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Can Rayleigh scattering be thought of as a special case of the Compton scattering?

Physics Asked by Solidification on October 29, 2021

The presence of the unmodified line (wavelength) in the scattered X-ray beam, in a Compton scattering experiment, is called the Rayleigh line. This line originates from the Compton scattering of the X-ray photons with the electrons so tightly bound that the entire atom recoils. Hence, the Compton shift is negligible. Does it mean that Rayleigh scattering can be thought of as the limiting case of the Compton scattering? If the answer is in negative, please explain why.

One Answer

Compton scattering is scattering of photons off the free electrons. Raylegh scattering is between a photon and an electron bound in an atom, when the photon energy is smaller than the transition energy: $$hbaromega < E_b - E_a = hbaromega_{ba}.$$ Resonant scattering when $$hbaromega = E_b - E_a = hbaromega_{ba}$$ (up to the precision of the level broadening). Finally, Raman scattering is scattering off the electrons bound in atoms when the photon length exceeds the transition energy: $$hbaromega > E_b - E_a = hbaromega_{ba}.$$

The problem with Compton scattering is that one cannot really have a static cloud of electrons - they are repelled by the coulomb forces - which is why one uses weakly bound electrons, which entails a bunch of effects typical for atomic scattering. Yet, Compton and Rayleigh scattering are very different things.

Recoil is another way to look at this difference: Compton scattering involves transfer of energy an momentum between between a photon and an electron, whereas Rayleigh scattering involve tsransfer of momentum (direction of photon changes) but not energy (re-emitted photon has the same energy).

Answered by Roger Vadim on October 29, 2021

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