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Unformatted text preview: 20 Solutions of NIaxwell’s Equations in
Free Space 20—1 Waves in free space; plane waves In Chapter 18 we had reached the point where we had the Maxwell equations
in complete form. All there is to know about the classical theory of the electric
and magnetic ﬁelds can be found in the four equations: I. vE=£ II. VXE=_§1_’
60 at
. (20.1)
III. VB=0 1v. c2vx3=L+E
60 at When we put all these equations together, a remarkable new phenomenon occurs:
ﬁelds generated by moving charges can leave the sources and travel alone through
space. We cons1dered a special example in which an inﬁnite current sheet is
suddenly turned on. After the current has been on for the time I, there are uniform
electric and magnetic ﬁelds extending out the distance ct from the source. Suppose
that the current sheet lies in the yzplane with a surface current density J going
toward positive y. The electric ﬁeld will have only a ycomponent, and the mag
netic ﬁeld, only a zcomponent. The magnitude of the ﬁeld components is given by __ J
260C ’ E,, = ch = (20.2)
for positive values of x less than ct. For larger x the ﬁelds are zero. There are,
of course, similar ﬁelds extending the same distance from the current sheet in the
negative xdirection. In Fig. 20—1 we show a graph of the magnitude of the ﬁelds
as a function of x at the instant I. As time goes on, the “wavefront” at ct moves
outward in x at the constant veIOCity c. Now conSider the following sequence of events. We turn on a current of unit
strength for a while, then suddenly increase the current strength to three units,
and hold it constant at this value. What do the ﬁelds look like then? We can see
what the ﬁelds will look like in the following way. First, we imagine a current of
unit strength that is turned on at t = 0 and left constant forever. The ﬁelds for
pOSitive x are then given by the graph in part (a) of Fig. 20—2. Next, we ask what
would happen if we turn on a steady current of two units at the time 11. The ﬁelds in this case will be twice as high as before, but will extend out in
x only the distance ('0 ~ II), as shown in part (b) of the ﬁgure. When we add
these two solutions, usmg the principle of superposition, we ﬁnd that the sum of
the two sources is a current of one unit for the time from zero to t1 and a current
of three units for times greater than 11. At the time t the ﬁelds will vary with x
as shown in part (c) of Fig. 20—2. Now let’s take a more complicated problem. ConSider a current which is
turned on to one unit for a while, then turned up to three units, and later turned
oﬁ" to zero. What are the ﬁelds for such a current? We can ﬁnd the solution in
the same way—by adding the solutions of three separate problems. First, we ﬁnd
the ﬁelds for a step current of unit strength. (We have solved that problem already.)
Next. we ﬁnd the ﬁelds produced by a step current of two units. Finally, we solve
for the ﬁelds of a step current of minus three units. When we add the three solutions,
we will have a current which is one unit strong from t = 0 to some later time,
say t1, then three units strong until a still later time t2, and then turned off—that 20~l 20—1 Waves in free space; plane
waves 20—2 Threedimensional waves
20—3 Scientiﬁc imagination 20—4 Spherical waves References: Chapter 47, Vol. 1: Sound:
T he Wave Equation
Chapter 28, Vol. 1: Electro
magnetic Radiation tEl=clB y t x l
n }
4
n Fig. 20—1. The electric and mag
netic ﬁeld as a function of x at the time t
after the current sheet is turned on. ct x (0) cit—m ct 'x
(C) Fig. 20—2. The electric field of a
current sheet. (a) One unit of current
turned on at t= O, (b) Two units of
current turned on at t = t]; (c) Super
position of (a) and (b). i, 2 , ° c(t 42) c(tt.) ct x
(a) (b) Fig. 20~3. If the current source strength varies as shown in (a), then at the
time t shown by the arrow the electric field as a function of x is as shown in (b). is, to zero. A graph of the current as a function of time is shown in Fig. 20—3(a).
When we add the three solutions for the electric ﬁeld, we ﬁnd that its variation
with x, at a given instant t, is as shown in Fig. 20—3(b). The ﬁeld is an exact
representation of the current. The ﬁeld distribution in space is a nice graph of
the current variation with time—only drawn backwards. As time goes on the whole
picture moves outward at the speed c, so there is a little blob of ﬁeld, travelling
toward positive x, which contains a completely detailed memory of the history of
all the current variations. If we were to stand miles away, we could tell from the
variation of the electric or magnetic ﬁeld exactly how the current had varied
at the source. You will also notice that long after all activity at the source has completely
stopped and all charges and currents are zero, the block of ﬁeld continues to travel
through space. We have a distribution of electric and magnetic ﬁelds that exist
independently of any charges or currents. That is the new effect that comes from
the complete set of Maxwell’s equations. If we want, we can give a complete
mathematical representation of the analysis we have just done by writing that the
electric ﬁeld at a given place and a given time is proportional to the current at the
source, only not at the same time, but at the earlier time t — x/c. We can write _ J(t — x/c). 26°C (20.3) E140) = We have, believe it or not, already derived this same equation from another
point of view in Vol. I, when we were dealing with the theory of the index of re
fraction. Then, we had to ﬁgure out what ﬁelds were produced by a thin layer of
oscillating dipoles in a sheet of dielectric material with the dipoles set in motion
by the electric ﬁeld of an incoming electromagnetic wave. Our problem was to
calculate the combined ﬁelds of the original wave and the waves radiated by the
oscillating dipoles. How could we have calculated the ﬁelds generated by moving
charges when we didn’t have Maxwell’s equations? At that time we took as our
starting point (without any derivation) a formula for the radiation ﬁelds produced
at large distances from an accelerating point charge. If you will look in Chapter
31 of Vol. I, you will see that Eq. (31.10) there is just the same as the Eq. (20.3)
that we have just written down. Although our earlier derivation was correct only
at large distances from the source, we see now that the same result continues to
be correct even right up to the source. We want now to look in a general way at the behavior of electric and magnetic
ﬁelds in empty space far away from the sources, i.e., from the currents and charges.
