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 absolutely 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
gravitational 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 (Fig. 19–1). 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 as shown in
Fig. 19–2 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.
 “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 $\tfrac{1}{2}m\,(dx/dt)^2$, 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 $t_1$ we started at some height and at the end of the time $t_2$ we
are definitely ending at some other place (Fig. 19–3).
 “Then the integral is
\begin{equation*}
\int_{t_1}^{t_2}\biggl[
\frac{1}{2}m\biggl(\ddt{x}{t}\biggr)^2-mgx\biggr]dt.
\end{equation*}
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 (Fig. 19–4). 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
uniform 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 shown in 
Fig. 19–5.
 “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 difference 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 (Fig. 19–6).
 “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.
 “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 complexities of life by proving
that it is so. The kind of mathematical problem we will have is very
difficult 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.
\begin{equation*}
\text{Action}=S=\int_{t_1}^{t_2}
(\text{KE}-\text{PE})\,dt.
\end{equation*}
Remember that the PE and KE are both functions of time. For each
different 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—Oh, 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 different 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 different 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
(Fig. 19–7).
 “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 (Fig. 19–8).
 “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 approximation, 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.
 “So we work it this way: We call $\underline{x(t)}$ (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 $\eta(t)$ (eta of $t$; Fig. 19–9).
 “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 $\underline{x(t)}$—to simplify the writing we
can call it $\underline{S}$—the difference of $\underline{S}$ and $S$
must be zero in the first-order approximation of small $\eta$. It can
differ in the second order, but in the first order the difference must
be zero.
 “And that must be true for any $\eta$ 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 $t_1$ and ends at a certain other point at $t_2$, and those points
and times are kept fixed. So the deviations in our $\eta$ have to be
zero at each end, $\eta(t_1)=0$ and $\eta(t_2)=0$. With that
condition, we have specified 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 $\eta$.
 “The idea is then that we substitute $x(t)=\underline{x(t)}+\eta(t)$
in the formula for the action:
\begin{equation*}
S=\int\biggl[
\frac{m}{2}\biggl(\ddt{x}{t}\biggr)^2-V(x)
\biggr]dt,
\end{equation*}
where I call the potential energy $V(x)$. The derivative $dx/dt$ is,
of course, the derivative of $\underline{x(t)}$ plus the derivative
of $\eta(t)$, so for the action I get this expression:
\begin{equation*}
S=\int_{t_1}^{t_2}\biggl[
\frac{m}{2}\biggl(
\ddt{\underline{x}}{t}+\ddt{\eta}{t}
\biggr)^2-V(\underline{x}+\eta)
\biggr]dt.
\end{equation*}
 “Now I must write this out in more detail. For the squared term I get
\begin{equation*}
\biggl(\ddt{\underline{x}}{t}\biggr)^2+
2\,\ddt{\underline{x}}{t}\,\ddt{\eta}{t}+
\biggl(\ddt{\eta}{t}\biggr)^2.
\end{equation*}
But wait. I’m not worrying about higher than the first order, so I
will take all the terms which involve $\eta^2$ 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
\begin{equation*}
\frac{m}{2}\biggl(\ddt{\underline{x}}{t}\biggr)^2+
m\,\ddt{\underline{x}}{t}\,\ddt{\eta}{t}+
(\text{second and higher order}).
\end{equation*}
 “Now we need the potential $V$ at $\underline{x}+\eta$. I consider
$\eta$ small, so I can write $V(x)$ as a Taylor series. It is
approximately $V(\underline{x})$; in the next approximation (from the
ordinary nature of derivatives) the correction is $\eta$ times the
rate of change of $V$ with respect to $x$, and so on:
\begin{equation*}
V(\underline{x}+\eta)=V(\underline{x})+
\eta V'(\underline{x})+\frac{\eta^2}{2}\,V''(\underline{x})+\dotsb
\end{equation*}
I have written $V'$ for the derivative of $V$ with respect to $x$ in
order to save writing. The term in $\eta^2$ 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, 
\begin{aligned}
S=\int_{t_1}^{t_2}\biggl[
&\frac{m}{2}\biggl(\ddt{\underline{x}}{t}\biggr)^2-V(\underline{x})+
m\,\ddt{\underline{x}}{t}\,\ddt{\eta}{t}\notag\\
&-\eta V'(\underline{x})+(\text{second and higher order})\biggr]dt.\notag
\end{aligned}
Now if we look carefully at the thing, we see that the first two terms
which I have arranged here correspond to the action $\underline{S}$
that I would have calculated with the true path $\underline{x}$. The
thing I want to concentrate on is the change in $S$—the difference
between the $S$ and the $\underline{S}$ that we would get for the
right path. This difference we will write as $\delta S$, called the
variation in $S$. Leaving out the ‘second and higher order’ terms, I
have for $\delta S$
\begin{equation*}
\delta S=\int_{t_1}^{t_2}\biggl[
m\,\ddt{\underline{x}}{t}\,\ddt{\eta}{t}-\eta V'(\underline{x})
\biggr]dt.
