## 14Work and Potential Energy (conclusion)

### 14–1Work

In the preceding chapter we have presented a great many new ideas and results that play a central role in physics. These ideas are so important that it seems worthwhile to devote a whole chapter to a closer examination of them. In the present chapter we shall not repeat the “proofs” or the specific tricks by which the results were obtained, but shall concentrate instead upon a discussion of the ideas themselves.

In learning any subject of a technical nature where mathematics plays a
role, one is confronted with the task of understanding and storing away
in the memory a huge body of facts and ideas, held together by certain
relationships which can be “proved” or “shown” to exist between
them. It is easy to confuse the proof itself with the relationship which
it establishes. Clearly, the important thing to learn and to remember is
the relationship, not the proof. In any particular circumstance we can
either say “it can be shown that” such and such is true, or we can
show it. In almost all cases, the particular proof that is used is
concocted, first of all, in such form that it can be written quickly and
easily on the chalkboard or on paper, and so that it will be as
smooth-looking as possible. Consequently, the proof may look deceptively
simple, when in fact, the author might have worked for hours trying
different ways of calculating the same thing until he has found the
neatest way, so as to be able to show that it can be shown in the
shortest amount of time! The thing to be remembered, when seeing a
proof, is not the proof itself, but rather that it *can be shown*
that such and such is true. Of course, if the proof involves some
mathematical procedures or “tricks” that one has not seen before,
attention should be given not to the trick exactly, but to the
mathematical idea involved.

It is certain that in all the demonstrations that are made in a course such as this, not one has been remembered from the time when the author studied freshman physics. Quite the contrary: he merely remembers that such and such is true, and to explain how it can be shown he invents a demonstration at the moment it is needed. Anyone who has really learned a subject should be able to follow a similar procedure, but it is no use remembering the proofs. That is why, in this chapter, we shall avoid the proofs of the various statements made previously, and merely summarize the results.

The first idea that has to be digested is *work done by a force*. The
physical word “work” is not the word in the ordinary sense of “Workers of the
world unite!,” but is a different idea. Physical work is expressed
as $\int\FLPF\cdot d\FLPs$, called “the line integral of $F$ dot $ds$,” which
means that if the force, for instance, is in one direction and the object on
which the force is working is displaced in a certain direction, then *only
the component of force in the direction of the displacement* does any work. If,
for instance, the force were constant and the displacement were a finite
distance $\Delta\FLPs$, then the work done in moving the object through that
distance is only the component of force along $\Delta\FLPs$ times $\Delta s$.
The rule is “force times distance,” but we really mean only the component of
force in the direction of the displacement times $\Delta s$ or, equivalently,
the component of displacement in the direction of force times $F$. It is evident
that no work whatsoever is done by a force which is at right angles to the
displacement.

Now if the vector displacement $\Delta\FLPs$ is resolved into components, in other words, if the actual displacement is $\Delta\FLPs$ and we want to consider it effectively as a component of displacement $\Delta x$ in the $x$-direction, $\Delta y$ in the $y$-direction, and $\Delta z$ in the $z$-direction, then the work done in carrying an object from one place to another can be calculated in three parts, by calculating the work done along $x$, along $y$, and along $z$. The work done in going along $x$ involves only that component of force, namely $F_x$, and so on, so the work is $F_x\,\Delta x + F_y\,\Delta y + F_z\,\Delta z$. When the force is not constant, and we have a complicated curved motion, then we must resolve the path into a lot of little $\Delta\FLPs$’s, add the work done in carrying the object along each $\Delta\FLPs$, and take the limit as $\Delta\FLPs$ goes to zero. This is the meaning of the “line integral.”

Everything we have just said is contained in the formula $W=\int\FLPF\cdot d\FLPs$. It is all very well to say that it is a marvelous formula, but it is another thing to understand what it means, or what some of the consequences are.

The word “work” in physics has a meaning so different from that of
the word as it is used in ordinary circumstances that it must be
observed carefully that there are some peculiar circumstances in which
it appears not to be the same. For example, according to the physical
definition of work, if one holds a hundred-pound weight off the ground
for a while, he is doing no work. Nevertheless, everyone knows that he
begins to sweat, shake, and breathe harder, as if he were running up a
flight of stairs. Yet running upstairs *is* considered as doing
work (in running *downstairs*, one gets work out of the world,
according to physics), but in simply holding an object in a fixed
position, no work is done. Clearly, the physical definition of work
differs from the physiological definition, for reasons we shall
briefly explore.

