# Does the field of a unifirmly moving charge satisfy Maxwell's wave equation?

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.

• 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) Commented Apr 22, 2016 at 15:53
• 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. Commented Apr 23, 2016 at 10:09

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 $c$.