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When does a charge radiate?

  1. Dec 25, 2004 #1
    The short answer is "when it accelerates". However, there are some very interesting controversies involved.

    One involves the equivalence principle and has been much argued over in the literature. I'm really asking about the current state of the arguments.

    Loosely the equivalence principle says if you are locked in an Einstein cage and cannot look outside, you cannot perform experiments inside the cage to tell whether you are at rest on a very large homogeneous, spherical planet with a downward gravitational acceleration g or whether you are lost in space with a rocket accelerating you upwards at g. If that is so, then why aren't you blinded by the radiation from a charged body sitting on a table in a uniform gravitational field. The short answer is that the radiation at 10 m/s^2 is too small to measure. But others have argued that the radiation is not even there!!! Any one know the current status of this?

    I apologize if this has been discussed before, but the search engine here could not find any postings involving 'radiation, accelerated charge'.
  2. jcsd
  3. Dec 25, 2004 #2

    radiates what exactly, may I ask ?

  4. Dec 25, 2004 #3
    I had in mind an electric charge and electromagnetic radiation.
  5. Dec 25, 2004 #4


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    It's not a "relativity" question so much as a "quantum theory" question. An accelerating charged particle does NOT radiate energy. It radiates energy only when the change in energy is an integral multiple of the quantum.
  6. Dec 25, 2004 #5
    The best theory so far on radiating charges is quantum electrodynamics. My question is not about this or any quantum theory of radiation. It is simply about classical electromagnetism and the equivalence principle. Once this is understood then you can look for quantum gravity by blending the equivalence principle and QED. I leave that to the HallsofIvy
  7. Dec 25, 2004 #6


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    I'm not up on the latest in the literature, but I had the impression that the answer was that notion of "radiating" was observer dependent.
  8. Dec 25, 2004 #7


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    Ron I believe I understand that the question is not a quantum mechanics issue but could be about a macroscopic charged ball, sitting on the table beside you (perched on an insulated stand)

    you and the ball are either in the cabin of an accelerating rocket ship or you are in your livingroom on earth in one gee.

    in neither case do you see any EM radiation because the ball is not
    accelerating relative to you

    you and the ball are either both in the one gee gravity field
    or you are both in the accelerating rocket ship

    somebody NOT on the rocket ship might be able to detect the EM waves made by the accelerating charged ball. assuming the cabin walls did not shield it. because they would see the charge accelerating

    there is a peculiar radiation associated with acceleration called "unruh radiation" after a guy at Univ. BC Vancouver named Bill Unruh
    but you dont need a CHARGE to be accelerated for that---it is something different

    anyway if you had a trillion extra electrons on a ball at the end of a wand and someone waved the wand back and forth it would make EM waves, like radio waves, because of the acceleration

    and as long as you were not also at the end of a wand being waved back and forth then you would be able to detect these waves.

    in principle, except for they're probably being very weak and low frequency so hard to detect.
  9. Dec 25, 2004 #8


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    there's been sporadic mention of the Caltech animation of
    radiation from an (accelerating) moving charge

    the Caltech MovingCharge applet

    for instance in this thread:

    I dont recall anyone raising just this very question tho
  10. Dec 26, 2004 #9
    Detecting the radiation should have nothing to do with whether you are in the same accelerating frame or not - the radiation is going to travel relative to space at c - and since the velocity of the comoving observer is nill compared to c, this observer will detect the radiation if it exists.

    In the case of a static G field, the question arises as to where the energy is derived to produce the radiation. Interesting question!
  11. Dec 26, 2004 #10
    The latest research that I know of on this subject is

    Radiation from a Uniformly Accelerated Charge, David G. Boulware, Annals of Physics: 124, 169-188 (1980)
    See also

    Classical Radiation from a Uniformly Accelerated Charge, Thomas Fulton, Fritz Rohrlich, Annals of Physics: 9, 499-517 (1960)
    Radiation Damping in a Gravitational Field, Bryce S. DeWitt, Robert W. Brehme, Annals of Physics: 9, 220-259 (1960)
    Principle of Equivalence, F. Rohrlich, Annals of Physics: 22, 169-191, (1963)
    Radiation from an Accelerated Charge and the Principle of Equivalence, A. Kovetz and G.E. Tauber, Am. J. Phys., Vol. 37(4), April 1969

  12. Dec 26, 2004 #11

    Andrew Mason

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    If a charge sitting in a gravitational field radiated energy, one could create a perpetual energy source without doing any work. The fact is that no one has ever detected radiation from a charge due to gravity.

    The notion that an accelerated charge should produce EM radiation is not required by Maxwell's equations. It is derived from the notion that an accelerated charge must interact with its own field.

    There have been controversial attempts to explain why gravitational acceleration of a charge should not produce radiation by showing that radiation is a function of the third time derivative of position (ie. non-uniform acceleration). The fact is, however, as anyone who tries to make electrons go in a circle using magnets knows, uniformly changing the direction of a moving electron with magnetic force produces radiation.

    It seems that uniform acceleration caused by electromagnetic force does produce radiation, whereas uniform acceleration caused by gravity does not. I think that the explanation for this is still a live issue. There is an interesting discussion in the "Speed of Gravity Controversy" thread relating to this.

