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Unformatted text preview: I2 Electrostatic Analogs 121 The same equations have the same solutions The total amount of information which has been acquired about the physical
world since the beginning of scientiﬁc progress is enormous, and it seems almost
impossible that any one person could know a reasonable fraction of it. But it is
actually quite possible for a physicist to retain a broad knowledge of the physical
world rather than to become a specialist in some narrow area. The reasons for
this are threefold: First, there are great principles which apply to all the different
kinds of phenomena—such as the principles of the conservation of energy and of
angular momentum. A thorough understanding of such principles gives an under—
standing of a great deal all at once. Second, there is the fact that many compli
cated phenomena, such as the behavior of solids under compression, really
basically depend on electrical and quantummechanical forces, so that if one
understands the fundamental laws of electricity and quantum mechanics, there is
at least some possibility of understanding many of the phenomena that occur
in complex situations. Finally, there is a most remarkable coincidence: The
equations for many diﬂerent physical situations have exactly the same appearance.
Of course, the symbols may be different—one letter is substituted for another—
but the mathematical form of the equations is the same. This means that having
studied one subject, we immediately have a great deal of direct and precise
knowledge about the solutions of the equations of another. We are now ﬁnished with the subject of electrostatics, and will soon go on to
study magnetism and electrodynamics. But before doing so, we would like to
show that while learning electrostatics we have simultaneously learned about a
large number of other subjects. We will ﬁnd that the equations of electrostatics
appear in several other places in physics. By a direct translation of the solutions
(of course the same mathematical equations must have the same solutions) it is
possible to solve problems in other ﬁelds with the same ease—or with the same
difﬁculty—as in electrostatics. The equations of electrostatics, we know, are v  (KE) = “I”, (12.1)
‘0
v x E = o. (12.2) (We take the equations of electrostatics with dielectrics so as to have the most
general situation.) The same physics can be expressed in another mathematical
form: E = — V¢, (12.3) V  (K W) = — “2:” (12.4) Now the point is that there are many physics problems whose mathematical
equations have the same form. There is a potential (11>) whose gradient multiplied
by a scalar function (K) has a divergence equal to another scalar function (—p/60).
Whatever we know about electrostatics can immediately be carried over into
that other subject, and vice versa. (It works both ways, of course—if the other
subject has some particular characteristics that are known, then we can apply
that knowledge to the corresponding electrostatic problem.) We want to consider
a series of examples from different subjects that produce equations of this form. 12—1 12—1 The same equations have the
same solutions 12—2 The ﬂow of heat; a point
source near an inﬁnite plane boundary
12—3 The stretched membrane 12—4 The diﬂ‘usion of neutrons; a
uniform spherical source in a
homogeneous medium 12—5 Irrotational ﬂuid ﬂow; the
ﬂow past a sphere 12—6 Illumination; the uniform
lighting of a plane 127 The “underlying unity” of
nature 122 The ﬂow of heat; a point source near an inﬁnite plane boundary We have discussed one example earlier (Section 3—4)—the ﬂew of heat.
Imagine a block of material, which need not be homogeneous but may consist of
different materials at different places, in which the temperature varies from point
to point. As a consequence of these temperature variations there is a ﬂow of heat,
which can be represented by the vector h. It represents the amount of heat energy
which ﬂows per unit time through a unit area perpendicular to the ﬂow. The di
vergence of h represents the rate per unit volume at which heat is leaving a region: V  h = rate of heat out per unit volume. (We could, of course, write the equation in integral form—just as we did in electro
statics with Gauss’ law—which would say that the ﬂux through a surface is equal
to the rate of change of heat energy inside the material. We will not bother to
translate the equations back and forth between the diﬂerential and the integral
forms, because it goes exactly the same as in electrostatics.) The rate at which heat is generated or absorbed at various places depends, of
course, on the problem. Suppose, for example, that there is a source of heat inside
the material (perhaps a radioactive source, or a resistor heated by an electrical
current). Let us call s the heat energy produced per unit volume per second by
this source. There may also be losses (or gains) of thermal energy to other internal
energies in the volume. If u is the internal energy per unit volume, —du/dt will
also be a “source” of heat energy. We have, then, du
V h—s—E (12.5)
We are not going to discuss just now the complete equation in which things
change with time, because we are making an analogy to electrostatics, where no
thing depends on the time. We will consider only steady heatﬂow problems, in which constant sources have produced an equilibrium state. In these cases,
V ' h = s. (12.6) It is, of course, necessary to have another equation, which describes how the
