# How to solve “EM wave equation” for the field of uniformly moving charge?

Is it possible to show that the field of a uniformly moving charge, which is according to Biot-Savart law is given by...

$${\bf E}({\bf r},t)=kq\left(\frac{1-v^2/c^2}{(1-v^2 \sin^2 \theta/c^2)^{3/2}}\right)\left(\frac{\hat{\bf r}}{\gamma^2(x-vt)^2 +y^2 + z^2}\right)$$

...satisfies the equation below (far from the charge itself)?

$$c^2\nabla^2\mathbf E = \dfrac{\partial^2}{\partial t^2}\mathbf E$$

Edit:

I know that uniform motion doesn't produce waves. I also know that this equation is valid only for empty space far from charges and currents.

But somehow this equation still has to hold for the electric field "dragged" inertially after the charge moving with constant velocity.

It's really difficult for me to grasp this idea. Any references would be most welcome. I tried to solve directly but failed.

• You should edit your post to make it clear that you know that a uniformly moving charge does not generate a wave, yet the fields seemingly have to satisfy this equation. I don't know the answer (up voting question), but I note that there is no source term in your wave equation... perhaps that makes a difference. – garyp Apr 26 '16 at 11:56
• One approach to this would be to first go to the rest frame of the charge, in this frame the electric field is particularly simple, namely $E=\frac{q}{4 \pi \epsilon_0 r^2}$, where it is straight forward to show (since there are no time derivatives) then to transform to a frame where the charge is moving and note that $\frac{\partial^2 }{\partial t^2} + \frac{1}{c^2}\nabla^2$ is invariant, see for example Purcell, Electricity and Magnetism. – jim Apr 27 '16 at 18:35
• Hi @jim , is it right to say the the so-called "EM wave equation" has obviously non-wavelike solutions (like an inertially moving field)? Why is it called wave equation after all? – Vitaly Korzhik Apr 28 '16 at 8:23
• @VitalyKorzhik I think because the EM equation can admit wave like solutions it is referred to as the EM wave equation. No particular distinction is made when you look at time independent equations. As well, you can have time dependent solutions that don't have to be sinusoidal, that people usually regard as a wave. – jim Apr 28 '16 at 10:28
• @jim Is there any obvious reason why the field function won't satisfy $v^2\nabla^2\mathbf E = \dfrac{\partial^2}{\partial t^2}\mathbf E$ (with $v$ instead of $c$)? Because it can't satisfy both at the same time. This probably has something to do with Lorentz transformation... – Vitaly Korzhik Apr 29 '16 at 7:25

People are not understanding your question. I think you want someone to verify explicitly that the fields produced in your special case do (or don't) obey the generally valid wave equation. After all, the field produced does not look like a wave. A general solution of the wave equation for a disturbance traveling in the $x$ direction is ${\bf{E}}({\bf r},t)= {\bf E_0}(x-ct, y, z)$. So I suppose $${\bf E}({\bf r},t) = \frac{1}{4\pi\epsilon_0}\frac{q}{\left[(x-ct)^2 +y^2 + z^2\right]}{\bf\hat{r}}$$ (where ${\bf\hat{r}}$ points away from the charge at any instant) is a solution of the wave equation.

Obviously this is good only for non-relativistic speeds. And my notation for the unit vector is not very good ... I hope it makes the point though.

Comment

I see that you have edited your question to include an explicit expression of the relativistic expression for the field. I think it's clear that what I did for the non-relativistic case works just as well for the relativistic case.

• I deleted this comment and added it to my answer. – garyp Apr 26 '16 at 14:32
• I have added time-dependence explicitly according to your suggestions, hope it looks right. I guess the relativistic part can be ignored for simplicity @Mathaholic – Vitaly Korzhik Apr 26 '16 at 16:49
• It's almost right ... you don't define ${\bf \hat{r}}$, and neither did I. With a correct definition of ${\bf \hat{r}}$, I think what you wrote is manifestly a solution to the wave equation! – garyp Apr 26 '16 at 16:52
• Problem is, I have $v$ instead of $c$ in my equation, so I don't see how it solves Maxwell's equation with $c$ in it. Can you elaborate a little? – Vitaly Korzhik Apr 26 '16 at 17:03
• Aha. I knew it looked too easy. :) – garyp Apr 26 '16 at 17:33