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Unformatted text preview: 26 Lorentz Transformations of the Fields 26—1 The fourpotential of a moving charge We saw in the last chapter that the potential A}. = (d), A) is a four—vector.
The time component is the scalar potential «iv, and the three space components are
the vector potential A. We also worked out the potentials of a particle movmg with
uniform speed on a straight line by using the Lorentz transformation (We had
already found them by another method in Chapter 21.) For a pomt charge whose
position at the time t is (Hf, 0, 0), the potentials at the point (x. y, z) are ‘1
¢ — _ 2
47reo\/l — v? [(11:2—12) + J12 + zﬂm
go
A; v . 7 ., . (26.1)
47reox/1 _ 02 [(11:12; + y2 + 211/2
— 1)
A1, = A2 = 0 Equations (26 1) give the potentials at x. y, and 2 at the time t, for a charge
whose “present” pos1tion (by which we mean the position at the time t) is at x = 2'!
Notice that the equations are in terms of (x — M), y, and 2, which are the coordi
nates measured from the current position P of the moving charge (see Fig. 26—1)
The actual inﬂuence we know really travels at the speed c, so it is the behavior of
the charge back at the retarded position P’ that really counts.1L The pomt P’ is at
x = rt’ (where. t’ = t — r’/c is the retarded time). But we said that the charge was
moving with uniform veloc1ty in a straight line, so naturally the behavior at P’ and
the current position are directly related. In fact, if we make the added assumptmn
that the potentials depend only upon the pOSition and the velocity at the retarded
moment, we have in equations (26.1) a complete formula for the potentials for a
charge moving any way. It works this way. Suppose that you have a charge
moving in some arbitrary fashion, say with the trajectory in Fig. 26—2, and you
are trying to ﬁnd the potentials at the pomt (x, y, z). First, you ﬁnd the retarded
position P’ and the veloc1ty I” at that pomt. Then you imgaine that the charge
would keep on moving with this veloc1ty during the delay time (t’ — I), so that
it would then appear at an imaginary position PM”, which we can call the “pro
jected position,” and would arrive there with the veloc1ty 12’. (Of course, it doesn’t
do that; its real posnion at t is at P.) Then the potentials at (x, y. z) are just what
equations (26 1) would give for the imaginary charge at the projected posnion
PM]. What we are saying is that since the potentials depend only on what the
charge is doing at the retarded time, the potentials will be the same whether the
charge continued movmg at a constant velocrty or whether it changed its veIOCity
after t’—that is. after the potentials that were going to appear at (x, y, 2) at the
time I were already determined. You know, of course, that the moment that we have the formula for the po
tentials from a charge movmg in any manner whatsoever, we have the complete
electrodynamics; we can get the potentials of any charge distribution by super T The primes used here to indicate the retarded posmons and times should not be confused
with the primes referring to a Lorentztransformed frame in the preceding chapter. 26—1 26—1 The fourpotential of a
moving charge 26—2 The ﬁelds of a point charge
with a constant velocity 26—3 Relativistic transformation
of the ﬁelds 264 The equations of motion in
relativistic notation In this chapter: c = 1 Review: Chapter 20. Vol. II. Solution
of Maxwel/‘i' Equations in
Free Space (My, Z)
l
RETARDED I
POSITION iy
.
/ I
l
PRESENT I
P, POSITION
O l n if,
I
I ~94 ., I ~, Fig. 26—]. Finding the fields at P due
to a charge q moving along the xoxis
with the constant speed v. The field
”now” at the point (x, y, 2) can be ex
pressed in terms of the “present” position
P, as well as in terms of P'. the ”retarded"
position (at t’ = t — r’/c). (mil) RETARDED
POSITION\
"PROJECT 0“
POSITIO
PRESENT
ePOSITION TRAJECTORY Fig. 26*2. A charge moves on an
arbitrary trajectory. The potentials at
(x,y,z) at the time t are determined by
the posnion P’ and velocny v' at the
retarded time t’ — r//c. They are con
veniently expressed in terms of the co
ordinates from the ”proiected” position PM”. (The actual position at t is P.) position. Therefore we can summarize all the phenomena of electrodynamics
either by writing Maxwell’s equations or by the following series of remarks.