heat ﬂows at various places. In many materials the heat current is approximately
proportional to the rate of change of the temperature with position: the larger the
temperature diﬂerence, the more the heat current. As we have seen, the vector
heat current is proportional to the temperature gradient. The constant of pro
portionality K, a property of the material, is called the thermal conductivity. h = ——KVT. (12.7) If the properties of the material vary from place to place, then K = K(x, y, z), a
function of position. [Equation (12.7) is not as fundamental as (12.5), which
expresses the conservation of heat energy, since the former depends upon a special
property of the substance] If now we substitute Eq. (12.7) into Eq. (12.6) we have V  (K VT) = —s, (12.8) which has exactly the same form as (12.4). Steady heatflow problems and electro
static problems are the same. The heat ﬂow vector h corresponds to E, and the
temperature T corresponds to :1». We have already noticed that a point heat source
produces a temperature ﬁeld which varies as l/r and a heat ﬂow which varies as
1/r2. This is nothing more than a translation of the statements from electrostatics
that a point charge generates a potential which varies as l/r and an electric ﬁeld
which varies as 1/r2. We can, in general, solve static heat problems as easily as
we can solve electrostatic problems. Consider a simple example. Suppose that we have a cylinder of radius a at the
temperature T1, maintained by the generation of heat in the cylinder. (It could be,
for example, a wire carrying a current, or a pipe with steam condensing inside.) 1 22 The cylinder is covered with a concentric sheath of insulating material which has a
conductivity K. Say the outside radius of the insulation is b and the outside is
kept at temperature T2 (Fig. 12—13). We want to ﬁnd out at what rate heat will
be lost by the wire, or steampipe, or whatever it is in the center. Let the total
amount of heat lost from a length L of the pipe be called G—which is what we are
trying to ﬁnd. How can we solve this problem? We have the differential equations, but since
these are the same as those of electrostatics, we have really already solved the
mathematical problem. The analogous problem is that of a conductor of radius a
at the potential (#1, separated from another conductor of radius b at the potential
4:2, with a concentric layer of dielectric material in between, as drawn in Fig.
12—1 (b). Now since the heat ﬂow h corresponds to the electric ﬁeld E, the quantity
G that we want to ﬁnd corresponds to the ﬂux of the electric ﬁeld from a unit
length (in other words, to the electric charge per unit length over so). We have
solved the electrostatic problem by using Gauss’ law. We follow the same pro
cedure for our heat—ﬂow problem. From the symmetry of the situation, we know that h depends only on the
distance from the center. So we enclose the pipe in a gaussian cylinder of length
L and radius r. From Gauss’ law, we know that the heat ﬂow h multiplied by
the area 21rrL of the surface must be equal to the total amount of heat generated
in51de, which is what we are calling G: G
21rth — G or h — ZTrL (12.9)
The heat ﬂow is proportional to the temperature gradient:
h = —K VT,
or, in this case, the magnitude of h is
dT
h — —K $
This, together with (12.9), gives
dT G
H7 _ _ 27rKLr' (12.10)
Integrating from r = a to r = b, we get
G b
Solving for G, we ﬁnd
G = —' T2) ' 1n (b/a) This result corresponds exactly to the result for the charge on a cylindrical conden
ser:
Q = 27760L(<i>1 — ¢2).
1n (b/a) The problems are the same, and they have the same solutions. From our knowledge
of electrostatics, we also know how much heat is lost by an insulated pipe. Let’s consider another example of heat ﬂow. Suppose we wish to know the
heat ﬂow in the neighborhood of a point source of heat located a little way beneath
the surface of the earth, or near the surface of a large metal block. The localized
heat source might be an atomic bomb that was set off underground, leaving an
intense source of heat, or it might correspond to a small radioactive source inside
a block of iron—there are numerous possibilities. We will treat the idealized problem of a point heat source of strength G at the
distance a beneath the surface of an inﬁnite block of uniform material whose
thermal conductivity is K. And we will neglect the thermal conductivity of the 1 2—3 (b) Fig. 12—]. (0) Heat ﬂow in a cylin
drical geometry. (b) The corresponding
electrical problem. SURFACE
TEMPERATURE p Fig. 12—2. The heat ﬂow and iso
thermals near a point heat source at the
distance 0 below the surface of a good
thermal conductor. An image source is
shown outside the material. air outside the material. We want to determine the distribution of the temperature
on the surface of the block. How hot is it right above the source and at various
places on the surface of the block? How shall we solve it? It is like an electrostatic problem with two materials
with different dielectric coefﬁcients K on opposite sides of a plane boundary. Aha!
Perhaps it is the analog of a point charge near the boundary between a dielectric
and a conductor, or something similar. Let’s see what the situation is near the
surface. The physical condition is that the normal component of h on the surface
is zero, since we have assumed there is no heat ﬂow out of the block. We should
ask: In what electrostatic problem do we have the condition that the normal
component of the electric ﬁeld E (which is the analog of h) is zero at a surface?
