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Let us consider a charged particle which is static with respect to a frame of reference $A$. Now suppose that I accelerate the charged particle to a speed $v$.

Now when I accelerated it to a speed $v$, during the course of acceleration the particle would radiate EM waves. Now these EM waves would keep propagating (at least the according variation of EM fields would go on) even when the particle has attained a uniform velocity. Now when the particle is moving with a uniform velocity, there would be time variations in E field and B field as per a particle with uniform velocity. Now let's bring this particle to rest by a retarding force. Again, during this course of acceleration the particle would radiate EM waves which will again start propagating all around.

But now the particle would have come to rest and its field should be static. But due to it's early origination there must be some points in space where the EM waves due to the first acceleration would have reached but about the subsequent events might not have . But this would lead to the implication there must be some points in space where the EM field due to a static electron is still undergoing variations and is not constant.

Can we express a static electric field as a superposition of two EM waves as in this case? Can we explain this sufficiently within the purview of classical theories?

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  • $\begingroup$ yes. special relativty + electrodynamics $\endgroup$
    – image357
    Mar 28, 2015 at 17:36

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The relativistically correct electromagnetic fields of any charge are given by the Lienard-Wiechert potentials.

The waves emitted during the course of acceleration do not belong to the field of the charge, since the waves are vacuum solutions of Maxwell's equations.

So, the presence or absence of radiation has nothing to do with the field being "static" or not.

In the fully relativistic treatment, the fields of the stopped charge are not static in time throughout all space though, but depend on the retarded time as is detailed in the Wikipedia article about Lienard-Wiechert. In the volume of space around the charge to which the "stop" has already propagated, the field is static, however.

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  • $\begingroup$ Could you elaborate the part "The waves do not belong to the field of the charge " $\endgroup$ Mar 28, 2015 at 18:24
  • $\begingroup$ @AgniveshSingh: EM waves are vacuum solutions (i.e. valid solutions of Maxwell's equations in a region free of charge and current) - they exist without need for the presence of a current or charge, and in this sense, they do not belong to the field of any object - they exist on their own. "The field" of an object is the part of the electromagnetic fields that is obtained by solving Maxwell's equation for the charge-current distribution of said object. $\endgroup$
    – ACuriousMind
    Mar 28, 2015 at 18:27
  • $\begingroup$ you mean only for the final configuration? $\endgroup$ Mar 28, 2015 at 18:29
  • $\begingroup$ And regardless of the categorization of the field the net electric field in the presence of a static charge can be time varying at some points ,i.e. at distant points ? $\endgroup$ Mar 28, 2015 at 18:31
  • $\begingroup$ @AgniveshSingh: If the charge was moving before it became static, then yes, that is what Lienard-Wiechert tell you - the information "the charge is static" propagates only at the speed of light, after all. $\endgroup$
    – ACuriousMind
    Mar 28, 2015 at 18:39

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