(Remember them in case you are ever on a desert island. From them, all can be
reconstructed. You will, of course, know the Lorentz transformation; you Will
never forget iliai on a desert island or anywhere else ) First, A” is a fourvector. Second, the Coulomb potential for a stationary
charge is q/47reor. Third, the potentials produced by a charge movmg in any way
depend only upon the veIOCity and position at the retarded time With those
three facts we have everything From the fact that A“ is a fourvector, we transform
the Coulomb potential, which we know, and get the potentials for a constant
veIOCity. Then. by the last statement that potentials depend only upon the past
veloc1ty at the retarded time, we can use the projected position game to ﬁnd them.
It is not a particularly useful way of doing things. but it is interesting to show that
the laws of physics can be put in so many diﬁerent ways It is sometimes said, by people who are careless, that all of electrodynamics
can be deduced solely from the Lorentz transformation and Coulomb’s law. Of
course. that is completely false. First, we have to suppose that there is a scalar
potential and a vector potential that together make a fourvector That tells us
how the potentials transform Then why is it that the effects at the retarded
time are the only things that count? Better yet, why is it that the potentials depend
only on the position and the velocity and not, for instance, on the acceleration"
The fields E and B do depend on the acceleration. If you try to make the same
kind of an argument with respect to them, you would say that they depend only
upon the position and velocity at the retarded time But then the ﬁelds from an
accelerating charge would be the same as the ﬁelds from a charge at the proiected
pOSition—which is false. Theﬁelds depend not only on the pOSition and the velomty
along the path but also on the acceleration. So there are several additional tacit
assumptions in this great statement that everything can be deduced from the
Lorentz transformation (Whenever you see a sweeping statement that a tremen
dous amount can come from a very small number of assumptions, you always
find that it is false. There are usually a large number of implied assumptions that
are far from obvious if you think about them sufﬁciently carefully.) 26—2 The ﬁelds of a point charge with a constant velocity Now that we have the potentials from a pomt charge movmg at constant
veloc1ty, we ought to ﬁnd the ﬁelds—for practical reasons There are many cases
where we have uniformly moving particles—for instance, cosmic rays going through
a cloud chamber, or even slowmoving electrons in a Wire. So let’s at least see
what the ﬁelds actually do look like for any speed—even for speeds nearly that
of light—assuming only that there is no acceleration. It is an interesting question. We get the ﬁelds from the potentials by the usual rules‘ First, for E E, _ ‘1 . _  (26.2) Similarly, for EU, ‘1 y
E, _<‘ . ﬂ *, 1 . (26.3)
47reom [L vt)1 —l— y2 + 22TH 1—02 The xcomponent is a little more work. The derivative of <1) is more complicated
262 and A, is not zero. First, _ , __ 2
:3): q r (V (f alga/(1. v A) 3/2” (26'4)
47r60\/l — 02 [—T—j— + y2 + 22J
Then, differentiating A, with respect to t, we ﬁnd
— l2 — l — 2
65:9” qh—ﬂ (xi—(3y WEI " ”in <26 5)
47T€0 l — 02 [W + yd + Z2J
And ﬁnally. taking the sum,
E” /——"q 76: — zit); — M 3/2” (26'6)
47T€0 l—Ugl>m~+y2+22l We’ll look at the phySlCS of E in a minute. let's ﬁrst ﬁnd B For the z—compo
nent, Since A,, is zero, we have Just one derivative to get. Notice, however, that A, is Just 1'45, and a/By of 17¢ is just ~—1'E,,. So B. ~ z'EU. (26.7)
Similarly,
_ 6A,, 6A,, _ '64)
By— 62 6x Fiﬁ/E,
and
3,, = —1)Ez (26.8) Finally, B, is zero. Since AU and A2 are both zero. We can write the magnetic ﬁeld
simply as B = v X E (26.9) Now let’s see what the ﬁelds look like. We Will try to draw a picture of the
ﬁeld at various pOSitions around the present pOSition of the charge. It is true that
the influence of the charge comes, in a certain sense. from the retarded posnion.