\end{equation*}
 “Now the problem is this: Here is a certain integral. I don’t know
what the $\underline{x}$ is yet, but I do know that no matter
what $\eta$ is, this integral must be zero. Well, you think, the only
way that that can happen is that what multiplies $\eta$ must be zero.
But what about the first term with $d\eta/dt$? Well, after all,
if $\eta$ can be anything at all, its derivative is anything also, so you
conclude that the coefficient of $d\eta/dt$ must also be zero. That
isn’t quite right. It isn’t quite right because there is a connection
between $\eta$ and its derivative; they are not absolutely
independent, because $\eta(t)$ must be zero at both $t_1$ and $t_2$.
 “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 $\eta$); 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 $(\eta)$,’ but with no other derivatives (no $d\eta/dt$). It
must be rearranged so it is always ‘something’ times $\eta$. 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 $d\eta/dt$ to make it have an $\eta$?
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 $\eta$
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 $d\eta/dt$ integrated with respect to $t$,
you write down the derivative of $\eta f$:
\begin{equation*}
\ddt{}{t}(\eta f)=\eta\,\ddt{f}{t}+f\,\ddt{\eta}{t}.
\end{equation*}
The integral you want is over the last term, so 
\begin{equation*}
\int f\,\ddt{\eta}{t}\,dt=\eta f-\int\eta\,\ddt{f}{t}\,dt.
\end{equation*}
 “In our formula for $\delta S$, the function $f$ is $m$
times $d\underline{x}/dt$; therefore, I have the following formula
for $\delta S$.

\begin{equation*}
\delta S=\left.m\,\ddt{\underline{x}}{t}\,\eta(t)\right|_{t_1}^{t_2}-
\int_{t_1}^{t_2}\ddt{}{t}\biggl(m\,\ddt{\underline{x}}{t}\biggr)\eta(t)\,dt-
\int_{t_1}^{t_2}V'(\underline{x})\,\eta(t)\,dt.
\end{equation*}


\begin{align*}
\delta S=\left.m\,\ddt{\underline{x}}{t}\,\eta(t)\right|_{t_1}^{t_2}-
\int_{t_1}^{t_2}\ddt{}{t}\biggl(m\,\ddt{\underline{x}}{t}\biggr)\eta(t)\,&dt\\[1ex]
-\int_{t_1}^{t_2}V'(\underline{x})\,\eta(t)\,&dt.
\end{align*}



The first term must be evaluated at the two limits $t_1$ and $t_2$. 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 $\eta$ 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 $\eta(t_1)=0$, and $\eta(t_2)=0$. So the integrated term is
zero. We collect the other terms together and obtain this:
\begin{equation*}
\delta S=\int_{t_1}^{t_2}\biggl[
-m\,\frac{d^2\underline{x}}{dt^2}-V'(\underline{x})
\biggr]\eta(t)\,dt.
\end{equation*}
The variation in $S$ is now the way we wanted it—there is the stuff
in brackets, say $F$, all multiplied by $\eta(t)$ and integrated from
$t_1$ to $t_2$.
 “We have that an integral of something or other times $\eta(t)$ is
always zero:
\begin{equation*}
\int F(t)\,\eta(t)\,dt=0.
\end{equation*}
I have some function of $t$; I multiply it by $\eta(t)$; and I
integrate it from one end to the other. And no matter what the $\eta$
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 $\eta(t)$ I took something which was zero for all $t$
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
(Fig. 19–10). When we do the integral of this $\eta$ times
any function $F$, the only place that you get anything other than zero
was where $\eta(t)$ 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 multiplied 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 $\eta$, then the
coefficient of $\eta$ must be zero. The action integral will be a
minimum for the path that satisfies this complicated differential
equation:
\begin{equation*}
\biggl[-m\,\frac{d^2\underline{x}}{dt^2}-V'(\underline{x})\biggr]=0.