Very near the sources—near enough so that during the delay in transmission, the
source has not had time to change much—the ﬁelds are very much the same as we
have found in what we called the electrostatic or magnetostatic cases. If we go out
to distances large enough so that the delays become important, however, the
nature of the ﬁelds can be radically different from the solutions we have found.
In a sense, the ﬁelds begin to take on a character of their own when they have
gone a long way from all the sources. So we can begin by discussing the behavior
of the ﬁelds in a region where there are no currents or charges. 20—2 Suppose we ask: What kind of ﬁelds can there be in regions where p and j are
both zero? In Chapter 18 we saw that the physics of Maxwell’s equations could also be expressed in terms of differential equations for the scalar and vector
potentials: 2 _ _1_ £22 _ _ P
V 4; 62 612 _ 6—0, (20.4)
2 _ 1 f4 _ _ _L.
V A — cg BIZ —— 6062 (20.5)
If p and j are zero, these equations take on the simpler form
1 62¢
v2 — _ __ =
CZ M2 0, (20.6)
va—iiz—ALo (207)
c2 6t2 — ' ' Thus in free space the scalar potential ¢ and each component of the vector potential
A all satisfy the same mathematical equation. Suppose we let p (psi) stand for any one of the four quantities «1», AI, A”, A2; then we want to investigate the general
solutions of the following equation: vztp—ii—o (208)
c2 at2 — ' '
This equation is called the threedimensional wave equation—three—dimensional,
because the function 4/ may depend in general on x, y, and z, and we need to worry
about variations in all three coordinates. This is made clear if we write out ex
plicrtly the three terms of the Laplacian operator:
6% 02¢ 62¢ 1 62¢ _
sic—2+3? 323—951—241 (209)
In free space, the electric ﬁelds E and B also satisfy the wave equation. For
example, since B = V X A, we can get a differential equation for B by taking
the curl of Eq. (207). Since the Laplacian is a scalar operator, the order of the
Laplacian and curl operations can be interchanged: v x (V2.4) = V2(V x A) = v23.
Similarly, the order of the operations curl and 6/62 can be interchanged: 162A 162 1923.
h c2 at? Using these results, we get the following differential equation for B: 2 _ 1 £12 _
V B a (“2 — O. (20.10)
So each component of the magnetic ﬁeld B satisﬁes the threedimensional wave
equatlon. Similarly, using the fact that E = —V¢ — dA/dt, it follows that the
electric ﬁeld E in free space also satisﬁes the threedimensional wave equation:
2 _ 1 6:5 _
V E 2—2 atz — 0. (20.11)
All of our electromagnetic ﬁelds satisfy the same wave equation, Eq. (20.8).
We might well ask: What is the most general solution to this equation? However,
rather than tackling that difﬁcult question right away, we will look ﬁrst at what
can be said in general about those solutions in which nothing varies in y and 2.
(Always do an easy case ﬁrst so that you can see what is going to happen, and
then you can go to the more complicated cases.) Let’s suppose that the magnitudes 203 of the ﬁelds depend only upon x—that there are no variations of the ﬁelds with
y and z. We are, of course, consrdermg plane waves again. We should expect to
get results something like those in the previous section. In fact, we will ﬁnd
precisely the same answers. You may ask: “Why do it all over again?” It 15 im
portant to do it again, ﬁrst, because we did not show that the waves we found were
the most general solutions for plane waves, and second, because we found the ﬁelds
only from a very particular kind of current source. We would like to ask now:
What is the most general kind of onedimensional wave there can be in free space?
We cannot ﬁnd that by seeing what happens for this or that particular source, but
must work with greater generality. Also we are going to work this time with differ
ential equations Instead of wrth integral forms. Although we will get the same re
sults, it is a way of practrcrng back and forth to show that it doesn’t make any
difference which way you go. Yea should know how to do things every which
way, because when you get a hard problem, you will often ﬁnd that only one of
the various ways is tractable. We could consider directly the solution of the wave equation for some elec
tromagnetic quantity. Instead, we want to start right from the beginning with
Maxwell’s equations in free space so that you can see their close relationship to
