According to the equivalence principle, a freely falling observer constitutes an inertial frame. Thus, locally, Maxwell's equations apply in their usual form. According to these equations, an accelerated charge should radiate electromagnetic energy. But in this frame, a stationary charge on the earth does indeed accelerate, meaning that it should radiate. So why don't we observe that stationary charges in gravitational fields, like that of the earth, radiate electromagnetic radiation? And do we observe that charges in free fall radiate (which they shouldn't according to the equivalence principle)?

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    $\begingroup$ Does this answer your question? Does a charged particle accelerating in a gravitational field radiate? $\endgroup$
    – Thorondor
    Oct 18, 2020 at 10:44
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    $\begingroup$ @Thorondor Well, I did see that question before I made my own, and the first answer did look promising, but it seemed like a lot of people didn't like it, so it wasn't clear whether the answer was correct. $\endgroup$ Oct 18, 2020 at 10:53
  • $\begingroup$ @Thorondor Oh, and another answer said that radiation had been detected, unlike the first. So again it isn't clear what the right answer is. $\endgroup$ Oct 18, 2020 at 10:57
  • $\begingroup$ Many years ago, when I was a grad student I asked Stephen Hawking this. He could still just about talk, and he gave the the same answer as @Alexey Bobrick: radiation is frame dependent. $\endgroup$
    – mike stone
    Jun 18, 2021 at 21:41

2 Answers 2


This is the answer I have given in The Large and the Small

The paradox concerning the radiation of a charge in a gravitational field can be resolved using either classical electromagnetism or in quantum electrodynamics. The resolution in quantum electrodynamics is interesting because it shows directly that so-called “virtual” photons can be observable. The paradox is this: it is a well recognised tenet of classical electromagnetism that an accelerated charge radiates, but we observe that a charged particle at rest at the Earth’s surface does not radiate, in spite of the fact that it has a proper upwards acceleration relative to an inertial reference frame. A charged particle, in free fall in a gravitational field, will be seen to radiate by a stationary observer on the ground. One would think it should lose energy, and slow down relative to a freely falling neutral particle. But if it did, then an observer in free fall with the charged particle would be able to distinguish free fall in a gravitational field from inertial motion in flat spacetime, both from a change in motion and from the presence of the radiation. If that were so, it would violate the equivalence principle.

This can be understood by thinking more carefully about how Maxwell’s equations describe radiation. An electromagnetic wave is a solution of Maxwell’s equations. What we see as radiation is actually a fluctuation in the electromagnetic field. If the observer and the charge are comoving, whether they are supported or in free fall, or even in oscillatory motion, no change in the electromagnetic field will be seen (to close approximation. A detailed analysis showing the transformations between these perspectives has been given by Rohrlich F., 1965, Classical Charged Particles, Addison-Wesley). So, no radiation will be seen. If the observer is supported by the Earth’s surface, and the charge is in free fall, he will see changes in the electromagnetic field. If the charge is supported and the observer is in free fall, the observer will see similar changes in the electromagnetic field.

In both cases the changes look like radiation, but although the appearance is the same, for the inertial charge the energy of the apparent “radiation” comes from the force accelerating the observer, not from the charge. The cause of the paradox is that to describe the detection of photons as “radiation” is misleading. The change in the electromagnetic field due to the acceleration of a single charge can be transformed away. This is fundamentally different from the radiation of a light bulb, where there are many charges moving differently. The motion of those charges cannot be transformed away through a suitable choice of reference frame.

Seen from the perspective of quantum electrodynamics, the paradox casts light on whether photons in the electromagnetic field are real or virtual. A charge is surrounded by photons containing the energy of the electromagnetic field generated by the charge. If the observer is not accelerating relative to the charge these photons are not observable. They are described as virtual by some authors. The acceleration of the observer relative to the charge changes the status of some photons in the field. They become observable. The observation of these photons in no way alters the behaviour of the charge, but it illustrates that the word “virtual” is misleading when applied to the photons which constitute the electromagnetic field.

  • $\begingroup$ How is quantum electrodynamics needed to explain this? This is a purely classical question, at least as far as QM is concerned. Also, I still don't understand how you can "transform away" electromagnetic radiation. If freely falling charges emit radiation, then they will slow down relative to a neutral particle, causing the distance between the two particles to increase, which is a violation of the EP. Emission of EM radiation causes the Abraham-Lorentz force, which is a very real force (unlike gravity), meaning I can't see how you would transform it away. $\endgroup$ Oct 18, 2020 at 12:59
  • $\begingroup$ QED is not needed. I gave both classical and quantum explanations. The explanations are complementary, but ultimately not distinct. The radiation is transformed away through choice of reference frame (since gravity is understood in general relativity). If the charge is not moving in the reference frame, no radiation is seen. As said, the freely falling charges in a non-inertial (supported) reference frame give changes to the e.m. field. These changes have the appearance of radiation, but are not radiation as any energy is supplied by the (supported) reference frame. $\endgroup$ Oct 18, 2020 at 13:58
  • $\begingroup$ I am sorry, it seems like I read your answer a bit fast, and I only noticed the QED part. But anyways, even after reading it more carefully, I still don't quite understand. On one hand, it makes sense that electromagnetic radiation only is observed if there is relative acceleration between the observer and the charge, as the equations of electromagnetism in accelerated frames are very similar to Maxwells equations. But on the other hand, it still seems like you couldn't just transform away the Abraham Lorentz force. $\endgroup$ Oct 18, 2020 at 14:13
  • $\begingroup$ Also, what do you mean when you say "for the inertial charge the energy of the apparent "radiation" comes from the force accelerating the observer, not from the charge". $\endgroup$ Oct 18, 2020 at 14:15
  • $\begingroup$ Put it this way. Just as in a non-inertial reference frame we observe a "fictitious" force, which appears to act on an inertial body, although there is in fact no active force, we also observe a "fictitious" radiation, which appears to radiate from an inertial charge, although it does not actually do so. $\endgroup$ Oct 18, 2020 at 14:20

But in this frame, a stationary charge on the earth does indeed accelerate, meaning that it should radiate. So why don't we observe that stationary charges in gravitational fields, like that of the earth, radiate electromagnetic radiation?

Three observations:

  1. in the free-falling frame, the earth-bound charge is moving with almost constant acceleration. Radiation field of so moving charge is very different from ordinary radiation. There are no oscillations, the field just has additional small component with interesting angular pattern (non-isotropic)

  2. charge having radiation field in free-falling frame by itself does not apparently imply it also has radiation field in the Earth frame. EM fields need to be transformed between accelerated frames and the result is not immediately obvious.

  3. even if it turns out there is some kind of radiation field in the Earth frame due to earth-bound charge, based on the standard retarded EM field of slow accelerated charge it is so weak it is very hard to measure. Especially due to number and mobility of other charges on the Earth.


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