\end{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 $z$ as functions of $t$; the action is more complicated.
For three-dimensional motion, you have to use the complete kinetic
energy—$(m/2)$ times the whole velocity squared. That is,
\begin{equation*}
\text{KE}=\frac{m}{2}\biggl[
\biggl(\ddt{x}{t}\biggr)^2\!\!+\biggl(\ddt{y}{t}\biggr)^2\!\!+
\biggl(\ddt{z}{t}\biggr)^2\,\biggr].
\end{equation*}
Also, the potential energy is a function of $x$, $y$, and $z$. 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 $\eta$?  Well, $\eta$ 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 $\eta$ 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 $\eta$ only in the
$x$-direction and say that coefficient 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
component. 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 $\eta$ 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 just add the kinetic energy of
both particles and take the potential energy of the mutual interaction.
And what do you vary? You vary the paths of both particles. Then,
for two particles moving in three dimensions, there are six equations.
You can vary the position of particle $1$ in the $x$-direction, in the
$y$-direction, and in the $z$-direction, and similarly for particle $2$;
so there are six equations. And that’s as it should be. There are the
three equations that determine the acceleration of particle $1$ in terms
of the force on it and three for the acceleration of particle $2$, from
the force on it. You follow the same game through, and you get Newton’s
law in three dimensions for any number of particles.
 “I have been saying that we get Newton’s law. That is not quite true,
because Newton’s law includes nonconservative forces like friction.
Newton said that $ma$ is equal to
any $F$. But the principle of least action only works for
conservative systems—where all forces can be gotten from a
potential function. You know, however, that on a microscopic level—on
the deepest level of physics—there are no nonconservative forces.
Nonconservative forces, like friction, appear only because we neglect
microscopic complications—there are just too many particles to
analyze. But the fundamental laws can be put in the form
of a principle of least action.
 “Let me generalize still further. Suppose we ask what happens if the
particle moves relativistically. We did not get the right relativistic
equation of motion; $F=ma$ is only right nonrelativistically. The
question is: Is there a corresponding principle of least action for
the relativistic case? There is. The formula in the case of relativity
is the following:

\begin{equation*}
S=-m_0c^2\int_{t_1}^{t_2}\sqrt{1-v^2/c^2}\,dt-
q\int_{t_1}^{t_2}[\phi(x,y,z,t)-\FLPv\cdot
\FLPA(x,y,z,t)]\,dt.
\end{equation*}


\begin{align*}
S=-m_0c^2&\int_{t_1}^{t_2}\sqrt{1-v^2/c^2}\,dt\\[1.25ex]
-q&\int_{t_1}^{t_2}[\phi(x,y,z,t)-\FLPv\cdot
\FLPA(x,y,z,t)]\,dt.
\end{align*}


The first part of the action integral is the rest mass $m_0$
times $c^2$ times the integral of a function of velocity,
$\sqrt{1-v^2/c^2}$. Then instead of just the potential energy, we have
an integral over the scalar potential $\phi$ and over $\FLPv$ times
the vector potential $\FLPA$. Of course, we are then including only
electromagnetic forces. All electric and magnetic fields are given in
terms of $\phi$ and $\FLPA$. This action function gives the complete
theory of relativistic motion of a single particle in an
electromagnetic field.
 “Of course, wherever I have written $\FLPv$, you understand that
before you try to figure anything out, you must substitute $dx/dt$
for $v_x$ and so on for the other components. Also, you put the point
along the path at time $t$, $x(t)$, $y(t)$, $z(t)$ where I wrote
simply $x$, $y$, $z$. Properly, it is only after you have made those
replacements for the $\FLPv$’s that you have the formula for the
action for a relativistic particle. I will leave to the more ingenious
of you the problem to demonstrate that this action formula does, in
fact, give the correct equations of motion for relativity. May I
suggest you do it first without the $\FLPA$, that is, for no magnetic
field?  Then you should get the components of the equation of motion,
$d\FLPp/dt=-q\,\FLPgrad{\phi}$, where, you remember,
$\FLPp=m_0\FLPv/\sqrt{1-v^2/c^2}$.
