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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. 193 “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 ﬁrstorder 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 ﬁrstorder 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 minimummaybe 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 ﬁrstorder
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 ﬁrstorder 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 threedimensional 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 ﬁrstorder variation has to be zero, we can do the
calculation by three successive shifts. We can shift 11 only in the xdirection and 19—6 say that coeﬁ‘icient must be zero. We get one equation. Then we shift it in the
ydirection and get another. And in the zdirection 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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 Spring '09
 LeeKinohara
 Physics

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