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I understand when wavelength is smaller than the atom interval, sound waves can't travel; hence, we need a frequency cutoff in the Debye Model.

But surely when it is the case, atoms are still oscillating; therefore, the oscillations must contribute some energy to the energy density.

I am left wondering what happens exactly beyond the frequency cutoff? Are those oscillation so small(?) that we can just ignore these very small energy contribution?

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"But surely when it is the case, atoms are still oscillating" I think not (if we don't count zero point energy). In the Debye model there are many frequencies corresponding to different modes of standing wave. When you reach the cut-off frequency, that's it: no more modes, no more ways of storing energy. You seem to want to revert to the Einstein model (individual atoms oscillating independently of each others' oscillations) beyond the Debye cut-off, but apart from the mixing of models, you may be forgetting that there's only one Einstein frequency, the supposed natural frequency of oscillation of independent atoms.

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  • $\begingroup$ Thanks for your answer, I am kind of new to Statistical Physics. so please forgive me if my confusion seems silly to you: since simple harmonic oscillation and coupling oscillation are kind of overlapping, so we can't superposite them. However, when there is no coupling oscillating (sound traveling) as beyond the cutoff frequency, won't natural oscillation "stands out"? (probably not, otherwise, the Debye model won't need that cutoff at the 1st place, but why not?) Or are you saying when sound can't travel, there is absolutely no any oscillation whatsoever (but I am not so sure how?)? $\endgroup$ – Shing Feb 28 '18 at 15:21
  • $\begingroup$ I'm saying the last thing you said. The Debye cut-off frequency for diamond is about $4 \times 10^{13} Hz$, whereas the Einstein frequency is about $3 \times 10^{13} Hz$, certainly not higher than the Debye cut-off, so even if you mix models, which I'm sure is quite impermissible, you ain't going to get vibrations of frequencies above the Debye cut-off. [You should check my calculations.] $\endgroup$ – Philip Wood Feb 28 '18 at 16:03
  • $\begingroup$ thanks, I have calculated the Debye frequency (I think it is $\approx 3 \times 10^{14}$?), but I am having hard time calculating the Einstein frequency, would you mind giving me some hints? $\endgroup$ – Shing Mar 2 '18 at 15:43
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    $\begingroup$ The Einstein temperature is defined as $\frac{hf}{k}$ in which $f$ is the supposed frequency of the atomic oscillators, $h$ is the Planck constant and $k$ is the Boltzmann constant. I found somewhere that the Einstein temperature for diamond is 12000 K. This is very high, but to be expected because of the stiff bonds between the (light) C atoms, making the Einstein frequency high. $\endgroup$ – Philip Wood Mar 2 '18 at 15:53

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