the electromagnetlc waves. So we Start With the equations In (20.1), setting the
charges and currents equal to zero. They become I. V ' E = 0
II. v x E = — 361—?
(20.12)
III. V B = 0
2 _ 2E
IV. c V X B — at
We write the ﬁrst equation out in components:
. _ m 6a 9% ,
V E _ j; + by + 62 ﬂ 0. (20.13) We are assuming that there are no variations with y and 2, so the last two terms are
zero. This equation then tells us that 6E,
6x = O. (20.14) Its solutlon is that E,“ the component of the electric ﬁeld in the xdirection. IS a
constant in space. If you look at IV in (20.12), supposmg no B—variation m y and
2 either, you can see that Ex 15 also constant in time. Such a ﬁeld could be the
steady DC ﬁeld from some charged condenser plates a long distance away. We are
not interested now in such an uninterestrng static ﬁeld; we are at the moment
interested only in dynamically varying ﬁelds. For dynamic ﬁelds, Ex = 0. We have then the Important result that for the propagation of plane waves
in any direction, the electric ﬁeld must be at right angles to the direction of propaga
(Ion. It can, of course, st111 vary in a complicated way with the coordinate x. The transverse E—ﬁeld can always be resolved into two components. say the
ycomponent and the zcomponent. So let’s ﬁrst work out a case in Which the elec
tric ﬁeld has only one transverse component. We’ll take ﬁrst an electric ﬁeld that
IS always in the ydirection, wrth zero zcomponent. Evrdently, if we solve this
problem we can also solve for the case where the electric ﬁeld IS always in the
zdirection. The general solution can always be expressed as the superposrtlon of
two such ﬁelds. How easy our equations now get. The only component of the electric ﬁeld
that is not zero is E,,, and all derivatives—except those With respect to x—are zero
The rest of Maxwell’s equations then become quite Simple. 20—4 Let’s look next at the second of Maxwell’s equations [11 of Eq. (20.12)].
Writing out the components of the curl E, we have 6E 6E ——.__£___7_’=
(vxr).—ay (,2 0,
6E. aEz_
(VXE)””E_E€“O’ _aE,,_aE,_a_E_,,
(“(5)2“5 '5; ex“ The xcomponent of V X E is zero because the derivatives with respect to y and
z are zero. The ycomponent is also zero; the ﬁrst term is zero because the derivative
with respect to z is zero, and the second term is zero because E2 is zero. The only
components of the curl ofE that is not zero 15 the z—component, which is equal to
aEy/ax. Setting the three components of V X E equal to the corresponding
components of ——aB/ar, we can conclude the following: 613,, _ 33,, at _ 0, 3t. 2 0_ (20.15)
332 _ _ 151:.
at _ 6x (20.16) Since the xcomponent of the magnetic ﬁeld and the ycomponent of the magnetic
ﬁeld both have zero time derivatives, these two components are just constant ﬁelds
and correspond to the magnetostatic solutions we found earlier. Somebody may
have left some permanent magnets near where the waves are propagating. We will
ignore these constant ﬁelds and set B, and By equal to zero. Incidentally, we would already have concluded that the x—component of B
should be zero for a different reason. Since the divergence of B is zero (from the
third Maxwell equation), applying the same arguments we used above for the
electric ﬁeld, we would conclude that the longitudinal component of the magnetic
ﬁeld can have no variation with x. Since we are ignoring such uniform ﬁelds in
our wave solutions, we would have set B, equal to zero. In plane electromagnetic
waves the Bﬁeld, as well as the E—ﬁeld, must be directed at right angles to the
direction of propagation. Equation (20.16) gives us the additional proposition that if the electric ﬁeld
has only a y—component, the magnetic ﬁeld will have only a z—component. So
E and B are at right angles to each other. This is exactly what happened in the
special wave we have already considered. We are now ready to use the last of Maxwell’s equations for free space [IV
of Eq. (20.12)]. Writing out the components, we have 2 632 2 613,, 6E, c(VXB)I=c ay—CbE': at,
2 _ 2aB,_ 2619,_a£,,
c(VXB),,—c 62 c (ax—:97 (20.17)
2 _ 2§_B_y_ 2033__6Ez
C(VXB)z—C 0x 6 6y’6t Of the six derivatives of the components of B, only the term aB,/ax is not equal
to zero. So the three equations give us simply
2 3B2 ~ 6E1,_ ax — w (20.18) —C The result of all our work is that only one component each of the electric and
magnetic ﬁelds is not zero, and that these components must satisfy Eqs. (20.16)
and (20.18). The two equations can be combined into one if we differentiate the
ﬁrst with respect to x and the second with respect to t; the lefthand sides of the 20—5 Fig. 204. The
represents a constant "shape" that travels
toward positive x with the speed c. function f(x — ct) two equations will then be the same (except for the factor c2). So we ﬁnd that
E,, satisﬁes the equation (20.19) We have seen the same differential equation before, when we studied the propaga
tion of sound. It is the wave equation for onedimensional waves. You should note that in the process of our derivation we have found something
more than is contained in Eq. (20 11). Maxwell’s equations have given us the
further information that electromagnetic waves have ﬁeld components only at
right angles to the direction of the wave propagation. Let’s review what we know about the solutions of the onedimensional wave
equation. If any quantity 1/1 satisﬁes the onedimensional wave equation 624/ 1 62¢ 672 “ 23 7972 0’ (2°20)
then one possible solution is a function ip(x, t) of the form tp(x, t) = f(x — ct), (20.21) that is, some function of the single variable (x — ct). The function f(x — ct)
represents a “rigid” pattern in x which travels toward posrtive x at the speed c
(see Fig. 20—4). For example, if the function f has a maxrmum when its argument
is zero, then for t = 0 the maximum of w Wlll occur at x = 0. At some later time,
say t = 10, t// will have its maximum at x = 10c. As time goes on, the maximum
moves toward positive x at the speed c. Sometimes it is more convenient to say that a solutlon of the onedimensxonal
wave equation is a function of (t — x/c). However, this is saying the same thing,
because any function of (t — x/c) is also a function of (x ~ ct): X—Ct
C F(t — x/c) = F[— ]= f(x — ct). Let’s show that f(x — ct) is indeed a solution of the wave equation. Since
