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We know that an accelerated electric charge produces electromagnetic radiation.

We also know that acceleration is relative to an observer.

Take electrically neutral observers A and B, who are accelerating relative to each other. B is at rest with C, a piece of electrically charged matter. This means C is accelerating with respect to A.

A would observe C emitting electromagnetic radiation, and thus, lose energy. And thus, A would condluce that the system consisting of A, B, and C is losing energy.

B would not observe electromagnetic radiation from C (since B is at rest with C). Nor would B observe A emitting electromagnetic radiation, as A is electrically neutral.

But both A and B must agree that the same amount of energy is being lost between them and C.

How is this energy loss observed? Does B observe A's acceleration to be slowing down?

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  • $\begingroup$ You assume a lot of questionable things. Emitting EM radiation does not always mean losing energy. But even if C loses energy, just because B and C share the same frame, this does not mean that C isn't radiating in the shared frame of B and C, or not losing energy in this frame. The frame of A isn't equivalent to frame of B; only one of them can be inertial, where Maxwell's equations hold. $\endgroup$
    – Ján Lalinský
    Commented Jan 26 at 3:15

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In special relativity, acceleration is absolute. Geometrically, the spacetime metric (the geometry of spacetime) is absolute. Straight wordines correspond to inertial observers while curved worldlines correspond to accelerating observers. This is the subject of the twin paradox.

Therefore, it makes absolute sense that a charge is accelerating, and all observers agree that the charge is radiating energy. The issue is that from the perspective of the accelerated observer, Maxwell's equations do not have the same form and an extra radiation term needs to be added. This is to be contrasted with inertial observers which all agree on Maxwell's equation. Therefore, depending on the perspective, the lost energy either correspond to an extra term in Maxwell's equation because the frame is non inertial or just corresponds to the usual radiation from the perspective of an inertial frame.

Hope this helps.

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    $\begingroup$ This gets at the crux of things. The OP's second sentence seems to imply that there's an equivalence between reference frames that are accelerating with respect to each other. But that's not true; it's only true for inertial frames that are moving at constant velocity with respect to each other. $\endgroup$
    – Michael Seifert
    Commented Jan 26 at 3:31
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"B would not observe electromagnetic radiation from C (since B is at rest with C)." In the instantaneous rest system of B and C, B and C would still be accelerating. There is no Lorentz transformation that can stop C from accelerating.

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If A is inertial, and C is an accelerating macroscopic charged body, there is radiation in technical sense (not periodic waves, but consistent acceleration field fading away as inverse of distance in the frame of A), and there is also outgoing flux of EM energy going away from C.

In non-inertial frame, like that of C, one has to define what one means by EM field, and radiation, because there are non-inertial forces which are not due to EM field. With appropriate definition, there can be radiation in this frame as well, despite the charge being at rest. Or there need not be, for another definition. This can be only resolved by formulating EM theory in non-inertial frame and carefully defining terms like electric and magnetic field, fictitious force, radiation. Nota bene in non-inertial frames, energy is not conserved in general, so don't rely on such arguments there.

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