me the following: Suppose you have a particle (in a gravita- tional field, for instance) which starts somewhere and moves to some other point by free motion—you throw it, and it goes up and comes down. —> It goes from the original place to the final place in a certain amount of time. Now, you try a different motion. Suppose that to get from here to there, it went like this _ but got there in just the same amount of time. Then he said this: If you calculate the kinetic energy at every moment on the path, take away the potential energy, and integrate it over the time during the whole path, you’ll find that the number you’ll get is bigger than that for the actual motion. * Later chapters do not depend on the material of this special lecture—which is in- tended to be for “entertainment.” 19—1 “In other words, the laws of Newton could be stated not in the form F = ma but in the form: the average kinetic energy less the average potential energy is as little as possible for the path of an object going from one point to another. “Let me illustrate a little bit better what it means. If you take the case of the gravitational field, then if the particle has the path x(t) (let’s just take one dimension for a moment; we take a trajectory that goes up and down and not sideways), where x is the height above the ground, the kinetic energy is %m (dx/dt) 2, and the potential energy at any time is mgx. Now I take the kinetic energy minus the potential energy at every moment along the path and integrate that with respect to time from the initial time to the final time. Let’s suppose that at the original time 11 we started at some height and at the end of the time t 2 we are definitely ending at some other place. ___.5 “Then the integral is ‘2 1 dx 2 ft! [5 m (a?) — mgx] dt. The actual motion is some kind of a curve—it’s a parabola if we plot against the time—and gives a certain value for the integral. But we could imagine some other motion that went very high and came up and down in some peculiar way. “ We can calculate the kinetic energy minus the potential energy and integrate for such a path . . . or for any other path we want. The miracle is that the true path is the one for which that integral is least. “Let’s try it out. First, suppose we take the case of a free particle for which there is no potential energy at all. Then the rule says that in going from one point to another in a given amount of time, the kinetic energy integral is least, so it must go at a uniform speed. (We know that’s the right answer—to go at a uniform speed.) Why is that? Because if the particle were to go any other way, the velocities would be sometimes higher and sometimes lower than the average. The average velocity is the same for every case because it has to get from ‘here’ to ‘there’ in a given amount of time. “As an example, say your job is to start from home and get to school in a given length of time with the car. You can do it several ways: You can accelerate like mad at the beginning and slow down with the brakes near the end, or you can go at a uniform speed, or you can go backwards for a while and then go forward, and so on. The thing is that the average speed has got to be, of course, the total distance that you have gone over the time. But if you do anything but go at a uni- form speed, then sometimes you are going too fast and sometimes you are going too slow. Now the mean square of something that deviates around an average, as you know, is always greater than the square of the mean; so the kinetic energy integral would always be higher if you wobbled your velocity than if you went at a uniform velocity. So we see that the integral is a minimum if the velocity is a constant (when there are no forces). The correct path is like this. _—_... “Now, an object thrown up in a gravitational field does rise faster first and then slow down. That is because there is also the potential energy, and we must have the least dtflerence of kinetic and potential energy on the average. Because the potential energy rises as we go up in space, we will get a lower difference if we can get as soon as possible up to where there is a high potential energy. Then we can take that potential away from the kinetic energy and get a lower average. So it is better to take a path which goes up and gets a lot of negative stuff from the potential energy. ——--D “On the other hand, you can’t go up too fast, or too far, because you will then have too much kinetic energy involved—you have to go very fast to get way up and come down again in the fixed amount of time available. So you don’t want to go too far up, but you want to go up some. So it turns out that the solution is some kind of balance between trying to get more potential energy with the least amount of extra kinetic energy—trying to get the difference, kinetic minus the potential, as small as possible. 1 9—2 “That is all my teacher told me, because he was a very good teacher and knew when to stop talking. But I don’t know when to stop talking. So instead of leaving it as an interesting remark, I am going to horrify and disgust you with the complexi- ties of life by proving that it is so. The kind of mathematical problem we will have is very diflicult and a new kind. We have a certain quantity which is called the action, S. It is the kinetic energy, minus the potential energy, integrated over time. Action = s = "(KB — PE) d1. ‘1 Remember that the PE and KB are both functions of time. For each difi‘erent possible path you get a different number for this action. Our mathematical problem is to find out for what curve that number is the least. . “You say—0h, that's just the ordinary calculus of maxima and minima. You calculate the action and just differentiate to find the minimum. “But watch out. Ordinarily we just have a function of some variable, and we have to find the value of that variable where the function is least or most. For instance, we have a rod which has been heated in the middle and the heat is spread around. For each point on the rod we have a temperature, and we must find the point at which that temperature is largest. But now for each path in space we have a number—quite a different thing—and we have to find the path in space for which the number is the minimum. That is a completely difi‘erent branch of mathematics. It is not the ordinary calculus. In fact, it is called the calculus of variations. “There are many problems in this kind of mathematics. For