The well known Larmor formula tells us that an accelerated classical charge radiates a power proportional to the square of the acceleration. It has been shown that a charge in free fall in a gravitational field does not radiate. Since gravity is locally equivalent to acceleration, one may wonder whether a charge at rest in a gravitational field radiates. The answer is that it does. However this radiation is not observable to an observer in the same reference frame! Apparently, the radiation does not carry energy or momentum, or it would be observable, in this reference frame and this would mean that charged matter behaves differently from neutral matter in gravity.

  • How does unobservable EM radiation without energy and momentum make sense (Q1)?

  • Does the physical origin of charge acceleration determine whether radiation is observable or not (Q2)?

  • Is the answer to the question in my title "It depends?"?

Note added:

I did not find the discussion at Does a charged particle accelerating in a gravitational field radiate? and For an accelerated charge to radiate, is an electromagnetic field as the source necessary? sufficiently enlightening.

  • $\begingroup$ You use phrases like "it has been shown" and "the answer is," which IMO is an overstatement. This stuff is complicated and does not lend itself to simple answers that everyone can agree on. There are fundamental ambiguities in defining what constitutes a radiation field, so you can get more than one answer, depending on what definition you like. $\endgroup$ – user4552 Mar 24 '19 at 16:22
  • $\begingroup$ @Ben Crowell I am quoting the wikipedia article, assuming that Rohrlich's results referenced there are generally accepted. If they are not, that is all the more interesting. To bring this out is exactly why I posted the question. Looking forward to your answer. $\endgroup$ – my2cts Mar 24 '19 at 16:33
  • $\begingroup$ Changed the title since it makes the question appear too broad. $\endgroup$ – my2cts Mar 24 '19 at 21:00
  • $\begingroup$ Thanks for both excellent answers. This gives a good level of understanding and the correct direction for further research. $\endgroup$ – my2cts Mar 24 '19 at 21:16

How does unobservable EM radiation without energy and momentum make sense (Q1)?

While an answer of probably_someone in terms of Rindler horizon offers a viewpoint in a global sense, we could also use local perspective. An example of local detector for EM radiation would be something that measures Poynting vector. If we surround the constantly accelerating charge with such detectors that are static in the frame of the charge, they would measure zero and we thus could conclude that there is no radiation. But that statement is only true in this specific reference frame, and since this reference frame covers only part of a spacetime (Rindler wedge) it is not true globally.

Does the physical origin of charge acceleration determine whether radiation is observable or not (Q2)?

Yes. There are two points to note. First, a point charge with electromagnetic field is not a local system. Because the field is long-range it can serve as a probe for other regions of spacetime and introduce effects observable even within an accelerated frame of reference. For example in a curved spacetime (as in nonzero curvature tensor) there would be an additional force term containing retarded Green's function acting on the charge and in principle depending on the whole history of motion of the charge and “global” structure of spacetime. Similar effects would be present even in flat spacetime if there are objects (dielectrics, conductors) interacting with EM field of a charge. Therefore we would have self-force acting on a charge, but also additional EM-fields measurable by detectors near the charge.

The second thing to keep in mind, is that a charge in a constant accelerating motion for all eternity is an impossibility. In a realistic physical situations the acceleration must have a beginning and an end (or at least its character must change). This would mean that “Rindler horizon” in the answer from probably_someone is not really a horizon but an approximation, and what really would be observable is dependent on the nature of accelerating force.

Is the answer to the question in my title "It depends?"?

Here is a statement true globally, independent of the choice of reference frame and the nature of accelerating force. A realistic setup would have the charge starting and ending its motion at some non-accelerating state. This means that charge should start at $i^-$ and end at $i^+$, timelike infinities. Then, if the charge undergoes a phase of acceleration, there would be an EM energy flux at future null infinity ($\mathscr{I}^+$). This mass-energy flux is the radiation of this charge.

An interesting perspective for this problem comes from noting that there is an additional contribution to the “bound momentum” of EM field of a moving charge that is proportional to the 4-acceleration, as first observed by Teitelboim:

  • Teitelboim, C. (1970). Splitting of the Maxwell tensor: radiation reaction without advanced fields. Physical Review D, 1(6), 1572, doi:10.1103/PhysRevD.1.1572.

A more recent reference:

  • Gal’tsov, D. (2009). Radiation Reaction and Energy–Momentum Conservation. In “Mass and Motion in General Relativity” (pp. 367-393). Springer, Dordrecht, arXiv:1012.2846.

contains the following passage:

Therefore the energy-momentum balance of the system consisting of the accelerated charge and its Maxwell field includes three, but not just two, ingredients: the particle momentum, the momentum carried by radiation, and the bound electromagnetic momentum. The radiation momentum can be extracted both from the particle momentum and indirectly from the bound momentum. This explains the origin of radiation of the uniformly accelerated charge, in which case the total reaction force is zero and thus the kinetic particle momentum is constant. While the charge is undergoing a constant acceleration, its bound electromagnetic momentum decreases and is transferred to radiation. Physically, however, the acceleration has to start at some moment and to finish at some moment, and during the stages of acquiring and loosing the acceleration the bound momentum is exchanged with the kinetic momentum. Therefore, the total energy-momentum loss of the charge will be equal to the momentum carried away by radiation. But an instantaneous balance is obscured by the presence of the Schott term. Another simple situation is periodic motion. Since the ambiguous Schott term is total derivative, its contribution vanishes if one integrates over the period, or, equivalently, averages over the period.

| cite | improve this answer | |

According to general relativity, any free-falling frame is an inertial frame. A free-falling observer would observe that an object sitting on the surface of the Earth would be accelerating uniformly upwards. Therefore, the surface of the Earth is a non-inertial frame (in particular, it's equivalent to a uniformly-accelerating frame).

A uniformly-accelerated reference frame in general relativity is described by Rindler coordinates (https://en.wikipedia.org/wiki/Rindler_coordinates). Rindler coordinates have a horizon, the spacetime beyond which is unobservable to an observer in this frame (though it can still be causally affected by the observer). In the uniformly-accelerating frame in which the charge is stationary relative to the surface of the Earth, the radiating components of the electromagnetic field are beyond that horizon, making them inaccessible to the uniformly-accelerating observer but visible to the inertial observer, for whom no horizons are present. The full details of this argument are laid out here: https://arxiv.org/abs/physics/0506049

| cite | improve this answer | |
  • $\begingroup$ I realise that this answer is correct, but it is not very intuitive. How can components of the field be beyond the horizon of the charge itself? And how do these (not) affect the state of motion of the charge? $\endgroup$ – my2cts Mar 24 '19 at 15:21
  • $\begingroup$ @my2cts I should probably clarify that "inaccessible" really means "unobservable." The uniformly-accelerated observer can still affect the spacetime beyond the horizon, but they cannot see anything beyond that horizon. This can be accomplished by, for example, letting go of a bowling ball you've been holding; in your uniformly-accelerating frame, the bowling ball begins to accelerate away until it passes beyond the horizon. This page might help to clarify: gregegan.net/SCIENCE/Rindler/RindlerHorizon.html $\endgroup$ – probably_someone Mar 24 '19 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.