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Equivalence principle implies uniformly accelerated charge doesn't radiate?

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johne1618
#1
Jan1-13, 06:13 PM
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Does a uniformly accelerated charged particle radiate em waves?

The equivalence principle says that a particle in uniform acceleration is equivalent to a particle at rest in a gravitational field.

A particle at rest in a gravitational field is clearly not going to radiate em waves therefore by the equivalence principle a uniformly accelerated charge won't either.

Is that right?
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Vanadium 50
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Jan1-13, 06:28 PM
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Quote Quote by johne1618 View Post
Does a uniformly accelerated charged particle radiate em waves?
Yes.

Quote Quote by johne1618 View Post
The equivalence principle says that a particle in uniform acceleration is equivalent to a particle at rest in a gravitational field.
False. The equivalence principle explains how to distinguish between a particle in a gravitational field and one in uniform acceleration - that is, a gravitational field will have tides.

Quote Quote by johne1618 View Post
A particle at rest in a gravitational field is clearly not going to radiate em waves therefore by the equivalence principle a uniformly accelerated charge won't either.

Is that right?
No. You have a history of pushing unconventional viewpoints. I sincerely hope this is not another attempt.
atyy
#3
Jan1-13, 06:30 PM
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No, an accelerating charge does radiate. The equivalence principle applies only locally. Radiation is inherently nonlocal, so the equivalence principle does not apply.

http://arxiv.org/abs/0806.0464

Andrew Mason
#4
Jan1-13, 10:20 PM
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Equivalence principle implies uniformly accelerated charge doesn't radiate?

This paper seems to say that a charge in free-fall in a uniform gravitational field (ie. accelerating) does not radiate:

Quote Quote by Ashok K. Singal, 1994;
4. CONCLUSIONS
We have shown that from the strong principle of equivalence the electric field of a freely falling charge in a static, uniform gravitational field would appear to fall along with the charge, remaining everywhere in a radial direction from the instantaneous position of the charge. Accordingly there will be no transverse fields (radiation!) from a freely falling charge in such a gravitational field. It is further shown that in the case of a charge supported in such a gravitational frame, the electric field energy, as measured by freely falling observers instantaneously at rest with respect to the charge, is equal to the Coulomb field energy of a charge permanently stationary in an inertial frame. It follows that in neither of the two cases will there be any electromagnetic radiation.
The OP also seems to have Richard Feynman on his side. This is from a paper "Does a Uniformly Accelerating charge radiate?":
Whether these solutions are realistic or not is an open question, but the predicted absence of radiation for hyperbolic motion has sometimes been cited as a way of reconciling the Equivalence Principle with the fact that a charged particle held stationary in a gravitational field (and therefore undergoing constant proper acceleration) does not radiate. For example, in Feynman's "Lectures on Gravitation" he says "we have inherited a prejudice that an accelerating charge should radiate", and then he goes on to argue that the usual formula giving the power radiated by an accelerating charge as proportional to the square of the acceleration "has led us astray" because it applies only to cyclic or bounded motions. ....
Thus the radiation reaction force (and therefore the radiated power) is proportional to the third derivative of position, so if the particle is undergoing constant acceleration it does not radiate (according to this formula).
AM
atyy
#5
Jan1-13, 10:24 PM
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Quote Quote by Andrew Mason View Post
This paper seems to say that a charge in free-fall in a uniform gravitational field (ie. accelerating) does not radiate:



The OP also seems to have Richard Feynman on his side. This is from a paper "Does a Uniformly Accelerating charge radiate?":
Whether these solutions are realistic or not is an open question, but the predicted absence of radiation for hyperbolic motion has sometimes been cited as a way of reconciling the Equivalence Principle with the fact that a charged particle held stationary in a gravitational field (and therefore undergoing constant proper acceleration) does not radiate. For example, in Feynman's "Lectures on Gravitation" he says "we have inherited a prejudice that an accelerating charge should radiate", and then he goes on to argue that the usual formula giving the power radiated by an accelerating charge as proportional to the square of the acceleration "has led us astray" because it applies only to cyclic or bounded motions. ....
Thus the radiation reaction force (and therefore the radiated power) is proportional to the third derivative of position, so if the particle is undergoing constant acceleration it does not radiate (according to this formula).
AM
Feynman was wrong on this point. (Try references 6 & 7 of http://arxiv.org/abs/0806.0464)

I think those references are not free, maybe try http://arxiv.org/abs/physics/0506049. I think this paper is technically right, but I prefer the resolution that the equivalence principle does not apply due to the nonlocal nature of radiation, rather than that a comoving observer does not measure radiation (but that might be splitting hairs - I agree that in flat spacetime a co-moving observer does not measure radiation.) The reason I prefer the non-locality resolution is that the equivalence principle should apply locally in curved spacetime. In that case, the charge is not "freely fallling" because in addition to the gravitational field, it is acted on by its own field. A similar point of view is expressed in http://arxiv.org/abs/1102.0529 (p144): "In the scalar and electromagnetic cases, the picture of a particle interacting with a free radiation field removes any tension between the nongeodesic motion of the charge and the principle of equivalence.", or http://arxiv.org/abs/0806.0464 (p2): "The principle of equivalence has a local character. The mentioned equivalence is only valid as far as the measurements does not reveal a possible curvature of space. So, if there is a non-local interaction between the charge and the curvature of spacetime, due to the non-local character of its electromagnetic field, this may modify the equation of motion of the charge."
the_emi_guy
#6
Jan2-13, 02:21 AM
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Quote Quote by Vanadium 50 View Post
Yes.



