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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.

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  • $\begingroup$ A moving charge does emit EM waves, in the form or EM-radiation. This is how radios work. $\endgroup$
    – D. W.
    Commented Apr 23, 2016 at 22:10
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    $\begingroup$ @D.W., shouldn't it be accelerating rather than moving? $\endgroup$
    – Alfred Centauri
    Commented Apr 23, 2016 at 22:26
  • $\begingroup$ 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. $\endgroup$
    – D. W.
    Commented Apr 23, 2016 at 22:32
  • $\begingroup$ @D.W., right but, and again, moving charge does not necessarily radiate. $\endgroup$
    – Alfred Centauri
    Commented Apr 23, 2016 at 22:37
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    $\begingroup$ 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). $\endgroup$
    – Vitaly Korzhik
    Commented Apr 23, 2016 at 22:38

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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.

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  • $\begingroup$ "energy stored in the fields" means standing waves? $\endgroup$
    – athanos lee
    Commented yesterday

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