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Feynman Physics Lectures V2 Ch26 1963-01-17 Lorentz Transformations of the Fields

Feynman Physics Lectures V2 Ch26 1963-01-17 Lorentz Transformations of the Fields

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Unformatted text preview: 26 Lorentz Transformations of the Fields 26—1 The four-potential 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 Lorentz-transformed frame in the preceding chapter. 26—1 26—1 The four-potential of a moving charge 26—2 The fields of a point charge with a constant velocity 26—3 Relativistic transformation of the fields 26-4 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 x-oxis 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 four-vector. 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 four-vector, 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 four-vector 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 slow-moving 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 x-component is a little more work. The derivative of <1) is more complicated 26-2 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 y-components 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 xA-ivi 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 [OW-velocity 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 right-hand Side of Eq (26.9) must have a factor l/cz: vXE c2. B = (26 12) For a slow-movmg 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 four-vector. 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 z-components is only a piece of something; there is also a l-component. 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 x-component is equal to a couple of terms that involve only y- and z-com- ponents. Suppose we call this combination of derivatives and components a “zy-thing,” 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 “xz-thing.” And 32 is, of course, the corresponding “yx-thing.” 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 z-component of E. Well, almost—there is a sign wrong But we forgot that in the four-dimensional gradient the t-dei‘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 four-vector 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 slow-movmg world (where we don’t worry about the speed of light) are not vectors in four-space. 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 three-dimensional 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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