There is none! That is one of the things that we have to watch out for. For physical reasons,
there may be certain restrictions in the kinds of mathematical conditions which
arise in any one subject. So if we have analyzed the differential equation only for
certain limited cases, we may have missed some kinds of solutions that can occur
in other physical situations. For example, there is no material with a dielectric
constant of zero, whereas a vacuum does have zero thermal conductivity. So there
is no electrostatic analogy for a perfect heat insulator. We can, however, still use
the same methods. We can try to imagine what would happen if the dielectric
constant were zero. (Of course, the dielectric constant is never zero in any real
Situation. But we might have a case in which there is a material with a very high
dielectric constant, so that we could neglect the dielectric constant of the air out
side.) How shall we ﬁnd an electric ﬁeld that has no component perpendicular to
the surface? That is, one which is always tangent at the surface? You will notice
that our problem is opposite to the one of a point charge near a plane conductor.
There we wanted the ﬁeld to be perpendicular to the surface, because the conductor
was all at the same potential. In the electrical problem, we invented a solution
by imagining a point charge behind the conducting plate. We can use the same
idea again. We try to pick an “image source” that will automatically make the
normal component of the ﬁeld zero at the surface. The solution is shown in
Fig. 12—2. An image source of the same sign and the same strength placed at the
distance a above the surface will cause the ﬁeld to be always horizontal at the sur
face. The normal components of the two sources cancel out. Thus our heat ﬂow problem is solved. The temperature everywhere is the
same, by direct analogy, as the potential due to two equal point charges! The
temperature T at the distance r from a single point source G in an inﬁnite medium is __ G _
_ 47rKr T (12.13) (This, of course, is just the analog of 4: = q/47reor.) The temperature for a point
source, together with its image source, is G G T = 41rKr1 + 47rKr2' (12.14) This formula gives us the temperature everywhere in the block. Several isothermal
surfaces are shown in Fig. 12—2. Also shown are lines of h, which can be obtained
from h = —K VT. We originally asked for the temperature distribution on the surface. For a
point on the surface at the distance p from the axis, r1 = r2 = \/p2 + 02, so T(surface) = Zi—K —72:——2. (12.15)
Vp a This function is also shown in the ﬁgure. The temperature is, naturally, higher
right above the source than it is farther away. This is the kind of problem that
geophysicists often need to solve. We now see that it is the same kind of thing we
have already been solving for electricity. 12—4 12—3 The stretched membrane Now let us consider a completely different physical situation which, never
theless, gives the same equations again. Consider a thin rubber sheet—a membrane
——which has been stretched over a large horizontal frame (like a drumhead).
Suppose now that the membrane is pushed up in one place and down in another;
as shown in Fig. 12—3. Can we describe the shape of the surface? We will show
how the problem can be solved when the deﬂections of the membrane are not too
large. There are forces in the sheet because it is stretched. If we were to make a
small cut anywhere, the two sides of the cut would pull apart (see Fig. 12—4). So
there is a surface tension in the sheet, analogous to the onedimensional tension
in a stretched string. We deﬁne the magnitude of the surface tension 1 as the force
per unit length which will just hold together the two sides of a out such as one of
those shown in Fig. 124. Suppose now that we look at a vertical cross section of the membrane. It
will appear as a curve, like the one in Fig. 12—5. Let u be the vertical displacement
of the membrane from its normal position, and x and y the coordinates in the
horizontal plane. (The cross section shown is parallel to the xaxis.) Consider a little piece of the surface of length Ax and width Ay. There will be
forces on the piece from the surface tension along each edge. The force along
edge 1 of the ﬁgure will be 71 Ay, directed tangent to the surface—that is, at the
angle 01 from the horizontal. Along edge 2, the force will be 12 Ay at the angle 02.
