How to show with Maxwells Equations that nonaccelerating charges don't radiate?

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    $\begingroup$ According to Maxwell's equations, does a charge at rest radiate? $\endgroup$ Commented Mar 1, 2014 at 23:54
  • $\begingroup$ true, the at rest situation is much more simple though (the current density is zero). But what about a uniformly moving charge (the current density is non-zero, and v times the dirac delta function)? Thats's when Maxwell's Equations are not so obvious anymore, or at least to me they are not. $\endgroup$ Commented Mar 2, 2014 at 0:02
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    $\begingroup$ But motion is relative. We can't say that the charge is absolutely uniformly moving. Consider the case that, according to you, the charge is at rest. Now, according to another reference frame, in relative motion with respect to you, the charge is uniformly moving. But according to you, the charge is motionless. All we can say is that the two reference frames have relative motion, i.e., that (uniform) motion is a relationship between reference frames, not a property of an object and its associated reference frame. $\endgroup$ Commented Mar 2, 2014 at 1:30

2 Answers 2


Alfred Centauri has almost answered the question for you (actually he has), but he's using knowledge about and properties of Maxwell's equations that it sounds as though you haven't yet met.

Maxwell's equations are covariant with respect to Lorentz transformations. That's a fancy way of saying that they keep their exact same form, and must foretell the same physics for all inertial observers. It doesn't matter whether or not I am moving uniformly relative to a charge: as long as I am not interacting with that charge (i.e. I am being an uncharged, passive observer and not part of the physics), Maxwell's equations must foretell the same physics. Therefore a charge uniformly moving with respect to me cannot radiate because one stationary relative to me does not.

Indeed historically this is what special relativity was all about. Einstein's famous 1905 relativity paper (there were several famous ones on vastly diverse fields of physics written by him that year) was called "Zur Elektrodynamik bewegter Körper" (on the electrodynamics of moving bodies), and he took as a beginning point this invariance of Maxwell's equations. His reasoning was that Maxwell's equations were really the only phsyics we knew at the time that accurately described something moving very fast, to wit: light, and therefore we should give them more weight than the assumed Galilean laws of relativity, whose validity for very swift things we had very few ways to check at the time.

So Einstein upheld the simple proposition that physics should be the same for all uniformly moving observers (as Alfred Centauri's other comment succinctly puts it the physics is a property of the charge, not of who observes it or their reference frames) and assumed Maxwell's equations were correct: thence derived the Lorentz transformations and special relativity to replace Galilean relativity from these assumptions, thus explaining the negative result for the Michelson-Morley experiment.

Edit in Response to Comment:

User Suresh makes the following point:

The principle of relativity is usually attributed to Galileo. One doesn't need SR to see that a charge moving with uniform velocity doesn't radiate. Choose the velocity to be so small so that SR effects can be neglected...

(actually I say more about Galileo and relativity here).

My response is:

Suresh, you are right, that is a good point. As you can say, you can just ignore Maxwell's equations and say by the principle of relativity, the uniformly moving one can't radiate if a stationary one doesn't, and this answers the OP's question in the sense that it gives the right physics. But then the OP would have to use special relativity, and not Galilean, to see that Maxwell's equations don't tell us anything different. His/her question was about Maxwell's equations, and using these to prove no radiation. He/she can't do this with Galilean relativity. The fact that Galilean relativity makes Maxwell's equation seem to foretell different physics for different observers throws up the interesting history.

  • $\begingroup$ except it wasn't Einstein who did this all, but Poincare who was friends with Einstein's advisor, Minkowsky. for some reason, it took about 30 years for Einstein to cite Poincare's work, of which he was told by Minkowsky before starting a work on special relativity $\endgroup$ Commented Mar 2, 2014 at 0:45
  • $\begingroup$ @Aksakal Yes I agree the principle of relativity and how it applied was thoroughly studied by Poincaré and others: my history knowledge is not detailed but I believe it was Einstein that recognised and embraced that the needed transformations would include time dilation. Have you read Poincaré's relevant papers (I haven't: would appreciate its name and links if you have one)? $\endgroup$ Commented Mar 2, 2014 at 1:00
  • $\begingroup$ i'll try to look up Poincare's work, it was in some philosophy journal, thus wasn't very technical $\endgroup$ Commented Mar 2, 2014 at 1:04
  • $\begingroup$ @Aksakal That would be great if you can find one, thanks! $\endgroup$ Commented Mar 2, 2014 at 1:05
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    $\begingroup$ @suresh Also, this may be interesting. $\endgroup$ Commented Mar 2, 2014 at 3:06

actually, they do radiate. prrof: Cherenkov radiation

  • $\begingroup$ Cherenkov radiation is a consequence of the medium and the finite speed of light; also the dispersive effects of the medium must be taken into account as well. This greatly convolutes the problem, especially if you are wanting to solve it with Maxwell's Equations. The question was directed more towards free space... $\endgroup$ Commented Mar 2, 2014 at 3:30
  • $\begingroup$ Are such Cherenkov radiating particles moving uniformly relative to the medium? $\endgroup$ Commented Mar 2, 2014 at 3:44
  • $\begingroup$ @AlfredCentauri, most likely not when Cherenkov found them, but as I recall that's not essential for them to radiate $\endgroup$ Commented Mar 2, 2014 at 3:45
  • $\begingroup$ See, for example: profmattstrassler.com/articles-and-posts/… $\endgroup$ Commented Mar 2, 2014 at 4:16

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