  13. Dec 26, 2004 #12


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    How would a charge know whether the field it was feeling was generated by itself or not? I don't see any problem with this assumption.

    I think Pete's link answered the question pretty well, but I have a bit more to say about it. Although showing things rigorously is difficult, the apparent paradox in this problem is due to a couple of misunderstandings. First of all, a charge is not a localized object. It is inevitably connected to its field, which makes it act like an extended body. The equivalence principle is therefore not applicable.

    Also, since it is effectively an extended body, its material portion doesn't necessarily fall on a geodesic in the absence of external electromagnetic fields.
  14. Dec 26, 2004 #13
    Sorry for dismissing your question, there is substance to it. What is electromagnetic radiation in Maxwell's theory (F=dA, d*F=*J)? Surely a photon is like an electron or an atom in that it exists independently of any observer's state of motion. Its existence begins at emission and ends in absorbtion. If a Maxwell radiation field is really a cloud of photons then we should expect the radiation to have an independent existence starting at emission and ending in absorbtion. The older literature defined e/m radiation only in the far field zone. Radiation left behind the bound fields of the emitter and was the only thing that survived at large distances. Work by Teitlebaum in the 70's split the Lennard Wiechert field of an arbitrarily accelerated charge into a bound field (associated with the charge) and a free field (radiation field). This permitted splitting to total field into bound and radiating pieces all the way down to the emitter itself. This still does not answer the question, I'll say more in my next replies.
    Last edited: Dec 26, 2004
  15. Dec 26, 2004 #14
    That is quite correct. However the equivalence principle will hold in a uniform gravitational field. At least when the field can be considered to be restricted within the region which is uniform.
    Unless the field is uniform.

  16. Dec 26, 2004 #15
    This is very disturbing, but possibly true. Real radiation, like light, ought to be observer independent. You may think you can red shift light out of existence, but its angular momentum is a Lorentz invariant. If Maxwell radiation is independent of the observer then there will be an invariant definition that does not involve the observer. The source free Maxwell equations (F=dA, d*F=0) produce two types of fields. When *F is closed and not exact, there is a topological charge found by integrating *F over a closed surface that encloses a missing point. This is how the charged black holes get their charge. Here r=0 is the missing point and the flat space-time analogue has a missing timelike line. There is another possibility. Instead of removing a timelike line in the flat space analogue, one could remove a null line and get a topological charge travelling at light speed. Like the magnetic monopole these have never been observed. But unlike the magnetic monopole these monsters are permitted by Maxwell's equations as they stand. To eliminate them one could require for radiation fields that *F was globally exact. In a similar way one bans magnetic monopoles by requiring F to be globally exact. Fooling with Maxwell's equations is not to be taken lightly. However I would suggest taking F=dA, d*F=0 , *F closed and not exact, for bound fields and F=dA, *F=dB for radiation fields with B an analogue of the vector potential. All the radiation fields I have checked have an F that is globally exact and globally coexact. Such a refinement nicely explains the duality rotation of one radiation field into another with A and B rotating into each other. This invariant characterization of radiation fields, leaves no room for the observer. I would be interested to see any observer dependent definition of a radiation field. I haven't yet checked this refinement against Teitlebaum's work. If this refinement is correct then there should be consistency between the two.
  17. Dec 26, 2004 #16
    Watch out! Acceleration is absolute, the inertial forces tell you so. Your argument leads to a charge in free fall with no acceleration radiating as it appears to accelerate past you at rest on earth. With no acceleration between a freely falling observer and freely falling charge, this observer would see no radiation. Clearly your radiation is observer dependant. I find this very hard to swallow.

    Here you are excavating particles out of the quantum background as you accelerate through it. The work done to make the particles is supplied by whatever is accelerating the detector. I may be wrong, but I don't think this observer dependent radiation is what we are talking about.
  18. Dec 26, 2004 #17
    Thank you Marcus. The moving charge applet is great. You can see the field lines bunching to make the radiation field. In MTW they have a great picture (due to JJ Thompson?) of the field lines from a charge given a short acceleration. One might think that the charge radiates in order to fight the acceleration directly. If so then the radiation would be in the forward direction, but it is perpendicular to the acceleration!!! In the applet at high speed you can see the bunched field lines leaving the charge directly opposite to the acceleration, again giving radiation perpendicular to the acceleration. Great video!!!
  19. Dec 26, 2004 #18
    Thanks Yogi, I share your belief that radiation ought to be observer independent

    Your second point is fascinating. In an accelerated cage, it is clearly the accelerating agent that supplies the energy for the radiation. How on earth could a static gravitational field produce the energy for the radiation???
  20. Dec 26, 2004 #19
    Pete thank you very much. I'll get these and have a look at them. I'm surprised that the most recent reference is almost 25 years old. May be people think this has been flogged to death.
  21. Dec 26, 2004 #20
    There are a few reasons for that. Here are two that I can think of

    (1) I couldn't find more recent ones. :rolleyes:
    (2) People don't think its worth redoing a calculation they can't find something wrong with.


    ps - The AJP article is short enough to scan and e-mail if you'd like?
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