(There will be similar forces on the other two edges of the piece, but we will forget
them for the moment.) The net upward force on the piece from edges 1 and 2 is AF = r2 Ay sin 02 — 7'1 Ay sin 61. We will limit our considerations to small distortions of the membrane, i.e., to
small slopes: we can then replace sin 0 by tan 0, which can be written as au/ax. The force is then
au Bu
AF — ‘ ’1 A” The quantity in brackets can be equally well written (for small Ax) as
.6. T 92) 6x 6x ’ e au
AF — 5(1' 5) Ax Ay. then There will be another contribution to AF from the forces on the other two
edges; the total is evidently a au 6 6a
The distortions of the diaphragm are caused by external forces. Let’s let
f represent the upward force per unit area on the sheet (a kind of “pressure")
from the external forces. When the membrane is in equilibrium (the static case), this force must be balanced by the internal force we have just computed, Eq
(12.16). That is (12.16) AF
f _ _ Ax Ay'
Equation (12.16) can then be written
f=  V ' (rVu), (12.17) where by V we now mean, of course, the twodimensional gradient operator
(a/ax, a/ay). We have the differential equation that relates u(x, y) to the applied 125 ‘i‘II“
he. 91“"‘a‘é  Fig. 123. A thin
stretched over a cylindrical frame (like rubber sheet a drumhead). 1f the sheet is pushed up
at A and down at B, what is the shape
of the surface? \\
s\\ Fig. 124. The surface tension 7 of
a stretched rubber sheet is the force per
unit length across a line. Fig. 12—5. Cross section of the de
ﬂected sheet. forces f(x, y) and the surface tension r(x, y), which may, in general, vary from
place to place in the sheet. (The distortions of a threedimensional elastic body are
also governed by similar equations, but we will stick to twodimensions.) We
will worry only about the case in which the tension 1' is constant throughout the
sheet. We can then write for Eq. (12.17), W = — (12.18) We have another equation that is the same as for electrostaticsl—only this
time, limited to twodimensions. The displacement u corresponds to ¢, and f/r
corresponds to p / so. So all the work we have done for inﬁnite plane charged sheets,
or long parallel wires, or charged cylinders is directly applicable to the stretched
membrane. Suppose we push the membrane at some points up to a deﬁnite height—that is,
we ﬁx the value of u at some places. That is the analog of having a deﬁnite potential
at the corresponding places in an electrical situation. So, for instance, we may
make a positive “potential” by pushing up on the membrane with an object having
the crosssectional shape of the corresponding cylindrical conductor. For example,
if we push the sheet up with a round rod, the surface will take on the shape shown
in Fig. 12—6. The height u is the same as the electrostatic potential as of a charged
cylindrical rod. It falls off as ln (1 /r). (The slope, which corresponds to the
electric ﬁeld E, drops off as l/r.) Fig. 12—6. Cross section of a
stretched rubber sheet pushed up by a
round rod. The function u(x, y) is the same
as the electric potential ¢(x, y) near a
very long charged rod. The stretched rubber sheet has often been used as a way of solving complicated
electrical problems experimentally. The analogy is used backwards! Various
rods and bars are pushed against the sheet to heights that correspond to the po
tentials of a set of electrodes. Measurements of the height then give the electrical
potential for the electrical situation. The analogy has been carried even further.
If little balls are placed on the membrane, their motion corresponds approximately
to the motion of electrons in the corresponding electric ﬁeld. One can actually
watch the “electrons” move on their trajectories. This method was used to design
the complicated geometry of many photomultiplier tubes (such as the ones used
for scintillation counters, and the one used for controlling the headlight beams on
Cadillacs). The method is still used, but the accuracy is limited. For the most
accurate work, it is better to determine the ﬁelds by numerical methods, using the
large electronic computing machines. 124 The diffusion of neutrons; a uniform spherical source in a homogeneous
medium We take another example that gives the same kind of equation, this time
having to do with diffusion. In Chapter 43 of Vol. I we considered the diffusion
of ions in a single gas, and of one gas through another. This time, let’s take a
different example—the diffusion of neutrons in a material like graphite. We choose
to speak of graphite (a pure form of carbon) because carbon doesn’t absorb slow
neutrons. In it the neutrons are free to wander around. They travel in a straight
line for several centimeters, on the average, before being scattered by a nucleus
and deﬂected into a new direction. So if we have a large block—many meters on
a side—the neutrons initially at one place will diffuse to other places. We want to
ﬁnd a description of their average behavior—that is, their average ﬂow. 12—6 Let N(x, y, z) AV be the number of neutrons in the element of volume AV
at the point (x, y, 2). Because of their motion, some neutrons will be leaving AV,
and others will be coming in. If there are more neutrons in one region than in a
nearby region, more neutrons will go from the ﬁrst region to the second than come
back; there will be a net ﬂow. Following the arguments of Chapter 43 in Vol. I,
we describe the ﬂow by a ﬂ0w vector J. Its xcomponent J, is the net number of
neutrons that pass in unit time a unit area perpendicular to the xdirection. We
found that (1219) where the diffusion constant D is given in terms of the mean velocity v, and the
meanfreepath [between scatterings is given by D: 111. l
3
The vector equation for J is J = —D VN. (12.20) The rate at which neutrons ﬂow across any surface element do is J  nda
(where, as usual, n is the unit normal). The net ﬂow out of a volume element is then
(following the usual gaussian argument) V  JdV. This flow would result in
a decrease with time of the number in AV unless neutrons are being created in
AV (by some nuclear process). If there are sources in the volume that generate S neutrons per unit time in a unit volume, then the net ﬂow out of AV will be equal
to (S — aN/at) AV. We have then that 6N
V 1— S—3; (12.21)
Combining (12.21) with (12.20), we get the neutron diﬂusion equation
v ~ (—1) VN) = s — (12.22) In the static case—where 6N/at = 0—we have Eq. (12.4) all over again!
