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Moving charged particle produces a magnetic field. If a gravitational wave passes by stationary charged particle and makes it oscillate together with the space, would that movement produce an EM wave?

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    $\begingroup$ +1 Have a look at the modes of distortion of a gravitational wave, en.wikipedia.org/wiki/Gravitational_wave and how the LIGO detector works, if you have time. Your question may well be a duplicate though.... $\endgroup$ – user154420 Jun 28 '17 at 10:19
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Light is a classical phenomenon, an electromagnetic wave described by sinusoidally varying electric and magnetic fields.

Photons are quantum mechanical elementary particles which are the building blocks of the classical wave, .

An interaction of photons and gravitons belongs to a quantization of gravity, which is in the research stage, but effective models do give photo production from graviton particle interactions, fig 4.

gravphot

Considering that a gravitational wave consists of gravitons an accumulation of graviton-charged_particle interaction will build up to classical electromagnetic waves.

So the answer is yes, depending on a definitive model of quantization of gravity.

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    $\begingroup$ Why does quantum mechanics, or photons or gravitons, need to be involved at all? The question makes perfect sense in the classical arena, and it should correspondingly have an answer in terms of classical electrodynamics over curved spacetime. $\endgroup$ – Emilio Pisanty Jun 28 '17 at 16:16
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    $\begingroup$ @EmilioPisanty because for me, a particle physicist, it answers the question without the full panoply of combining gravitation with maxwell's equations, is the simplest solution. Go ahead and answer for the classical case $\endgroup$ – anna v Jun 28 '17 at 17:00
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    $\begingroup$ @EmilioPisanty not the least reason, that quantum mechanics is the basic underlying framework of nature, as far as our knowledge goes. $\endgroup$ – anna v Jun 28 '17 at 17:21
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    $\begingroup$ And yeah, I look forward to a full classical solution. As mentioned out in chat, the problem is nontrivial, because the proper accelerations are zero, so it requires some rather careful analysis to work out correctly. $\endgroup$ – Emilio Pisanty Jun 28 '17 at 17:39
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    $\begingroup$ @EmilioPisanty that is why I answered from the bottom up. It is simpler to know that yes, there will be radiation. $\endgroup$ – anna v Jun 28 '17 at 18:27

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