how a wave looks in a different frame 453
The key to our problem is the way two particular scalar quantities transform, namely, E⋅B and E2−B2. Let us use Eq. (9.49) to calculate E'⋅B' and see how it is related to E⋅B:
E'⋅B' = E'xB'x + E'yB'y + E'zB'z,
= ExBx + γ2EyBy + (v/c2)EyEz − vByBz − (v/c)2EzBz +
+ γ2EzBz − (v/c2)EyEz + vByBz − (v/c)2EyBy
= ExBx + γ2(1−β2)(EyBy + EzBz) = E⋅B. (9.50)
The scalar product E⋅B is not changed in the Lorentz transformation of the fields; it is an invariant. A similar calculation, which will be left to the reader as Exercise 9.32, shows that E2x + E2y + E2z − c2(B2x + B2y + B2z) is also unchanged by the Lorentz transformation. We therefore have
E'⋅B' = E⋅B and E'2 − c2B'2 = E2 − c2B2 (9.51)
The invariance of these two quantities is an important general property of any electromagnetic field, not just the field of an electromagnetic wave with which we are concerned at the moment. For the wave field, its implications are especially simple and direct. We know that the plane wave has B perpendicular to E, and cB = E. Each of our two invariants, E⋅B and E2 − c2B2, is therefore zero. And if an invariant is zero in one frame, it must be zero in all frames. We see that any Lorentz transformation of the wave will leave E and cB perpendicular and equal in magnitude. A light wave looks like a light wave in any inertial frame of reference. That should not surprise us. It could be said that we have merely come full circle, back to the postulates of relativity, Einstein's starting point. Indeed, according to Einstein's own autobiographical account, he had begun 10 years earlier (at age 16!) to wonder what one would observe if one could "catch up" with a light wave. With the transformations in Eq. (9.49), which were given in Einstein's 1905 paper, the question can be answered. Consider a traveling wave with amplitudes given by Ey = E0, Ex = Ez = 0, Bz = E0/c, Bx = By = 0. This is a wave traveling in the x^ direction, as we can tell from the fact that E×B points in that direction. Using Eq. (9.49) and the relation γ = 1/√1−β2, we find that
E'y = E0√(1−β)/(1+β), B'z = (E0/c)√(1−β)/(1+β). (9.52)
As observed in F' the amplitude of the wave is reduced. The wave velocity, of course, is c in F', as it is in F. The electromagnetic wave has no rest frame. In the limit β = 1, the amplitudes E'y and B'z observed in F' are reduced to zero. The wave has vanished!
454 Maxwell's equations and EM waves