It is a fact that when one holds a weight he has to do “physiological”
work. Why should he sweat? Why should he need to consume food to hold
the weight up? Why is the machinery inside him operating at full
throttle, just to hold the weight up? Actually, the weight could be held
up with no effort by just placing it on a table; then the table, quietly
and calmly, without any supply of energy, is able to maintain the same
weight at the same height! The physiological situation is something like
the following. There are two kinds of muscles in the human body and in
other animals: one kind, called *striated* or
*skeletal* muscle, is the type of muscle we have in our arms, for
example, which is under voluntary control; the other kind, called
*smooth* muscle,
is like the muscle in the intestines or, in the clam, the greater
adductor muscle that closes the shell. The smooth muscles work very slowly, but they
can hold a “set”; that is to say, if the clam tries to close its shell
in a certain position, it will hold that position, even if there is a
very great force trying to change it. It will hold a position under load
for hours and hours without getting tired because it is very much like a
table holding up a weight, it “sets” into a certain position, and the
molecules just lock there temporarily with no work being done, no effort
being generated by the clam. The fact that we have to generate effort to
hold up a weight is simply due to the design of striated muscle. What
happens is that when a nerve impulse reaches a muscle fiber, the fiber
gives a little twitch and then relaxes, so that when we hold something
up, enormous volleys of nerve impulses are coming in to the muscle,
large numbers of twitches are maintaining the weight, while the other
fibers relax. We can see this, of course: when we hold a heavy weight
and get tired, we begin to shake. The reason is that the volleys are
coming irregularly, and the muscle is tired and not reacting fast
enough. Why such an inefficient scheme? We do not know exactly why, but
evolution has not been able to develop *fast* smooth muscle. Smooth muscle would be much more effective
for holding up weights because you could just stand there and it would
lock in; there would be no work involved and no energy would be
required. However, it has the disadvantage that it is very
slow-operating.

Returning now to physics, we may ask *why* we want to calculate
the work done. The answer is that it is interesting and useful to do
so, since the work done on a particle by the resultant of all the
forces acting on it is exactly equal to the change in kinetic energy
of that particle. That is, if an object is being pushed, it picks up
speed, and
\begin{equation*}
\Delta(v^2)=\frac{2}{m}\,\FLPF\cdot\Delta\FLPs.
\end{equation*}

### 14–2Constrained motion

Another interesting feature of forces and work is this: suppose that
we have a sloping or a curved track, and a particle that must move
along the track, but without friction. Or we may have a pendulum with
a string and a weight; the string constrains the weight to move in a
circle about the pivot point. The pivot point may be changed by having
the string hit a peg, so that the path of the weight is along two
circles of different radii. These are examples of what we call
*fixed, frictionless constraints*.

In motion with a fixed frictionless constraint, no work is done by the constraint because the forces of constraint are always at right angles to the motion. By the “forces of constraint” we mean those forces which are applied to the object directly by the constraint itself—the contact force with the track, or the tension in the string.

The forces involved in the motion of a particle on a slope moving
under the influence of gravity are quite complicated, since there is a
constraint force, a gravitational force, and so on. However, if we
base our calculation of the motion on conservation of energy and
*the gravitational force alone*, we get the right result. This
seems rather strange, because it is not strictly the right way to do
it—we should use the *resultant* force. Nevertheless, the work
done by the gravitational force alone will turn out to be the change
in the kinetic energy, because the work done by the constraint part of
the force is zero (Fig. 14–1).

The important feature here is that if a force can be analyzed as the sum of two or more “pieces” then the work done by the resultant force in going along a certain curve is the sum of the works done by the various “component” forces into which the force is analyzed. Thus if we analyze the force as being the vector sum of several effects, gravitational plus constraint forces, etc., or the $x$-component of all forces and the $y$-component of all forces, or any other way that we wish to split it up, then the work done by the net force is equal to the sum of the works done by all the parts into which we have divided the force in making the analysis.

