# Can the question of a gravitationally accelerated charge radiation be tested experimentally?

I know that the question of radiation from a gravitationally accelerated charge has been discussed extensively at Does a charged particle accelerating in a gravitational field radiate?. Yet the experimental aspect hasn't been touched up on. My question strictly on this topic can appear dumb, yet for a layman like me it appears as a self-evident proposition: why can't it be verified by a test or some observation - whether a freely falling charge radiate photons, how strongly and relative to which frame of reference it does or does not radiate? If it cannot be tested practically, could you please explain why, if possible. Many thanks.

• Can we detect the motion of a freely falling charge by its electromagnetic emissions? Yes, of course we can. Why would you think otherwise? If you want to do the experiment, there is nothing to stop you from dropping a few charged metal balls and measuring the electric field from a distance. May 26, 2016 at 2:42
• @CuriousOne, the question is about detection of radiation, not detection of motion of charged particle. Detection of changing electric field does not mean detection of radiation. Nov 23, 2018 at 14:55

There is some misunderstanding here.

whether a freely falling charge radiate photons, how strongly and relative to which frame of reference it does or does not radiate

if you mean a charge in free fall.

In this calculation:, from the conclusion

It is found that the "naive" conclusion from the principle of equivalence - that a freely falling charge does not radiate, and a charge supported at rest in a gravitational field does radiate - is a correct conclusion, and one should look for radiation whenever a relative acceleration exists between an electric charge and its electric field. The electric field which falls freely in the gravitational field is accelerated relative to the static charge. The field is curved, and the work done in overcoming the stress force created in the curved eld, is the source of the energy carried by the radiation. This work is done by the gravitational field on the electric field, and the energy carried by the radiation is created in the expense of the gravitational energy of the system.

italics mine

So it is not a charge in free fall that radiates but a charge supported and stationary. To measure experimentally on earth the electromagnetic radiation of a stationary charge is not possible

a) because the gravitational constant is so small that any radiation will have such small energy that it cannot be detected.

Look at the formula

which is equivalent to the power radiated by an accelerated charged particle (Larmor formula), where the acceleration is replaced by g.

b) accumulating charges as in van der Graaff accelerators induce a number of electromagnetic interactions which will radiate, not to ignore also the black body radiation, even in vacuum, and the electromagnetic coupling is orders of magnitude larger than the gravitational, which will inundate any signal

It is only in cosmological observations that one might need a contribution from such a mechanism , as discussed in the paper:

Motz suggested that the huge radiation emerging from quasars may be created by charges located in the strong gravitational fields close to the surface of the quasars. Although the current explanation for this phenomenon is different, radiation from charges located in strong gravitational fields can still play a role in certain cosmological phenomena.

Edit with some more quotes for clarification:

A freely falling charge in a uniform GF follows a geodetic line in this system, and it is not subject to any external force. The electric field of the charge follows similar geodesics. The charge and its field both are located in the same frame of reference, and in that frame their relative situation is similar to the one existing between a static charge and its field in a free space. No relative acceleration exists between the charge and its electric field, and we conclude that a freely falling charge does not radiate.

......

The electric field of a charge supported at rest in the lab against GF seems static, but it is not. The electric field, which is an independent physical entity, is not supported with the charge, and it falls freely in the gravitational field. There is a relative acceleration between the charge and its electric field, the field is curved (both in the lab system and in the freely falling system), and a stress force exists between the charge and its field. The (freely falling) electric field follows the system of reference characterized by the geodesics.

The italics (mine) explain the difference between the resting mass and the electric field. The mass reacts to the gravitational force, the field is freely falling.

The electric field is detached from the supported charge, and it is not supported against gravity as the charge is. Hence the electric field falls in a free fall, and it has an acceleration g relative to the supported charge. In the freely falling system, which also has an acceleration g relative to the supported charge, the charge is accelerated upward with an acceleration g

They go on to demonstrate a non vanishing Poynting vector, i.e. electromagnetic radiation.

Hand waving, I see it as the mass/charge_carrying part of the particle settling to a lower gravitational level as the energy is being radiated by the distorted electric field, and as I said it is a very small effect , significant maybe for cosmological dimensions.