We can use our knowledge of electrostatics to solve problems about the diffusion
of neutrons. So let’s solve a problem. (You may wonder: Why do a problem if
we have already done all the problems in electrostatics? We can do it faster this
time because we have done the electrostatic problems!) Suppose we have a block of material in which neutrons are being generated——
say by uranium ﬁssion—uniformly throughout a spherical region of radius a
(Fig. 12—7). We would like to know: What is the density of neutrons everywhere?
How uniform is the density of neutrons in the region where they are being gen
erated ? What is the ratio of the neutron density at the center to the neutron density
at the surface of the source region? Finding the answers is easy. The source
density SO replaces the charge density p, so our problem is the same as the problem
of a sphere of uniform charge density. Finding N is just like ﬁnding the potentlal
4:. We have already worked out the ﬁelds inside and outside of a uniformly charged
sphere; we can integrate them to get the potential. Outside, the potential is
Q/47reor, with the total charge Q given by 47m 3p/ 3. So 3 . _ £11..
¢outsrde “ 360’. For points inside, the ﬁeld is due only to the charge Q(r) inside the sphere of radius
r, Q(r) = 47rr3p/3, so (12.24)
12—7 NEUTRON
FLOW VECTOR o l'
(a)
. I ,
ELECTRIC. /"\ FIELD /—’..r‘.\
W 4‘51.»
\' . 1
\./ /1 l
I
x l I \
. I
1 I
I
4' I
1
I
I
I
o 0 r
(b) Fig. 12—7. (0) Neutrons are produced
uniformly throughout a sphere of radius a
in a large graphite block and diﬁuse
outward. The neutron density N is found
as a function of r, the distance from the
center of the source. (b) The analogous
electrostatic situation: a uniform sphere of
charge, where N corresponds to (ﬁt and J
corresponds to E. The ﬁeld increases linearly with r. Integrating E to get ¢, we have pr2
¢inside =  —— + aconstant.
660 At the radius a, dam,“ must be the same as ¢outside, so the constant must be
pa 2/2<eo. (We are assuming that d; is zero at large distances from the source, which
will correspond to N being zero for the neutrons.) Therefore, 2 2
mm. = L — (12,25) We know immediately the neutron density in our other problem. The answer
is S 3
Noumde = 3—3,, (12.26)
and
2 2
Ninside = % — (12.27) N is shown as a function of r in Fig. 12—7. Now what is the ratio of density at the center to that at the edge? At the center
(r = 0), it is proportional to 3a2/2. At the edge (r = a) it is proportional to
2a 2 / 2, so the ratio of densities is 3/2. A uniform source doesn’t produce a uniform
density of neutrons. You see, our knowledge of electrostatics gives us a good start
on the physics of nuclear reactors. There are many physical circumstances in which diffusion plays a big part.
The motion of ions through a liquid, or of electrons through a semiconductor,
obeys the same equation. We ﬁnd again and again the same equations. 12—5 Irrotational ﬂuid ﬂow; the ﬂow past a sphere Let’s now consider an example which is not really a very good one, because
the equations we will use will not really represent the subject with complete
generality but only in an artiﬁcial idealized situation. We take up the problem
of water ﬂow. In the case of the stretched sheet, our equations were an approxima
tion which was correct only for small deﬂections. For our consideration of water
ﬂow, we will not make that kind of an approximation; we must make restrictions
that do not apply at all to real water. We treat only the case of the steady ﬂow of
an incompressible, nonviscous, circulationfree liquid. Then we represent the ﬂow
by giving the velocity v(r) as a function of position r. If the motion is steady
(the only case for which there is an electrostatic analog) v is independent of time.
If p is the density of the ﬂuid, then pv is the amount of mass which passes per unit
time through a unit area. By the conservation of matter, the divergence of pv will
be, in general, the time rate of change of the mass of the material per unit volume.
We will assume that there are no processes for the continuous creation or destruc
tion of matter. The conservation of matter then requires that V pv = 0. (It
should, in general, be equal to —6p/at, but since our ﬂuid is incompressible, p
cannot change.) Since p is everywhere the same, we can factor it out, and our equa
tion is simply V  v = 0. Good! We have electrostatics again (with no charges); it’s just like V ' E = 0.