### 14–3Conservative forces

In nature there are certain forces, that of gravity, for example,
which have a very remarkable property which we call “conservative”
(no political ideas involved, it is again one of those “crazy
words”). If we calculate how much work is done by a force in moving
an object from one point to another along some curved path, in general
the work depends upon the curve, but in special cases it does not. If
it does not depend upon the curve, we say that the force is a
conservative force. In other words, if the integral of the force times
the distance in going from position $1$ to position $2$ in
Fig. 14–2 is calculated along curve $A$ and then along $B$,
we get the same number of joules, and if this is true for this pair of
points on *every curve*, and if the same proposition works *no
matter which pair of points* we use, then we say the force is
conservative. In such circumstances, the work integral going from $1$
to $2$ can be evaluated in a simple manner, and we can give a formula for
the result. Ordinarily it is not this easy, because we also have to
specify the curve, but when we have a case where the work does not
depend on the curve, then, of course, the work depends only upon the
*positions* of $1$ and $2$.

To demonstrate this idea, consider the following. We take a “standard” point $P$, at an arbitrary location (Fig. 14–2). Then, the work line-integral from $1$ to $2$, which we want to calculate, can be evaluated as the work done in going from $1$ to $P$ plus the work done in going from $P$ to $2$, because the forces are conservative and the work does not depend upon the curve. Now, the work done in going from position $P$ to a particular position in space is a function of that position in space. Of course it really depends on $P$ also, but we hold the arbitrary point $P$ fixed permanently for the analysis. If that is done, then the work done in going from point $P$ to point $2$ is some function of the final position of $2$. It depends upon where $2$ is; if we go to some other point we get a different answer.

We shall call this function of position $-U(x,y,z)$, and when we wish
to refer to some particular point $2$ whose coordinates
are $(x_2,y_2,z_2)$, we shall write $U(2)$, as an abbreviation
for $U(x_2,y_2,z_2)$. The work done in going from point $1$ to point $P$
can be written also by going the *other way* along the integral,
reversing all the $d\FLPs$’s. That is, the work done in going from $1$
to $P$ is *minus* the work done in going from the point $P$ to $1$:
\begin{equation*}
\int_1^P\FLPF\cdot d\FLPs=\int_P^1\FLPF\cdot(-d\FLPs)=
-\int_P^1\FLPF\cdot d\FLPs.
\end{equation*}
Thus the work done in going from $P$ to $1$ is $-U(1)$, and from $P$
to $2$ the work is $-U(2)$. Therefore the integral from $1$ to $2$ is
equal to $-U(2)$ plus [$-U(1)$ backwards], or $+U(1) - U(2)$:
\begin{gather}
U(1)=-\int_P^1\FLPF\cdot d\FLPs,\quad
U(2)=-\int_P^2\FLPF\cdot d\FLPs,\notag\\[1.5ex]
\label{Eq:I:14:1}
\int_1^2\FLPF\cdot d\FLPs=U(1)-U(2).
\end{gather}
The quantity $U(2) - U(1)$ is called the change in the potential energy,
and we call $U$ the potential energy. We shall say that when the object
is located at position $2$, it has potential energy $U(2)$ and at
position $1$ it has potential energy $U(1)$. If it is located at
position $P$, it has zero potential energy. If we had used any other
point, say $Q$, instead of $P$, it would turn out (and we shall leave it
to you to demonstrate) that the *potential energy is changed only
by the addition of a constant*. Since the conservation of energy depends
only upon *changes*, it does not matter if we add a constant to the
potential energy. Thus the point $P$ is arbitrary.

Now, we have the following two propositions: (1) that the work done by
a force is equal to the change in kinetic energy of the particle, but
(2) mathematically, for a conservative force, the work done is minus
the change in a function $U$ which we call the potential energy. As a
consequence of these two, we arrive at the proposition that *if
only conservative forces act, the kinetic energy $T$ plus the
potential energy $U$ remains constant:*
\begin{equation}
\label{Eq:I:14:2}
T+U=\text{constant}.
\end{equation}

Let us now discuss the formulas for the potential energy for a number
of cases. If we have a gravitational field that is uniform, if we are
not going to heights comparable with the radius of the earth, then the
force is a constant vertical force and the work done is simply the
force times the vertical distance. Thus
\begin{equation}
\label{Eq:I:14:3}
U(z)=mgz,
\end{equation}
and the point $P$ which corresponds to zero potential energy happens to
be any point in the plane $z=0$. We could also have said that the
potential energy is $mg(z - 6)$ if we had wanted to—all the results
would, of course, be the same in our analysis except that the value of
the potential energy at $z = 0$ would be $-6mg$. It makes no difference,
because only *differences* in potential energy count.

