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.
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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. – CuriousOne May 26 '16 at 02:42
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1@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. – Ján Lalinský Nov 23 '18 at 14:55
3 Answers
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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.
anna v
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@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. – anna v May 26 '16 at 05:33
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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. – CuriousOne May 26 '16 at 05:46
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@CuriousOne yes Gen.Rel.Grav. 36 (2004) 315-330 , I added the arxiv link . If the mathematics tells me that a freely falling body radiates NOT , because the mathematics says it is like a uniformly moving one ( things do cancel in mathematical solutions after all), then I will believe the formulae and not my "field" ( note not radiation) intuition – anna v May 26 '16 at 05:49
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In physics it doesn't matter what the mathematics tells you. I can put a battery with tens of thousands of Coulomb of charge on your table. Do you expect it to drain because it is in a gravitational field? If it doesn't, then where does the energy come from to cause the radiation that you expect? From the gravity field? Does that mean Earth loses mass because we charge it a little? – CuriousOne May 26 '16 at 05:52
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Yes the paper says the energy comes from the gravity field, but if you look at the formula the power is tiny. Whys should the earth not lose a bit of its invariant mass due to this? It is the total energy momentum vector of the earth after all that will be affected – anna v May 26 '16 at 06:14
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So you are saying that the radiated power comes from the gravitational field? Does this mean that each atom of Earth is shrinking in mass a little bit or are individual atoms disappearing once in a while? – CuriousOne May 26 '16 at 06:32
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But @CuriousOne we are talking relativity here. Even if some angles of the atoms change the energy can be supplied, if the speed of the earth changes ditto. Just energy momentum changes a tiny bit because of the small couplings. mass is not a conserved quantity in S and GR – anna v May 26 '16 at 09:46
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So you are saying that a sitting charge can radiate because of some thermal movements of atoms in a body the size of a planet and that can go on forever? OK. And general relativity (which doesn't know anything about the structure of matter) says that? :-) – CuriousOne May 26 '16 at 11:03
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@CuriousOne not thermal, thermal is electromagnetic. gravitational. General relativity has four vectors, i.e. invariant masses and not additive masses. – anna v May 26 '16 at 11:39
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So you are saying that the only reason why sitting charges radiate is because the planet is warm? Even though temperature is nowhere defined in GR? And space changes? GR assigns a structure to space that can emit energy forever? Is that how this works? I am confused. – CuriousOne May 26 '16 at 16:25
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@CuriousOne No, I am not saying that. I am saying that the gravitational field exists and in conjunction with the electromagnetic one the result is that stationary charges radiate, while free falling ones do not. The g acceleration of the sitting charge creates the em wave. The energy comes from the gravitational field. Isupose if one goes to gravitons, energy and momentum are exchanged with gravitons in some complicated diagrams (one more reason will be very improbable) with the stationary charge. – anna v May 26 '16 at 16:32
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1Look, 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. – CuriousOne May 26 '16 at 16:41
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@CuriousOne I am trusting on the paper I am quoting, not on your intuition. It is not the charged body that is the source of energy, but the gravitational field. Similar to black body radiation, it is not the body that is the source of energy but the stored electromagnetic energy in the body. – anna v May 26 '16 at 18:09
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1I 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. – CuriousOne May 26 '16 at 18:16
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@CuriousOne Gravitation is distortion of space. Distortion of space has energy momentum consequences that affect all matter/energy(as seen at LIGO) . The energy is taken from distortion of space, which will fall to a lower potential energy by emitting a photon . quantum mechanics and its feynman graphs for the interaction have to enter here for a consistent theory. – anna v May 27 '16 at 04:16
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Philipp see my answer above to your question. "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? –" It is obvious to me that one has to have a solid quantum mechanical model including quantization of gravity to be able to solve for specific boundary conditions and frames. It will depend on the solutions and the boundary values for a given framework, as for all answers to electromagnetic radiation problems. – anna v May 27 '16 at 04:21
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Nothing is falling in the case of a resting charge. It just sits there. General relativity doesn't make dW=Fds and energy conservation go away. Your paper claims that a resting charge is constantly sending out em radiation. That energy has to come from somewhere... and I can't see where it is supposed to be coming from. There is no quantum theory of gravity and the paper makes a perfectly classical claim, so, please, tell me where the energy is coming from. What is the source? – CuriousOne May 27 '16 at 04:50
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@CuriousOne exactly where the energy recorded in LIGO electromagnetically is coming from, the distortion of space of the gravitational field due to the electromagnetic charge's energy. Effective quantization of gravity models are used continuously. I agree that the situation is not satisfactory. We accept for LIGO that as the gravitational wave passes the distance changes and the electromagnetic wave changes. I cannot see why the reverse is not true, even classically, though I do not have the math tools to show it. – anna v May 27 '16 at 05:28
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1LIGO 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? – CuriousOne May 27 '16 at 05:49
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@CuriousOne let us stop this. I cannot convince you and you cannot make me doubt. The energy is coming from released gravitational potential energy, in my books, and the original supplier of energy, as for all energies of the universe, comes from the big bang. The sitting charge is very very slowly sinking in space time, due to g, and the energy is released in em waves. That is my intuitive understanding which clashes with your understanding. – anna v May 27 '16 at 06:50
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But you can convince me. Tell me where the energy is coming from in case of a resting charge. There is no change in the gravitational field. There is no change in the geometry. Everything is perfectly static and the big bang plays no role in this. Now, I think we have been around this block often enough to know that you don't have an answer and that you just can't admit that. Have a good day, then. – CuriousOne May 27 '16 at 06:53