Not so! Electrostatics is not simply V  E = 0. It is a pair of equations. One
equation does not tell us enough; we need still an additional equation. To match
electrostatics, we should have also that the curl of v is zero. But that is not generally
true for real liquids. Most liquids will ordinarily develop some circulation. So
we are restricted to the situation in which there is no circulation of the ﬂuid. Such
ﬂow is often called irrotational. Anyway, if we make all our assumptions, we can 12—8 imagine a case of ﬂuid ﬂow that is analogous to electrostatics. So we take V  v = 0 (12.28)
and V X 0 = O. (12.29) We want to emphasize that the number of circumstances in which liquid
ﬂow follows these equations is far from the great majority, but there are a few.
They must be cases in which we can neglect surface tension, compressibility, and
viscosity, and in which we can assume that the ﬂow is irrotational. These assump
tions are valid so rarely for real water that the mathematician John von Neumann
said that people who analyze Eqs. (12.28) and (12.29) are studying “dry water”!
(We take up the problem of ﬂuid ﬂow in more detail in Chapters 40 and 41.) Because V X v = 0, the velocity of “dry water” can be written as the
gradient of some potential: v = —w. (12.30) What is the physical meaning of up? There isn’t any very useful meaning. The
velocity can be written as the gradient of a potential simply because the ﬂow is
irrotational. And by analogy with electrostatics, w is called the velocity potential,
but it is not related to a potential energy in the way that d: is. Since the divergence
of v is zero, we have v  (w) = vzw = 0. (12.31) The velocity potential rp obeys the same differential equation as the electrostatic
potential in free space (p = 0). Let’s pick a problem in irrotational ﬂow and see whether we can solve it by
the methods we have learned. Consider the problem of a spherical ball falling
through a liquid. If it is going too slowly, the viscous forces, which we are dis
regarding, will be important. If it is going too fast, little Whirlpools (turbulence)
will appear in its wake and there will be some circulation of the water. But if the
ball is going neither too fast nor too slow, it is more or less true that the water ﬂow
will ﬁt our assumptions, and we can describe the motion of the water by our
simple equations. It is convenient to describe what happens in a frame of reference ﬁxed in the
sphere. In this frame we are asking the question: How does water ﬂow past a sphere
at rest when the ﬂow at large distances is uniform? That is, when, far from the
sphere, the ﬂow is everywhere the same. The flow near the sphere will be as shown
by the streamlines drawn in Fig. 12—8. These lines, always parallel to v, correspond
to lines of electric ﬁeld. We want to get a quantative description for the velocity
ﬁeld, i.e., an expression for the velocity at any point P. We can ﬁnd the velocity from the gradient of up, so we ﬁrst work out the po
tential. We want a potential that satisﬁes Eq. (12.31) everywhere, and which
also satisﬁes two restrictions: (1) there is no ﬂow in the spherical region inside
the surface of the ball, and (2) the ﬂow is constant at large distances. To satisfy
(1), the component of 0 normal to the surface of the sphere must be zero. That
means that ail/6r is zero at r = a. To satisfy (2), we must have aw/az = ya at
all points where r >> a. Strictly speaking, there is no electrostatic case which
corresponds exactly to our problem. It really corresponds to putting a sphere of
dielectric constant zero in a uniform electric ﬁeld. If we had worked out the
solution to the problem of a sphere of a dielectric constant K in a uniform ﬁeld,
then by putting K = 0 we would immediately have the solution to this problem. We have not actually worked out this particular electrostatic problem in de tail, but let’s do it now. (We could work directly on the ﬂuid problem with v and 14, but we will use E and 45 because we are so used to them.)
The problem is: Find a solution of vzqs = 0 such that E = ~V¢ is a con
stant, say E 0, for large r, and such that the radial component of E is equal to zero
at r = a. That is,
as _
5; r=a — 0. (12.32) 12—9 >>—
_.———
._’——. Fig. 12—8. The velocity ﬁeld of ir
rotational ﬂuid ﬂow past a sphere. Our problem involves a new kind of boundary condition, not one for which
«1: is a constant on a surface, but for which 6¢/6r is a constant. That is a
little different. It is not easy to get the answer immediately. First of all, without
the sphere, ¢ would be —Eoz. Then E would be in the zdirection and have the con
stant magnitude E0, everywhere. Now we have analyzed the case of a dielectric
sphere which has a uniform polarization inside it, and we found that the ﬁeld
inside such a polarized sphere is a uniform ﬁeld, and that outside it is the same as
the ﬁeld of a point dipole located at the center. So let’s guess that the solution we
want is a superposition of a uniform ﬁeld plus the ﬁeld of a dipole. The potential
of a dipole (Chapter 6) is pz/41reor3. Thus we assume that _ P2
_ —E02 + 41reor3  Since the dipole ﬁeld falls off as l/r3, at large distances we have just the ﬁeld E 0.
