2012-03-04 06:56:22 8 Comments

A charged particle undergoing an acceleration radiates photons.

Let's consider a charge in a freely falling frame of reference. In such a frame, the local gravitational field is necessarily zero, and the particle does not accelerate or experience any force. Thus, this charge is free in such a frame. But, a free charge does not emit any photons. There seems to be a paradox. Does a freely falling charge in a gravitational field radiate?

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## 7 comments

## @Alexey Bobrick 2012-03-16 21:46:20

The paradox is resolved as follows: the number of photons changes when you switch between non-inertial frames. This is actually a remarkable fact and it holds also for quantum particles, which can be created in pairs of particles and antiparticles, and whose number depends on the frame of reference.

Now, a step back. Forget about gravity for a moment, as it is irrelevant here (we are still in GR, though). Imagine a point charge, which is accelerating with respect to a flat empty space. If you switch to the rest frame of the charge, you observe a constant electric field. When you switch back to the inertial frame, you see the field changing with time at each point and carrying away radiation from the charge.

In the presence of gravity the case is absolutely similar. To conclude, switching between non-inertial frames makes a static electric field variable and corresponds to a radiation flow.

Another relevant point: When moving with charge, no energy is emitted, but when standing in the lab frame, there is a flux observed. However, there is no contradiction here as well, as the energy as a quantity is not defined for noninertial frames.

## @Ján Lalinský 2014-01-10 21:57:26

" If you switch to the rest frame of the charge, you observe a constant electric field. " Could you please explain why? It is not clear what electric field will be, because it is no clear how the charge is accelerated and what is your definition of electric field in accelerating frame (ordinarily electric field is part of force due to other charges in inertial frame).

## @Alexey Bobrick 2014-01-10 22:18:50

@JánLalinský: It is beacuase when you switch to an accelerated frame as an observer, metric will be constant with your proper time. If you write down Einstein-Maxwell equations, you will have equations for electric and magnetic fields, which do not depend on time. There may possibly be some flux components, though, I don't have a proof at hand.

## @Jerry Schirmer 2014-06-03 19:12:16

I am not at all convinced by this answer. It feels heuristic at best.

## @Alexey Bobrick 2014-06-04 14:39:58

Dear @JerrySchirmer, many thanks for pointing this out!

## @Alexey Bobrick 2014-06-04 14:42:39

@JerrySchirmer, and yet, could you please kindly be specific in your statements? Is there any particular argument that you find wrong in my answer? Do you think some aspect of the question was not touched upon? Or in which way could in general the answer be better in your opinion?

## @Jerry Schirmer 2014-06-04 14:58:49

@AlexeyBobrick: your answer doens't feel wrong to me, it just doesn't feel convincing -- in particular, I'd have trouble making this argument quantitative, to show how much radiation a charge $q$ will produce in a schwarzschild spacetime of mass $m$. (I didn't downvote). To me, the essential source of the problem is self-force, which I guess could be interpreted as a particle accelerating relative to its field.

## @Alexey Bobrick 2014-06-04 15:04:04

@JerrySchirmer: I would say, the answer is also qualitative, it just asks why does the paradox appear. The paradox is resolved by two statements: 1) The charge in its own frame does not radiate only locally, 2) Additionally, static fields may turn into radiation when the frames are switched. So I just didn't go into calculations. And as you show in your answer, it takes an article size derivation to get the field estimates.

## @Alexey Bobrick 2014-06-04 15:05:44

@JerrySchirmer, by the way, even non-charged point masses may have self-force in GR, similarly to Abraham-Lorenz force.

## @Jerry Schirmer 2014-06-04 15:08:53

@AlexeyBobrick: fair enough :)

## @sure 2014-12-19 23:07:10

To radiate is like to accelerate : it is an absolute. Therefore, if you radiate for someone, it must be possible to detect it in any other frame (inertial or not). A free falling charge does not radiate because it is inertial. A charge at rest on earth do radiate for a free falling observer. Yet, if you're also at rest with respect to the charge, even though you won't see it radiates, the electric field is curved because of strong equivalence principle ! Thus, the field of the charge acts on the charge itself in such a way that its all consistent, even from an energy point of view.

## @Alexey Bobrick 2014-12-21 16:16:18

@sure, sorry, but it is not absolute. Consider a charge in vacuum - it certainly does not radiate for an observer, for whom the charge is at rest. However, an accelerated observer will see the energy flux. Therefore, the notion "radiate" is dependent on the observer. And therefore, some of your statements need to be restated with observers being specified.