it is a function of only one variable—the variable (x — ct)—we will let f’ represent
the derivatlve of f with respect to its variable and f” represent the second derivative
off. Differentiating Eq. (20.21) with respect to x, we have 6P__ , _
5;_f(x Ct), since the derivative of (x — ct) with respect to x is l. The second derivative of
ip With respect to x is clearly 2
37‘: = f”(x — ct). (20.22)
Taking derivatives of \p with respect to t, we ﬁnd
6
g; = f’(x ~ czx—c),
62¢ 2 //
51—2 2 +c f (x — C!) (2023) We see that .p does indeed satisfy the onedimensmnal wave equation. You may be wondering: “If I have the wave equation, how do I know that
I should take f(x — ct) as a solution? I don’t like this backward method. Isn’t
there some forward way to ﬁnd the solution?” Well, one good forward way is
to know the solution. It is possible to “cook up” an apparently forward mathe
matical argument, expecially because we know what the solution is supposed to
be, but with an equation as simple as this we don’t have to play games. Soon
you will get so that when you see Eq. (20.20), you nearly simultaneously see 20—6 ¢ = f(x — ct) as a solution. (Just as now when you see the integral of x2 dx, you
know right away that the answer is x3/ 3.) Actually you should also see a little more. Not only is any function of (x — ct)
a solution, but any function of (x + ct) is also a solution. Since the wave equation
contains only c2, changing the sign of c makes no difference. In fact, the most
general solution of the onedimensional wave equation is the sum of two arbitrary
functions, one of (x — ct) and the other of (x + ct): up = f(x — ct) + g(x + ct). (20.24) The ﬁrst term represents a wave travelling toward positive x, and the second term
an arbitrary wave travelling toward negative x. The general solution is the super—
position of two such waves both existing at the same time. We will leave the following amusrng question for you to think about. Take a function
\b of the following form: up = cos kx cos kct. This equation isn’t in the form of a function of (x — ct) or of (x + ct). Yet you can
easily show that this function 15 a solution of the wave equation by direct substitution into
Eq. (20.20). How can we then say that the general solution is of the form of Eq. (20.24)? Applying our conclusions about the solution of the wave equation to the
y—component of the electric ﬁeld, E,,, we conclude that E, can vary with x in any
arbitrary fashion. However, the ﬁelds which do exist can always be considered as
the sum of two patterns. One wave is sailing through space in one direction with
speed c, with an associated magnetic ﬁeld perpendicular to the electric ﬁeld;
another wave is travelling in the opposite direction with the same speed. Such
waves correspond to the electromagnetic waves that we know about—light, radio
waves, infrared radiation, ultraviolet radiation, xrays, and so on. We have already
discussed the radiation of light in great detail in Vol. I. Since everything we learned
there applies to any electromagnetic wave, we don’t need to consider in great detail
here the behavior of these waves. We should perhaps make a few further remarks on the question of the polariza
tion of the electromagnetic waves. In our solution we chose to consider the special
case in which the electric ﬁeld has only a ycomponent. There is clearly another
solution for waves travelling in the plus or minus xdirection, with an electric
ﬁeld which has only a zcornponent. Since Maxwell’s equations are linear, the
general solution for onedimensional waves propagating in the xdirection is the
sum of waves of E1, and waves of E. This general solution is summarized in the
following equations: E = (0, Eu: E2)
E1; = f(x — ct) + g(x + ct)
E = F(x — ct) + G(x + ct) (20.25)
B = (0, B1,, 3,.)
c8, = f(x—ct) —g(x + ct) cBy = —F(x — ct) + G(x + ct). Such electromagnetic waves have an E—vector whose direction is not constant but
which gyrates around in some arbitrary way in the yzplane. At every point
the magnetic ﬁeld is always perpendicular to the electric ﬁeld and to the direction
of propagation. 20—7 If there are only waves travelling in one direction, say the positive xdirection,
there is a simple rule which tells the relative orientation of the electric and mag
netic ﬁelds. The rule is that the cross product E X B—which is, of course, a
vector at right angles to both E and B——points in the direction in which the wave is
travelling. If E is rotated into B by a righthand screw, the screw points in the
direction of the wave velocity. (We shall see later that the vector E X B has a
special physical signiﬁcance: it is a vector which describes the ﬂow of energy in an
electromagnetic ﬁeld.) 20—2 Threedimensional waves We want now to turn to the subject of three—dimensional waves. We have
already seen that the vector E satisﬁes the wave equation. It is also easy to arrive
at the same conclusion by arguing directly from Maxwell’s equations. Suppose we
start with the equation 6B
V X E — — 37
and take the curl of both sides:
v x (v x E) = — gw x B). (20.26) You will remember that the curl of the curl of any vector can be written as the sum
of two terms, one involving the divergence and the other the Laplacian, VX (VXE)= V(V~E)—V2E. ln free space, however, the divergence of E is zero, so only the Laplacian term
remains. Also, from the fourth of Maxwell’s equations in free space [Eq. (20.12)]
the time derivative of (‘2 V X B is the second derivative of E With respect to t: ,2 a _ 62E
‘5‘VX3)—‘at2'
Equation (20.26) then becomes
2 _ i 62E
V E _ c2 >6t2 ’ which is the threedimensional wave equation. Written out in all its glory, this
equation is, of course, 62E 62E 62E 1 62E
2 + 2 + 7 2 _ 2 v _
6x By 62 c at 0. (20.27) How shall we ﬁnd the general wave solution” The answer is that all the solu
tions of the threedimensional wave equation can be represented as a superposition
of the onedimensional solutions we have already found. We obtained the equation
for waves which move in the xdirection by supposing that the ﬁeld did not depend
on y and 2. Obviously, there are other solutions in which the ﬁelds do not depend
on x and 2, representing waves going in the ydirection. Then there are solutions
which do not depend on x and y, representing waves travelling in the zdirection.
