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EM wave phenomenon is usually described as "changing electric field creates changing magnetic field which creates a changing electric field etc."

But fields of a uniformly moving (not accelerating) charge also change, and as far as I can see, all the second-order derivatives in time and space required by the wave equation are in place (because of the 1/r^2 dependence).

So, why doesn't the field of a uniformly moving charge create EM waves all by itself, because of all the second-order derivatives? It seems to me that mathematics requires it.

Volker Siegel
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Vitaly Korzhik
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  • A moving charge does emit EM waves, in the form or EM-radiation. This is how radios work. – D. W. Apr 23 '16 at 22:10
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    @D.W., shouldn't it be accelerating rather than moving? – Alfred Centauri Apr 23 '16 at 22:26
  • Yes, that's true. Charges emit radiation when accelerated. But charges produce electric fields, and their motion (even when moving at constant speed) counts as a changes in the electric field. Jefimenko's equations, will tell the corresponding change in the vector potential, and therefor the EM-fields. – D. W. Apr 23 '16 at 22:32
  • @D.W., right but, and again, moving charge does not necessarily radiate. – Alfred Centauri Apr 23 '16 at 22:37
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    Thanks for posting everyone. I do mean "moving at constant speed", we are not talking about acceleration here. Just imagine the field of a uniformly moving charge - it will be changing both in space and in time. So the question is why those changes do not start the waves (in every point everywhere, actually). – Vitaly Korzhik Apr 23 '16 at 22:38
  • Vitaly, the field of a uniformly moving point charge has terms that go as $1/r^2$ where as radiation goes as $1/r$. To get $1/r$ terms requires non-uniform motion. Remarkably, I have my Jackson book open to page 664 (for a different reason altogether) where equation (14.14) clearly shows this. – Alfred Centauri Apr 23 '16 at 22:41
  • True, but I don't see how this solves the problem. The EM wave equation doesn't require any specific dependence on $r$ or $r^2$, does it? And my question is - why it isn't working (i.e. producing waves) in case of $1/r^2$. – Vitaly Korzhik Apr 23 '16 at 22:45
  • "The EM wave equation doesn't require any specific dependence on r or r2, does it?" - Vitaly, I'm not sure what your thinking. The field of a point charge at rest is not a wave solution to the wave equation. Transforming the field (correctly) to a uniformly moving frame will not turn the field into a propagating wave solution. Please try to edit your question with more details on what you've tried. As it is, I'm just not sure what your thinking is. – Alfred Centauri Apr 23 '16 at 22:51
  • @AlfredCentauri So, just to be clear: the function of the field of a uniformly moving charge (for a stationary observer) is not a solution to the wave equation? – Vitaly Korzhik Apr 23 '16 at 22:57
  • A duplicate indeed. That clears it up. Thanks a lot for your help! – Vitaly Korzhik Apr 24 '16 at 11:53

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A uniformly moving charge creates electromagnetic waves. But the electric and magnetic field contain only velocity dependent terms. If it's accelerating, then the field terms contain acceleration term also and in addition to the energy stored in the fields, there will be radiation also which causes considerable decrease in the energy stored in the fields.

In your case, there is electromagnetic wave. The energy will be then stored in the fields as there is no radiation.

UKH
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