but because the motion is exactly spec1ﬁed, the retarded pos1tion is uniquely given
in terms of the present pos1tion For uniform veloc1ties, it’s nicer to relate the
ﬁelds to the current position, because the ﬁeld components at (x,y. z) depend
only on (x — vi), y. and z——which are the components of the displacements
rp from the present pos1tion to (x, y, 2) (see Fig. 26—3). ConSider ﬁrst a pomt with Z = 0. Then E has only x— and y—components.
From Eqs. (26.3) and (26.6), the ratio of these components is Just equal to the
ratio of the x and ycomponents of the displacement. That means that E is in
the same (ll/'eCIIOfl as rp, as shown in Fig. 26—3. Since E2 is also proportional to 2.
it is clear that this result holds in three dimenSions. In short, the electric ﬁeld is
radial from the charge, and the ﬁeld lines radiate directly out of the charge, jll§l
as they do for a stationary charge. Of course, the ﬁeld isn't exactly the same as
for the stationary charge. because of all the extra factors of (1 —~ v2) But we
can show something rather interesting. The difference is just what you would get
if you were to draw the Coulomb ﬁeld With a peculiar set of coordinates in which
the scale of x was squashed up by the factor V1 — v2. If you do that, the ﬁeld
lines will be spread out ahead and behind the charge and will be squeezed together
around the Sides, as shown in Fig. 26—4. lfwe relate the strength of E to the density of the ﬁeld lines in the conventional
way, we see a stronger ﬁeld at the sides and a weaker ﬁeld ahead and behind.
which is just what the equations say. First, if we look at the strength of the ﬁeld
at right angles to the line of motion, that is, for (x — at) = 0, the distance from 26—3 H
E)' E
(tn/ﬁlls),
l
/ l
’P/ I
l

l
_77 7 7 P xAivi l r);
’ q\
t ‘v’ ’* \PRESENT
POSITION
Fig. 26—3. For a charge moving with constant speed, the electric ﬁeld points
radially from the “present" position of
the charge. The electric ﬁeld of a Fig. 26—4.
charge moving with the constant speed
v = 0.9c, part (b), compared with the
ﬁeld of a charge at rest, part (a). Fig. 26—5. The magnetic ﬁeld near a moving charge is v X E.
With Fig. 26—4.) (Compare the charge is (y2 + 22). Here the total ﬁeld strength is \/E§ +—E_%, which is _ q 1 .
E — 47r60\/1——— ”2 y2 + 22 (26m) The ﬁeld is proportional to the inverse square of the distance—Just like the Cou
lomb ﬁeld except increased by the constant, extra factor 1/\/1 —— 02, which is
always greater than one. So at the sides of a moving charge, the electric ﬁeld is
stronger than you get from the Coulomb laW. In fact, the ﬁeld in the Sidew1se
direction is bigger than the Coulomb potential by the ratio of the energy of the
particle to its rest mass. Ahead of the charge (and behind), y and z are zero and _ _ M.
E _ Er — 47reo(x — at)2 (2611) The ﬁeld again varies as the inverse square of the distance from the charge but is
now reduced by the factor (1 — U2), in agreement with the picture of the ﬁeld lines.
