Let us consider two charged particles travelling at a uniform velocity, V, as seen from a frame of reference A . Now let us consider a frame of reference, B, which is also travelling at a uniform velocity V with respect to A. Now the interaction of the two charged particles as seen from A would give rise to EM waves. (Initially, I thought that acceleration of charged particles is necessary for EM waves, but an answer to my last question shows a system of equations for this interaction giving an EM wave). Now let us consider the interaction of the two particles as seen from B. As seen from B, this interaction would be an electrostatic interaction. But the EM waves once generated will propagate everywhere in space and hence should also be detected in the reference frame B. And at the same time there doesn't seem to be any way in which the waves can evade the reference frame B. Although I admit I am unaware of the way field transformation takes place, I still guess that if a field is changing in a wave-like manner in one frame of reference then the field to which it gets transformed must also change in such a manner. So, what exactly happens in such a scenario ?
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$\begingroup$ Yes, and I am with your question. Now, from the beginning I am telling you that I believe (but not yet checked thoroughly) that the answer to the question is that a charge travelling with constant velocity emits an e.m. field in its own direction of movement. This is what I want to check. $\endgroup$– SofiaCommented Feb 27, 2015 at 19:18
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$\begingroup$ @sofia : But , in this case I have extended the question to electrostatic interactions too. Have you checked the first answer that I received to my last question which claimed that a steady magnetic field is produced by uniform velocity charge ?I guess that answer can be refuted ... $\endgroup$– Agnivesh SinghCommented Feb 27, 2015 at 19:25
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$\begingroup$ @sofia: Except for my doubt of validity of Maxwell's equation in case of induced fields I would like to ask if you have completely solved those equations to obtain a solution to it, in order to be thorough . $\endgroup$– Agnivesh SinghCommented Feb 27, 2015 at 19:28
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$\begingroup$ That wasn't a good idea. You don't solve a problem by complicating it. The problem should have remained with a single travelling charge. Two charges attract or repel one another and their movement is accelerated. Thus, you ask another problem that was left unexplained in your former question. With accelerated charges, no doubt one has e.m. waves, just, the acceleration can be small, therefore the e.m. waves emission negligible, or big acceleration, in which case the e.m. waves can't be neglected, and one has retarded Lienard-Wiechert waves. $\endgroup$– SofiaCommented Feb 27, 2015 at 19:57
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$\begingroup$ @Sofia : I am not actually intending to make it complicated. I am trying to explore it from every possible angle.In order to understand things truly one has to endure all the complicacies coming on the way. $\endgroup$– Agnivesh SinghCommented Feb 28, 2015 at 3:32
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