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In most contexts in which I've come across Mie theory so far, authors do not even mention inelastic scattering but just seem to imply that all scattering is elastic. The only exception I've found is Bohren and Huffman who say

We restrict our treatment to elastic scattering: the frequency of the scattered light is the same as that of the incident light.

in their introductory chapter and later, when they compare classical and quantum mechanical concepts:

It is also possible for a photon to interact in a solid by exciting a phonon of lesser energy, the energy difference being carried off by another photon; such processes, called inelastic, include Raman and Brillouin scattering and are beyond the scope of this book.

That is better than not mentioning inelastic scattering at all, but I'm still a bit confused.

Mie theory is based on classical electrodynamics. It seems to me, however, that inelastic scattering requires a quantum mechanical treatment, since I think of it exactly as described in the above quote, i.e. as the absorption of an incident photon and prompt emission of a secondary photon. I.e. Mie theory, or any classical theory for that matter, should be incapable of describing inelastic scattering.

  1. Is this correct?

Elastic scattering must be strongly dominant in many phenomena involving small particles, since Mie theory is quite successfully used in a number of areas.

  1. Is there an intuitive way to understand why elastic scattering is dominant?
  2. Can you think of examples where it is not dominant?

I hope I was able to provide clear questions despite my confusion.

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Inelastic scattering may be understood using a classical theory. Of course, this is a limit of the more fundamental quantum theory, but it basically goes like this:

The electric field excites a linear polarization $\mathbf{P}=\chi \mathbf{E}$ in the substance through which the light propagates. This polarization has the same time-dependence as $\mathbf{E}$. Since $\mathbf{P}$ is oscillating, it re-emits light (accelerating charges emit), and that light will have the same frequency as $\mathbf{E}$. This is linear, elastic scattering.

Now consider a vibration of the molecules/atoms of the medium (frequency $\omega_p)$. These vibrations will only occur at particular frequencies which correspond to resonances of the medium (depending on the arrangement of atoms and coupling strengths). Due to a coupling between the electron distributions and the atomic placements, a distortion of the atoms can cause a slight polarization in them (coupling constant $q$).

$$ P=\chi E\cos(\omega t)+qE\cos(\omega t)\cos(\omega_p t) $$

The second term on the right is the inelastic scattering term, which has frequency components at $\omega+\omega_p$ and $\omega-\omega_p$. So the polarization $P$ will emit colors shifted by the vibration frequency. But this is a relatively small effect because the coupling is typically very small ($<10^{-5})$. In principle, it could be dominant, but as far as I know it never is.

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  • $\begingroup$ Nice, thank you! But could you explain "Due to a coupling between the electron distributions and the atomic placements, a distortion of the atoms can cause a slight polarization in them (coupling constant q)." a little better, please? Is it that the vibration of the atoms causes them to acquire a polarization independently of the incident light? So if there were no incident light this would be thermal emission? $\endgroup$ – user35915 Dec 19 '18 at 7:24
  • $\begingroup$ @user35915 yes, you’re right. The coupling is (to first order) independent of the incident light, and there can be thermal emission at the vibration frequency. The coupling is the special sauce of this process, and depends on a lot of details (including the atomic species, the symmetries involved, the phonon spectrum, etc.). These details are where quantum mechanics really becomes necessary. But as this answer shows, a classical treatment can give you the broad strokes of inelastic scattering. $\endgroup$ – Gilbert Dec 19 '18 at 17:51

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