## @sure 2014-12-21 16:32:04

By absolute, I mean that if the charges radiate in an inertial frame, then there must exists a way to

knowit does. The same happens with acceleration. The non preservation of the syntaxic form of a quantity between frames does not mean that the quantity you're looking at isn't in some sense "absolute". That is, something non trivial happens somewhere, so something equivalent that is non trivial happens everywhere.## @Alexey Bobrick 2014-12-21 16:41:49

@sure: Well, I mean quite precisely, that if you take an observer and consider all the physical quantities (metric, fields, curvature) in his/her vicinity, you will get what observer 'sees'. This way, accelerated observer will 'see' that if it releases a test particle, the particle will get accelerated (cf to what you said). Similarly, accelerated observer can take a detector and measure the flux from a field of an inertial charge. Note, the charge didn't radiate for being inertial, but still can learn that the energy was extracted by a non-inertial observer (again, cf to what you stated).

## @sure 2014-12-21 16:53:00

An accelerated charge, seen from its proper frame, has curved field line and a non zero self-force. This is precisely what I mean by "you know you're accelerating even if you're in the accelerated frame". Or maybe I didn't understand something?

## @Alexey Bobrick 2014-12-21 17:12:58

@sure: No, this is definitely fine, though I thought you meant acceleration in general. Yes, I do agree that an accelerated charge radiates non-locally to infinity, hence feels self-force. However, the radiation as a locally measured quantity is dependent on observer. I guess this settles the discussion.

## @sure 2014-12-21 17:32:31

Yes, we agree indeed.

## @thermomagnetic condensed boson 2018-02-08 20:16:28

How does this answer reconciles with the accepted answer of physics.stackexchange.com/questions/241522/…?

## @Alexey Bobrick 2018-02-09 00:12:03

Lorentz boost is not the same as acceleration

## @Ján Lalinský 2019-04-20 21:49:47

>

"you see the field changing with time at each point. This naturally corresponds to appearing magnetic fields, and hence radiation"-- Changing electric field is not sufficient to conclude there is radiation. Radiation means electric field disturbance spreads to infinity and decays with distance no faster than $C/r$.## @Alexey Bobrick 2019-04-20 22:12:01

Thanks a lot for your comment! You are correct, radiation does correspond to a C/r fall-off, and it requires more than just a changing electric field. However, my point was only that accelerated charges do radiate for inertial observers. I will edit the answer a bit to avoid confusion.

## @Jerry Schirmer 2014-06-03 19:10:55

The charge accelerates. This is proven in a paper written by Bryce DeWitt and Robert Brehme in the '60s, cited in the paper at this link:

https://www.sciencedirect.com/science/article/pii/0003491660900300

The article is out of print, and I had to look it up at a university library to read it. The interesting part of the result is that the acceleration of the particle picks up a non-local term that depends on a path integral over the particle's path.

## @Alexey Bobrick 2014-06-04 14:48:01

By the way, correct answer, but doesn't expicitly address the question. Indeed, the word 'locally' is the resolution of the paradox. The charge does not emit radiation locally in its comoving frame. And yet does so generally for accelerated or non-comoving observers.

## @Virgo 2018-11-08 20:53:51

There is a more recent article on this at link.springer.com/article/10.12942/lrr-2011-7

## @Elio Fabri 2018-10-22 13:22:38

I can't follow most of the answers and comments. I would like to keep the question as simple as possible (which does not mean easy). First of all, I find quite improper to involve QM, still worse QFT. It is well known that relations between GR and QM are far from being settled... So I will insist on keeping the discourse strictly classical. I would like to see an answer to the following question (not original, but not answered here, as far I can remember.)

We are on a lonely, non-rotating planet (or cold star it you like better). At some distance from its surface there is a

stationarycharged body. E.g. the body is kept fixed by a solid vertical bar raised from planet's surface. It seems to me that someone stated that in such situation, thanks to equivalence principle, the charged body is radiating. The obvious objection is: where does the radiated energy come from?Let me explain better. The physical situation, as far planet and charged body are concerned, is stationary. Moreover, there is no doubt that spacetime is asymptotically flat. To me radiation means an energy flow whose total flux through a sphere of radius $r$ has a finite non-zero limit as $r\to\infty$. Since GR says that energy is globally conserved, I find it to forbid a continuous energy flux.

Please show where the error is in my reasoning.

## @A.V.S. 2018-03-06 20:18:59

A recent answer by John Rennie linked this question as 'definitive' yet there are issues with the accepted answer by Alexey Bobrick.

The 'number of photons' mentioned makes one think that this is about a purely quantum effect. It is not. Motion and radiation of a point charge in curved space could be handled classically.