The energy needed to compress a linear spring a distance $x$ from an equilibrium point is \begin{equation} \label{Eq:I:14:4} U(x)=\tfrac{1}{2}kx^2, \end{equation} and the zero of potential energy is at the point $x=0$, the equilibrium position of the spring. Again we could add any constant we wish.

The potential energy of gravitation for point masses $M$ and $m$, a distance $r$ apart, is \begin{equation} \label{Eq:I:14:5} U(r) = -GMm/r. \end{equation} The constant has been chosen here so that the potential is zero at infinity. Of course the same formula applies to electrical charges, because it is the same law: \begin{equation} \label{Eq:I:14:6} U(r) = q_1q_2/4\pi\epsO r. \end{equation}

Now let us actually use one of these formulas, to see whether we
understand what it means. *Question*: How fast do we have to
shoot a rocket away from the earth in order for it to leave?
*Solution*: The kinetic plus potential energy must be a constant;
when it “leaves,” it will be millions of miles away, and if it is
just barely able to leave, we may suppose that it is moving with zero
speed out there, just barely going. Let $a$ be the radius of the
earth, and $M$ its mass. The kinetic plus potential energy is then
initially given by $\tfrac{1}{2}mv^2 - GmM/a$. At the end of the
motion the two energies must be equal. The kinetic energy is taken to
be zero at the end of the motion, because it is supposed to be just
barely drifting away at essentially zero speed, and the potential
energy is $GmM$ divided by infinity, which is zero. So everything is
zero on one side and that tells us that the square of the velocity
must be $2GM/a$. But $GM/a^2$ is what we call the acceleration of
gravity, $g$. Thus
\begin{equation*}
v^2=2ga.
\end{equation*}

At what speed must a satellite travel in order to keep going around
the earth? We worked this out long ago and found that $v^2 =
GM/a$. Therefore to go *away* from the earth, we need
$\sqrt{2}$ times the velocity we need to just go *around* the earth near
its
surface. We need, in other words, *twice as much energy* (because
energy goes as the square of the velocity) to leave the earth as we do
to go around it. Therefore the first thing that was done historically
with satellites was to get one to go around the earth, which requires
a speed of five miles per second. The next thing was to send a
satellite away from the earth permanently; this required twice the
energy, or about seven miles per second.

Now, continuing our discussion of the characteristics of potential energy, let us consider the interaction of two molecules, or two atoms, two oxygen atoms for instance. When they are very far apart, the force is one of attraction, which varies as the inverse seventh power of the distance, and when they are very close the force is a very large repulsion. If we integrate the inverse seventh power to find the work done, we find that the potential energy $U$, which is a function of the radial distance between the two oxygen atoms, varies as the inverse sixth power of the distance for large distances.

If we sketch the curve of the potential energy $U(r)$ as in Fig. 14–3, we thus start out at large $r$ with an inverse sixth power, but if we come in sufficiently near we reach a point $d$ where there is a minimum of potential energy. The minimum of potential energy at $r = d$ means this: if we start at $d$ and move a small distance, a very small distance, the work done, which is the change in potential energy when we move this distance, is nearly zero, because there is very little change in potential energy at the bottom of the curve. Thus there is no force at this point, and so it is the equilibrium point. Another way to see that it is the equilibrium point is that it takes work to move away from $d$ in either direction. When the two oxygen atoms have settled down, so that no more energy can be liberated from the force between them, they are in the lowest energy state, and they will be at this separation $d$. This is the way an oxygen molecule looks when it is cold. When we heat it up, the atoms shake and move farther apart, and we can in fact break them apart, but to do so takes a certain amount of work or energy, which is the potential energy difference between $r = d$ and $r=\infty$. When we try to push the atoms very close together the energy goes up very rapidly, because they repel each other.

The reason we bring this out is that the idea of *force* is not
particularly suitable for quantum mechanics; there the idea of
*energy* is most natural. We find that although forces and
velocities “dissolve” and disappear when we consider the more
advanced forces between nuclear matter and between molecules and so
on, the energy concept remains. Therefore we find curves of potential
energy in quantum mechanics books, but very rarely do we ever see a
curve for the force between two molecules, because by that time people
who are doing analyses are thinking in terms of energy rather than of
force.