 “It is much more difficult to include also the case with a vector
potential. The variations get much more complicated. But in the end,
the force term does come out equal to $q(\FLPE+\FLPv\times\FLPB)$, as
it should. But I will leave that for you to play with.
 “I would like to emphasize that in the general case, for instance in
the relativistic formula, the action integrand no longer has the form of
the kinetic energy minus the potential energy. That’s only true in the
nonrelativistic approximation. For example, the
term $m_0c^2\sqrt{1-v^2/c^2}$ is not what we have called the kinetic
energy. The question of what the action should be for any particular
case must be determined by some kind of trial and error. It is just the
same problem as determining what are the laws of motion in the first
place. You just have to fiddle around with the equations that you know
and see if you can get them into the form of the principle of least
action.
 “One other point on terminology. The function that is integrated over
time to get the action $S$ is called the Lagrangian,
$\Lagrangian$,
which is a function only of the velocities and positions of particles.
So the principle of least action is also written
\begin{equation*}
S=\int_{t_1}^{t_2}\Lagrangian(x_i,v_i)\,dt,
\end{equation*}
where by $x_i$ and $v_i$ are meant all the components of the positions
and velocities. So if you hear someone talking about the ‘Lagrangian,’
you know they are talking about the function that is used to
find $S$. For relativistic motion in an electromagnetic field
\begin{equation*}
\Lagrangian=-m_0c^2\sqrt{1-v^2/c^2}-q(\phi-\FLPv\cdot\FLPA).
\end{equation*}
 “Also, I should say that $S$ is not really called the ‘action’ by the
most precise and pedantic people. It is called ‘Hamilton’s first
principal function.’ Now I hate to give a lecture on
‘the-principle-of-least-Hamilton’s-first-principal-function.’ So I call
it ‘the action.’ Also, more and more people are calling it the action.
You see, historically something else which is not quite as useful was
called the action, but I think it’s more sensible to change to a newer
definition. So now you too will call the new function the action, and
pretty soon everybody will call it by that simple name.
 “Now I want to say some things on this subject which are similar to the
discussions I gave about the principle of least time. There is quite a
difference in the characteristic of a law which says a certain integral
from one place to another is a minimum—which tells something about the
whole path—and of a law which says that as you go along, there is a
force that makes it accelerate. The second way tells how you inch your
way along the path, and the other is a grand statement about the whole
path. In the case of light, we talked about the connection of these two.
Now, I would like to explain why it is true that there are differential
laws when there is a least action principle of this kind. The reason is
the following: Consider the actual path in space and time. As before,
let’s take only one dimension, so we can plot the graph of $x$ as a
function of $t$. Along the true path, $S$ is a minimum. Let’s suppose
that we have the true path and that it goes through some point $a$ in
space and time, and also through another nearby point $b$
(Fig. 19–11). Now if the entire integral from $t_1$ to $t_2$
is a minimum, it is also necessary that the integral along the little
section from $a$ to $b$ is also a minimum. It can’t be that the part
from $a$ to $b$ is a little bit more. Otherwise you could just fiddle
with just that piece of the path and make the whole integral a little
lower.
 “So every subsection of the path must also be a minimum. And this is
true no matter how short the subsection. Therefore, the principle that
the whole path gives a minimum can be stated also by saying that an
infinitesimal section of path also has a curve such that it has a
minimum action. Now if we take a short enough section of
path—between two points $a$ and $b$ very close together—how the
potential varies from one place to another far away is not the
important thing, because you are staying almost in the same place over
the whole little piece of the path. The only thing that you have to
discuss is the first-order change in the potential. The answer can
only depend on the derivative of the potential and not on the
potential everywhere. So the statement about the gross property of the
whole path becomes a statement of what happens for a short section of
the path—a differential statement. And this differential statement
only involves the derivatives of the potential, that is, the force at
a point. That’s the qualitative explanation of the relation between
the gross law and the differential law.
 “In the case of light we also discussed the question: How does the
particle find the right path? From the differential point of view, it
is easy to understand. Every moment it gets an acceleration and knows
only what to do at that instant. But all your instincts on cause and
effect go haywire when you say that the particle decides to take the
path that is going to give the minimum action. Does it ‘smell’ the
neighboring paths to find out whether or not they have more action? In
the case of light, when we put blocks in the way so that the photons
could not test all the paths, we found that they couldn’t figure out
which way to go, and we had the phenomenon of diffraction.