Our guess will automatically satisfy condition (2) above. But what do we take for
the dipole strength p ? To ﬁnd out, we may use the other condition on :15, Eq. (12.32).
We must diﬂ‘erentiate ¢ with respect to r, but of course we must do so at a constant
angle 9, so it is more convenient if we ﬁrst express d: in terms of r and 0, rather than
of z and r. Since z = r cos 9, we get 0
¢ = ——Eor cos a + {:20er  (12.34) The radial component of E is _ 6_¢ _ p cos 6_
6r — +Eo cos 0 + zmora (12.35) This must be zero at r = a for all 0. This will be true if
p = —27r60a3E0. Note carefully that if both terms in Eq. (12.35) had not had the same 0depen
dence, it would not have been possible to choose p so that (12.35) turned out to be
zero at r = a for all angles. The fact that it works out means that we have guessed
wisely in writing Eq. (12.33). Of course, when we made the guess we were looking
ahead; we knew that we would need another term that (a) satisﬁed V245 = 0 (any
real ﬁeld would do that), (b) dependent on cos 6, and (c) fell to zero at large r.
The dipole ﬁeld is the only one that does all three. Using (12.36), our potential is 3
¢ = —E0 cos a (r + . (12.37) The solution of the ﬂuid ﬂow problem can be written simply as 1P 3
—v0 cos 0 (r + . (12.38) It is straightforward to ﬁnd 0 from this potential. We will not pursue the matter
further. 126 Illumination; the uniform lighting of a plane In this section we turn to a completely different physical problem—we want
to illustrate the great variety of possibilities. This time we will do something that
leads to the same kind of integral that we found in electrostatics. (If we have a
mathematical problem which gives us a certain integral, then we know something
about the properties of that integral if it is the same integral that we had to do for
another problem.) We take our example from illumination engineering. Suppose
there is a light source at the distance a above a plane surface. What is the illumina
tion of the surface? That is, what is the radiant energy per unit time arriving at a
unit area of the surface? (See Fig. 129.) We suppose that the source is spherically 1210 Fig. 129. The illumination In of a
surface is the radiant energy per unit
1 time arriving at a unit area of the surface. symmetric, so that light is radiated equally in all directions. Then the amount of
radiant energy which passes through a unit area at right angles to a light ﬂow varies
inversely as the square of the distance. It is evident that the intensity of the light in
the direction normal to the ﬂow is given by the same kind of formula as for the
electric ﬁeld from a point source. If the light rays meet the surface at an angle 9 to
the normal, then I, the energy arriving per unit area of the surface, is only cos 0 as
great, because the same energy goes onto an area larger by l/cos 0. If we call the
strength of our light source S, then In, the illumination of a surface, is 1,, = — e,  n, (12.39) where e, is the unit vector from the source and n is the unit normal to the surface.
The illumination In corresponds to the normal component of the electric ﬁeld from
a point charge of strength 41rEOS. Knowing that, we see that for any distribution of
light sources, we can ﬁnd the answer by solving the corresponding electrostatic
problem. We calculate the vertical component of electric ﬁeld on the plane due to
a distribution of charge in the same way as for that of the light sources.* Consider the following example. We wish for some Special experimental
situation to arrange that the top surface of a table will have a very uniform illumina
tion. We have available long tubular ﬂuorescent lights which radiate uniformly
along their lengths. We can illuminate the table by placing the ﬂuorescent tubes
in a regular array on the ceiling, which is at the height 2 above the table. What is
the widest spacing b from tube to tube that we should use if we want the surface
illumination to be uniform to, say, within one part in a thousand? Answer; ( 1)
Find the electric ﬁeld from a grid of wires With the spacing b, each charged uni
formly; (2) compute the vertical component of the electric ﬁeld; (3) ﬁnd out what
b must be so that the ripples of the ﬁeld are not more than one part in a thousand. In Chapter 7 we saw that the electric ﬁeld of a grid of charged wires could be
represented as a sum of terms, each one of which gave a sinusoidal variation of
the ﬁeld with a period of b/n, where n is an integer. The amplitude of any one of
these terms is given by Eq. (7.44): Fn ___ Ane—Zr’nzlb We need consider only n = 1, so long as we only want the ﬁeld at points not too
close to the grid. For a complete solution, we would still need to determine the
coefﬁcients A.” which we have not yet done (although it is a straightforward
calculation). Since we need only A 1, we can estimate that its magnitude is roughly
the same as that of the average ﬁeld. The exponential factor would then give us
directly the relative amplitude of the variations. If we want this factor to be 103,
we ﬁnd that b must be 0.912. If we make the spacing of the ﬂuorescent tubes 3/4 * Since we are talking about incoherent sources whose intensities always add linearly,
the analogous electric charges will always have the same sign. Also, our analogy applies
only to the light energy arriving at the top of an opaque surface, so we must include in
our integral only the sources which shine on the surface (and, naturally, not sources
located below the surface l). 12—11 of the distance to the ceiling, the exponential factor is then 1/4000, and we have a
