# Why doesn't finite field propagation speed contradict Gauss's law? [duplicate]

 Not sure why this was closed. The answers there do not answer my question, and are not even correct...

Imagine a charge sitting in space. It causes an electric field everywhere, with magnitude $$\propto r_{old}^{-2}$$

But now let's say we move the charge a little. This will change the electric field everywhere to be $$\propto r_{new}^{-2}$$. However, according to Maxwell's other equations (I'm told), this propagation is not instant, but happens with a finite speed $$c$$.

Let's draw a box that intersects the boundary of this wavefront. One end of the box will be have electric field vectors with magnitude $$\propto r_{old}^{-2}$$, while the other end will be $$\propto r_{new}^{-2}$$. These are not equal, so $$\nabla \cdot E \neq 0$$. But since there's no charge in the box, this should be impossible according to Gauss's law!

What's going on?

• "These are not equal, so $\nabla\cdot E\ne0$." Why do you say this? Jul 14, 2020 at 3:27
• @Sandejo: Because there would be more electric-flux on one side of the box than the other, since one side feels the old field while the other feels the new one. Jul 14, 2020 at 5:17
• Note that it's not just the magnitude of the field that changes. The old field points away from the old position, while the new field points away from the new position. Jul 14, 2020 at 5:41
• @Sandejo: The difference in angle shouldn't be nearly enough to make up for the difference in magnitude, though. For example, on the mostly-perpendicular faces, if $r_{old}$ is far enough away that $\cos(\theta) > 0.5$, then the contribution to the flux from the vector's angle can at most be doubled, since it's always true that $\cos(\theta) \leq 1$. But doubling the distance will reduce the magnitude of the vectors by a factor of 4. Jul 14, 2020 at 21:35
• Jul 16, 2020 at 21:31

You're right: if there were just a Coulomb field outside some expanding shell, and a different Coulomb field inside the shell, then Gauss's law wouldn't hold, as can plainly be seen by drawing a Gaussian surface that straddles the shell.

However, the shell itself contains an additional, transverse electric field. This is the pulse of radiation produced by accelerating the charge, and it ensures that the flux through the Gaussian surface is zero. To see this visually, note that having zero flux through a Gaussian surface is equivalent to an equal number of electric field lines enter and exit.

Now consider the Gaussian surface drawn in red. Four field lines enter it radially and only one exits radially. But three extra field lines exit transversely, so the radiation field ensures that Gauss's law keeps working. (And it keeps working no matter how quickly you kick the charge: kicking it faster makes the shell narrower, but the radiation field larger as well.)

In fact, this is one of the nicest ways of deriving the radiation field; see Appendix H of Purcell and Morin, Electricity and Magnetism for a full derivation using this method.

• This is probably the missing piece. Could you include a simple example that shows the math balances out with the inclusion of transverse waves? Jul 17, 2020 at 17:46

Gauss' law holds for all classical electromagnetism, including moving sources and electromagnetic waves. Your key mistake is here:

These are not equal, so ∇⋅E≠0.

The mere fact that the two are not equal in no way implies, by itself, a violation of Gauss' law. You must actually evaluate the divergence of the field to find out if it is non-zero. It depends on the details of how it transitions from one to the other. In this case, the fact that these waves are solutions to Maxwell's equations ensures that the transition is such that Gauss' law is satisfied everywhere.

Note that while Gauss’ law is satisfied at all times, it is not the only law involved. Both Ampere’s law and Faraday’s law are also involved. Due to those as the wave actually traverses the box there are large (same size as the change in the field) transverse fields generated. These can be calculated explicitly using the Lienard Wiechert potentials.

• It's almost trivial to see that the flux through the two parts of the box will not balance out. For example, consider a box that's almost entirely on one side of the wavefront, except for one face. Then the only difference between the flux of this box and one entirely in a static field will be the flux through that one face; thus in order for it to balance, the flux through that face must be the same in both cases. But the flux through that face decreases with $𝑟^{−2}$, so for drastically different $r_{old}$ & $r_{new}$, it will be drastically different. Jul 14, 2020 at 22:29
• You are forgetting the other parts of the box and the angles of the vectors through the surfaces.
– Dale
Jul 14, 2020 at 22:35
• The other parts of the box are irrelevant since we can make the portion outside the wave arbitrarily small. As for angles, it should be obvious that the increase to $\cos(\theta)$ won't be able to contribute an eg. 4x increase when doubling $r$, but for a proof see my response to @Sandejo above. Jul 14, 2020 at 23:24
• They are relevant. When doubling r, for example, the flux through the sides will greatly increase. It cannot be arbitrarily small for both the before and after configuration. Any time you have a scenario that you think violates some well known law and your justification involves something that is negligible, chances are it is actually not negligible
– Dale
Jul 14, 2020 at 23:42
• You are right. I was describing before vs after. During the passage of the wave there are large transverse fields due to Ampere’s law and Faraday’s law. These are not negligible in any box configuration since they are the same size as the change in the field.
– Dale
Jul 15, 2020 at 2:22

Let's draw a box that intersects the boundary of this wavefront. One end of the box will be have electric field vectors with magnitude $$\propto r_{old}^{-2}$$, while the other end will be $$\propto r_{new}^{-2}$$. These are not equal, so $$\nabla \cdot E \neq 0$$. But since there's no charge in the box, this should be impossible according to Gauss's law!

This analysis doesn't take into account that the electric field vectors are also changing along the sides of the box.

Gauss's law would only be violated if the point charge was sending out electric field vectors that were parallel to the sides of the box, in which case OP would be correct, we could just reduce the total flux through the box surface to (far end flux) - (near end flux). But the electric field vectors actually extend radially outward from the point charge, so there will be nonzero flux through every side of the box, not just the far end and the near end. The net flux through all sides of the box remains zero when the charge is shifted as a result.

Here's a picture of the difference. The electric field vectors are perpendicular to the electric potential lines shown in the picture. • "This analysis doesn't take into account that the electric field vectors are also changing along the sides of the box" - It does. We're allowed to move the box s.t. the surface area of the box-sides resting on the $r_{new}$ half of the wavefront is arbitrarily small. In that case the sides will have the same flux as the case where the charge never moved. (cont.) Jul 17, 2020 at 23:19
• (cont.) "Gauss's law would only be violated if the [..] electric field vectors were parallel to the sides of the box" - We can also make the box as small as we want, or wait for the wavefront to be far away, so the contour lines you've drawn (which are not the same as field lines, btw) can be as arbitrarily-close to parallel as we please. Jul 17, 2020 at 23:26
• If you’re so far away that the field lines are basically parallel, you’re also so far away that the difference in field strength between the two ends of the box (from shifting the charge) is basically negligible. And they’re never actually parallel, so you still get nonzero flux through the other sides, which is what makes the net flux equal to zero. The stuff you’re trying to ignore is exactly the stuff that is making Gauss’s law continue to hold in this situation. Jul 17, 2020 at 23:33
• You don't need to deform the box. You move it so that the flux through the sides on one half of the wavefront is negligible, leaving only the flux through the surface area of the back-face on that half of the wavefront. It should now be trivial to see that the total flux is non-zero (without transverse waves, see above answers), since the total flux is zero when the surface is fully embedded on one side of the wavefront, and the flux through the back-face changes when it moves to the other side. Jul 17, 2020 at 23:40
• The issue is that your wavefront’s intersection with one end of the box isn’t a flat plane, it’s a circle. You’re mistakenly taking the limit as the box gets infinitely far (when the wavefront would be a line parallel to the end) and treating it as if it was the same as the actual case when the wavefront is spherical. Jul 17, 2020 at 23:46