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Why do photons/electrons scatter at angles in compton scattering?

  1. Apr 8, 2012 #1
    This is something that I always just took for granted, but I have no idea how a photon scatters off of an electron at an angle other than 0 or 180 degrees. I haven't seen this mentioned in a modern physics or nuclear engineering textbook either, so I assuming its a pretty complicated reason?
     
  2. jcsd
  3. Apr 8, 2012 #2
    An incident photon has a well-defined momentum. According to the Heisenberg's Uncertainty principle, its position is unknown. Particularly, the impact parameter, as defined in classical scattering theory, is not defined and may take any value. But, different impact parameters correspond to different scattering angles, so you get a whole spectrum of scattering angles.

    More formally, the scattering matrix element between two plane-wave states is non-zero. The square of the modulus, multiplied by the available phase space around the final state, gives the scattering cross section.
     
  4. Apr 8, 2012 #3
    Dickfore,

    Thanks for the quick reply! So essentially all comes down to the uncertainty principle? I understand classical scattering and understood the connection you made to an impact parameter.
     
  5. Apr 8, 2012 #4
    The inelastic scattering of a photon off of a free stationary electron is described classically by Thomson scattering, and relativistically by Compton scattering. The two solutions agree for photon energies up to roughly 100 keV.
    The kinematics requires conservation of energy, and conservation of momentum in both longitudinal and transverse planes (3 unknowns, and 3 equations total). This is very similar to billiard ball kinematics.

    See Section 4 in http://farside.ph.utexas.edu/teaching/em/lectures/node92.html for a discussion of Thomson scattering.
     
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