While it is generally true that in a curved background it is difficult to define invariantly what is radiated, there is a whole class of setups where it could be done

globallyand with ease (at least at conceptual level). Let us consider asymptotically flat spacetimes with time-like Killing vector field. Now consider the point charge that starts moving from infinity with constant velocity at $t=-\infty$ interacts with the nontrivial part of the metric around $t=0$ and flies away at $t=+\infty$. Killing v.f. gives us conservation of energy for the system 'charge ${}+{}$ electromagnetic field' and so the difference between initial and final kinetic energy would be well defined, and it had to be the energy radiated away. (Of course, if there is black hole horizon, 'away' may also mean into the black hole). Even for a charge in a bound motion the decay of bound orbits is such a case also would be observer independent. So if a charge's radius of orbit after, say, $10^{20}$ revolutions in a Schwartzschild metric becomes a half of its initial radius then no coordinate transform could erase the radiation at infinity.One more note: this question is quite different from the question whether a uniformly accelerating charge radiates: there the motion is given, the charge is being dragged by an external force, while here we are dealing with a charge and its field interacting with gravitational field without interference from additional forces, how the charge moves is unknown beforehand. And while for uniformly accelerated charge the main problem is how to extract radiation from the known EM field, in the current question we could sidestep this issue by restricting ourselves to asymptotically flat spacetimes. The main problem now would be to find the motion of the charge and by applying conservation of energy we would then know the energy of that was radiated away.

Jerry Schirmer's answer does link to a correct paper but does not provide an detailed explanation for the 'paradox'.

So here is the paradox resolution in terms of the question: in a freely falling frame gravitational field acting on a charge is indeed zero. However the charge would not be moving along a geodesic. Instead charge would be moving under the DeWitt-Brehme radiation reaction force: $$ m{a^\mu} = f_{{\rm{ext}}}^\mu + {e^2}(\delta _{\;\nu}^\mu + {u^\mu}{u_\nu})\left({{2 \over {3m}}{{Df_{{\rm{ext}}}^\nu} \over {d\tau}} + {1 \over 3}R_{\;\lambda}^\nu {u^\lambda}} \right) + \\{} + 2{e^2}{u_\nu}\int\nolimits_{- \infty}^{{\tau ^ -}} {{\nabla ^{\left[ \mu \right.}}} G_{+ \,\lambda^{'}}^{\left. {\;\nu} \right]}(z(\tau),z(\tau^{'})){u^{\lambda^{'}}}\,d\tau^{'}, \tag{*} $$ where $G_{+ \,\lambda^{'}}^{ {\;\nu}}(x,x^{'})$ is (retarded) Green's function of EM field (it is a bitensor: so indices $\lambda^{'}$ and $\nu$ correspond to (co)tangent bundles at different points), and integration is carried along the past motion of the charge. In the case of absent external force ($f_{\rm ext}=0$) and in a Ricci-flat metric this force is given only by a non-local integral. Incidentally, original 1960 DeWitt & Brehme paper did not include the term with Ricci tensor. This was corrected in 1968 by Hobbs.

Green's functions in GR for massless fields have a richer structure than in flat space: it is generally nonzero

insidethe future lightcone of a point $x^{'}$, and so the integral would be nonzero. This property reflects the fact thatin curved spacetime, electromagnetic waves propagate not just at the speed of light, but at all speeds smaller than or equal to the speed of light, the delay is caused by an interaction between the radiation and the spacetime curvature. So the integral would generally be nonzero and the charge would radiate EM waves.There is nothing mysterious in the non-local character of the force (*). It is the result of going from the system with infinite degrees of freedom 'charge ${}+{}$ elecromagnetic field' to a finite dimensional description in terms of charge motion alone.

Locallypoint charge is acted upon by electromagnetic field. This electromagnetic field originated on the same charge in the past, been scattered by a gravitational field some distance away from it and produced a potentially non-zero force in the present.Situation might be easier to understand if we consider the following flat space situation: a point charge and a small dielectric ball at some distance $d$ from it. The ball gains dipole moment in the field of a charge and exerts a certain force on it. Now let us wiggle the ball a little around the moment $t=0$, then perturbations of EM field from this wiggle would be propagating and the original charge would feel additional force at time $t=d/c$. This is a usual Lorentz force, but since the only source of EM field is our charge we can write it as an integral of Green's function over the past worldline of the charge. This Green's function encodes all information about ball and its wiggle. And since there is diffraction on the ball, this function would be nonzero inside the future lightcone of its argument $x^{'}$. The force felt by the charge (both constant contribution and signal from wiggling) now would be written as an integral over the charge past, an expression similar to the DeWitt-Brehme force.

