1
$\begingroup$

The so-called "electromagnetic wave equation" is so general that it should obviously hold for every point in space and for all field configurations (if there are no charges).

Here it is: $$ \left(c^2 \nabla^2 - \dfrac{\partial^2}{\partial t^2}\right)\mathbf E = 0$$

Now, for a uniformly moving (not accelerating) charge, the field configuration "moves" with it, with velocity $v$ (let's ignore relativistic effects).

Although this is no "wave", any function that "moves" should satisfy its own "wave equation", but with $v^2$ instead of $c^2$.

(is that right?)

But now we have a contradiction, because in my example the electic field is supposed to satisfy two different equations at once.

Where is the mistake? I'll appreciate any input.

$\endgroup$
2
  • $\begingroup$ have a look at Motl's answer here physics.stackexchange.com/q/3580 . I do not agree that ""any function that "moves" should satisfy its own "wave equation"" . where did you see this? . A uniformly moving charge does not radiate. depending on the inertial frame of observation the shape of the field will be distorted according to the motion (not 1/r^2 for the stationary observer) $\endgroup$
    – anna v
    Apr 22, 2016 at 15:53
  • $\begingroup$ Hi @annav, thanks for your answer. The general solution to a wave equation is usually described as a "propagating disturbance of arbitrary shape". That's why I think a function of any form at all, including a field of a moving charge, should satisfy the wave equation. Maybe my mistake is here? Motl's answer includes virtual photons and a bit hard for me to grasp, but thanks anyway. $\endgroup$ Apr 23, 2016 at 10:09

1 Answer 1

0
$\begingroup$

The speed of light in a vacuum, $c$, refers to the phase velocity of an EM wave. It is the speed of a pure sine wave.

What you are talking about for the speed of the disturbance that propagates with a moving charge is the group velocity; the velocity of the interference pattern of many individual sine waves with different frequencies. This is not constrained to move at $c$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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