Or in general, since we have written our equations in vector form, the three
dimensional wave equation can have solutions which are plane waves movmg in
any direction at all. Again, Since the equations are linear, we may have Simultane
ously as many plane waves as we wish, travelling in as many different directions.
Thus the most general solution of the threedimenSional wave equation 18 a
superposition of all sorts of plane waves moving in all sorts of directions. Try to imagine what the electric and magnetic ﬁelds look like at present in
the space in this lecture room. First of all, there is a steady magnetic ﬁeld; it comes
from the currents in the interior of the earth—that is, the earth’s steady magnetic
ﬁeld. Then there are some irregular, nearly static electric ﬁelds produced perhaps
by electric charges generated by friction as various people move about in their m—S chairs and rub their coat sleeves against the chair arms. Then there are other
magnetic ﬁelds produced by oscillating currents in the electrical wiring—ﬁelds
which vary at a frequency of 60 cycles per second, in synchronism with the genera
tor at Boulder Darn. But more interesting are the electric and magnetic ﬁelds vary
ing at much higher frequencies. For instance, as light travels from window to
ﬂoor and wall to wall, there are little wiggles of the electric and magnetic ﬁelds
moving along at 186,000 miles per second. Then there are also infrared waves
travelling from the warm foreheads to the cold blackboard. And we have forgotten
the ultraviolet light, the xrays, and the radiowaves travelling through the room. Flying across the room are electromagnetic waves which carry music of a jazz
band. There are waves modulated by a series of impulses representing pictures of
events going on in other parts of the world, or of imaginary aspirins dissolving in
imaginary stomachs. To demonstrate the reality of these waves it is only necessary
to turn on electronic equipment that converts these waves into pictures and sounds. If we go into further detail to analyze even the smallest wiggles, there are
tiny electromagnetic waves that have come into the room from enormous distances.
There are now tiny oscillations of the electric ﬁeld, whose crests are separated by
a distance of one foot, that have come from millions of miles away, transmitted
to the earth from the Mariner 11 space craft which has just passed Venus. Its
signals carry summaries of information it has picked up about the planets (infor
mation obtained from electromagnetic waves that travelled from the planet to
the space craft). There are very tiny wiggles of the electric and magnetic ﬁelds that are waves
which originated billions of light years away—from galaxies in the remotest corners
of the universe. That this is true has been found by “ﬁlling the room with wires”—
by building antennas as large as this room. Such radiowaves have been detected
from places in space beyond the range of the greatest optical telescopes. Even they,
the optical telescopes, are simply gatherers of electromagnetic waves. What we
call the stars are only inferences, inferences drawn from the only physical reality
we have yet gotten from them—from a careful study of the unendingly complex
undulations of the electric and magnetic ﬁelds reaching us on earth. There is, of course, more: the ﬁelds produced by lightning miles away, the
ﬁelds of the charged cosmic ray particles as they zip through the room, and more,
and more. What a complicated thing is the electric ﬁeld in the space around you!
Yet it always satisﬁes the threedimensional wave equation. 20—3 Scientiﬁc imagination I have asked you to imagine these electric and magnetic ﬁelds. What do you
do? Do you know how? How do I imagine the electric and magnetic ﬁeld? What
do I actually see? What are the demands of scientiﬁc imagination? Is it any
different from trying to imagine that the room is full of invisible angels? No, it is
not like imagining invisible angels. It requires a much higher degree of imagination
to understand the electromagnetic ﬁeld than to understand invisible angels. Why?