example, the circle is usually defined as the locus of all points at a constant distance from a fixed point, but another way of defining a circle is this: a circle is that curve of given length which encloses the biggest area. Any other curve encloses less area for a given perimeter than the circle does. So if we give the problem: find that curve which encloses the greatest area for a given perimeter, we would have a problem of the calculus of variations—a difierent kind of calculus than you’re used to. “So we make the calculation for the path of an object. Here is the way we are going to do it. The idea is that we imagine that there is a true path and that any other curve we draw is a false path, so that if we calculate the action for the false path we will get a value that is bigger than if we calculate the action for the true path. _ “Problem: Find the true path. Where is it? One way, of course, is to calculate the action for millions and millions of paths and look at which one is lowest. When you find the lowest one, that’s the true path. “That’s a possible way. But we can do it better than that. When we have a quantity which has a minimum—for instance, in an ordinary function like the temperature—one of the properties of the minimum is that if we go away from the minimum in the first order, the deviation of the function from its minimum value is only second order. At any place else on the curve, if we move a small distance the value of the function changes also in the first order. But at a minimum, a tiny motion away makes, in the first approximation, no difference. _ “That is what we are going to use to calculate the true path. If we have the true path, a curve which differs only a little bit from it will, in the first approxima- tion, make no difference in the action. Any difference will be in the second approximation, if we really have a minimum. “That is easy to prove. If there is a change in the first order when I deviate the curve a certain way, there is a change in the action that is proportional to the deviation. The change presumably makes the action greater; otherwise we haven’t got a minimum. But then if the change is proportional to the deviation, reversing the sign of the deviation will make the action less. We would get the action to increase one way and to decrease the other way. The only way that it could really be a minimum is that in the first approximation it doesn’t make any change, that the changes are proportional to the square of the deviations from the true path. 19-3 “So we work it this way: We call it) (with an underline) the true path—the one we are trying to find. We take some trial path x(t) that differs from the true path by a small amount which we will call 770) (eta of t). — “Now the idea is that if we calculate the action S for the path x(t), then the difference between that S and the action that we calculated for the path x_(t)—to simplify the writing we can call it S—the difierence of S and S must be— zero in the first-order approximation of small 17. It can differ in the second order, but in the first order the difierence must be zero. “And that must be true for any '0 at all. Well, not quite. The method doesn’t mean anything unless you consider paths which all begin and end at the same two points—each path begins at a certain point at t1 and ends at a certain other point at t2, and those points and times are kept fixed. So the deviations in our 17 have to be zero at each end, 1700 = 0 arid n(t2) = 0. With that condition, we have speci- fied our mathematical problem. “If you didn’t know any calculus, you might do the same kind of thing to find the minimum of an ordinary function f(x). You could discuss what happens if you take f(x) and add a small amount h to x and argue that the correction to f(x) in the first order in h must be zero at the minimum. You would substitute x + h for x and expand out to the first order in h . . . just as we are going to do with 71- “The idea is then that we substitute x(t) = x(t) + 17(t) in the formula for the action: m dx 2 where I call the potential energy V(x). The derivative dx/dt is, of course, the derivative of it) plus the derivative of 1,0), so for the action I get this expression: ’2 _ mdl (1172 ] S—/;I[—2-(217+d—t>—V(&+fl)dt- “Now I must write this out in more detail. For the squared term I get d; dx dn d1; (3)2 + 2711th + <dt)2 But wait. I’m not worrying about higher than the first order, so I will take all the terms which involve 112 and higher powers and put them in a little box called ‘second and higher order.’ From this term I get only second order, but there will be more from something else. So the kinetic energy part is 2 g (g?) + m % (:11? + (second and higher order). “Now we need the potential V at a: + 17- I consider 11 small, so I can write V(x) as a Taylor series. It is approximately V(x); in the next approximation (from the ordinary nature of derivatives) the correction is 17 times the rate of change of V with respect to x, and so on: V(gc + n)= V(x) + nV’OC) +— 2V"Q) +- I have written V’ for the derivative of V with respect to x in order to save writing. The term in 172 and the ones beyond fall into the ‘second and higher order’ category and we don’t have to worry about them. Putting it all together, ‘2 _ 2 dz _ dx d7, S‘/,, [2 (d7) VU+md7d7 — nV’Q) + (second and higher order)] dt 19—4 Now if we look carefully at the thing, we see that the first two terms which I have arranged here correspond to the action S that I would have calculated with the true path 5. The thing I want to concentrate on is the change in S—the difference between the S and the S that we would get for the right path. This difference we will write as as, called the variation in S. Leaving out the ‘second and higher order’ terms, I have for as ‘2 dx d as = f [m E a; — nV’(J_c)]dt. “Now the problem is this: Here is a certain integral. I don’t know what the f is yet, but I do know that no matter what n is, this integral must be zero. Well, you think, the only way that that 'can happen is that what multiplies 77 must be zero. But what about the first term with dn/dt? Well, after all, if n can be anything at all, its derivative is anything also, so you conclude that the coefficient of dn/dt must also be zero. That isn’t quite right. It isn’t quite right because there is a connection between 17 and its derivative; they are not absolutely independent, because n(t) must be zero at both t1 and