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To radiate or not to radiate

  1. Jan 14, 2010 #1
    An isolated charge, accelerated by a constant force, theoretically radiates (Larmor). Does the same charge, held at rest in a gravitational field, constantly radiate?
     
  2. jcsd
  3. Jan 14, 2010 #2

    Dale

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    How do you intend to detect the radiation?
     
  4. Jan 14, 2010 #3
    It does not radiate because the large distance behavior of the fields in the latter case do not correspond to what you would get from the former case by applying the equivalence principle.
     
  5. Jan 14, 2010 #4

    atyy

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  6. Jan 14, 2010 #5
    If you refer to the radiation emitted (or not emitted) in the gravitational field, I envisioned placing the charge in a blackened container and monitoring to see if the container's temperature increases (up to a limit).
     
  7. Jan 14, 2010 #6

    sylas

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    Great question! [strike]The equivalence principle DOES apply, and the charge does radiate.[/strike] Reference given by atyy.
    Caveat: This paper is new to me. I am going to check up further and make sure I understand it better; hence this answer should be understood simply as the answer of the reference, and not of me personally. I do not have the expertise to confirm it independently. But I'll check around.

    Cheers -- sylas

    Postscript. I am speaking above out of turn; and so I have struck out the sentence above for which I do not have sufficient confidence to give such a definite answer. Apologies, I shall focus for now on reading rather than giving answers.
     
    Last edited: Jan 14, 2010
  8. Jan 14, 2010 #7
    I think this paper is better:

    http://arxiv.org/abs/gr-qc/9303025

    Put simply, if you're going to apply the equivalence principle, you have to do it correctly, and thus also look at the proper boundary conditions/initial conditions on the fields.
     
  9. Jan 14, 2010 #8
    And that paper can now be updated as the ancient problem of the electromagnetic self force which is responsible for the radiation of accelerated charges, has now been solved rigorously:

    http://arxiv.org/abs/0905.2391
     
  10. Jan 14, 2010 #9

    atyy

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    Yes, I like Parrott's paper, which is one of those I had in mind when I said the EP does not apply to charged particles (but you seem to have drawn a different conclusion?). In my understanding the Harpaz and Soker paper is just an amusing case where by accident the EP "applies" to a charged particle.
     
    Last edited: Jan 14, 2010
  11. Jan 14, 2010 #10

    sylas

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    Thank you! I withdraw my previous answer, and shall look further.

    Cheers -- sylas
     
  12. Jan 14, 2010 #11

    atyy

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    Some more references, in addition to Parrott's given above by Count Iblis.

    The qualification "uncharged" in the statement of the EP in http://arxiv.org/abs/0707.2748

    The EP does not apply to charged particles (unless it's a gravitational charge):
    http://relativity.livingreviews.org/Articles/lrr-2004-6/ [Broken]
    http://arxiv.org/abs/0806.0464
    http://arxiv.org/abs/gr-qc/0008065

    Also in Rindler's and J L Martin's GR texts (though Fredrik has now made me suspicious of Rindler ... :rolleyes:)
     
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  13. Jan 14, 2010 #12
    If the charge is held at rest, meaning mgh = constant, then where would the radiated energy come from? Wouldn't conservation of energy apply?
    Bob S
     
  14. Jan 14, 2010 #13
    Why the restriction 'isolated' charge?

    I think it does the same for all charges in any configuration subject to acceleration.

    And as all matter of the Universe is accelerated at all times ....
     
  15. Jan 14, 2010 #14
    the conservation of energy...
    the energy has to be of such a tiny value that I think is not measurable, and all instruments of measure also vary in the same way (the reference atom at lab) giving a null result.
     
  16. Jan 15, 2010 #15
    Those were questions I had in mind when I submitted the thread. The majority consensus seems to be that the EP doesn't apply to electric charge. I'll give some thought to an analogous situation that doesn't involve electric charge. Thanks to all for responding, and for the links.
     
  17. Jan 18, 2010 #16
    Is it just coincidence that the article http://www.maxwellsociety.net/Charge%20and%20the%20Equivalence%20Principle.html" [Broken] that you linked to in the other thread on the same subject, was authored by someone called G.R.Dixon?
     
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  18. Jan 18, 2010 #17

    bcrowell

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    I've spent some time studying this, and although the mathematical treatment of charged particles moving in curved spacetime is heinously complex, and I haven't dug into it, I think I've gotten to the point where I understand the issue reasonably well at the conceptual level. Of course this stuff is very subtle, so most likely I'm making mistakes :-), but I felt confident enough to incorporate an explanation in my book, http://www.lightandmatter.com/html_books/genrel/ch01/ch01.html#Section1.5 [Broken] .