The actual calculations of Green's function in GR are quite complicated and for a sample I would recommend the review

The motion of point particles in curved spacetime. Living Reviews in Relativity, 14(1), 7, open access web.Once that has been done the work done by the DeWitt-Brehme force allows us to calculate the energy of radiated EM wave. This has been demonstrated by Quinn & Wald:

who have proven that the net energy radiated to infinity equals to minus the net work done on the particle by the DeWitt-Brehme radiation reaction force.

## @anna v 2012-03-04 10:57:12

We have $F=m_1a$ where $m_1$ is the mass of the charged particle and $a$ the acceleration.

The gravitational force is $F=Gm_1m_2/r^2$.

Hence $a=m_2/r^2$ where $m_2$ is the mass of the large body (earth) towards which the charged particle is falling and $r$ is the distance from the center of gravity and $G$ the gravitational constant. There is always an acceleration, though when $r$ becomes very large the acceleration is very small and the photons emitted will be very low energy.

What is happening to the freely falling charged particle is that part of the potential energy it is giving up by falling, turns into radiated photon energy, rather than totally to velocity of fall towards the center of gravity, which will happen to an uncharged particle.

Here is a relevant theoretical study of charge and acceleration.

A more recent link is here. . It shows that a freely falling charge should not radiated after all. Only a stationary one. See my answer to a newer relevant question here.

## @Sergio 2012-03-04 11:08:46

The quastion is in the plane of General Relativity. The free falling charge stay at rest in a local frame of reference. Does the observer, which is in the same frame, detect the radiation?

## @Siyuan Ren 2012-03-04 11:16:45

You are using classical mechanics here. OP asks to interpret in terms of general relativity.

## @anna v 2012-03-04 19:22:27

@Sergio you should be asking the question differently then, stating in the body of the question that you are talking of GR transformations. Still, your assumption that in the rest frame of the particle the gravitational potential is zero is wrong. It will be something complicated by the transformations to reach the rest frame of m1, but still there. I expect that an observer at rest in the rest frame of m1 will not be seeing the radiation, in the same way as he/she would not know that the particle is falling and increasing its velocity in the total center of mass.

## @anna v 2012-03-04 19:28:41

continued: the physical photons observed in the overall CMS would be cooled into the infrared by the same transformations to the point of being virtual, once one utilizes quantum field theory.

## @Sergio 2012-03-04 22:31:53

@anna v Dear Anna, thanks for your answer. My quastion was is in the frame of General Relativity. I find that the quastion is no agreement among physicist for today. It is clearly describe in this article xxx.lanl.gov/abs/gr-qc/0006037

## @anna v 2012-03-05 05:27:23

@Sergio I looked at the paper you linked. You have to realize that paradoxes arise in physics when mathematical frameworks are mixed. The mathematical framework of GR must not be mixed with the classical newtonian gravity, which happens if you imply GR and want to see gravitational fields. There are no gravitational fields in GR , just distortions of space-time. The statement " the local gravitational field is necessarily zero" is wrong because it mixes two such mathematical frames if you are implying that you are thinking in the GR framework. Solutions need one consistent framework.

## @anna v 2012-03-05 05:28:40

p.s The paper you linked is interesting and comes to the same conclusion as my newtonian argument in the comments.

## @innisfree 2013-12-17 18:46:46

Bit late, but doesn't this answer violate the weak equivalence principle? You suggest that a charged particle in a gravitational field accelerates slower than a neutral one. But gravity knows nothing of quantum numbers, charges etc because of the WEP.

## @anna v 2013-12-17 19:42:00

@innisfree Gravity at the moment knows nothing of quantum mechanics in a consistent framework. A charged particle necessarily is a quantum mechanical entity. We have to wait for the unified q uantum theory to be able to see what the weak equivalence principle will look in it. Till then semiclassical arguments have to hold, and it is an experimental fact that accelerating or decelerating charges emit radiation, and this energy will come from somewhere.

## @innisfree 2013-12-17 23:59:44

What experiment demonstrates that charge accelerated by gravity radiates? I agree that charges accelerated in EM fields radiate.

## @anna v 2013-12-18 04:31:28

@innisfree our intuition that equivalence should hold from the quantum mechanical side too? here is a preprint that calculates this cds.cern.ch/record/403583/files/9910019.pdf , to be consistent with the equivalence principle. Maybe some day black hole observations might confirm this. It would be a small effect to the radiation emitted by by the hot gases , which is the standard explanation I find for xray sources being black holes.

