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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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  • $\begingroup$ possible duplicate of For an accelerated charge to radiate, is an electromagnetic field as the source necessary? $\endgroup$ – Mark Eichenlaub Mar 4 '12 at 7:26
  • $\begingroup$ No claims on correctness; there are some interesting points here: physicsforums.com/archive/index.php/t-72035.html $\endgroup$ – Manishearth Mar 4 '12 at 7:38
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    $\begingroup$ I am sorry but "Let's consider a charge in a freely falling frame of reference." If it is falling it falls to some gravitational center, thus the statement " In such a frame, the local gravitational field is necessarily zero," is wrong . If the particle is in orbit, there is the angular acceleration and it will radiate. If it is free falling there is the acceleration of the gravitational field. The 1/r of a gravitational potential becomes 0 when r=infinity. $\endgroup$ – anna v Mar 4 '12 at 10:20
  • $\begingroup$ @annav That deserves to be expanded a bit and put in an answer :D $\endgroup$ – Manishearth Mar 4 '12 at 10:36
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    $\begingroup$ @annav this is true, but then the question could have been asked like this: "Imagine a static charge. It has an electric field but there is no magnetic field and hence no radiation. However, now let's transform into an accelerating reference frame. In this new reference frame we have an accelerating charge, so it should emit radiation. How can the charge emit photons in one reference frame but not in the other? And since an accelerating reference frame is the same as a gravitational field (Einstein's principle), does a gravitationally accelerated charge radiate or not?" $\endgroup$ – Nathaniel Mar 4 '12 at 13:40
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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.

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  • $\begingroup$ " 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). $\endgroup$ – Ján Lalinský Jan 10 '14 at 21:57
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    $\begingroup$ I am not at all convinced by this answer. It feels heuristic at best. $\endgroup$ – Jerry Schirmer Jun 3 '14 at 19:12
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    $\begingroup$ @JerrySchirmer, by the way, even non-charged point masses may have self-force in GR, similarly to Abraham-Lorenz force. $\endgroup$ – Alexey Bobrick Jun 4 '14 at 15:05
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    $\begingroup$ 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. $\endgroup$ – sure Dec 19 '14 at 23:07
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    $\begingroup$ @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. $\endgroup$ – Alexey Bobrick Dec 21 '14 at 16:16
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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

Radiation Damping in a Gravitational Field, Bryce S. DeWitt, Robert W. Brehme, Annals of Physics: 9, 220-259 (1960) The charged particle tries to do its best to satisfy the equivalence principle, and on a local basis, in fact, does so. In the absence of an externally applied electromagnetic field the motion of the particle deviates from geodetic motion only because of the unavoidable tail in the propagation function of the electromagnetic field, which enters into the picture nonlocally by appearing in an integral over the past history of the particles.

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.

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    $\begingroup$ 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. $\endgroup$ – Alexey Bobrick Jun 4 '14 at 14:48
  • $\begingroup$ There is a more recent article on this at link.springer.com/article/10.12942/lrr-2011-7 $\endgroup$ – Virgo Nov 8 '18 at 20:53
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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 globally and 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 inside the future lightcone of a point $x^{'}$, and so the integral would be nonzero. This property reflects the fact that in 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. Locally point 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

  • Poisson, E., Pound, A., & Vega, I. (2011). 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:

Quinn, T. C., & Wald, R. M. (1999). Energy conservation for point particles undergoing radiation reaction. Physical Review D, 60(6), 064009, doi, arXiv.

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.

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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.

The conditions in which electromagnetic radiation is formed are discussed. It is found that the main condition for the emission of radiation by an electric charge is the existence of a relative acceleration between the charge and its electric field. Such a situation exists both for a charge accelerated in a free space, and for a charge supported at rest in a gravitational field. Hence, in such situations, the charges radiate. It is also shown that relating radiation to the relative acceleration between a charge and its electric field, solves several difficulties that existed in earlier approaches, like the “energy balance paradox,” and the “relativistic” nature of the observation of the emitted radiation

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.

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    $\begingroup$ 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? $\endgroup$ – Sergio Mar 4 '12 at 11:08
  • $\begingroup$ You are using classical mechanics here. OP asks to interpret in terms of general relativity. $\endgroup$ – Siyuan Ren Mar 4 '12 at 11:16
  • $\begingroup$ @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. $\endgroup$ – anna v Mar 4 '12 at 19:22
  • $\begingroup$ 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. $\endgroup$ – anna v Mar 4 '12 at 19:28
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    $\begingroup$ @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 $\endgroup$ – Sergio Mar 4 '12 at 22:31
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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 not accelerate 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.

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  • $\begingroup$ Principle of equivalence works everywhere you apply it. Check the article gave by anna v $\endgroup$ – sure Dec 19 '14 at 23:09
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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 stationary charged 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.

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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.

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  • $\begingroup$ There's nothing left to be resolved. General relativity predicts that the particle will radiate. This has been known since the '60s. $\endgroup$ – Jerry Schirmer Jun 3 '14 at 19:28
  • $\begingroup$ What about the equivalence principle. Is it valid in all the situation ? Can you apply the equivalence principle to an electron ? $\endgroup$ – Sijo Joseph Jun 3 '14 at 19:38
  • $\begingroup$ Not globally. And the electromagnetic self force breaks the premise of the equivalence princple. $\endgroup$ – Jerry Schirmer Jun 3 '14 at 19:40
  • $\begingroup$ 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 ? $\endgroup$ – Sijo Joseph Jun 3 '14 at 19:44
  • $\begingroup$ What does it have to do with quantum mechanics? $\endgroup$ – Jerry Schirmer Jun 3 '14 at 19:58

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