Because to make invisible angels understandable, all I have to do is to alter their
properties a little bit—I make them slightly visible, and then I can see the shapes
of their wings, and bodies, and halos. Once I succeed in imagining a visrble angel,
the abstraction required—which is to take almost invisible angels and imagine
them completely invisible—is relatively easy. So you say, “Professor, please give
me an approximate description of the electromagnetic waves, even though it may
be slightly inaccurate, so that I too can see them as well as I can see almost invisible
angels. Then I W!“ modify the picture to the necessary abstraction.” I'm sorry I can’t do that for you. I don’t know how. I have no picture of this
electromagnetic ﬁeld that is in any sense accurate. I have known about the electro
magnetic ﬁeld a long time—I was in the same position 25 years ago that you are
now, and I have had 25 years more of experience thinking about these wiggling
waves. When I start describing the magnetic ﬁeld moving through space, I speak
of the E and B ﬁelds and wave my arms and you may imagine that I can see them. 20—9 I’ll tell you what I see. I see some kind of vague shadowy, wiggling lines—here
and there is an E and B written on them somehow, and perhaps some of the lines
have arrows on them—an arrow here or there which disappears when I look too
closely at it. When I talk about the ﬁelds swishing through space, I have a terrible
confusion between the symbols I use to describe the objects and the objects them
selves. I cannot really make a picture that is even nearly like the true waves. So
if you have some difﬁculty in making such a picture, you should not be worried
that your difﬁculty is unusual. Our science makes terriﬁc demands on the imagination. The degree of
imagination that is required is much more extreme than that required for some of
the ancient ideas. The modern ideas are much harder to imagine. We use a lot
of tools, though. We use mathematical equations and rules, and make a lot of
pictures. What I realize now is that when I talk about the electromagnetic ﬁeld in
space, I see some kind of a superposition of all of the diagrams which I’ve ever
seen drawn about them. I don’t see little bundles of ﬁeld lines running about be
cause it worries me that if I ran at a different speed the bundles would disappear.
I don’t even always see the electric and magnetic ﬁelds because sometimes I think
I should have made a picture with the vector potential and the scalar potential,
for those were perhaps the more phySically signiﬁcant things that were wiggling. Perhaps the only hope, you say, is to take a mathematical View. Now what is
a mathematical View? From a mathematical view, there is an electric ﬁeld vector
and a magnetic ﬁeld vector at every pomt in space; that is, there are six numbers
associated with every point. Can you imagine six numbers associated with each
point in space? That’s too hard. Can you imagine even one number associated
With every point? I cannot! I can imagine such a thing as the temperature at every
point in space. That seems to be understandable. There is a hotness and coldness
that varies from place to place. But I honestly do not understand the idea of a
number at every point. So perhaps we should put the question: Can we represent the electric ﬁeld by
something more like a temperature, say like the displacement of a piece of jello?
Suppose that we were to begin by imagining that the world was ﬁlled With thin
jello and that the ﬁelds represented some distortion—say a stretching or tw1sting—
of the jello. Then we could visualize the ﬁeld. After we “see” what it is like we
could abstract the jello away. For many years that’s what people tried to do.
Maxwell, Ampere, Faraday, and others tried to understand electromagnetism
this way. (Sometimes they called the abstract jello “ether.”) But it turned out that
the attempt to imagine the electromagnetic ﬁeld in that way was really standing in
the way of progress. We are unfortunately limited to abstractions, to using in
struments to detect the ﬁeld, to using mathematical symbols to describe the ﬁeld,
etc. But nevertheless, in some sense the ﬁelds are real, because after we are all
ﬁnished ﬁddling around With mathematical equations—With or without making
pictures and drawings or trying to visualize the thing—we can still make the instru
ments detect the Signals from Mariner II and ﬁnd out about galaxies a billion miles
away, and so on. The whole question of imagination in science is often misunderstood by people
in other disciplines. They try to test our imagination in the following way. They
say, “Here is a picture of some people in a situation. What do you imagine will
happen next?” When we say, “I can’t imagine,” they may think we have a weak
imagination. They overlook the fact that whatever we are allowed to imagine in
science must be conszstent Wlth everything else we know: that the electric ﬁelds and
the waves we talk about are not just some happy thoughts which we are free to
make as we wish, but ideas which must be consistent with all the laws of phySics
we know. We can’t allow ourselves to seriously imagine things which are obviously
in contradiction to the known laws of nature. And so our kind of imagination is
quite a difﬁcult game. One has to have the imagination to think of something that
has never been seen before, never been heard of before. At the same time the
thoughts are restricted in a strait jacket, so to speak, limited by the conditions that
come from our knowledge of the way nature really is. The problem of creating 20—10 something which is new, but which is consistent With everything which has been
seen before, IS one of extreme difﬁculty. While I’m on this subject I want to talk about whether it Will ever be poss1ble
to imagine beauty that we can't see It is an interesting question. When we look
at a rainbow, it looks beautiful to us. Everybody says, “Ooh, a rainbow.” (You
see how scientiﬁc I am. I am afraid to say something is beautiful unless I have an
experimental way of deﬁning it.) But how would we describe a rainbow if we were
blind? We are blind when we measure the infrared reﬂection coefﬁcient of sodium
chloride, or when we talk about the frequency of the waves that are coming from
some galaxy that we can‘t see—we make a diagram,‘we make a plot. For instance,
for the rainbow, such a plot would be the intenSity of radiation vs. wavelength
measured With a spectrophotometer for each direction in the sky. Generally, such
measurements would give a curve that was rather ﬂat. Then some day, someone
would discover that for certain conditions of the weather, and at certain angles in
the sky, the spectrum of intens1ty as a function of wavelength would behave
strangely; it would have a bump. As the angle of the instrument was varied only a
little bit, the maXimum of the bump would m0ve from one wavelength to another.
