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If a charged particle is free falling in the gravitational field-

  1. Mar 13, 2007 #1
    If a "charged particle" is free falling in the gravitational field----------

    If a "charged particle" is free falling in the earth`s gravitational field, it will emit electromagnetic waves????

    (1)If a person stands on the ground, he will say:Yes ! because,the charged particle is accelerated by the gravitational field!

    (2)If another person is free falling along with that particle,he will say:No! because according to the "equivalent principle" he and the particle have the same amount of acceleration, from his point of view,the particle has no acceleration,it can not emit electromagnetic waves!

    (3)Paradox!!! I want to ask: who is right???

    . Thank you
     
  2. jcsd
  3. Mar 13, 2007 #2

    Demystifier

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    Acceleration is not relative, it is absolute. A free fall is actually a motion along a geodesic, which is locally the same as motion with uniform velocity without presence of the gravitational field. An analogous question is: Will a charge at rest (without a gravitational force) radiate from the point of view of an accelerated observer? The answer is - no.
     
  4. Mar 13, 2007 #3

    pervect

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    This problem has created a surprising amount of controversy. One paper that I rather like on the topic is http://xxx.lanl.gov/abs/gr-qc/0006037. As the authors note

    and this one paper shouldn't necessarily be regarded as definitive. A key issue seems to be what one means when one says "emits electromagnetic waves" or "radiates".
     
  5. Mar 13, 2007 #4

    Haelfix

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    Yea this is a complicated question. The resolution is still controversial, but people generally agree that the paradox dissappears.

    The first thing that helps is to forget ones intuition about what radiation is and instead focus on say, its effects. Namely does an accelerating particle heat up say, a glass of water? When properly formulated, a stationary observer and say someone in an elevator passing by, will both agree that this happens.

    The root of the nastiness and the cause of the paradox is that radiation is defined with respect to moving a test particle away to infinity. So there is implicitly a prefered inertial frame where 'classical radiation' is defined with respect to a far field. This definition is slightly problematic b/c you can see its somewhat clashy with relativity. Making it mesh, is the technical problem people clash about.
     
  6. Mar 13, 2007 #5

    Stingray

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    Regardless of how you define radiation, a freely-falling charge does not necessarily move on a geodesic (even in the absence of other charges). It violates the equivalence principle by virtue of being inseparable from its electromagnetic field. This makes the particle act as though it had a large radius no matter how small it actually is. The equivalence principle only holds (as far as it goes) for experiments which can only sample an infinitesimally small region of space.
     
  7. Mar 13, 2007 #6
    Remember: a charged particle in an EM field is not freely falling.
     
  8. Mar 13, 2007 #7

    Garth

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    Does a supported electron sitting stationary on a laboratory bench radiate?

    According to the equivalence principle it is accelerating at g upwards and therefore according to Maxwell's equations it should radiate, however, if so, where is the power for this radiation coming from?

    Garth
     
    Last edited: Mar 14, 2007
  9. Mar 13, 2007 #8

    robphy

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    This question has been bothering me for while.... especially when I have received convincing-at-the-time answers from knowledgeable professors which I now cannot reconstruct [or even recall whether it's yes or no, let alone any of the caveats].

    As pervect's reply suggests, one must first find a clear and precise mathematical formulation of the problem [intrepreting the physics mathematically], prove a resulting mathematical theorem, then interpret the mathematics physically. Assuming the mathematics is done correctly, there's nothing there to argue over. The key issues must lie in the translation to or from the mathematics... e.g.,
    a particle is said to "radiate" when... <insert a mathematical condition regarding spacetime curves and tensor fields>,
    the equivalence principle states.... <insert a mathematical statement>.

    Is there such a formulation?
    (The paper referenced by pervect looks like a good starting point to review the literature.... time-permitting, of course.)

    Underlying my reply is the question:
    Is the question being posed even well-defined?
    Or does it need to be better formulated before a resolution can be obtained?
    Does the question capture the underlying issues involved?
     
