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Does a co-accelerated charge radiate?

  1. Aug 10, 2009 #1
    Is there a consensus on whether or not a co-accelerated observer measures radiation from an accelerated charge?
     
  2. jcsd
  3. Aug 10, 2009 #2

    clem

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    I think it is clear that such radiation is measured.
     
  4. Aug 10, 2009 #3

    atyy

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  5. Aug 10, 2009 #4
    Last edited by a moderator: Apr 24, 2017
  6. Aug 11, 2009 #5
    Thanks. So do you have a link to a reference to an experiment that measured the radiation from an accelerating charge when the receiver was co-accelerating with it?
     
  7. Aug 11, 2009 #6
    Last edited by a moderator: Apr 24, 2017
  8. Aug 11, 2009 #7
    Look at the equations in the link. They imply that with no relative acc(and no change in rel acc) there is no radiation.
     
    Last edited: Aug 11, 2009
  9. Aug 12, 2009 #8

    Vanadium 50

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    The short answer is "yes, there is radiation". If you hold up a radiation measuring device, it indicates radiation.

    One can then quibble endlessly about this. The longer answer is that locally, you can't tell if you have radiation or just near-field electric and magnetic fields. You can have electric and magnetic fields acting as if they are radiation, but until you measure the 1/r fall off, you can't tell radiation from near-field. So you need an extended object to do the measurement, and this extended object doesn't form a unique reference frame. So the question is actually ill-defined: one strictly speaking cannot build a radiation detector that is in a comoving frame with a point charge.

    Let me reiterate the short answer, because it's more important than the quibbling: "yes, there is radiation". If you hold up a radiation measuring device, it indicates radiation.
     
  10. Aug 12, 2009 #9
    It is clearly written in the link that a charge on the surface of the earth, being actually accelerating, doess not radiate or you'd have a perpetual energy souce.
    The extended object is in the same comoving frame in which the charge is accelerated, so according to the equivalence principle the situation is the same as an observer with the charge on the ground which does not radiate
     
  11. Aug 12, 2009 #10

    Vanadium 50

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    A charge resting on the surface of the earth does not accelerate.

    The equivalence principle does not apply to extended objects - or, more properly, the equivalence principle allows us to distinguish between uniform acceleration and gravity by seeing where they are not equivalent - gravity has tides.

    If you hold up a radiation measuring device, it indicates radiation.
     
  12. Aug 12, 2009 #11
    Sorry, I meant subject to acceleration as writtenhere
    "One of the most familiar propositions of elementary classical electrodynamics is that "an accelerating charge radiates". In fact, the power (energy per time) of electromagnetic radiation emitted by a charged particle is often said to be strictly a function of the acceleration of that particle. However, if we accept the strong Equivalence Principle (i.e., the equivalence between gravity and acceleration), the simple idea that radiation is a function of acceleration becomes problematic, because in this context an object can be both stationary and accelerating. For example, a charged object at rest on the Earth's surface is stationary, and yet it's also subject to a (gravitational) acceleration of about 9.8 m/sec2. It seems safe to say (and it is evidently a matter of fact) that such an object does not radiate electromagnetic energy, at least from the point of view of co-stationary observers. If it did, we would have a perpetual source of free energy."
     
  13. Aug 12, 2009 #12
    The charge on the earth with the observer is the same case as he meant here(the earth also changes acceleration as the latter changes direction)
     
  14. Aug 12, 2009 #13

    atyy

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    The equivalence principle does not apply to electrically charged particles. A charged particle cannot fall freely (ie. under the influence of gravity alone), because as it accelerates in the gravitational field it will give off electromagnetic radiation, which will act on the charge itself - so the charge is no longer falling under the influence of gravity alone.

    In the case of a gravitationally charged particle, the charge gives off gravitational radiation (very roughly speaking), so the charge does not fall freely with respect to the "background" spacetime, but does fall freely with respect to the combined "background" + "gravitational radiation" spacetime.

    http://relativity.livingreviews.org/Articles/lrr-2004-6/ [Broken]
     
    Last edited by a moderator: May 4, 2017
  15. Aug 12, 2009 #14

    Vanadium 50

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    This is exactly what I meant by quibbling.

    If you hold up a radiation measuring device, it indicates radiation.
     
  16. Aug 12, 2009 #15
    Nice answer. But a measuring device co accelerating with the charge in the far field won't detect radiation, right?
     
  17. Aug 12, 2009 #16

    Vanadium 50

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    I'm arguing that there is no such thing as "a measuring device co accelerating". This device has to be an extended object, and relativity of simultaneity means its not in a single frame at all, much less the same frame as the source.
     
  18. Aug 12, 2009 #17
    That's wrong. The equivalence principle does apply to electrically charged particles. The effect on a charged particle accelerating under earth's gravity(and note, a change in acc can only produce radiation) is the same as in a frame with corresponding opposite acc, look at the first two equations in the link provided in this threadhttp://www.mathpages.com/HOME/kmath528/kmath528.htm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  19. Aug 13, 2009 #18
    OK, so let's minimise the effects of relativity of simultaneity by using an infinitesimally small measuring device and velocity.
     
  20. Aug 14, 2009 #19

    Vanadium 50

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    Already answered.

     
  21. Aug 14, 2009 #20
    What's your problem?
    Did you read the links? I already said "and note, a change in acc can only produce radiation"
    You use an extended object, an infinitesimally small object or whatever, you don't measure radiation The equation is this-
    dW/dt= -(2e^2/3c^2)(dx/dt) (d^3x/dt^3)
     
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