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I was in Solid State Physics lecture yesterday and we BRIEFLY went over what causes phonons to collide with one another. Things such as crystal imperfections, boundaries, Temperature, but I was wondering:

Do Optical mode phonons interact differently than acoustic ones? Also, is there a way I can quantify this?

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IIRC phonons of any type are non-interacting only if the potential is quadratic (i.e. harmonic). At low temperatures this is a good approximation so you tend to get scattering only at defects. At higher temperatures the potential becomes significantly anharmonic and you can get scattering in the bulk. –  John Rennie Oct 17 '13 at 15:28

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Do Optical mode phonons interact differently than acoustic ones?

YES.

Look at the dispersion curve for any material and you can understand this better. Let us take silicon for example. enter image description here

You can see that the gradient of the frequency as a function of wave vector (aka, group velocity) is what determines the amount of energy that can be carried by phonons. You can see that the optical branches have a much flatter profile than the acoustic branches. Therefore, they do not participate heavily in energy transfer and storage (thermal conductivity and specific heat capacity).

You can also see that their frequency is much higher than acoustic branches. This means that they interact with EM waves of similar frequcencis. This is why, for example, CO2 is a greenhouse gas.

Also, is there a way I can quantify this?

YES.

Thier interaction is different from acoustic phonons and were first successfully described by Einstein where he assumed a dispersion relations at a single frequency, $\omega_0$, and independent single frequency oscillator at each atom. The density of states is given by $D(\omega) = N\delta(\omega - \omega_0)$

The acoustic phonons are described by Debye model and they have a liner dispersion relation $\omega= v_s k$ and are responsible for sound wave. The density of states is given by $$D(\omega) = \frac{V \omega^2}{2 \pi^2 v_s^3}$$

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