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Suppose a charge is at rest in the laboratory frame. Also, say two observers in two vehicles A and B are approaching towards it. Let A be accelerating towards it and B be moving uniformly towards it. Will A observe the charge radiate and B observe it not radiate?

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  • $\begingroup$ Can we also ask how the presence of a gravitation field might affect this? For example, if a charge in in "free fall" it will accelerate due to the gravity and if "C" is supported and not falling, but still in the gravitation field, does this count as the charge accelerating and therefore radiating? $\endgroup$ – Paul Young May 31 '19 at 15:12
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    $\begingroup$ @PaulYoung: The Wiki article on paradox of radiation of charged particles in a gravitational field is worth reading. $\endgroup$ – Michael Seifert May 31 '19 at 15:18
  • $\begingroup$ If the lab frame is inertial, the charge does not radiate. Therefore no observer sees it radiate. Why would you ever think otherwise? $\endgroup$ – WillO Jun 4 '19 at 3:28
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According to relativity, "B" will observe the charge in uniform motion, not radiating. "A" will see the charge accelerating, but not radiating. Indeed, the laws are the same for "inertial" reference systems, but are different when the phenomena are seen from non-inertial reference systems. "A" does not have an inertial reference system, thus "A" will not see the particle radiating, as would happen for a particle accelerating in an inertial reference system.

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    $\begingroup$ So the statement that an accelerating charged particle emits radiation is true only from the perspective of an inertial observer? I never saw that qualifier attached. Any reference? $\endgroup$ – mithusengupta123 Jun 5 '19 at 12:56
  • $\begingroup$ The sentence you wrote is slightly ambigous. I stated that a particle radiates only if it is accelerating with respect to an inertial system. Typically this point is discussed when the Maxwell equations are given, saying that they are valid for inertial systems. The relativistic formulation is useful for this, see en.wikipedia.org/wiki/… $\endgroup$ – Doriano Brogioli Jun 5 '19 at 13:21
  • $\begingroup$ Thanks. So Maxwell's equations change when transformed to a non-inertial frame and that's under the jurisdiction of general relativity. And a proper treatment would reveal that if a change accelerates w.r.t a non-inertial observer, it would not radiate. Do I get it? $\endgroup$ – mithusengupta123 Jun 5 '19 at 13:28
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    $\begingroup$ Yes, the Maxwell's equations change in an accelerated reference system. Writing the transformed equations is not difficult and in principle does not require the general relativistic treatment (which however helps to keep them compact). So it is possible that a charge is accelerating with respect to an accelerated reference system, but it does not radiate. This happens when it is in uniform motion with respect to an inertial reference system. Btw, "inertiality" is something absolute, also in relativity... $\endgroup$ – Doriano Brogioli Jun 5 '19 at 13:46
  • $\begingroup$ Thanks. This answer cured my headache. :-) $\endgroup$ – mithusengupta123 Jun 5 '19 at 13:48
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B observes it not radiate, and for A we need a longer story.

Before treating A, I'll treat a related but easier case: an inertial observer observing the fields of an accelerating charge. In this case the charge puts energy continuously into a term in the electromagnetic field energy tensor which propagates away from it at the speed of light, and therefore it radiates continuously. However, the radiative nature of the field only becomes fully unambiguous once the charge stops accelerating.

But the question concerned another case, namely an inertialy moving charge observed by a non-inertial observer A. The first step is to say what you mean by "observe" for an accelerating observer. To do it thoroughly and carefully we can adopt the methods of general relativity. For example, to find an observed energy one takes the 4-momentum $\bf p$ of the entity being observed and the instantaneous 4-velocity $\bf u$ of the observer and forms the combination $-\bf p \cdot u$. In this way one can determine how the electromagnetic field of the charge will appear to observer A: one can determine its effects. But the question whether or not A considers the field to be a radiative field has no single answer until one agrees on a definition of a radiative field in this context. I have read enough about this to discover that this is a non-trivial issue, and it usually gets solved by making an appeal to the far field, and inquiring whether energy is carried away to null infinity, and in order to do this we need to let there be nice simple asymptotic conditions, such as both parties moving inertially. If we do this then I think A has to conclude that the charge did not radiate.

Now suppose A is standing still on a gravitating planet and the charged particle is in free fall. It seems to be the same situation, doesn't it? But I think it is not! The interpretation is different. Now the particle does radiate according to a distant inertial observer, and it radiates according to A too. This conclusion doesn't break the equivalence principle, because it is a conclusion about how we interpret the situation at null infinity, rather than about the local forces and energy movements. However I will admit that I am a little hazy on this matter and will gratefully accept the correction of anyone else who can write in and explain it more fully.

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