    Here's a slightly edited version of what I'm saying in the book:

    ----------------

    The equivalence principle is not a single, simple, mathematically well defined statement. As an example of an ambiguity that is still somewhat controversial, 90 years after Einstein first proposed the principle, consider the question of whether or not it applies to charged particles. Raymond Chiao http://arxiv.org/abs/quant-ph/0601193v7 proposes the following thought experiment. Let a neutral particle and a charged particle be set, side by side, in orbit around the earth. If the equivalence principle applies regardless of charge, then these two particles must go on orbiting amicably, side by side. But then we have a violation of conservation of energy, since the charged particle, which is accelerating, will radiate electromagnetic waves (with very low frequency and amplitude). It seems as though the particle's orbit must decay.

    The resolution of the paradox, as demonstrated by detailed calculations by Gron and Naess http://arxiv.org/abs/0806.0464 is interesting because it exemplifies the local nature of the equivalence principle. When a charged particle moves through a gravitational field, in general it is possible for the particle to experience a reaction from its own electromagnetic fields. This might seem impossible, since an observer in a frame momentarily at rest with respect to the particle sees the radiation fly off in all directions at the speed of light. But there are in fact several different mechanisms by which a charged particle can be reunited with its long-lost electromagnetic offspring. An example (not directly related to Chiao's scenario) is the following.

    Bring a laser very close to a black hole, but not so close that it has strayed inside the event horizon at rH. It turns out that at r=(3/2)RH, a ray of light can have a circular orbit around the black hole. Since this is greater than RH, we can, at least in theory, hold the laser stationary at this value of r using a powerful rocket engine. If we point the laser in the azimuthal direction, its own beam will come back and hit it.

    Since matter can experience a back-reaction from its own electromagnetic radiation, it becomes plausible how the paradox can be resolved. The equivalence principle holds locally, i.e., within a small patch of space and time. If Chiao's charged and neutral particle are released side by side, then they will obey the equivalence principle for at least a certain amount of time --- and "for at least a certain amount of time" is all we should expect, since the principle is local. But after a while, the charged particle will start to experience a back-reaction from its own radiated electromagnetic fields. Since Chiao's particles are orbiting the earth, and the earth is not a black hole, the mechanism clearly can't be as simple as the one described above, but Gron and Naess show that there are similar mechanisms that can apply here, e.g., scattering of light waves by the nonuniform gravitational field.

    [later...]

    The equivalence principle says that electromagnetic waves have gravitational mass as well as inertial mass, so it seems clear that the same must hold for static fields. In Chiao's paradox (p. 26), the orbiting charged particle has an electric field that extends out to infinity. When we measure the mass of a charged particle such as an electron, there is no way to separate the mass of this field from a more localized contribution. The electric field "falls" through the gravitational field, and the equivalence principle, which is local, cannot guarantee that all parts of the field rotate uniformly about the earth, even in distant parts of the universe. The electric field pattern becomes distorted, and this distortion causes a radiation reaction which back-reacts on the particle, causing its orbit to decay.

    ---------------

    So if the question is whether the equivalence principle applies to charged particles or not, I think the answer is not a yes/no answer. You have to keep in mind that the equivalence principle only applies locally, and this can be tricky to translate into actual experiments. A lot of thought experiments that might seem purely local are actually nonlocal, so they appear to violate the equivalence principle. If the e.p. were a mathematically well defined statement, then we'd be able to define "local" and "nonlocal" in a rigorous way, and probably everyone would have to agree on a yes/no answer. However, nobody has ever succeeded in stating the e.p. in a mathematically rigorous way.

    I like Chiao's thought experiment a lot better than the ones involving linear acceleration. I think it's conceptually simpler. For one thing, there's been some debate over the significance of horizons in the linear case. Boulware says that the radiation disappears behind the event horizon of the accelerated observer, where it can't be observed, but Parrott argues that this is wrong.
     
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  19. Jan 18, 2010 #18

    Dale

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    Well, I don't know about using heat to measure it. I was thinking about measuring it with an antenna. My guess is that a charge accelerating past a stationary antenna will induce the same signal in the antenna as an antenna falling past a stationary charge.
     
  20. Jan 18, 2010 #19
    Very interesting comment. My guess is that a charge moving past a stationary antenna at constant velocity will induce the same signal in the antenna as an antenna falling past a stationary charge at constant velocity. Both are Faraday induction of an electric field.

    So how does the charge then radiate energy?

    My guess is that a charge accelerating past a stationary antenna will radiate the same signal in the antenna as an antenna accelerating past a stationary charge.