Next we note that if several conservative forces are acting on an object at the same time, then the potential energy of the object is the sum of the potential energies from each of the separate forces. This is the same proposition that we mentioned before, because if the force can be represented as a vector sum of forces, then the work done by the total force is the sum of the works done by the partial forces, and it can therefore be analyzed as changes in the potential energies of each of them separately. Thus the total potential energy is the sum of all the little pieces.

We could generalize this to the case of a system of many objects interacting with one another, like Jupiter, Saturn, Uranus, etc., or oxygen, nitrogen, carbon, etc., which are acting with respect to one another in pairs due to forces all of which are conservative. In these circumstances the kinetic energy in the entire system is simply the sum of the kinetic energies of all of the particular atoms or planets or whatever, and the potential energy of the system is the sum, over the pairs of particles, of the potential energy of mutual interaction of a single pair, as though the others were not there. (This is really not true for molecular forces, and the formula is somewhat more complicated; it certainly is true for Newtonian gravitation, and it is true as an approximation for molecular forces. For molecular forces there is a potential energy, but it is sometimes a more complicated function of the positions of the atoms than simply a sum of terms from pairs.) In the special case of gravity, therefore, the potential energy is the sum, over all the pairs $i$ and $j$, of $-Gm_im_j/r_{ij}$, as was indicated in Eq. (13.14). Equation (13.14) expressed mathematically the following proposition: that the total kinetic energy plus the total potential energy does not change with time. As the various planets wheel about, and turn and twist and so on, if we calculate the total kinetic energy and the total potential energy we find that the total remains constant.

### 14–4Nonconservative forces

We have spent a considerable time discussing conservative forces; what
about nonconservative forces? We shall take a deeper view of this than
is usual, and state that there are no nonconservative forces! As a
matter of fact, all the fundamental forces in nature appear to be
conservative. This is not a consequence of Newton’s
laws. In fact, so far as
Newton himself knew, the forces
could be nonconservative, as friction apparently is. When we say
friction *apparently* is, we are taking a modern view, in which it
has been discovered that all the deep forces, the forces between the
particles at the most fundamental level, are conservative.

If, for example, we analyze a system like that great globular star cluster that we saw a picture of, with the thousands of stars all interacting, then the formula for the total potential energy is simply one term plus another term, etc., summed over all pairs of stars, and the kinetic energy is the sum of the kinetic energies of all the individual stars. But the globular cluster as a whole is drifting in space too, and, if we were far enough away from it and did not see the details, could be thought of as a single object. Then if forces were applied to it, some of those forces might end up driving it forward as a whole, and we would see the center of the whole thing moving. On the other hand, some of the forces can be, so to speak, “wasted” in increasing the kinetic or potential energy of the “particles” inside. Let us suppose, for instance, that the action of these forces expands the whole cluster and makes the particles move faster. The total energy of the whole thing is really conserved, but seen from the outside with our crude eyes which cannot see the confusion of motions inside, and just thinking of the kinetic energy of the motion of the whole object as though it were a single particle, it would appear that energy is not conserved, but this is due to a lack of appreciation of what it is that we see. And that, it turns out, is the case: the total energy of the world, kinetic plus potential, is a constant when we look closely enough.

When we study matter in the finest detail at the atomic level, it is not
always *easy* to separate the total energy of a thing into two
parts, kinetic energy and potential energy, and such separation is not
always necessary. It is *almost* always possible to do it, so let
us say that it *is* always possible, and that the
potential-plus-kinetic energy of the world is constant. Thus the total
potential-plus-kinetic energy inside the whole world is constant, and if
the “world” is a piece of isolated material, the energy is constant if
there are no external forces. But as we have seen, some of the kinetic
and potential energy of a thing may be internal, for instance the
internal molecular motions, in the sense that we do not notice it. We
know that in a glass of water everything is jiggling around, all the
parts are moving all the time, so there is a certain kinetic energy
inside, which we ordinarily may not pay any attention to. We do not
notice the motion of the atoms, which produces heat, and so we do not
call it kinetic energy, but heat is primarily kinetic energy. Internal
potential energy may also be in the form, for instance, of chemical
energy: when we burn gasoline energy is liberated because the potential
energies of the atoms in the new atomic arrangement are lower than in
the old arrangement. It is not strictly possible to treat heat as being
pure kinetic energy, for a little of the potential gets in, and vice
versa for chemical energy, so we put the two together and say that the
total kinetic and potential energy inside an object is partly heat,
partly chemical energy, and so on. Anyway, all these different forms of
internal energy are sometimes considered as “lost” energy in the sense
described above; this will be made clearer when we study thermodynamics.