    Last edited: Mar 13, 2007
  10. Mar 13, 2007 #9
    Hmm, perhaps I miss some essential element here, but, for instance an electron radiating, simply means that one or more photons are emitted, right?

    So, to me that seems like a clear space-time event. So how do you conclude that that event does not exist for an accelerated observer?

    To make it even more specific, what if the electron emits a photon in the direction of the observer, and assuming the case that there is no Rindler horizon here, are you saying it won't hit him since he is accelerating? :confused:
     
    Last edited: Mar 13, 2007
  11. Mar 13, 2007 #10
    I'm not qualified to answer the question but it's to interesting a paradox not to. It seems to me that the very reason why it wouldn't be visible is the reason why it can exist. The during laspse in time before the radiation or effects of radiation reach you will have moved.
     
  12. Mar 13, 2007 #11

    pervect

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    In the past, I've argued that particle number is not conserved under acceleration, pointing to the Unruh effect as an example. "Rindler" particles are not the same as "Minkowski" particles, in particular it turns out that Rindler particles correspond to Minkowski particles with a negative frequency. (This much is mentioned in Wald).

    Thus, by this explanation, there is not any reason to expect the number of photons to agree in the accelerated and non-accelerated frame.

    I see that at least one paper on the topic, by Demystifier aka Hrojve Nikolic, http://arxiv.org/abs/gr-qc/9909035, disagrees with this explanation, however.

    (This might be another good paper to read with the one I first suggested (ideally, along with all of the referenced papers in both articles)
     
  13. Mar 14, 2007 #12
    So far ,is there any relevant "experiment test" have conducted??
     
    Last edited: Mar 14, 2007
  14. Mar 14, 2007 #13

    Demystifier

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    Yes, you miss an essential element. More precisely, you do not read carefully. I was talking about a charge at rest, which certainly does not radiate from the point of view of an inertial observer. Thus it should not be surprising that it does not radiate from the point of view of an accelerated observer as well.
     
  15. Mar 14, 2007 #14

    Demystifier

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    Thanks for mentioning it. For a more explicit critique of the Rindler particles see also
    http://arxiv.org/abs/gr-qc/0103108
     
  16. Mar 14, 2007 #15
    Now you still got me interested here. :smile:

    So you say a charge at rest does not radiate, right?
    But you also say that an observer who accelerates away from such a charge does not measure radiation, right?

    So then I would like to know what you consider a charge at rest. At rest relative to what?
     
    Last edited: Mar 14, 2007
  17. Mar 14, 2007 #16

    Demystifier

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    Now you are right, I was not sufficiently precise.
    At rest with respect to an inertial frame in flat spacetime. OK?
     
  18. Mar 14, 2007 #17

    Garth

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    And what happens with a supported charge at rest relative to the Centre of Mass in a gravitational field with 'curved' space-time?

    Garth
     
  19. Mar 14, 2007 #18

    pervect

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    I have a few questions:

    Would you agree with Wigner's field-based defintion of a particle in flat space-time as an irreducible representation of the Poincare group?

    Is your conclusion basically that we have no good defintion of a particle at all in curved space-times, or am I misreading your conclusion? (I.e. to use Wigner's defintion, no Poincare group -> no particles).
     
  20. Mar 14, 2007 #19

    pervect

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    Some more comments:

    As I understand it, an inertial observer will see an accelerated charge as emitting particles. A co-accelerating observer will see the same charge not as emitting particles, but as absorbing them.

    I think this implies that particle number is not a tensor quantity, but I probably need to think about this more.
     
    Last edited: Mar 14, 2007
  21. Mar 14, 2007 #20

    Haelfix

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    Again the key thing is that both observers will agree on the measurement of any experiment, its just the interpretation that will change (for instance one person sees 'radiation' the other comoving observer sees inductance)

    Particle number is most definitely not a conserved quantity in curved space, as that only makes sense when the relevant isometry group is Poincare. Only the Smatrix is conserved, given suitably nice asymptotics and time like killing vectors
     
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