## @innisfree 2013-12-18 11:15:38

The abstract you quote is interesting, however; it says that charges supported at rest in a GF radiate! That means an electron sitting on the earth's surface. That is controversial. Without reading more, I can't follow how energy is conserved from the back reaction.

## @anna v 2013-12-18 11:45:12

@innisfree I think it is a theoreticians idea of not bothering how it will be at rest. They want to stress that it is the distortion in the field that generates radiation A free electron at rest in vacuum would take the energy from the gravitational potential. Sitting on the surface electrons are bound, Bound electrons do not radiate, which is the point of quantum mechanics.

## @Jerry Schirmer 2014-06-03 19:07:55

@innisfree: Adding general relativity or quantum mechanics to the answer will not change the result. There is no reason to complicate the answer with these things, which will cloud the physics. +1

## @sure 2014-12-19 23:01:10

The paper of Harpaz and Soker is beautiful, simple, and should be taught to any student in general relativity. It clearly close the question (I'm now totally convinced that there's no problem with electrodynamics and equivalence principle), and one can even conclude that the earth on which a charge lies on the ground Hawking's radiate as a result. Really beautiful piece of physics there.

## @Sijo Joseph 2014-01-10 19:46:55

This is an interesting topic, but I don't think the issue is fully resolved. Suppose there is a stationary charge in a gravitational field will it radiate ? If we apply equivalence principle, it seems that it will radiate, since the gravitational field is equivalent to an accelerating frame. Hence the stationary charge in a gravitational field is equivalent to an accelerating charge in a zero-gravitational field. Hence it should emit radiation.Some physicists says, that won't happen. This is a true paradox which challenge the principle of equivalence. Secondly consider an uncharged rocket which is accelerated by burning the fuel. Suppose, to get an acceleration of 1g we have used 1kg of fuel. How much fuel we have to use to accelerate a charged rocket ? less than 1kg of fuel or greater than 1kg of fuel ? If the charged particle emit radiation, then the second rocket need more fuel. But some physicist calculations shows that we need the same amount of fuel? Then there is a paradox !,What about the energy ? what is the difference between charged and uncharged rocket ? I think this is the topic we physicists have to explore to uncover the mystery ! Actually there is a book titled $\textbf{Uniformly Accelerating Charged Particles Threat to the Equivalence Principle}$ by $\textbf{Stephen N. Lyle}$ dedicated to explain this paradox.

## @Jerry Schirmer 2014-06-03 19:28:43

There's nothing left to be resolved. General relativity predicts that the particle will radiate. This has been known since the '60s.

## @Sijo Joseph 2014-06-03 19:38:32

What about the equivalence principle. Is it valid in all the situation ? Can you apply the equivalence principle to an electron ?

## @Jerry Schirmer 2014-06-03 19:40:08

Not globally. And the electromagnetic self force breaks the premise of the equivalence princple.

## @Sijo Joseph 2014-06-03 19:44:56

OK, If this is true, how can you say that there is nothing left to be resolved ! If the equivalence principle cannot be applied in the quantum situation how can you surely say that we answered this question completely ?

## @Jerry Schirmer 2014-06-03 19:58:24

What does it have to do with quantum mechanics?

## @Sijo Joseph 2014-06-04 10:59:03

Radiation is a pure quantum phenomena.

## @Jerry Schirmer 2014-06-04 14:55:48

It has a classical explanation. Single photon states aren't eigenstates of the hamiltonian, anyway.

## @Sijo Joseph 2014-06-05 01:48:29

Classical theories are the special cases of quantum theory when $\hbar\to0$ or $m\to\infty$ and photon itself is a quantum concept !

## @Jerry Schirmer 2014-06-06 15:25:52

I understand the classical limit. Show me a "virtual photon" state. It's nonphysical nonsense that can clarify some things, but is also not observable and dependent on perturbation theory. The field is what is real, the photon states are abstractions of the field. Talking about virtual photons in answering this question is more misleading than just using the classical picture.

## @Cloudswrest 2014-03-25 19:51:21

A charge is surrounded by an electric field, which can be considered "attached" to the charge, moves with it, and stretches off to infinity. It is just as much a "physical" object as the charge itself and has mass/energy and momentum density if moving. Gravitationally accelerating a charge also gravitationally accelerates the local electric field around it, but it does

notaccelerate the parts of the electric field that are far away from the source of gravity. These far way parts of the field will exert some drag on the charge and represents the energy lost to radiation. When you realize the electric field stretches off to infinity you realize that a charge is a non-local object, and hence it is inappropriate to apply the Principle of Equivalence.## @sure 2014-12-19 23:09:22

Principle of equivalence works everywhere you apply it. Check the article gave by anna v