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@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. – Philipp Jun 01 '16 at 01:26
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@anna 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? – Philipp Jun 01 '16 at 01:27
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@Philipp You summary is correct. Non uniformity in the path of a free falling charge brick will have an effect, but one would have to solve for the specific situation and find whether the field and the mass retain their symmetry or there are distortions, imo. It is the change that induces radiation – anna v Jun 01 '16 at 03:15
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@anna Yep, I gotcha. Yet, talking about the charge resting on the earth, I still can't get rid of some vague 'doubts' whether this case is in all aspects analogous to the charge just moving with constant acceleration. – Philipp Jun 01 '16 at 23:20
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@anna Yep, I gotcha. Yet, speaking about th charge resting on th earth, I still can't get rid of some vague 'doubts' whether this case is in all aspects analogous to a charge just moving with constant acceleration. You see, th problem is, that as long as th charge is accelerated, say, by an engine, it not only distorts its field (creating just stress), but it brings about a never-stopping - continuous - distortion, 'compressing' th field more and more as long as th acceleration continues, cos there is a constant time lag from th getgo between th charge & very distant front edge of th field..? – Philipp Jun 01 '16 at 23:27
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From the given example, unless I missed something important, th field of th charge resting on th Earth is just distorted & this distortion geometrically stays constant - th field tries 2 fall w th same acceleration at all its points but againt the supported charge. But it doesn't seem 2 be th same with just an engine-accelerated charge: the infinite edges of its field first 'don't know' that th charge started 2 move, but even after that they can't fully 'catch up' in their velocity with th charge as long as it continues to constantly accelerate. But it doesn't seem to be so in uniform gravity. – Philipp Jun 01 '16 at 23:33
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But the field is constantly accelerated. The image is for time t, but at time t1 it will have radiated and still be distorted with respect to the mass/charge – anna v Jun 02 '16 at 03:06
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@anna But does that mean that the field with time gets more and more distorted - that it continues to 'fall' relative to the charge basically indefinitely? Or will it at some moment time 'stabilize' at some 'statically deformed' state, like on the picture you showed? Cos if this is th case (electron will radiate indefinitely), it really invokes serious implications:
- Either a distant observer can make an eternal engine based on using that radiation - complete no-no
- Or the Earth really looses energy (and mass too) through this phenomenon. If it is so, it is a Nobel Prize huge discovery
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The energy lost is infinitessimally small within the lifetime of the universe , just substitute the numbers in the power formula. And this is for free charges sitting on the surface, which is neutral. Quantum mechanically bound charges do not radiate. The earth continually loses electromagnetic radiation energy through black body radiation anyway. About the shape of the field, I cannot answer, whatever happens will be very slow because of the infinitessimally small power lost. – anna v Jun 06 '16 at 03:24
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@anna My suspicion is that those waves can be seen only from the reference frame falling freely relative to that charge lying on the support, whereas the observer standing still close to the charge won't see anything, and the observer far from earth where the space is linear won't see (won't detect) anything either. But maybe I'm wrong – Philipp Jul 09 '16 at 00:50
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I highly doubt the conclusions drawn here. Feynmann himself states in his Lectures on Gravitation, that the radiated power is proportional to the third derivative of position, which collapses to your equation (9) for sinusiodal motion. I found a very mathematical but convincing approach here: http://www.mathpages.com/home/kmath528/kmath528.htm – Dschoni Mar 30 '17 at 12:55
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How can the charge "sink" when it is resting on a (relatively) inflexible solid surface? Is that surface somehow slowly becoming more and more flexed? If so, wouldn't that translate to a gradually increasing "weight" being applied to it? To what do we attribute this? It is attracted more strongly than a neutral object? But then why doesn't a suitably heavier normal object also inevitably sink in - don't you get an equilibrium of forces? What prevents it? – The_Sympathizer Sep 18 '18 at 06:05
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@The_Sympathizer It is the field that sinks, in an appropriate frame, not the particle, – anna v Sep 18 '18 at 06:15
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We already have a question that asks about the theory. This question is solely about experiment. This answer starts off with an overly confident theoretical answer on a difficult and controversial topic, and references a single paper, which appears to be an unpublished report. – Nov 23 '18 at 13:36
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@BenCrowell there is a publication behind a paywall https://www.worldscientific.com/doi/abs/10.1142/S0217751X05024547 ; In addition I hope, as an experimental physicist, I am allowed to decide whether an experiment can be possibly done, relying on the theoretical predictions for the system under consideration. – anna v Nov 23 '18 at 13:51
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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.
S. McGrew
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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.
Count Iblis
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@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. – Philipp May 26 '16 at 23:37
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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? – Philipp May 26 '16 at 23:45
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A double slit is a very imprecise experiment, as it is. If anything this would require a very precise atomic or nuclear physics experiment. – CuriousOne May 27 '16 at 04:53
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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. – Nov 23 '18 at 13:38
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@BenCrowell It causes decoherence, you just detect the decoherence caused by the soft photons, not the soft photons themselves directly. – Count Iblis Jan 05 '19 at 04:19
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Perhaps, but this sounds unlikely to me because the softer a photon gets, the less it decoheres a superposition (since for large wavelengths it cannot resolve the two slits). Decoherence by photon emission has been observed, but typically requires a very delicate double slit experiment with the object heated to extreme temperatures, so that it emits photons of sufficiently short wavelength. – knzhou May 05 '20 at 19:07