This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 14 Work and Potential Energy (conclusion) 14—1 Work 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 worth while 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 cir
cumstance 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 smoothlooking 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 fresh
man 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
unitel,” but is a different idea. Physical work is expressed as f F  ds, 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 As, then the work done in moving the constant force through that
distance is only the component of force along As times As. The rule is “force times
distance,” but we really mean only the component of force in the direction of the
displacement times As 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 As is resolved into components, in other
words, if the actual displacement is As and we want to consider it effectively as a
component of displacement Ax in the xdirection, Ay in the ydirection, and A2 14—1 14—1 Work
14—2 Constrained motion
14—3 Conservative forces 14—4 Nonconservative forces
14—5 Potentials and ﬁelds in the zdirection, 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 2. The work done in going along x involves only that component
of force, namely Fm, and so on, so the work is FI Ax + F1, Ay + F, A2. When
the force is not constant, and we have a complicated curved motion, then we must
resolve the path into a lot of little As’s, add the work done in carrying the object
along each As, and take the limit as As goes to zero. This is the meaning of the
“line integral.” Everything we have just said is contained in the formula W = jF  ds. 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 \ex
ample, according to the physical deﬁnition of work, if one holds a hundredpound
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 deﬁni
tion 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 2 A(v2)= EFAs 14—2 i
g
l
l
, 14—2 Constrained 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 xcomponent
of all forces and the ycomponent 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—3 Conservative 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 conserva
tive 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 goingfrom l 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, ‘f course, the work depends only upon the positions of 1 and 2. To demonstjate this idea, consider the following. We take a “standard”
point P, at an arbitrary location (Fig. 14—2). Then, the work lineintegral 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 con
servative 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 pomt 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 diﬁerent 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 (x2, y2, 22), we shall write U(2), 14—3 FORCE OF
CONSTRAINT gong: OF GRAVlTY DIRECTION OF MOTION Fig. 14—1. Forces acting on a sliding body (no friction). Fig. 142. Possible paths
two points in a ﬁeld of force. between as an abbreviation for U(xg, y2, 22). 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 ds’s. That is, the work done in going from 1 to P is minus the work done
in going from the point P to l: [IPFds =fPlF(—ds) = —/PlFds. Thus the work done in going from P to l is — U(l), and from P to 2 the work is
— U(2). Therefore the integral from 1 to 2 is equal to — U(2) plus [— U(l) back
wards], or +U(l) — U(2): U(l) = —fPIFds, U(2) = —/:Fds, leds = U(l) — U(2). (14.1)
The quantity U(l) — U(2) 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(l).
If it is located at posrtion 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 demon
strate) 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 followmg 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: T + U = constant. (14.2) 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 U(2) = mgz, (14.3) and the point P which corresponds to zero potential energy happens to be any
point in the plane 2 = 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
—mg6. It makes no diﬂerence, because only differences in potential energy count.
The energy needed to compress a linear spring a distance x from an equ11ibrium point is
U(X) = %kx2, (144) 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
U(r) = —GMm/r. (14.5) 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: U(r) = q1q2/47reor. (14.6) 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 14—4 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 %m2)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 ZGM/a. But GM/a2 is what we call the acceleration of
gravity, g. Thus 02 = 2ga. At what speed must a satellite travel in order to keep going around the earth?
We worked this out long ago and found that 212 = GM /a. Therefore to go
away from the earth, we need \/§ 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 sulﬁciently near
we reach a point (1 where there is a minimum of potential energy. The minimum of
potential energy at r = (1 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 = 00. 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 14—5 \—U(r)~ l/rs (IF r >>a) Fig. 14—3. The potential energy be
tween two atoms as a function of the
distance between them. l
i
l 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 gravita
tion, 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 1' and j,
of —Gm,m,/r,,, 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—4 Nonconservative 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
pluskinetic energy of the world is constant. Thus the total potentialpluskinetic
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 14—6 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—5 Potentials 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 F. We have already noted in Chapter 12 that the gravitational force on a
particle can be written as its mass, m, times another vector, C, which is dependent
only upon the position of the particle: F=mC. We can analyze gravitation, then, by imagining that there is a certain vector C 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.