Then one day the physical reView of the blind men might publish a technical article
with the title “The IntenSity of Radiation as a Function of Angle under Certain
Conditions of the Weather.” In this article there might appear a graph such as
the one in Fig. 20—5 The author would perhaps remark that at the larger angles
there was more radiation at long wavelengths, whereas for the smaller angles the
maXimum in the radiation came at shorter wavelengths. (From our point of viCW,
we would say that the light at 40° is predominantly green and the light at 42° is
predominantly red.) Fig. 20—5. The intensity of electro
magnetic waves as a function of wave
length for three angles (measured from
the direction opposite the sun), observed
Only with certain meteorological con Wovelength ditions. lntensrty Now do we ﬁnd the graph of Fig. 20—5 beautiful? It contains much more de
tail than we apprehend when we look at a rainbow, because our eyes cannot see
the exact details in the shape of a spectrum. The eye, however, ﬁnds the rainbow
beautiful. Do we have enough imagination to see in the spectral curves the same
beauty we see when we look directly at the rainbow? I don’t know. But suppose I have a graph of the reﬂection coefﬁcient of a sodium chloride
crystal as a function of wavelength in the infrared, and also as a function of angle.
I would have a representation of how it would look to my eyes if they could see
in the infrared~perhaps some glowing, shiny “green,” mixed with reﬂections from
the surface in a “metallic red.” That would be a beautiful thing, but I don’t know
whether I can ever look at a graph of the reﬂection coeﬂicrent of NaCl measured
with some instrument and say that it has the same beauty. On the other hand. even if we cannot see beauty in particular measured results,
we can already claim to see a certain beauty in the equations which describe general
phySical laws. For example, in the wave equation (20.9), there’s something nice
about the regularity of the appearance of the x, the y, the z, and the t. And this
nice symmetry in appearance of the x. y, z, and 1 suggests to the mind still a greater
beauty which has to do with the four dimenSions, the possibility that space has
fourdimensional symmetry, the poss1bility of analyzrng that and the developments
of the speCial theory of relatiVity. So there is plenty of intellectual beauty asso
Ciated With the equations. 20—11 20—4 Spherical waves We have seen that there are solutions of the wave equation which corre
spond to plane waves, and that any electromagnetic wave can be described as a
superposition of many plane waves. In certain special cases, however, it is more
convenient to describe the wave ﬁeld in a ditferent mathematical form. We would
like to discuss now the theory of spherical waves—waves which correspond to
spherical surfaces that are spreading out from some center. When you drop a
stone into a lake, the ripples spread out in circular waves on the surface—they are
twodimensional waves. A spherical wave is a similar thing except that it spreads
out in three dimensions. Before we start describing spherical waves, we need a little mathematics.
Suppose we have a function that depends only on the radial distance r from a
certain origindin other words, a function that is spherically symmetric. Let’s
call the function Mr), where by r we mean r= Vx2+y2+;. the radial distance from the origin. In order to ﬁnd out what functions Mr) satisfy
the wave equation, we will need an expression for the Laplacian of 1/1. So we want
to ﬁnd the sum of the second derivatives of 11/ with respect to x, y, and z. We will
use the notation that 11/(r) represents the derivative of 1/2 with respect to r and ll/"(r)
represents the second derivative of it with respect to r. First, we ﬁnd the derivatives with respect to x. The ﬁrst derivative is 61100) 6x 6r
= I __— u
i0 (r) ax
The second derivative of ip with respect to x is 2 2 2
a¢=¢n<ﬂ> +War 672 6x 5;?" We can evaluate the partial derivatives of r with respect to x from 59r_x’ 62r_ll_x2
fix—r 6x2 r r2 So the second derivative of it with respect to x is 624’ _ x2 H 1 x2 ’ ax2_ﬁ¢ +;(1_ﬁ w (20.28)
Likewise, a2 2 1 2 6;: = {7 W, + 7 (I _ W, (20.29) 621p Z2 II 1 22 ’ (723251,, +;(1_7§ ¢_ (20.30) The Laplacian is the sum of these three derivatives. Remembering that
x2 + y2 + z2 = r2, we get 2
VZW) = W’tr) + ; Mr). (20.31)
It is often more convenient to write this equation in the following form:
2 _ 1 d2
v 1p _ 7 2]? (All). (20.32) If you carry out the differentiation indicated in Eq. (20.32), you will see that the
right—hand Side is the same as in Eq. (20.31). If we Wish to consider spherically symmetric ﬁelds which can propagate as
spherical waves, our ﬁeld quantity must be a function of both r and t. Suppose 20—12 we ask, then, what functions Mr, I) are solutions of the threedimensional wave
equation 2
We, 0 — i 3 Mr, 0 = 0. c, at, (20.33) Since 11/(r, t) depends only on the spatial coordinates through r, we can use the equa
tion for the LaplaCian we found above, Eq. (20.32). To be precise, however, since
i; is also a function of t, we should write the derivatives with respect to r as partial
derivatives. Then the wave equation becomes We must now solve this equation, which appears to be much more complicated
than the plane wave case. But notice that if we multiply this equation by r, we get 162 a2 (no) — ~— W) = o. C, at, (20.34) This equation tells us that the function rip satisﬁes the onedimensional wave equa
tion in the variable r. Using the general prinCiple which we have emphaSized so
often, that the same equations always have the same solutions, we know that if
rip is a function only of (r — ct) then it will be a solution of Eq. (20.34). 50 we
know that spherical waves must have the form rip(r, t) = f(r —— ct). Or, as we have seen before, we can equally well say that rip can have the form r112 = f(t — r/C) Dividing by r, we ﬁnd that the ﬁeld quantity 4/ (whatever it may be) has the follow
ing form: 10 — r/c). l’ ¢= (2035) Such a function represents a general spherical wave travelling outward from the
