My old "classical" understanding of how electromagnetic waves propagates is this: varying electric field generates a varying magnetic filed which in turns generates a varying electric field .. and so on the wave will travel by it's own.

But when thinking about it from Special Relativity point of view this description does not fit for me!

For instance in special relativity the magnetic force is a consequence of length contraction as observed in a different frames of reference, so there will be only an electric field and no magnetic filed in the travelling wave frame of reference (is that correct?).

So how exactly EM waves travel in SR from an observer frame of reference? and from the travelling wave frame of reference?


  • $\begingroup$ The problem with your argument is that there is no inertial reference frame in which the magnetic field vanishes. We can prove this by contradiction. Assume you find an inertial reference frame in which the magnetic field vanishes. The electric field in this new reference frame must be nonzero and still vary with time. Then Maxwell's equation tells us that there must be a magnetic field present which is a contradiction. $\endgroup$ – honey.mustard Mar 23 '17 at 11:32
  • 2
    $\begingroup$ A briefer statement of @honey.mustard's point: there is no "traveling wave frame of reference". $\endgroup$ – WillO Mar 23 '17 at 13:09
  • $\begingroup$ @honey.mustard good point $\endgroup$ – DeepBlue Mar 23 '17 at 18:18
  • $\begingroup$ @honey.mustard What about the case where EM wave travels at speed lower than c like inside a medium? We can talk then about a frame of reference where B field vanishes! $\endgroup$ – DeepBlue Mar 23 '17 at 19:41
  • $\begingroup$ This question cannot be answered as asked. There is no frame of reference for the travelling wave and, thus, we cannot tell you how it travels in its frame of reference $\endgroup$ – Jim Mar 21 '19 at 13:14

In relativistic electrodynamics there is really only one "electromagnetic field". However, if one wishes to maintain the distinction between electric and magnetic fields, these concepts are not frame-invariant (ordinary vectors cannot be Lorentz covariant).

What are invariant are the scalar quantities $\vec{E} \cdot \vec{B}$ and $E^2 - c^2B^2$ (in SI units).

For a transverse electromagnetic wave in vacuum, both of these invariants equal zero.

Thus in any other frame of reference, the (transformed) electric and magnetic fields are perpendicular and the E-field strength is always $c$ times the B-field strength. There is no frame in which the electric or magnetic fields disappear. What is not so immediately obvious (but is true nonetheless) is that a Lorentz transformation of $\vec{k}\cdot r - \omega t$ (which is the argument of the specific electromagnetic wave functional form) transforms to $\vec{k'}\cdot r' - \omega' t'$, such that the wave speed $\omega/k = \omega'/k' =c$ and the wave travels at $c$ according to observers in any frame of reference.

In fact your question does not make sense because there is no "travelling frame" for light. All frames of reference "see" light travelling locally at $c$.

  • $\begingroup$ Good answer. Can you expand your answer to explain how the EM wave travel (instead of self generating fields)? Regarding frames of reference what about EM wave travelling at speed less than c like inside a medium? $\endgroup$ – DeepBlue Mar 23 '17 at 18:16
  • $\begingroup$ @DeepBlue I don't understand your problem. The E-field does not "generate" the B-field, or vice-versa. They coexist. There are lots of similar questions on this site. $\endgroup$ – ProfRob Mar 23 '17 at 18:33
  • $\begingroup$ @DeepBlue Maybe what you're trying to understand is the Doppler effect. $\endgroup$ – honey.mustard Mar 23 '17 at 18:51
  • $\begingroup$ I am trying to understand how EM wave propagates in a vacuum, my old understanding was wrong (self generating E & B fields), so I am left with 2 options to explain it: space is the medium that carry the wave and it is not empty (I read something about that), or that photons truly travel as particles. $\endgroup$ – DeepBlue Mar 23 '17 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.