 “Is the same thing true in mechanics? Is it true that the particle
doesn’t just ‘take the right path’ but that it looks at all the other
possible trajectories? And if by having things in the way, we don’t
let it look, that we will get an analog of diffraction? The miracle of
it all is, of course, that it does just that. That’s what the laws of
quantum mechanics say. So our principle of least action is
incompletely stated. It isn’t that a particle takes the path of least
action but that it smells all the paths in the neighborhood and
chooses the one that has the least action by a method analogous to the
one by which light chose the shortest time. You remember that the way
light chose the shortest time was this: If it went on a path that took
a different amount of time, it would arrive at a different phase. And
the total amplitude at some point is the sum of contributions of
amplitude for all the different ways the light can arrive. All the
paths that give wildly different phases don’t add up to anything. But
if you can find a whole sequence of paths which have phases almost all
the same, then the little contributions will add up and you get a
reasonable total amplitude to arrive. The important path becomes the
one for which there are many nearby paths which give the same phase.
 “It is just exactly the same thing for quantum mechanics. The
complete quantum mechanics (for the nonrelativistic case and
neglecting electron spin) works as follows: The probability that a
particle starting at point $1$ at the time $t_1$ will arrive at
point $2$ at the time $t_2$ is the square of a probability amplitude. The
total amplitude can be written as the sum of the amplitudes for each
possible path—for each way of arrival. For every $x(t)$ that we
could have—for every possible imaginary trajectory—we have to
calculate an amplitude. Then we add them all together. What do we take
for the amplitude for each path? Our action integral tells us what the
amplitude for a single path ought to be. The amplitude is proportional
to some constant times $e^{iS/\hbar}$, where $S$ is the action for
that path. That is, if we represent the phase of the amplitude by a
complex number, the phase angle is $S/\hbar$. The action $S$ has
dimensions of energy times time, and 
Planck’s constant $\hbar$ has the
same dimensions. It is the constant that determines when quantum
mechanics is important.
 “Here is how it works: Suppose that for all paths, $S$ is very large
compared to $\hbar$. One path contributes a certain amplitude. For a
nearby path, the phase is quite different, because with an enormous $S$
even a small change in $S$ means a completely different phase—because
$\hbar$ is so tiny. So nearby paths will normally cancel their effects
out in taking the sum—except for one region, and that is when a path
and a nearby path all give the same phase in the first approximation
(more precisely, the same action within $\hbar$). Only those paths will
be the important ones. So in the limiting case in which Planck’s
constant $\hbar$ goes to zero, the
correct quantum-mechanical laws can be summarized by simply saying:
‘Forget about all these probability amplitudes. The particle does go on
a special path, namely, that one for which $S$ does not vary in the
first approximation.’ That’s the relation between the principle of least
action and quantum mechanics. The fact that quantum mechanics can be
formulated in this way was discovered in 1942 by a student of that same
teacher, Bader, I spoke of at the beginning of this lecture. [Quantum
mechanics was originally formulated by giving a differential equation
for the amplitude (Schrödinger) and also by some other matrix mathematics
(Heisenberg).]
 “Now I want to talk about other minimum principles in physics. There
are many very interesting ones. I will not try to list them all now
but will only describe one more. Later on, when we come to a physical
phenomenon which has a nice minimum principle, I will tell about it
then. I want now to show that we can describe electrostatics, not by
giving a differential equation for the field, but by saying that a
certain integral is a maximum or a minimum. First, let’s take the case
where the charge density is known everywhere, and the problem is to
find the potential $\phi$ everywhere in space. You know that the
answer should be
\begin{equation*}
\nabla^2\phi=-\rho/\epsO.
\end{equation*}
But another way of stating the same thing is this: Calculate the
integral $U\stared$, where
\begin{equation*}
U\stared=\frac{\epsO}{2}\int(\FLPgrad{\phi})^2\,dV-
\int\rho\phi\,dV,
\end{equation*}
which is a volume integral to be taken over all space. This thing is a
minimum for the correct potential distribution $\phi(x,y,z)$.
 “We can show that the two statements about electrostatics are
equivalent. Let’s suppose that we pick any function $\phi$. We want to
show that when we take for $\phi$ the correct
potential $\underline{\phi}$, plus a small deviation $f$, then in the first
order, the change in $U\stared$ is zero. So we write
\begin{equation*}
\phi=\underline{\phi}+f.