As another example, when friction is present it is not true that kinetic energy is lost, even though a sliding object stops and the kinetic energy seems to be lost. The kinetic energy is not lost because, of course, the atoms inside are jiggling with a greater amount of kinetic energy than before, and although we cannot see that, we can measure it by determining the temperature. Of course if we disregard the heat energy, then the conservation of energy theorem will appear to be false.

Another situation in which energy conservation appears to be false is when we study only part of a system. Naturally, the conservation of energy theorem will appear not to be true if something is interacting with something else on the outside and we neglect to take that interaction into account.

In classical physics potential energy involved only gravitation and electricity, but now we have nuclear energy and other energies also. Light, for example, would involve a new form of energy in the classical theory, but we can also, if we want to, imagine that the energy of light is the kinetic energy of a photon, and then our formula (14.2) would still be right.

### 14–5Potentials and fields

We shall now discuss a few of the ideas associated with potential
energy and with the idea of a *field*. Suppose we have two large
objects $A$ and $B$ and a third very small one which is attracted
gravitationally by the two, with some resultant force $\FLPF$. We have
already noted in Chapter 12 that the gravitational force
on a particle can be written as its mass, $m$, times another vector,
$\FLPC$, which is dependent only upon the *position* of the
particle:
\begin{equation*}
\FLPF=m\FLPC.
\end{equation*}
We can analyze gravitation, then, by imagining that there is a certain
vector $\FLPC$ at every position in space which “acts” upon a mass
which we may place there, but which is there itself whether we
actually supply a mass for it to “act” on or not. $\FLPC$ has three
components, and each of those components is a function of $(x,y,z)$, a
function of position in space. Such a thing we call a *field*,
and we say that the objects $A$ and $B$ *generate* the field,
i.e., they “make” the vector $\FLPC$. When an object is put in a
field, the force on it is equal to its mass times the value of the
field vector at the point where the object is put.

We can also do the same with the potential energy. Since the potential
energy, the integral of $(-\textbf{force})\cdot(d\FLPs)$ can be
written as $m$ times the integral
of $(-\textbf{field})\cdot(d\FLPs)$, a mere change of scale, we see
that the potential energy $U(x,y,z)$ of an object located at a
point $(x,y,z)$ in space can be written as $m$ times another function
which we may call the *potential* $\Psi$. The
integral $\int\FLPC\cdot d\FLPs=-\Psi$, just as $\int\FLPF\cdot
d\FLPs=-U$; there is only a scale factor between the two:
\begin{equation}
\label{Eq:I:14:7}
U=-\int\FLPF\cdot d\FLPs=-m\int\FLPC\cdot d\FLPs=m\Psi.
\end{equation}
\begin{align}
\label{Eq:I:14:7}
U&=-\int\FLPF\cdot d\FLPs\notag\\
&=-m\int\FLPC\cdot d\FLPs=m\Psi.
\end{align}
By having this function $\Psi(x,y,z)$ at every point in space, we can
immediately calculate the potential energy of an object at any point
in space, namely, $U(x, y, z) = m\Psi(x, y, z)$—rather a trivial
business, it seems. But it is not really trivial, because it is
sometimes much nicer to describe the field by giving the value
of $\Psi$ everywhere in space instead of having to give $\FLPC$. Instead
of having to write three complicated components of a vector function,
we can give instead the scalar function $\Psi$. Furthermore, it is
much easier to calculate $\Psi$ than any given component of $\FLPC$
when the field is produced by a number of masses, for since the
potential is a scalar we merely add, without worrying about
direction. Also, the field $\FLPC$ can be recovered easily
from $\Psi$, as we shall shortly see. Suppose we have point masses $m_1$,
$m_2$, … at the points $1$, $2$, … and we wish to know the
potential $\Psi$ at some arbitrary point $p$. This is simply the sum
of the potentials at $p$ due to the individual masses taken one by
one:
\begin{equation}
\label{Eq:I:14:8}
\Psi(p)=\sum_i-\frac{Gm_i}{r_{ip}},\quad
i=\text{$1$, $2$, $\ldots$}
\end{equation}