• @CuriousOne if you read my answer you would see that I say it is not possible in laboratory conditions due to the constants. One is talking of radiation, not of a field, and the paper seems legitimate to me. In space even with uniform velocity a charge's field will be changing. The thing is to look at the equations, and they have done it. May 26, 2016 at 5:33
• Let's start with the obvious: has the paper been peer reviewed and published in a journal? There is no physical difference between "the field" and "radiation". Neither the charge, nor the field, nor the gravitating body, nor your test equipment care about whether you are looking at the near field or the far field. The field is the field. Can you measure the changing field of a freely falling body? Yes. What does it depend in? Your relative state of motion, no matter what causes it. May 26, 2016 at 5:46
• Look, you are saying no more and no less that a charged gravitating body is an eternal source of energy but you can't identify the mechanism by which that energy is generated. Currently you are simply grasping for ever more straws, none of which exist in the theory rather than examining whether maybe you are resolving the paradox in the wrong way. A charge sitting in a gravity field does not radiate. That may not agree with a naive interpretation of the equivalence principle, so one has to work on a non-naive one, which you can not offer, you rather violate energy conservation. May 26, 2016 at 16:41
• I am not asking you to trust anything, I am asking you to explain where the energy comes from. The "gravitational field is the source of the energy" is not a sufficient explanation. How is the gravitational field the source of the energy? How does it change while the charge is bleeding this energy into space? How does gravity convert fermions into bosons? Is this a mechanism that violates lepton number conservation? How does it work? These are important questions and you have to be able to answer them if you want me to believe that a charge sitting on a table sends out em radiation. May 26, 2016 at 18:16
• LIGO gets its signals from two times twenty solar masses falling into each other. I can clearly see where the energy is coming from in that case. Here you have a few electrons sitting on the table, doing exactly nothing. You know how it is in physics... when nothing happens, no work is being done, either. So where is the energy coming from when nothing happens? May 27, 2016 at 5:49

I think it can be tested experimentally. Absent a gravitational field, it is possible to detect EM radiation from a moving charge: if the electrical influence of the charge at a distance is delayed relative to the position of the charge, as is expected from the finite propagation speed of disturbances in the EM field, then I'd say radiation is being detected.

A charged sphere could be dropped on a hard plate so it bounces. The bounce would be "chirped", increasing in frequency until the bouncing stops. The frequency would be low, perhaps as low as a few hundred hertz at its maximum. However, that signal should be detectable with a sufficiently sensitive apparatus.

There would be three possible components of the signal: one that would occur at the moment of bounce when the sphere is accelerated, and two components corresponding to the upward motion and the downward motion, both of which would be in free fall.

I'm not entirely sure that the two kinds of signals would be distinguishable, but considering that the frequency of the free-fall components would be somewhat independent of the details of the brief acceleration at the moment of bounce, I suspect that the components could be examined separately.

Testing whether a charged mass stationary in a gravitational field radiates is much more difficult, because there would not be a particular frequency to measure.

Detecting the emitted photons from a charge accelerating in Earth's gravity looks like a hopeless task, given that the radiated power is tiny (see formula 9 in Anna's answer). But we can try to exploit the fact that the number of emitted photons always diverges no matter how small the acceleration. While very low energy photons (so-called "soft photons") cannot be detected directly, the fact that such photons have been emitted can be detected indirectly by measuring decoherence effects.

For example, one can consider doing a two slit interference experiment with electrons where one slit is put above the other slit. Electrons on paths that end on some spot on the screen will emit soft photons, paths that pass through the upper slit will do so slightly differently than paths that pass through the lower slit. The state of the electromagnetic field is similar but not exactly the same for the two paths. The overlap between the two states will depend on where on the screen the electron ends up, the squared modulus of this function multiples the interference pattern that you would get without this effect. The larger the distance between the two slits, the larger this decoherence effect due to the emission of soft photons will be.

• @anna_v Many thanks for the thorough answer. So, to recap, if my layman's understanding is correct, an electrically charged brick that falls freely towards the Earh doesn't radiate the slightest iota, because the electric field itself falls 'with' the brick alike, accelerating in the same way - equivalence principle stands 'iron-clad'. A stationary, electrically charged brick on the surface of the Earth DOES RADIATE. But both of those things, alas, at present point of history can't be tested experimentally cos the (non)expected radiation is prohibitively weak. May 26, 2016 at 23:37
• Yet one aspect still looks a bit unclear to me: because in reality there are no such things as uniform gravitational field, and the tidal forces are always there, how would (or not) electric field of the freely falling charge (towards the Earth or Moon) interact, if that's a proper word, with the non-uniform gravity field? Will it cause any radiation or not? Obviously, th gravity acceleration at different 'outskirts' of the electric field of the charge at any given time will be different to a greater or lesser degree. Does it have any consequences in terms of any radiation? May 26, 2016 at 23:45
• A double slit is a very imprecise experiment, as it is. If anything this would require a very precise atomic or nuclear physics experiment. May 27, 2016 at 4:53
• This doesn't seem very plausible. If there were some special technique for detecting an infinitesimal flux of ultra-low-energy photons, that would be incredible. I don't see how a double slit is going to do that.
– user4552
Nov 23, 2018 at 13:38
• @BenCrowell It causes decoherence, you just detect the decoherence caused by the soft photons, not the soft photons themselves directly. Jan 5, 2019 at 4:19