C 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 C. 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 (force)  (ds) can be written as in times the integral of the (ﬁeld)  (ds),
a mere change of scale, we see that the potential energy U(x, y, z) of an object
located at a point (x, y, 2,) in space can be written as In times another function which
we may call the potential \I/. The integral fC  ds = —\II, just as fF  ds = — U;
there is only a scale factor between the two: U=—/Fds=—m/cds=mxiz. (14.7) By having this function \Il(x, y, 2) 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\I/(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 14—7 ——w~r————a—uq ._.'_ 44. Mr) =CONSTANT = —Gm/a Fig. 14—4. Potential due to a spher—
ical shell of radius a. of \II everywhere in space instead of having to give C. Instead of having to write
three complicated components of a vector function, we can give instead the scalar
function \I/. Furthermore, it is much easier to calculate \II than any given component
of C 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 C can be
recovered easily from ‘11, as we shall shortly see. Suppose we have point masses
m1, m2, . . . at the points 1, 2, . . . and we wish to know the potential \I/ at some
arbitrary point p. This is simply the sum of the potentials at P due to the individual
masses taken one by one: \I/(p) = Z — Gmi, 1': 1,2,... (14.8) rip In the last chapter we used this formula, that the potential is the sum of the
porentials 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 = 00 and varying as l/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 xcomponent of the force? (If we can do this, of course, we can also
ﬁnd the y and zcomponents, and we will then know the whole force.) Now, if
we were to move the object a small distance Ax, the work done by the force on the
object would be the xcomponent of the force times Ax, if Ax is sufficiently small,
and this should equal the change in potential energy in going from one point to
the other: AW = —AU = Fl Ax. (14.9) We have merely used the formula IF  ds = —AU, but for a very short path.
Now we divide by Ax and so ﬁnd that the force is Fr = —AU/Ax. (14.10) Of course this is not exact. What we really want is the limit of (14.10) as Ax
gets smaller and smaller, because it is only exactly right in the limit of inﬁnitesimal
Ax. 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 con
sidering that only x varies, and y and 2 do not vary. Instead of a d they simply
make a “backwards 6,” or a. (A a should have been used in the beginning of
calculus because we always want to cancel that d, but we never want to cancel a 6!)
So they write aU/ax, 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 (aU/axlw), 14—8 which means “Take the derivative of U with respect to x, keeping y and zconstant.”
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
2. However, always use a 6 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 deriva
tive in which we vary only x. Therefore, we ﬁnd that the force in the xdirection is minus the partial deriva
tive of U with respect to x: F, = —aU/6x. (14.11) In a similar way, the force in the ydirection 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: F, = —6U/6y, F, = —aU/az. (14.12) 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: C, = —6\II/6x, C, = —a\I//ay, C, = —6\I//az. (14.13) Incidentally, we shall mention here another notation, which we shall not
actually use for quite a while: Since C is a vector and has x, y, and zcomponents,
the symbolized a/ax, a/ay, and 6/62 which produce the x, y, and zcompo
nents are something like vectors. The mathematicians have invented a glorious
new symbol, V, called “grad” or “gradient” which is not a quantity but an operator
which makes a vector from a scalar. It has the following “components”: The
xcomponent of this “grad” is 6/6x, the y—component is a/ay, and the zcomponent
is 6/62, and then we have the fun of writing our formulas this way: F = —VU, C = —v\1/. (14.14) Using V gives us a quick way of testing whether we have a real vector equation or
not, but actually Eq. (14.14) means precisely the same as Eqs. (14.11) and (14.12);
it is just another way of writing them, and since we do not want to write three
equations every time, we just write VU 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: F = qE. (In general, of course, the xcomponent of force in an
electrical problem has also a part which depends on the magnetic ﬁeld. It is easy
to show from Eq. (12.10) 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 (15 which is minus the integral of E  ds,
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 <1): W) = /E'ds,
U = qd>. Let us take, as an example, the case of two parallel metal plates, each with a
surface charge of to per unit area. This is called a parallelplate 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
o/eo (Fig. 14—5). We would like to know how much work would be done in 149 Fig. 1 4—5.
plates. Field between parallel carrying a charge from one plate to the other. The work would be the (force) ~ (ds)
integral, which can be written as charge times the potential value at plate 1 minus
that at plate 2: 2
W=f Fds = qwl — 4»).
1 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: 60 1 50 The difference in potential, A4» = ad/eo, is called the voltage dzﬂerence, and ¢
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
andso many volts. For a capacitor made of two parallel plates carrying a surface
charge iv, the voltage, or diﬁerence in potential, of the pair of plates is (rd/60. 14—10 ...
View
Full Document
 Spring '09
 LeeKinohara
 Physics, Energy, Potential Energy, Work

Click to edit the document details