In the last chapter we used this formula, that the potential is the
sum of the potentials from all the different objects, to calculate the
potential due to a spherical shell of matter by adding the
contributions to the potential at a point from all parts of the
shell. The result of this calculation is shown graphically in
Fig. 14–4. It is negative, having the value zero at $r =
\infty$ and varying as $1/r$ down to the radius $a$, and then is
constant inside the shell. Outside the shell the potential is $-Gm/r$,
where $m$ is the mass of the shell, which is exactly the same as it
would have been if all the mass were located at the center. But it is
not *everywhere* exactly the same, for inside the shell the
potential turns out to be $-Gm/a$, and is a constant! *When the
potential is constant, there is no field*, or when the potential energy
is constant there is no force, because if we move an object from one
place to another anywhere inside the sphere the work done by the force
is exactly zero. Why? Because the work done in moving the object from
one place to the other is equal to minus the change in the potential
energy (or, the corresponding field integral is the change of the
potential). But the potential energy is the *same* at any two
points inside, so there is zero change in potential energy, and
therefore no work is done in going between any two points inside the
shell. The only way the work can be zero for all directions of
displacement is that there is no force at all.

This gives us a clue as to how we can obtain the force or the field,
given the potential energy. Let us suppose that the potential energy
of an object is known at the position $(x,y,z)$ and we want to know
what the force on the object is. It will not do to know the potential
at only this one point, as we shall see; it requires knowledge of the
potential at neighboring points as well. Why? How can we calculate the
$x$-component of the force? (If we can do this, of course, we can also
find the $y$- and $z$-components, and we will then know the whole
force.) Now, if we were to move the object a small distance $\Delta
x$, the work done by the force on the object would be the
$x$-component of the force times $\Delta x$, if $\Delta x$ is
sufficiently small, and this should equal the change in potential
energy in going from one point to the other:
\begin{equation}
\label{Eq:I:14:9}
\Delta W=-\Delta U=F_x\,\Delta x.
\end{equation}
We have merely used the formula $\int\FLPF\cdot d\FLPs=-\Delta U$, but
for a *very short* path. Now we divide by $\Delta x$ and so find
that the force is
\begin{equation}
\label{Eq:I:14:10}
F_x=-\Delta U/\Delta x.
\end{equation}

Of course this is not exact. What we really want is the limit
of (14.10) as $\Delta x$ gets smaller and smaller, because it
is only *exactly* right in the limit of infinitesimal $\Delta x$.
This we recognize as the derivative of $U$ with respect to $x$, and we
would be inclined, therefore, to write $-dU/dx$. But $U$ depends on
$x$, $y$, and $z$, and the mathematicians have invented a different symbol to
remind us to be very careful when we are differentiating such a
function, so as to remember that we are considering that *only $x$
varies*, and $y$ and $z$ do not vary. Instead of a $d$ they simply make
a “backwards $6$,” or $\partial$. (A $\partial$ should have been used
in the beginning of calculus because we always want to cancel that $d$,
but we never want to cancel a $\partial$!) So they write $\ddpl{U}{x}$,
and furthermore, in moments of duress, if they want to be *very*
careful, they put a line beside it with a little $yz$ at the bottom
($\ddpl{U}{x}|_{yz}$), which means “Take the derivative of $U$ with
respect to $x$, keeping $y$ and $z$ constant.” Most often we leave out
the remark about what is kept constant because it is usually evident
from the context, so we usually do not use the line with the $y$
and $z$. However, *always* use a $\partial$ instead of a $d$ as a
warning that it is a derivative with some other variables kept constant.
This is called a *partial derivative*; it is a derivative in
which we vary only $x$.

Therefore, we find that the force in the $x$-direction is minus the
partial derivative of $U$ with respect to $x$:
\begin{equation}
\label{Eq:I:14:11}
F_x=-\ddpl{U}{x}.
\end{equation}
In a similar way, the force in the $y$-direction can be found by
differentiating $U$ with respect to $y$, keeping $x$ and $z$ constant,
and the third component, of course, is the derivative with respect
to $z$, keeping $y$ and $x$ constant:
\begin{equation}
\label{Eq:I:14:12}
F_y=-\ddpl{U}{y},\quad
F_z=-\ddpl{U}{z}.
\end{equation}
This is the way to get from the potential energy to the force. We get
the *field* from the *potential* in exactly the same way:
\begin{equation}
\label{Eq:I:14:13}
C_x=-\ddpl{\Psi}{x},\quad
C_y=-\ddpl{\Psi}{y},\quad
C_z=-\ddpl{\Psi}{z}.
\end{equation}
\begin{equation}
\begin{gathered}
C_x=-\ddpl{\Psi}{x},\\[.5ex]
C_y=-\ddpl{\Psi}{y},\\[.5ex]
C_z=-\ddpl{\Psi}{z}.
\end{gathered}
\label{Eq:I:14:13}
\end{equation}