safety factor of 4, so we are fairly sure that we will have the illumination constant
to one part in a thousand. (An exact calculation shows that A 1 is really twice the
average ﬁeld, so the exact answer is b = 0.82.) It is somewhat surprising that for
such a uniform illumination the allowed separation of the tubes comes out so large. 12—7 The “underlying unity” of nature In this chapter, we wished to show that in learning electrostatics you have
learned at the same time how to handle many subjects in physics, and that by
keeping this in mind, it is possible to learn almost all of physics in a limited number of years. However, a question surely suggests itself at the end of such a discussion:
Why are the equations from different phenomena so similar? We might say: “It is
the underlying unity of nature.” But what does that mean? What could such a
statement mean? It could mean simply that the equations are similar for different
phenomena; but then, of course, we have given no explanation. The “underlying
unity” might mean that everything is made out of the same stuff, and therefore
obeys the same equations. That sounds like a good explanation, but let us
think. The electrostatic potential, the diﬁusion of neutrons, heat ﬂow—are we
really dealing with the same stuff? Can we really imagine that the eleCtrostatic po
tential is physically identical to the temperature, or to the density of particles?
Certainly d) is not exactly the same as the thermal energy of particles. The displace
ment of a membrane is certainly not like a temperature. Why, then, is there “an
underlying unity” ? A closer look at the physics of the various subjects shows, in fact, that the
equations are not really identical. The equation we found for neutron diffusion is
only an approximation that is good when the distance over which we are looking
is large compared with the mean free path. If we look more closely, we would see
the individual neutrons running around. Certainly the motion of an individual
neutron is a completely diﬂerent thing from the smooth variation we get from
solving the differential equation. The differential equation is an approximation,
because we assume that the neutrons are smoothly distributed in space. Is it possible that this is the clue? That the thing which is common to all the
phenomena is the space, the framework into which the physics is put? As long as
things are reasonably smooth in space, then the important things that will be
involved will be the rates of change of quantities with position in space. That is
why we always get an equation with a gradient. The derivatives must appear in
the form of a gradient or a divergence; because the laws of physics are independent
of direction, they must be expressible in vector form. The equations of electro
statics are the simplest vector equations that one can get which involve only the
spatial derivatives of quantities. Any other simple problem—or simpliﬁcation of a
complicated problem—must look like electrostatics. What is common to all our
problems is that they involve space and that we have imitated what is actually a
complicated phenomenon by a simple differential equation. That leads us to another interesting question. Is the same statement perhaps
also true for the electrostatic equations? Are they also correct only as a smoothed
out imitation of a really much more complicated microscopic world? Could it be
that the real world consists of little Xons which can be seen only at very tiny dis
tances? And that in our measurements we are always observing on such a large
scale that we can’t see these little Xons, and that is why we get the differential
equations? Our currently most complete theory of electrodynamics does indeed have its
difﬁculties at very short distances. So it is possible, in principle, that these equations
are smoothedout versions of something. They appear to be correct at distances
down to about 10—14 cm, but then they begin to look wrong. It is possible that
there is some as yet undiscovered underlying “machinery,” and that the details of
an underlying complexity are hidden in the smoothlooking equations—as is so 12—12 in the “smooth” diffusion of neutrons. But no one has yet formulated a successful
theory that works that way. Strangely enough, it turns out (for reasons that we do not at all understand)
that the combination of relativity and quantum mechanics as we know them seems
to forbid the invention of an equation that is fundamentally different from Eq.
(12.4), and which does not at the same time lead to some kind of contradiction.
Not simply a disagreement with experiment, but an internal contradiction. As, for
example, the prediction that the sum of the probabilities of all possible occurrences
is not equal to unity, or that energies may sometimes come out as complex numbers,
or some other such idiocy. No one has yet made up a theory of electricity for which
V24> = —p/eo is understood as a smoothedout approximation to a mechanism
underneath, and which does not lead ultimately to some kind of an absurdity.
But, it must be added, it is also true that the assumption that V24: = —p/eo is
valid for all distances, no matter how small, leads to absurdities of its own (the
electrical energy of an electron is inﬁnite)——absurdities from which no one yet
knows an escape. 12—13 ...
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This note was uploaded on 06/18/2009 for the course PHYSICS none taught by Professor Leekinohara during the Spring '09 term at Uni. Nottingham  Malaysia.
 Spring '09
 LeeKinohara
 Physics

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