# Does a charged particle accelerating in a gravitational field radiate?

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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– Mark Eichenlaub Mar 4 '12 at 7:26
No claims on correctness; there are some interesting points here: physicsforums.com/archive/index.php/t-72035.html – Manishearth Mar 4 '12 at 7:38
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. – anna v Mar 4 '12 at 10:20
@annav That deserves to be expanded a bit and put in an answer :D – Manishearth Mar 4 '12 at 10:36
@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?" – Nathaniel Mar 4 '12 at 13:40

The paradox is, sort of, resolved as follows: the number of photons changes when you switch between non-inertial frames. This is actually a remarkable fact, and holds also for quantum particles, which can be created in pairs-antipairs, 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. This naturally corresponds to appearing magnetic fields, and hence radiation.

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

Another point might be: 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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" 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). – Ján Lalinský Jan 10 '14 at 21:57
@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. – Alexey Bobrick Jan 10 '14 at 22:18
I am not at all convinced by this answer. It feels heuristic at best. – Jerry Schirmer Jun 3 '14 at 19:12
Dear @JerrySchirmer, many thanks for pointing this out! – Alexey Bobrick Jun 4 '14 at 14:39
@JerrySchirmer, by the way, even non-charged point masses may have self-force in GR, similarly to Abraham-Lorenz force. – Alexey Bobrick Jun 4 '14 at 15:05

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

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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? – Sergio Mar 4 '12 at 11:08
You are using classical mechanics here. OP asks to interpret in terms of general relativity. – Siyuan Ren Mar 4 '12 at 11:16
@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 Mar 4 '12 at 19:22
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. – anna v Mar 4 '12 at 19:28
@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 – Sergio Mar 4 '12 at 22:31

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:

http://home.comcast.net/~peter.m.brown/ref/falling_charge.htm

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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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. – Alexey Bobrick Jun 4 '14 at 14:48

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

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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Principle of equivalence works everywhere you apply it. Check the article gave by anna v – sure Dec 19 '14 at 23:09