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How is the frequency unchanged in elastic wave scattering with a crystal?

  1. Aug 13, 2015 #1
    it makes sense that with an inelastic scattering process, a wave approaches a crystal, and then some energy is imparted to the phonons, so that the outgoing wave has a different frequency from the incoming wave. However, with an elastic scattering process, the frequency is unchanged-- how does the crystal not affect the wave? What exactly is happening? I don't understand the precise steps of what is going on, that causes an elastic vs inelastic result. Thanks!
     
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  3. Aug 13, 2015 #2

    Orodruin

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    Before trying scattering on a crystal, have you considered how light scattering on a single atom works with respect to elastic/inelastic scattering?
     
  4. Aug 14, 2015 #3
    Thanks, Orodruin. I looked into it, and please tell me if I understand correctly: for elastic scattering, the light induces a dipole on the atom, which in turn radiates secondary light at the same frequency as the incident light. When the light is incident on a large molecule, there is induced secondary light from several parts of the molecule, and these interfere with each other, causing phase shifts, and constructive and destructive interference toward elastic behavior.
    So, would I extend this picture to a crystal by saying that the incident light is inducing dipoles all over the crystal??
    Thanks!
     
  5. Aug 14, 2015 #4

    Orodruin

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    I would say the main point in order to have elastic scattering is to not excite internal degrees of freedom. In the case of the single atom, this would be corresponding to the atom entering an excited state. For the crystal case, exciting internal degrees of freedom is essentially exciting the phonon states.
     
  6. Aug 14, 2015 #5
    If we are not exciting the phonons, then what steps are actually happening? Thanks!
     
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