False. The equivalence principle explains how to distinguish between a particle in a gravitational field and one in uniform acceleration - that is, a gravitational field will have tides.



No. You have a history of pushing unconventional viewpoints. I sincerely hope this is not another attempt.
I have not followed this OPs other activity, but it sure seems like, for this particular question, he deserves more than a one word answer followed by chiding.


This is a very famous paradox, and had been called a "perpetual problem".

Quoting from Alemida (attachment):

"The long-standing paradox about the electromagnetic
radiation emitted by a uniformly accelerated charge has received considerable attention. Eminent figures
such as Pauli, Born, Sommerfeld, Schott, von Laue, and many others have contributed to this debate with different
answers. The relevant questions we consider are: Does a uniformly accelerated charge actually radiate? In a constant
gravitational field should free-falling observers detect any radiation emitted by free-falling charges? Is the equivalence
principle valid for such situations?"


I have posted some papers:

Boulware: Annals of Physics 124, 169-188 (1980) - Probably the most well known paper on this subject.
Singal: General Relativity and Gravitation, Vol. 27 No. 9, 1995 - Argues that no radiation occurs (mentioned by Andrew Mason).
Alemida, arXiv:physics/0506049v5, 2005 - fairly recent, includes some history of the debate.
Ch52: http://rickbradford.co.uk/ChoiceCutsCh52.pdf - This is a very good survey of work on this topic, from Born in 1909 through Feynman in 1999.

The concensus right now is that uniformly accelerating charges probably radiate, but not everyone can see the radiation.

The various papers on this topic make it clear that part of the problem is in definition.

For example, what do we mean by "radiate".

According to the Boulware paper, and further in Alemida, a co-moving observer will not see any radiation from a uniformly accelerating charge,
even though the radiation exists, because it is outside of his event horizon and unobservable to him.
Attached Files
File Type: pdf boulware.pdf (1.00 MB, 5 views)
File Type: pdf almeida.pdf (166.3 KB, 3 views)
File Type: pdf singal.pdf (564.5 KB, 3 views)
andrien
#7
Jan2-13, 05:48 AM
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feynman said that radiation is proportional to(dx/dt)(d3x/dt3) but it can be written as (d2x/dt2)2- d/dt(dx/dt d2x/dt2) and the last term vanishes for periodic motions so it is also written only with first term present which applies only to periodic motion.In fact a very careful treatment is needed which is given in book of schwinger's 'classical electrodynamics'.
Jano L.
#8
Jan2-13, 01:37 PM
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john,
this is notoriously difficult question. Although it is easy-to-state, nobody has provided equally easy-to-understand answer. There are these long papers published for decades about it and still no revelation.

I think this has happened because the question addressed is formulated in rather unclear concepts. It uses expressions such as "accelerated charge radiates", "accelerated charge is equivalent to stationary one", which are too vague.

It may help you to think about this:

Acceleration is a relative concept - with respect to what should the charge accelerate?

Production of radiation can mean very different things to different people - does it mean that the particle blows energy away? That the flux of the Poynting vector through the surface in large distance distance is positive? That the EM field in the vicinity of the particle is time-dependent? That the EM field is given by the retarded solution of Maxwell's equations ? That the latter contains non-zero 1/r component? That photons are produced?

Also, it is important to keep in mind that the EM field is relative concept as well.

What is meant by the word "particle in sit. A is equivalent to the particle in sit. B"? That the trajectory of the particle is the same? Or that the state of the fields around the particle is the same?

I can see many combinations of choices which lead to many different questions with possibly different answers; unless the question is made clear, there is no hope for useful answer.

I am sure that if you propose clear formulation of the question, it will meet will much better reception and we may even find some good answer.
Andrew Mason
#9
Jan3-13, 12:06 AM
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Perhaps it depends on how the charge is accelerated. If it is accelerated by gravity, there is no interaction of electrical fields. If the charge is accelerated by electrical force, there is an interaction of electrical fields. In the first case there may be no radiation but in the latter there would be.

In other words, radiation is not caused by the charge's acceleration. Rather it is a consequence of an electrical interaction, which necessarily also causes the charge to accelerate. If we could find a particle that possessed charge but no mass, we might be able to determine this experimentally. But it appears that charge and mass cannot be separated.