t2. “The method of solving all problems in the calculus of variations always uses the same general principle. You make the shift in the thing you want to vary (as we did by adding 11); you look at the first-order terms; then you always arrange things in such a form that you get an integral of the form ‘some kind of stuff times the shift (n),’ but with no other derivatives (no dn/dt). It must be rearranged so it is always ‘something’ times 17- You will see the great value of that in a minute. (There are formulas that tell you how to do this in some cases without actually calculating, but they are not general enough to be worth bothering about; the best way is to calculate it out this way.) “How can I rearrange the term in dn/dt to make it have an n? I can do that by integrating by parts. It turns out that the whole trick of the calculus of variations consists of writing down the variation of S and then integrating by parts so that the derivatives of 1? disappear. It is always the same in every problem in which derivatives appear. “You remember the general principle for integrating by parts. If you have any function f times dn/dt integrated with respect to t, you write down the derivative ofnf: a; d(nf) = n:{+ ftz—z The integral you want is over the last term, so [fis—2' dt= nf— fd—fdt. “In our formula for 6S, the function f is m‘times die/dt; therefore, I have the following formula for «SS. lg 52 t2 — / d(m d") (t) dt / V’(x)n(t)dt. l ‘1 dt 11 .- The first term must be evaluated at the two limits t1 and t2. Then I must have the integral from the rest of the integration by parts. The last term is brought down without change. “Now comes something which always happens—the integrated part disappears. (In fact, if the integrated part does not disappear, you restate the principle, adding conditions to make sure it does!) We have already said that 11 must be zero at both ends of the path, because the principle is that the action is a minimum provided that the varied curve begins and ends at the chosen points. The condition is that 19—5 BS= m—nU) 11(11) = 0, and 1:02) = 0. So the integrated term is zero. We collect the other terms together and obtain this: :3 2 as = A [—m ‘27? — my] n(t)dt. The variation in S is now the way we wanted it—there is the stuff in brackets, say F, all multiplied by n(t) and integrated from t1 to t2. “We have that an integral of something or other times n(t) is always zero: /F(t) ”(0dr = o. I have some function of t; I multiply it by n(t); and I integrate it from one end to the other. And no matter what the 71 is, I get zero. That means that the function F(t) is zero. That’s obvious, but anyway I’ll show you one kind of proof. “Suppose that for 17(t) I took something which was zero for all 1 except right near one particular value. It stays zero until it gets to this t, ————> then it blips up for a moment and blips right back down. When we do the integral of this 11 times any function F, the only place that you get anything other than zero was where 11(1) was blipping, and then you get the value of F at that place times the integral over the blip. The integral over the blip alone isn’t zero, but when multi- plied by F it has to be; so the function F has to be zero where the blip was. But the blip was anywhere I wanted to put it, so F must be zero everywhere. “We see that if our integral is zero for any n, then the coefficient of 1! must be zero. The action integral will be a minimum for the path that satisfies this compli- cated diflerential equation: It’s not really so complicated; you have seen it before. It is just F = ma. The first term is the mass times acceleration, and the second is the derivative of the potential energy, which is the force. “So, for a conservative system at least, we have demonstrated that the principle of least action gives the right answer; it says that the path that has the minimum action is the one satisfying Newton’s law. “One remark: I did not prove it was a minimum-maybe it’s a maximum. In fact, it doesn’t really have to be a minimum. It is quite analogous to What we found for the ‘principle of least time’ which we discussed in optics. There also, we said at first it was ‘least’ time. It turned out, however, that there were situations in which it wasn’t the least time. The fundamental principle was that for any first-order variation away from the optical path, the change in time was zero; it is the same story. What we really mean by ‘least’ is that the first-order change in the value of S, when you change the path, is zero. It is not necessarily a ‘minimum.’ “Next, I remark on some generalizations. In the first place, the thing can be done in three dimensions. Instead of just x, I would have x, y, and 2 as functions of t; the action is more complicated. For three-dimensional motion, you have to use the complete kinetic energy—(m/Z) times the whole velocity squared. That is, _ m dx 2 dy 2 dz)2]_ KB ‘ 7 [(27) + (a) + (3 Also, the potential energy is a function of x, y, and 2. And what about the path? The path is some general curve in space, which is not so easily drawn, but the idea is the same. And what about the 11? Well, 7i can have three components. You could shift the paths in x, or in y, or in z—or you could shift in all three directions simultaneously. So 77 would be a vector. This doesn’t really complicate things too much, though. Since only the first-order variation has to be zero, we can do the calculation by three successive shifts. We can shift 11 only in the x-direction and 19—6 say that coefi‘icient must be zero. We get one equation. Then we shift it in the y-direction and get another. And in the z-direction and get another. Or, of course, in any order that you want. Anyway, you get three equations. And, of course, Newton’s law is really three equations in the three dimensions—one for each com- ponent. I think that you can practically see that it is bound to work, but we will leave you to show for yourself that it will work for three dimensions. Incidentally, you could use any coordinate system you want, polar or otherwise, and get Newton’s laws appropriate to that system right off by seeing what happens if you have the shift 71 in radius, or in angle, etc. “Similarly, the method can be generalized to any number of particles. If you have, say, two particles with a force between them, so that there is a mutual potential energy, then you ju...
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