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Radiation of an accelerated charge

  1. Nov 1, 2013 #1
    I am having a lot of trouble understanding this concept. It seems to be the intersection of many theories: electrodynamics, special relativity, and quantum theory. Classical electrodynamics and quantum theory apparently have two different conceptions of what an EM wave is; in classical electrodynamics, an EM wave is a continuous change in the continuous EM field surrounding an electron undergoing continuous acceleration. In quantum theory, the EM wave is a photon. But a photon has to have a certain frequency, but a classical EM wave could be a complex, possibly non-periodic wave-form, having a whole spectrum of frequencies. A photon doesn't have a whole spectrum of frequencies, so if quantum theory is right, does this mean there is a whole spectrum of photons being emanated from the accelerating electron, with probability of finding a photon of a particular frequency equivalent to the amplitude of that frequency according to the Fourier decomposition of the EM field of the electron? But if each photon comes in a discrete amount, does this mean that according to quantum theory, there is no such thing as smooth acceleration? It is said that EM energy is transmitted by photon particles, but I don't understand this if the only thing that approximates a photon with a particular frequency in classical EM is the electric field of a uniformly oscillating charged particle, and the particle does not necessarily have to follow such a path. The more I think about this, the less I understand.
     
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  3. Nov 1, 2013 #2

    UltrafastPED

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    This should help your mental picture: Photons are what you get by quantizing the electromagnetic field modes.

    The other thing to bear in mind is that classical particle trajectories don't make sense in quantum theories ... because everything has a wave function, which is the origin of the uncertainty principle. So "smooth acceleration of the electron" is not a meaningful statement in QM.
     
  4. Nov 2, 2013 #3

    Jano L.

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    Your confusion is probably caused by trying to understand light with what Jaynes called "the buckshot theory" of light. In this conception, light is made of "photons, which are discrete particles of light with energy ##\hbar \omega##".

    Teachers and textbooks still teach this simple concept from 10's, because it looks so simple and apparently helps to understand many phenomena.

    The problem with the buckshot theory is that all the wave phenomena appear as miracles, and also that absorption/emission of a buckshot light particle violate energy and momentum conservation, not to mention that probability that the buckshot will find the electron is 0 if they are both point-like.

    The modern picture of light is little bit more complicated but also more satisfying. In the modern theory of light, classical or quantum, the basic object is the EM field, which is continuously varying thing in space and time. The basic tool of the theory are Maxwell equations, which are differential equations. (In the quantum version, there are also additional conditions, like commutation relations.)

    But discrete particles of light with energy ##\hbar \omega## play no basic role whatsoever. The word photon is still used, but in different meanings. In the theory, it is most often used as a name for any state of the field from the subset of possible states that are eigenstates of the number operator ##N =\sum_{\mathbf{m}} a_{\mathbf{m}}^+ a_{\mathbf{m}}## with eigenvalue 1. For example for the first excited state of mode ##\mathbf m##, it is customary to say that there is 1 photon in the mode ##\mathbf m##.

    The simplest example of such state is harmonically oscillating stationary wave in a cubical reflective box. The wave oscillates continuously with angular frequency ##\omega## determined by the mode ##\mathbf m##, and is associated with energy ##\hbar \omega##. But it is not a localized particle and does not just pop in and out of existence. The evolution of the field is continuous.
     
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