\end{equation*}
The $\underline{\phi}$ is what we are looking for, but we are making a
variation of it to find what it has to be so that the variation
of $U\stared$ is zero to first order. For the first part of $U\stared$,
we need
\begin{equation*}
(\FLPgrad{\phi})^2=(\FLPgrad{\underline{\phi}})^2+
2\,\FLPgrad{\underline{\phi}}\cdot\FLPgrad{f}+
(\FLPgrad{f})^2.
\end{equation*}
The only first-order term that will vary is
\begin{equation*}
2\,\FLPgrad{\underline{\phi}}\cdot\FLPgrad{f}.
\end{equation*}
In the second term of the quantity $U\stared$, the integrand is
\begin{equation*}
\rho\phi=\rho\underline{\phi}+\rho f,
\end{equation*}
whose variable part is $\rho f$. So, keeping only the variable parts,
we need the integral
\begin{equation*}
\Delta U\stared=\int(\epsO\FLPgrad{\underline{\phi}}\cdot\FLPgrad{f}-
\rho f)\,dV.
\end{equation*}
 “Now, following the old general rule, we have to get the darn thing
all clear of derivatives of $f$. Let’s look at what the derivatives
are. The dot product is
\begin{equation*}
\ddp{\underline{\phi}}{x}\,\ddp{f}{x}+
\ddp{\underline{\phi}}{y}\,\ddp{f}{y}+
\ddp{\underline{\phi}}{z}\,\ddp{f}{z},
\end{equation*}
which we have to integrate with respect to $x$, to $y$, and to $z$. Now
here is the trick: to get rid of $\ddpl{f}{x}$ we integrate by parts
with respect to $x$. That will carry the derivative over onto
the $\underline{\phi}$. It’s the same general idea we used to get rid of
derivatives with respect to $t$. We use the equality
\begin{equation*}
\int\ddp{\underline{\phi}}{x}\,\ddp{f}{x}\,dx=
f\,\ddp{\underline{\phi}}{x}-
\int f\,\frac{\partial^2\underline{\phi}}{\partial x^2}\,dx.
\end{equation*}
The integrated term is zero, since we have to make $f$ zero at infinity.
(That corresponds to making $\eta$ zero at $t_1$ and $t_2$. So our
principle should be more accurately stated: $U\stared$ is less for the
true $\phi$ than for any other $\phi(x,y,z)$ having the same values at
infinity.) Then we do the same thing for $y$ and $z$. So our
integral $\Delta U\stared$ is
\begin{equation*}
\Delta U\stared=\int(-\epsO\nabla^2\underline{\phi}-\rho)f\,dV.
\end{equation*}
In order for this variation to be zero for any $f$, no matter what,
the coefficient of $f$ must be zero and, therefore,
\begin{equation*}
\nabla^2\underline{\phi}=-\rho/\epsO.
\end{equation*}
We get back our old equation. So our ‘minimum’ proposition is correct.
 “We can generalize our proposition if we do our algebra in a little
different way. Let’s go back and do our integration by parts without
taking components. We start by looking at the following equality:
\begin{equation*}
\FLPdiv{(f\,\FLPgrad{\underline{\phi}})}=
\FLPgrad{f}\cdot\FLPgrad{\underline{\phi}}+f\,\nabla^2\underline{\phi}.
\end{equation*}
If I differentiate out the left-hand side, I can show that it is just
equal to the right-hand side. Now we can use this equation to integrate
by parts. In our integral $\Delta U\stared$, we replace
$\FLPgrad{\underline{\phi}}\cdot\FLPgrad{f}$
by $\FLPdiv{(f\,\FLPgrad{\underline{\phi}})}-f\,\nabla^2\underline{\phi}$,
which gets integrated over volume. The divergence term integrated over
volume can be replaced by a surface integral:
\begin{equation*}
\int\FLPdiv{(f\,\FLPgrad{\underline{\phi}})}\,dV=
\int f\,\FLPgrad{\underline{\phi}}\cdot\FLPn\,da.
\end{equation*}
Since we are integrating over all space, the surface over which we are
integrating is at infinity. There, $f$ is zero and we get the same
answer as before.
