Gauss's Law with Moving Charges

Question

My text claims that Gauss's Law has been proven to work for moving charges experimentally, is there a non-experimental way to verify this?

Answer 1

Any physics equations you write down could be wrong, so you need to verify them experimentally, or you're just doing math. I can imagine a universe in which the Gauss's law doesn't work for moving charges; and I have to test to see if we live in such a universe. In that sense, there is no non-experimental way to verify it.

On the other hand, if you mean to ask if there's a mathematical way to prove it from the experimentally verified equations of electromagnetism, then sure! But you need to be able to calculate the electric field of a moving charge. The derivation is a bit complicated, and I'll leave it to you to find your favorite version of it, but the answer for a single charge moving at constant velocity is \begin{equation} \vec{E} = \frac{k\, q\, \vec{r}}{r^3}\, \frac{1-v^2}{(1-v^2\, \sin^2\theta)^{3/2}} \end{equation} It's enough to show this for just one charge because the integral is linear, so you can just add up contributions from different charges. Also, we can take the integral over a sphere centered on this charge, since the divergence theorem tells us that moving the surface of integration won't change anything (as long as we keep the charge inside).

So, we do \begin{align} \oint \vec{E} \cdot d\vec{a} &= \oint \frac{k\, q\, \vec{r} \cdot \hat{r}}{r^3}\, \frac{1-v^2}{(1-v^2\, \sin^2\theta)^{3/2}}\, r^2\, \sin \theta\, d\theta\, d\phi \\ &= k\, q\, \int_0^{2\pi} \int_0^\pi \frac{1-v^2}{(1-v^2\, \sin^2\theta)^{3/2}}\, \sin \theta\, d\theta\, d\phi\\ &= 2\pi\, k\, q\, \int_0^\pi \frac{1-v^2}{(1-v^2\, \sin^2\theta)^{3/2}}\, \sin \theta\, d\theta \\ &= 4\pi\, k\, q~, \end{align} which is just the usual Gauss's law.

This, of course, is only for constant velocities, as Peter Kravchuk pointed out below. To see it more generally, it would probably be easier to go over to the differential form of Gauss's law, which is one of Maxwell's equations. Then, you can note that Maxwell's equations are relativistically covariant.