    Bob S
     
  21. Jan 18, 2010 #20
    I have not read the linked papers yet. (I'm in the middle of "A Rigorous Derivation of Electromagnetic Self-force". So far, so good.) But here is my question.

    If we are asking, "Does a charged particle radiate when it accelerates?" we have to specify: accelerate relative to what?

    If it has to be relative to an inertial frame, ... well, the only (locally) inertial frame we have in a (non-uniform) gravitational field is the freely falling one. So I would say that the equivalence principle has to hold because .. what else is there?

    This business about local vs non-local.. the equivalence principle is local, Maxwell's equations are local (curved spacetime or flat), so what is there that is nonlocal?
     
  22. Jan 18, 2010 #21

    Dale

    Staff: Mentor

    Well, that is not exactly what I was saying, but that may be correct also. I really don't know.

    I was thinking of a non-inertial charge and an inertial antenna in both cases, one where the charge was accelerating in the absence of gravity (e.g. on a rocket), and the other where the charge was stationary on the earth and the antenna was in free fall. But it could be that for EM it doesn't matter which is inertial and which is non-inertial. I honestly don't have a good enough feel for it without working out the equations in detail.

    In the absence of gravity case you would probably say that the energy comes from the work done by the rocket, and in the gravity case you would probably say that the energy comes from the lost gravitational potential energy of the antenna. But again, I don't have a good feel for it.

    I expect that in both cases a co-accelerating antenna will detect no radiation. That should be easier to analyze than a relatively moving antenna.
     
  23. Jan 18, 2010 #22

    atyy

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    The EP does not apply to charged particles (unless the charge is a gravitational charge). Is there such a thing as a freely falling charge? No. Because the charge is in its own electric field (roughly speaking).
     
  24. Jan 18, 2010 #23
    No, it's not a coincidence. I authored the article and linked to it rather than type in a voluminous reply here. I hope that's an OK thing to do. Best regards, George.
     
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  25. Jan 18, 2010 #24

    atyy

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    Yes, the orbiting case is simpler, and Rindler uses that in his textbook too. I think Boulware is right about the horizons and the detectability of the radiation. Gron, with Eriksen, is one of those who agree with Boulware's conclusion (ref [7] in Gron and Naess), also Almeida and Saa's http://arxiv.org/abs/physics/0506049. Nonetheless, I think the horizons are irrelevant to saving the EP for charged particles. I would agree with Gron and Naess's "As measured by this observer the charge is at rest in an inertial frame, and hence it does not radiate. However, a thorough analysis proves that this is not the solution to the paradox. It has been shown that, as measured by an observer that is not falling freely, a freely falling charge radiates with a power given by Larmor’s formula.... 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." Try also section 5.5.3 in http://relativity.livingreviews.org/Articles/lrr-2004-6/ [Broken] "In the scalar and electromagnetic cases, the picture of a particle interacting with a radiative field removes any tension between the nongeodesic motion of the charge and the principle of equivalence."
     
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  26. Jan 18, 2010 #25

    bcrowell

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    There are different ways of defining "local." After all, you can determine the Riemann tensor by "local" measurements, but the e.p. is specifically supposed to be talking about things that are "more local" than that. The e.p. basically says you can do a local linear approximation to the structure of spacetime, just like the freshman calculus idea of approximating a function with a line. But that doesn't mean that the second derivative isn't also a "local" quantity.

    AFAICT the reason that we see this long string of papers all disagreeing with one another is completely because nobody knows the right way to define "local" where it occurs in the equivalence principle. A good example of this is Parrott's criticism of Boulware's idea about how the radiation gets dumped behind a horizon where you can't see it. He says that the distinction between the near-field region and the radiation region is only an approximate one. He argues that any analytic function can be determined everywhere from knowledge of all its derivatives at a point, so an observer in the "near field" region can actually extrapolate to find the radiation field. IMO this is kind of over-reaching, because by the same argument you could do local measurements of all the derivatives of the metric (well, modulo gauge) and determine the metric everywhere in space. The equivalence principle would then be meaningless, because local experiments would all come out different from one another. In real measurements, if you try to determine the 37th derivative of something by local measurements, you're going to get killed by random errors. If this general idea were easy to formalize in the case of the e.p., I think someone would have done it by now, and we wouldn't see people like Parrott politely saying that people like Boulware are idiots, and vice versa.

    I think the basic answer is: accelerate relative to what Isaac Newton would have considered to be a good, God-fearing inertial frame. This seems to give the right answer for all the cases I have in mind (charge lying on a table, charge in an accelerating elevator, charge orbiting the earth). The other rule of thumb that seems to work is just to check conservation of energy. All of these answers are completely inconsistent with how we usually think about the e.p., but I think they just go to show that the e.p. has fuzzy boundaries.
     
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