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Radiation by accelerating charge

  1. May 1, 2010 #1
    Why according to QFT accelerating electron radiates? What is the vertex in Feynman diagram for this proces.
     
  2. jcsd
  3. May 1, 2010 #2
    electron -> electron + photon
     
  4. May 6, 2010 #3
    Could anyone give me a reference to a book where I can find explanation why
    accelerating electron radiates. I'm interested in argument based on QED.
    I know the classical derivation of thise effect. Thanks in advance.
     
  5. May 6, 2010 #4
    What is the amplitude of emission of a photon with momentum
    k and polarization lambda by the accelerated charge in
    the Minkowski vacuum. Let's consider accelerated charge as
    a current.
     
  6. May 6, 2010 #5
    I don't see why it requires QFT or even QM to explain why an electron radiates when it accelerates, radiation is just a change in position of the electric force termed the electromagnetic wave.

    QFT is the study of the force itself weather it is radiating or not right?
     
  7. May 6, 2010 #6
    It is explained in detail in the book: "Quantum Field Theory" by Claude Itzykson and Jean-Bernard Zuber. In the introduction, they also explain in detail why the classical treatment is problematic.
     
  8. May 6, 2010 #7
    The conventional view (also explained in the book by Itzykson & Zuber) is that within classical electromagnetism one cannot give a fully rigorous derivation of the radiation emitted by an accelerated charge.

    The source of the problem is actually quite easy to understand. If a charge is accelerated by an electric field and it radiates, it loses the energy and momentum in the radiation. So, the force on the charge is not given by the applied electric field times the charge. But, by assumption, only electromagnetic fields can interact with the charge. So, to correctly describe the motion of the charge from first principles, one would have to consider the interaction of the charge with the total electromagnetic fields, which consists of the applied field plus the fields generated by the charge itself.

    Now, in principle, one would always have to include the intereaction of the charge with its own field, but this, of course, leads to infinities. But one can ignore this problem in case of non accelerating charges as the net effect of the self-interaction should vanish.

    For accelerating charges this is no longer true (indeed, it is the very source of the energy and momentum radiated by the charge). But in the 19th century and early 20th century, this problem was never satisfactory solved. The theory leads to either equations that allow a charge to undergo runaway accelerations in the absense of an applied field. Or, you end up with pre-acceleration. I.e. if you suddenly switch on a field at t = 0, the charge starts to accelerate slightly before t = 0.

    Recently a solution of this problem was proposed here:

    http://arxiv.org/abs/0905.2391
     
  9. May 6, 2010 #8
    Thanks for your answer.
    I have one more question. Do you understand why the radiation isn't seen in the
    frame of reference of the charge. I heard that the energy of emmited quanta is
    equal 0 in this frame so they cannot be detected.
     
  10. May 6, 2010 #9
    Keep in mind that QFT provides only the S-matrix coefficients for you. For example, QED can calculate the amplitude for the process in which two electrons collide and produce two electrons plus one photon (this is called bremsstrahlung). Apparently, at the point of collision the two electrons were accelerated, which was the cause for the photon emission. However, QED S-matrix does not tell you how exactly this emission occurred. It only tells you the final result (emitted photon) and the probability for it.

    Eugene.
     
  11. May 6, 2010 #10
    I vaguely remember reading something like that a long time ago. I'm not sure this is really correct, though. I think it is really an issue on how you define what you mean by "radiation". You sometime hear arguments like this when discussing whether or not the equivalence prinicple is valid when a charge is accelerated by a gravitational field.

    Naively a charge accelerated by a gravitational field should should look the same as a stationary charge floating in space viewed by an accelerated observer. But this doesn't sound right as in the former case, the charge should emit radiation while in the latter case, just because the obserer is accelerating cannot mean that the charge emits radiation.

    But from what I've read, the paradox is resolved by taking into account the fact that applying the equivalence principle doesn't yield the corect boundary conditons at infinity here and those boundary conditions are important if you want to compute the radiation. So, the accelerated observer does see an accelerated charge, just like stationary observer sees an accelerated charge when it falls in a gravitational field, but the asymptotic behavior of the fields is not the same in both cases.
     
  12. May 7, 2010 #11
    Last edited by a moderator: Apr 25, 2017
  13. May 7, 2010 #12

    Hans de Vries

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    Science Advisor
    Gold Member

    What you calculate in QFT (from the vertex) is the transition current, the interference
    between the initial momentum state and the final momentum state. This pattern is a
    sinusoidal charge/current density + spin density. It is this pattern which is the source
    of the electromagnetic radiation.

    Lienard Wiechert is used to determine the propagation from the source as expressed
    by the propagator 1/q^2. The transverse components stem from the sinusoidal spin
    density (magnetization) components of the transition current.


    Regards, Hans
     
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