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I am currently rather uneducated on the subject, but I was thinking of a general relativity thought experiment. Say I take a charge from infinity and give it velocity to orbit a planet in a circle. Now from the earth's perspective the charge is accelerating thus it should radiate EM waves. But Einstein says there is no experiment you can perform (locally) to determine if you are in free fall in a gravitational field as opposed to floating in free space. But if what I described was the case then there would be an experiment you could perform; you could check for the emission of a photon. Where is the problem?

  • This reminds me of the apparent paradox in atomic physics where if electrons are orbitting a positive nucleus they are being accelerated and should emit EM waves, but they don't. This question was discussed at the time when quantum mechanics was being developed for small objects like atoms. – tom Nov 07 '14 at 22:14
  • There is an experiment you can do to determine if you're falling in a gravitational field versus floating in space. Your feet are being pulled on harder than your head (tidal force) and this is detectable. For example, this comet knew it was falling in an gravitational field: http://en.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9 – Brandon Enright Nov 07 '14 at 22:57
  • yes but locally it is not possible, one can get arbitrarily close to the point charge to measure this photon –  Nov 07 '14 at 23:04
  • Are you sure that your charge will orbit the planet? It is rather likely to surround the planet and return to infinity. Unless the charge hits the earth, which you say it's not the case, its movement in the gravitational field, which is a classical field, should be symmetrical in time. The charge should return to the infinity with the same energy as it had in the beginning. – Sofia Nov 08 '14 at 11:38
  • By the way, the comparison with an electron in the electric field of the nucleus is not so good. The electron movement has a wavelength and interfere with itself. I a naïve atomic model, the electron moves on an orbit on which the interference is constructive, and is not allowed to move on any other orbit where the interference is destructive. This is why the electron orbits are discrete. – Sofia Nov 08 '14 at 11:46
  • Imagine a point P on the electron orbit. In a naïve atomic model, the radius of the orbit is exactly so that the wave just leaving the point P interferes IN PHASE with the wave just returning from surrounding the nucleus. Orbits of radii such that the interference is destructive, are, of course, impossible. However, if surrounding the earth, the radius is so big and energy needed for the object renders the wave-length so small, that interference with the wave returning after surrounding the earth is not possible. – Sofia Nov 08 '14 at 12:12

2 Answers2

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In a consistent theory the electromagnetic potential should be coupled to gravity. Then the body does not radiate on geodesics.

It can be easily shown. Consider the usual Minkowski space and a charge at rest. Surely it doesn't radiate. But there are coordinates in which its worldline is curved. The radiation is still absent, because affine connection components appear in the Maxwell equations. Same goes for arbitrary spacetimes.

So no, the orbiting charge does not emit radiation.

Prof. Legolasov
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  • But in other reference frames magnetic fields come and go, right? These are just static, or move with the charge? – Alan Rominger Nov 08 '14 at 02:42
  • Yes, it does. So does the electric field. You can evaluate it by transforming $F_{\mu \nu}$ from one frame to another. But it does not change the fact that there is no radiation. – Prof. Legolasov Nov 08 '14 at 03:10
  • It would be nice if the resolution of this issue were as simple as this, but it's not. See the answers to the question that this question duplicates. –  Nov 10 '14 at 15:41
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The equivalence principle does not apply to the local rest frame of the orbiting charge, at least not in the way you are applying it, because such an application of the equivalence principle clearly involves taking into account not only the charge but also the electromagnetic field it carries. However the EM field is not an object that is localized to some neighborhood of the charge worldline, it extends out to spatial infinity. The equivalence principle does not apply to such a configuration-it only applies to those which are confined to local regions of space-time.

FenderLesPaul
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  • This is a nice answer. How about moving it to the question that this question duplicates? –  Nov 10 '14 at 15:41