 “Only now we see how to solve a problem when we don’t know
where all the charges are. Suppose that we have conductors with
charges spread out on them in some way. We can still use our minimum
principle if the potentials of all the conductors are fixed. We carry
out the integral for $U\stared$ only in the space outside of all
conductors. Then, since we can’t vary $\underline{\phi}$ on the
conductor, $f$ is zero on all those surfaces, and the surface integral
\begin{equation*}
\int f\,\FLPgrad{\underline{\phi}}\cdot\FLPn\,da
\end{equation*}
is still zero. The remaining volume integral
\begin{equation*}
\Delta U\stared=\int(-\epsO\,\nabla^2\underline{\phi}-\rho)f\,dV
\end{equation*}
is only to be carried out in the spaces between conductors. Of course,
we get Poisson’s equation again,
\begin{equation*}
\nabla^2\underline{\phi}=-\rho/\epsO.
\end{equation*}
So we have shown that our original integral $U\stared$ is also a minimum if
we evaluate it over the space outside of conductors all at fixed
potentials (that is, such that any trial $\phi(x,y,z)$ must equal the
given potential of the conductors when $(x,y,z)$ is a point on the
surface of a conductor).
 “There is an interesting case when the only charges are on
conductors. Then
\begin{equation*}
U\stared=\frac{\epsO}{2}\int(\FLPgrad{\phi})^2\,dV.
\end{equation*}
Our minimum principle says that in the case where there are conductors
set at certain given potentials, the potential between them adjusts
itself so that integral $U\stared$ is least. What is this integral? The
term $\FLPgrad{\phi}$ is the electric field, so the integral is the
electrostatic energy. The true field is the one, of all those coming
from the gradient of a potential, with the minimum total energy.
 “I would like to use this result to calculate something particular to
show you that these things are really quite practical. Suppose I take
two conductors in the form of a cylindrical condenser
(Fig. 19–12). The inside conductor has the potential $V$,
and the outside is at the potential zero. Let the radius of the inside
conductor be $a$ and that of the outside, $b$. Now we can suppose
any distribution of potential between the two. If we use the
correct $\underline{\phi}$, and
calculate $\epsO/2\int(\FLPgrad{\underline{\phi}})^2\,dV$, it should be
the energy of the system, $\tfrac{1}{2}CV^2$. So we can also
calculate $C$ by our principle. But if we use a wrong distribution of
potential and try to calculate the capacity $C$ by this method, we will
get a capacity that is too big, since $V$ is specified. Any assumed
potential $\phi$ that is not the exactly correct one will give a
fake $C$ that is larger than the correct value. But if my false $\phi$
is any rough approximation, the $C$ will be a good approximation,
because the error in $C$ is second order in the error in $\phi$.
 “Suppose I don’t know the capacity of a cylindrical condenser. I can
use this principle to find it. I just guess at the potential
function $\phi$ until I get the lowest $C$. Suppose, for instance, I pick a
potential that corresponds to a constant field. (You know, of course,
that the field isn’t really constant here; it varies as $1/r$.) A
field which is constant means a potential which goes linearly with
distance. To fit the conditions at the two conductors, it must be
\begin{equation*}
\phi=V\biggl(1-\frac{r-a}{b-a}\biggr).
\end{equation*}
This function is $V$ at $r=a$, zero at $r=b$, and in between has a
constant slope equal to $-V/(b-a)$. So what one does to find the
integral $U\stared$ is multiply the square of this gradient by $\epsO/2$
and integrate over all volume. Let’s do this calculation for a
cylinder of unit length. A volume element at the radius $r$ is $2\pi
r\,dr$. Doing the integral, I find that my first try at the capacity
gives
\begin{equation*}
\frac{1}{2}\,CV^2(\text{first try})=\frac{\epsO}{2}
\int_a^b\frac{V^2}{(b-a)^2}\,2\pi r\,dr.
\end{equation*}
The integral is easy; it is just
\begin{equation*}
\pi V^2\biggl(\frac{b+a}{b-a}\biggr).
\end{equation*}
So I have a formula for the capacity which is not the true one but is
an approximate job:
\begin{equation*}
\frac{C}{2\pi\epsO}=\frac{b+a}{2(b-a)}.
\end{equation*}
It is, naturally, different from the correct
answer $C=2\pi\epsO/\ln(b/a)$, but it’s not too bad. Let’s compare it
with the right answer for several values of $b/a$. I have computed out
the answers in Table 19–1. Even when $b/a$ is as big
as $2$—which gives a pretty big variation in the field compared with a
linearly varying field—I get a pretty fair approximation. The answer
is, of course, a little too high, as expected. The thing gets much worse
if you have a tiny wire inside a big cylinder. Then the field has
enormous variations and if you represent it by a constant, you’re not
doing very well. With $b/a=100$, we’re off by nearly a factor of two.