If P/C is small, 1*2/c2 is still smaller, and the effect of the (l — 212) terms is very
small; we get back to Coulomb‘s law. But if a particle is movmg very close to
the speed of light, the ﬁeld in the forward direction is enormously reduced, and
the ﬁeld in the sidewise direction is enormously increased. Our results for the electric ﬁeld of a charge can be put this way: Suppose
you were to draw on a piece of paper the ﬁeld lines for a charge at rest, and then
set the picture to travelling with the speed 2;. Then, of course, the whole picture
would be compressed by the Lorentz contraction; that is, the carbon granules
on the paper would appear in diﬁerent places The miracle of it is that the picture
you would see as the page ﬂies by would still represent the ﬁeld lines of the point
charge. The contraction moves them closer together at the sides and spreads them
out ahead and behind, just in the right way to give the correct line densities. We
have emphasized before that ﬁeld lines are not real but are only one way of repre
senting the ﬁeld. However, here they almost seem to be real. In this particular
case, if you make the mistake of thinking that the ﬁeld lines are somehow really
there in space, and transform them, you get the correct ﬁeld. That doesn’t, however,
make the ﬁeld lines any more real All you need do to remind yourself that they
aren’t real is to think about the electric ﬁelds produced by a charge together With
a magnet; when the magnet moves, new electric ﬁelds are produced, and destroy
the beautiful picture So the neat idea of the contracting picture doesn‘t work in
general. It is, however, a handy way to remember what the ﬁelds from a fast
movmg charge are like. The magnetic ﬁeld is v X E [from Eq. (26.9)]. lfyou take the velomty crossed
into a radial E—ﬁeld, you get a B which circles around the line of motion, as shown
in Fig. 26—5. If we put back the c’s, you Will see that it’s the same result we had
for [OWvelocity charges. A good way to see where the c’s must go is to refer back
to the force law, F = q(E + I) X B). You see that a velocity times the magnetic ﬁeld has the same dimensions as an
electric ﬁeld. So the righthand Side of Eq (26.9) must have a factor l/cz: vXE
c2. B = (26 12) For a slowmovmg charge (71 << 6), we can take for E the Coulomb ﬁeld: then _ 47re..c2 w (26.13) This formula corresponds exactly to equations for the magnetic ﬁeld of a current
that we found in Section 1477. 2641 We would like to point out, in passing, something interesting for you to think
about. (We will come back to discuss it again later.) Imagine two electrons with
velocities at right angles, so that one will cross over the path of the other, but in
front of it, so they don’t collide. At some instant, their relative positions Will be
as in Fig 26—6(a). We look at the force on ql due to qg and Vice versa. On qg
there is only the electric force from ([1, since ql makes no magnetic ﬁeld along its
line of motion. On ql, however, there is again the electric force but, in addition,
a magnetic force, since it is moving in a B—ﬁeld made by qg. The forces are as drawn
in Fig. 26—6(b). The electric forces on ql and q2 are equal and opposite. However,
there is a sidewise (magnetic) force on q; and no szdewise force on q2. Does action
not equal reaction? We leave it for you to worry about. 26—3 Relativistic transformation of the ﬁelds In the last section we calculated the electric and magnetic ﬁelds from the
transformed potentials. The ﬁelds are important, of course, in spite of the argu
ments given earlier that there is physical meaning and reality to the potentials.
The ﬁelds, too, are real. It would be convenient for many purposes to have a way
to compute the ﬁelds in a moving system if you already know the ﬁelds in some
“rest” system. We have the transformation laws for ¢ and A, because A“ is a
fourvector. Now we would like to know the transformation law of E and B.
Given E and B in one frame, how do they look in another frame movmg past?
It is a convenient transformation to have. We could always work back through the
potentials, but it is useful sometimes to be able to transform the ﬁelds directly.