From what I can gather, the contention that an electron radiates when accelerated is based on a theory that the electron interacts with its own field when accelerated. Feynman appears to have taken different positions on that. He may have ended up believing that it all depends on how you want to look at it, suggesting that there may be more than one 'correct' equivalent explanation. He talks about this in his Nobel lecture, which is well worth reading.

AM
andrien
#10
Jan3-13, 05:45 AM
P: 1,020
there is no confusion with it.Accelerating charges radiate.In case of discrepancy,stick to larmor law(generalised)
the_emi_guy
#11
Jan3-13, 11:06 AM
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Quote Quote by andrien View Post
there is no confusion with it.Accelerating charges radiate.In case of discrepancy,stick to larmor law(generalised)
You figured this out yourself?


@atyy
Your references do not address the OPs question. Your references address radiation from a charge that is in orbit around a gravitating body. As Jano L pointed out, there are a multitude of scholarly works addressing exactly the OPs question (I have posted some). Wouldn't it make sense to begin the discussion with those?
atyy
#12
Jan3-13, 11:20 AM
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Quote Quote by the_emi_guy View Post
@atyy
Your references do not address the OPs question. Your references address radiation from a charge that is in orbit around a gravitating body. As Jano L pointed out, there are a multitude of scholarly works addressing exactly the OPs question (I have posted some). Wouldn't it make sense to begin the discussion with those?
Did you see my post #5?
Andrew Mason
#13
Jan3-13, 02:39 PM
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I have to agree with the_emi_guy and Janos L. There has been a long-standing debate among eminent physicists about whether a uniformly accelerated charge (ie a charge in gravitational orbit) radiates. I don't think we are going to resolve this debate here.
Without the right experimental data there is going to remain a controversy.

Remarkably, it appears that this is still an unresolved problem in physics. Tying gravity into the rest of physics remains an intractable problem.

AM
Vanadium 50
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Jan3-13, 05:42 PM
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Andrew, I disagree. Whether an accelerated charge radiates is settled. This is section 14.3 of Jackson (2nd). An accelerated charge doesn't know that its acceleration is or is not gravitational, so it can't "decide" whether or not to radiate, as you suggest in Post #9.

A full GR treatment agrees, insofar as "radiation" is well-defined. (Observers, especially nearby accelerating ones, may disagree about what is near-field and what is far-field, but that's a quibble.) Where people get confused is when they try and find a shortcut or simplification, like the initial misunderstanding of the equivalence principle.
Andrew Mason
#15
Jan3-13, 11:25 PM
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Quote Quote by Vanadium 50 View Post
Andrew, I disagree. Whether an accelerated charge radiates is settled. This is section 14.3 of Jackson (2nd). An accelerated charge doesn't know that its acceleration is or is not gravitational, so it can't "decide" whether or not to radiate, as you suggest in Post #9.
I don't think a charge has to "know" anything. But surely there is a difference between a charge interacting with an electrical field and interaction with a gravitational field. Why should they be the same?

A full GR treatment agrees, insofar as "radiation" is well-defined. (Observers, especially nearby accelerating ones, may disagree about what is near-field and what is far-field, but that's a quibble.) Where people get confused is when they try and find a shortcut or simplification, like the initial misunderstanding of the equivalence principle.
Perhaps you can explain to me, then, how the charge in circular gravitational orbit radiates without losing energy. One of the papers stated: "To explain this puzzle, we need to recognize that the concept of radiation has no absolute meaning, and that it depends both on the radiation field and the state of motion of the observer." While that may well be true, I don't see how radiation can exist without some kind of loss of energy.

It is a very difficult experiment to do, so it may be a while before anyone can experimental data to test the theory. So it may be unresolved for some time.

AM
K^2
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Jan3-13, 11:53 PM
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Don't we have direct evidence of orbiting charges radiating from black hole X-Ray radiation? The metric is the same, so I don't see why it should be much different on Earth's orbit, other than in magnitude, of course.

And surely, for static charge in gravitational field, it cannot be that hard to solve Maxwell's equations with four-current being just j0 using covariant derivatives and Shwarzschild metric.
andrien
#17
Jan4-13, 12:13 AM
P: 1,020
You figured this out yourself?
you need to see schwinger's book 'classical electrodynamics' for this.He shows there that if a charge is accelerating and you don't care about the starting of motion and other factors,then you will find that accelerating charges don't radiate as opposed to larmor's formula.
the_emi_guy
#18
Jan4-13, 12:14 AM
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Quote Quote by K^2 View Post
Don't we have direct evidence of orbiting charges radiating from black hole X-Ray radiation?
Could supply a scholarly reference to this radiation that connects it to our classic problem?


Quote Quote by K^2 View Post
And surely, for static charge in gravitational field, it cannot be that hard to solve Maxwell's equations with four-current being just j0 using covariant derivatives and Shwarzschild metric.
If it is easy then it surely has been done. Can you supply a reference?


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