origin at the speed c. If we forget about the r in the denominator for a moment,
the amplitude of the wave as a function of the distance from the origin at a given
time has a certain shape that travels outward at the speed c. The factor r in the
denominator, however, says that the amplitude of the wave decreases in proportion
to l/r as the wave propagates. In other words, unlike a plane wave in which the
amplitude remains constant as the wave runs along, in a spherical wave the ampli—
tude steadily decreases, as shown in Fig. 20—6. This effect is easy to understand
from a simple physical argument. 0 r I’ 2
F———c(t2—tl) —A
(a) (b) Fig. 20—6. A spherical wove \l/ = fl! — r/cl/r. ((1)312 as a function of r for 1‘
some wave for the later time #2. (b) 11/ as a function of t for r = r, and the some wove seen at r2. 2043 ll h and the We know that the energy density in a wave depends on the square of the wave
amplitude. As the wave spreads, its energy is spread over larger and larger areas
proportional to the radial distance squared. If the total energy is conserved, the
energy density must fall as 1/r2, and the amplitude of the wave must decrease as
l/r. So Eq. (20.35) is the “reasonable” form for a spherical wave. We have disregarded the second possible solution to the onedimenSional
wave equation: rr = g(1 + r/c),
or
: 80 + r/c)_ , 1r This also represents a spherical wave, but one which travels inward from large r
toward the origin. We are now going to make a speCial assumption. We say, without any demon
stration whatever, that the waves generated by a source are only the waves which
go outward. Since we know that waves are caused by the motion of charges, we
want to think that the waves proceed outward from the charges. It would be
rather strange to imagine that before charges were set in motion, a spherical wave
started out from inﬁnity and arrived at the charges just at the time they began to
move. That is a pOSSible solution, but experience shows that when charges are
accelerated the waves travel outward from the charges. Although Maxwell’s
equations would allow either possibility, we will put in an additional fact—based
on experience—that only the outgoing wave solution makes “physical sense.” We should remark, however, that there is an interesting consequence to this
additional assumption: we are removing the symmetry with respect to time that
exists in Maxwell’s equations. The original equations for E and B, and also the
wave equations we derived from them, have the property that if we change the Sign
of t, the equation is unchanged. These equations say that for every solution
corresponding to a wave going in one direction there is an equally valid solution
for a wave travelling in the opposite direction. Our statement that we will consider
only the outgoing spherical waves is an important additional assumption. (A
formulation of electrodynamics in which this additional assumption is avoided has
been carefully studied. Surprismgly, in many circumstances it does not lead to
physically absurd conclusions, but it would take us too far astray to discuss these
ideas just now. We Will talk about them a little more in Chapter 28.) We must mention another important point. In our solution for an outgoing
wave, Eq. (20.35), the function tp is inﬁnite at the origin. That is somewhat peculiar.
We would like to have a wave solution which is smooth everywhere. Our solution
must represent physically a situation in which there is some source at the origin.
In other words, we have inadvertently made a mistake. We have not solved the
free wave equation (20.33) everywhere; we have solved Eq. (20.33) with zero on
the right everywhere, except at the origin. Our mistake crept in because some of
the steps in our derivation are not “legal” when r = 0. Let’s show that it is easy to make the same kind of mistake in an electrostatic
problem. Suppose we want a solution of the equation for an electrostatic potential
in free space, V24: = 0. The Laplacian is equal to zero, because we are assuming
that there are no Charges anywhere. But what about a spherically symmetric
solution to this equation—that is, some function qb that depends only on r. Using
the formula of Eq. (20.32) for the Laplacian, we have 1d2
;m (’45) = 0 Multiplying this equation by r, we have an equation which is readily integrated: d2
W (W) = 0 If we integrate once With respect to r, we ﬁnd that the ﬁrst derivative of r¢ is a
20—14 constant, which we may call a: d E —— (1.
Integrating again, we ﬁnd that r4: is of the form r¢=ar+b, where b is another constant of integration. So we have found that the following ¢
is a solution for the electrostatic potential in free space: b
¢—a+;' Something is evidently wrong. In the region where there are no electric
charges, we know the solution for the electrostatic potential: the potential is
everywhere a constant. That corresponds to the ﬁrst term in our solution. But we
also have the second term, which says that there is a contribution to the potential
that varies as one over the distance from the origin. We know, however, that such
a potential corresponds to a pOint charge at the origin. So, although we thought
we were solving for the potential in free space, our solution also gives the ﬁeld
for a point source at the origin. Do you see the Similarity between what happened
now and what happened when we solved for a spherically symmetric solution to
the wave equation? If there were really no charges or currents at the origin, there
would not be spherical outgoing waves. The spherical waves must, of course, be
produced by sources at the origin. In the next chapter we w111 investigate the con
nection between the outgoing electromagnetic waves and the currents and voltages
which produce them. 20—15 ...
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This note was uploaded on 03/26/2010 for the course PHYSICS V85.0093.0 taught by Professor Chaikin during the Spring '10 term at NYU.
 Spring '10
 Chaikin
 Physics

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