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Compton scattering elastic?

  1. Sep 14, 2008 #1
    In the usual calculaltion of the Compton effect we assume an elastic collision between a photon and an electron. Energy and momentum of the particles are conserved before and after the collision.

    In fact, the electron is subject to a high acceleration during the collision, and classically it should therefore radiate. Is this accounted for in the usual treatment of the subject? Or should there be a correction of some kind for radiative losses?
  2. jcsd
  3. Sep 14, 2008 #2


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    The collision itself is elastic. Anything that happens afterwards (radiation losses) is irrelevant.
  4. Sep 14, 2008 #3
    How do I know when a question is irrelevant?
  5. Sep 14, 2008 #4
    Yes, the definition of Compton scattering is elastic. But the OP is referring to the fact that there is a radiative resistance to acceleration (so by definition this occurs during any collision, not afterwards).

    In this case ("collision" with a "photon" of all things) it seems like folly to factor in classical radiative losses. Either consider a completely classical treatment (in which, you will find, what you considered the "scattered photon" really is the radiation due to an acceleration of the particle, albeit the transverse acceleration, and the corresponding radiative losses are crucial to the mechanism of the longitudinal acceleration) or a completely quantum treatment (sure, the acceleration may be high, but the time period is small).
  6. Sep 15, 2008 #5


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    Applying classical effects to a quantum process leads to confusion.
    The emission of one or more additional photons in the collision (making it inelastic) can be, and has to be, treated in QED.
  7. Sep 16, 2008 #6


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    Marty -- Great question. You are absolutely right that the electron radiates because of its acceleration. For many scattering experiments, the so-called radiative corrections are important -- they come from Feynman diagrams in which the scattering charged particles emit photons. For example, one process involves 1. emit a photon, then 2. absorb the incoming photon, and 3. emit the final Compton photon.

    The computations are tricky, they involve the specific experimental detection set-up, and both infrared and ultraviolet divergences -- the first solved by Bloch and Nordseick, the second by Schwinger. The basic idea is to subtract out the effects of the acceleration induced radiation; the topic is called radiative corrections.

    Strictly speaking, there is no such thing as elastic scattering of charged particles. However, there are plenty of circumstances for which the scattering is elastic to a very good approximation.

    My dissertation dealt with radiative corrections to various electron-nucleon scattering experiments -- which experiments helped explore the electromagnetic structure of protons and neutrons in the early 1960s. A very exciting time it was.

    Reilly Atkinson
  8. Sep 19, 2008 #7
    Wouldn't the radiated energy be much less than a single quantum, because the electron doesn't go through even a single cycle of oscillation? How do we reconcile this with the quantization of radiation?
  9. Sep 19, 2008 #8


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    What oscillation? The old idea of quantization implying discrete energy levels is out the door. In fact, depending on the boundary conditions, photons can have continuous energies or discrete energies. The quantization refers to the (canonically) quantized fields used in the basic theory.

    The radiated photon can have any momentum, and associated energy, subject to conservation of total momentum and energy.

    Reilly Atkinson
  10. Sep 19, 2008 #9
    Are we even talking about the same photon? We all know that the Compton photon exchanges energy and momentum with the electron. That's not the issue. I was talking about the EXTRA photon which you said we needed to account for the radiation due to the acceleration of the electron.

    Are you saying that I can have a photon of frequency f whose energy is a very small fraction of hf? Because that's what I need to balance the Compton interaction if I must account for radiative losses due to the acceleration of the electron.
  11. Sep 21, 2008 #10


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    Photons have energy hw, with w the frequency. This holds for all frequencies from 0 to infinity. Write down the conservation of energy and momentum equations for, say,

    electron + photon -> electron + photon + photon

    to see how things balance. The actual restriction of the radiating photon energies are due to the properties of the detection apparatus. Typically, one detects over a small area or volume, which means that an electron that radiated a little bit can be detected as well as one that does not radiate.

    Radiative corrections simply factor out the contribution of radiation for the detection scheme used..

    Reilly Atkinson
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