Things are much better for small $b/a$. To take the opposite extreme,
when the conductors are not very far apart—say $b/a=1.1$—then the
constant field is a pretty good approximation, and we get the correct
value for $C$ to within a tenth of a percent.
 “Now I would like to tell you how to improve such a calculation. (Of
course, you know the right answer for the cylinder, but the
method is the same for some other odd shapes, where you may not know
the right answer.) The next step is to try a better approximation to
the unknown true $\phi$. For example, we might try a constant plus an
exponential $\phi$, etc. But how do you know when you have a better
approximation unless you know the true $\phi$? Answer: You
calculate $C$; the lowest $C$ is the value nearest the truth. Let us try this
idea out. Suppose that the potential is not linear but say quadratic
in $r$—that the electric field is not constant but linear. The most
general quadratic form that fits $\phi=0$ at $r=b$ and $\phi=V$
at $r=a$ is
\begin{equation*}
\phi=V\biggl[1+\alpha\biggl(\frac{r-a}{b-a}\biggr)-
(1+\alpha)\biggl(\frac{r-a}{b-a}\biggr)^2
\biggr],
\end{equation*}
where $\alpha$ is any constant number. This formula is a little more
complicated. It involves a quadratic term in the potential as well as
a linear term. It is very easy to get the field out of it. The field
is just
\begin{equation*}
E=-\ddt{\phi}{r}=-\frac{\alpha V}{b-a}+
2(1+\alpha)\,\frac{(r-a)V}{(b-a)^2}.
\end{equation*}
Now we have to square this and integrate over volume. But wait a moment.
What should I take for $\alpha$? I can take a parabola for the $\phi$;
but what parabola? Here’s what I do: Calculate the capacity with
an arbitrary $\alpha$. What I get is
\begin{equation*}
\frac{C}{2\pi\epsO}=\frac{a}{b-a}
\biggl[\frac{b}{a}\biggl(\frac{\alpha^2}{6}+
\frac{2\alpha}{3}+1\biggr)+
\frac{1}{6}\,\alpha^2+\frac{1}{3}\biggr].
\end{equation*}
It looks a little complicated, but it comes out of integrating the
square of the field. Now I can pick my $\alpha$. I know that the truth
lies lower than anything that I am going to calculate, so whatever I put
in for $\alpha$ is going to give me an answer too big. But if I keep
playing with $\alpha$ and get the lowest possible value I can, that
lowest value is nearer to the truth than any other value. So what I do
next is to pick the $\alpha$ that gives the minimum value for $C$.
Working it out by ordinary calculus, I get that the minimum $C$ occurs
for $\alpha=-2b/(b+a)$. Substituting that value into the formula, I
obtain for the minimum capacity
\begin{equation*}
\frac{C}{2\pi\epsO}=\frac{b^2+4ab+a^2}{3(b^2-a^2)}.
\end{equation*}
 “I’ve worked out what this formula gives for $C$ for various values
of $b/a$. I call these numbers $C (\text{quadratic})$.
Table 19–2 compares $C (\text{quadratic})$ with the
true $C$.
 “For example, when the ratio of the radii is $2$ to $1$, I
have $1.444$, which is a very good approximation to the true answer,
$1.4427$. Even for larger $b/a$, it stays pretty good—it is much,
much better than the first approximation. It is even fairly
good—only off by $10$ percent—when $b/a$ is $10$ to $1$. But when
it gets to be $100$ to $1$—well, things begin to go wild. I get that
$C$ is $0.347$ instead of $0.217$. On the other hand, for a ratio of
radii of $1.5$, the answer is excellent; and for a $b/a$ of $1.1$, the
answer comes out $10.492063$ instead of $10.492059$. Where the answer
should be good, it is very, very good.
 “I have given these examples, first, to show the theoretical value of
the principles of minimum action and minimum principles in general
and, second, to show their practical utility—not just to calculate a
capacity when we already know the answer. For any other shape, you can
guess an approximate field with some unknown parameters like $\alpha$
and adjust them to get a minimum. You will get excellent numerical
results for otherwise intractable problems.”