Incidentally, we shall mention here another notation, which we shall
not actually use for quite a while: Since $\FLPC$ is a vector and has
$x$-, $y$-, and $z$-components, the symbolized $\ddpl{}{x}$, $\ddpl{}{y}$,
and $\ddpl{}{z}$ which produce the $x$-, $y$-,
and $z$-components are something like vectors. The mathematicians have
invented a glorious new symbol, $\FLPnabla$, called “grad” or
“gradient”, which is not a quantity but an *operator* that
makes a vector from a scalar. It has the following “components”: The
$x$-component of this “grad” is $\ddpl{}{x}$, the $y$-component
is $\ddpl{}{y}$, and the $z$-component is $\ddpl{}{z}$, and then we have
the fun of writing our formulas this way:
\begin{equation}
\label{Eq:I:14:14}
\FLPF=-\FLPgrad{U},\quad
\FLPC=-\FLPgrad{\Psi}.
\end{equation}
Using $\FLPnabla$ gives us a quick way of testing whether we have a
real vector equation or not, but actually Eqs. (14.14)
mean precisely the same as Eqs. (14.11), (14.12)
and (14.13); it is just another way of writing them, and
since we do not want to write three equations every time, we just
write $\FLPgrad{U}$ instead.

One more example of fields and potentials has to do with the
electrical case. In the case of electricity the force on a stationary
object is the charge times the electric field: $\FLPF = q\FLPE$. (In
general, of course, the $x$-component of force in an electrical
problem has also a part which depends on the magnetic field. It is
easy to show from Eq. (12.11) that the force on a
particle due to magnetic fields is always at right angles to its
velocity, and also at right angles to the field. Since the force due
to magnetism on a moving charge is at right angles to the velocity,
*no work is done* by the magnetism on the moving charge because
the motion is at right angles to the force. Therefore, in calculating
theorems of kinetic energy in electric and magnetic fields we can
disregard the contribution from the magnetic field, since it does not
change the kinetic energy.) We suppose that there is only an electric
field. Then we can calculate the energy, or work done, in the same way
as for gravity, and calculate a quantity $\phi$ which is minus the
integral of $\FLPE\cdot d\FLPs$, from the arbitrary fixed point to the
point where we make the calculation, and then the potential energy in
an electric field is just charge times this quantity $\phi$:
\begin{gather*}
\phi(\FLPr)=-\int\FLPE\cdot d\FLPs,\\[1ex]
U=q\phi.
\end{gather*}

Let us take, as an example, the case of two parallel metal plates, each
with a surface charge of $\pm\sigma$ per unit area. This is called a
parallel-plate capacitor. We found previously
that there is zero force outside the plates and that there is a constant
electric field between them, directed from $+$ to $-$ and of
magnitude $\sigma/\epsO$ (Fig. 14–5). We would like to know
how much work would be done in carrying a charge from one plate to the
other. The work would be the $(\textbf{force})\cdot(d\FLPs)$
integral, which can be written as charge times the potential value at
plate $1$ minus that at plate $2$:
\begin{equation*}
W=\int_1^2\FLPF\cdot d\FLPs=q(\phi_1-\phi_2).
\end{equation*}
We can actually work out the integral because the force is constant,
and if we call the separation of the plates $d$, then the integral is
easy:
\begin{equation*}
\int_1^2\FLPF\cdot d\FLPs=\frac{q\sigma}{\epsO}\int_1^2dx=
\frac{q\sigma d}{\epsO}.
\end{equation*}
The difference in potential, $\Delta\phi=\sigma d/\epsO$, is called
the *voltage difference*, and $\phi$ is measured in volts. When
we say a pair of plates is charged to a certain voltage, what we mean
is that the difference in electrical potential of the two plates is
so-and-so many volts. For a capacitor made of two parallel
plates carrying a surface charge $\pm\sigma$, the voltage, or difference
in potential, of the pair of plates is $\sigma d/\epsO$.