We Will now see how that goes. How can we ﬁnd the transformation laws of the ﬁelds? We know the trans
formation laws of the <15 and A, and we know how the ﬁelds are given in terms of
(1; and A—it should be easy to ﬁnd the transformation for the B and E. (You
might think that With every vector there should be something to make it a four
vector, so With E there’s got to be something else we can use for the fourth com—
ponent. And also for B. But it’s not so. It’s quite different from what you would
expect.) To begin with, let’s take just a magnetic ﬁeld B, which is, of course
V X A. Now we know that the vector potential with its x, y, and zcomponents
is only a piece of something; there is also a lcomponent. Also we know that for
derivatives like V, besides the x, y, 2 parts, there is also a derivative with respect to
I. So let’s try to ﬁgure out what happens if we replace a “y" by a “t", or a “z"
by a “I,” or something like that. First, notice the form of the terms in V X A when we write out the com
ponents: _ 6A2 6A,, B. _ 6A., _ 9A2 62 6x ~6Ay 3A,
’ BZ—‘J ay‘ 6y 02 , 3,, = (26.14) The xcomponent is equal to a couple of terms that involve only y and zcom
ponents. Suppose we call this combination of derivatives and components a
“zything,” and give it a shorthand name, F2”. We simply mean that F2y 2 ~— — —. (26.15) Similarly. By is equal to the same kind of “thing,” but this time it is an “xzthing.”
And 32 is, of course, the corresponding “yxthing.” We have Bx = F2”, By = F12, B2 = F1/I‘ (26.16) Now what happens if we simply try to concoct also some “t”type things
like F“ and Ft; (since nature should be nice and symmetric in x, y, z, and I)? For
instance, what is F”? It is, of course, 6A; 6A2 . 62 at 26—5 Q,"_’ Q2
(0) Iva
F, ___(1l:’l ‘ 8
(b) i. Vi q2E2=F2 Fig. 26—6. The forces between two
moving charges are not always equal and
opposite. It appears that “action" is not
equal to ”reaction." Table 26—1 The components of Fm, l Fl” = _Fvu I Fm, = 0 J Fr); — _Bz Fxt : Ex
§ F,” : ~13; F”, : EU
1 Fa —‘ —By F“ = E; But remember that A, 2 ¢, so it is also 0d) 0A2 02 8t You’ve seen that before. It is the zcomponent of E. Well, almost—there is a
sign wrong But we forgot that in the fourdimensional gradient the tdei‘ivative
comes With the oppos1te Sign from x, y, and : So we should really have taken the
more COHSistent extension of F12 as _ 3ﬂ 0/2.
F12 — #62 + Vat (2617)
Then it is exactly equal to —E: Trying also F” and F,,,, we ﬁnd that the three pOSSibilities give
F”, = 5“ F”, = —Eu~ F22 = ~E2. (26.18) What happens if both subscripts are 1" Or, for that matter, if both are x?
We get things like 6A; 8A;
F = — '— a
” at at
and
_ 0A,, 6A,,
F” — 6x — 797’ Which give nothing but zero.
We have then six of these F—things. There are six more which you get by
reversing the subscripts, but they give nothing really new, since F,,, = —F,,,. and so on. So, out of sixteen poss1ble combinations of the four subscripts taken
in pairs, we get only Six different physical objects; and the]; are the components
of B and E. To represent the general term of F, we Will use the general subscripts ,u and 1’,
where each can stand for 0, 1, 2, or 3~meaning in our usual fourvector notation
I, x, y, and 2 Also, everything Wlll be consistent with our four~vector notation if
we deﬁne FW by F,” : VHA, — VVA,“ (2619) remembering that V,, = (6/6]. —6/6x, ——6/6y, —6/02) and that A,, : ((15, At, A,,.
AZ) What we have found is that there are six quantities that belong together in
nature~that are different aspects of the same thing. The electric and magnetic
ﬁelds which we have conSidered as separate vectors in our slowmovmg world
(where we don’t worry about the speed of light) are not vectors in fourspace.
They are parts ofa new “thing." Our phySical “ﬁeld” is really the six—component
object FM. That is the way we must look at it for relatiVity We summaiize our
results on F,” in Table 26—1 You see that what we have done here is to generalize the cross product We
began With the curl operation, and the fact that the transformation properties of
the curl are the same as the transformation properties of two vectors—the ordinary
threedimensional vector A and the gradient operator which we know also behaves
like a vector Let’s look for a moment at an ordinary cross product in three di
menSions, for ...
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 Spring '09
 LeeKinohara
 Physics

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