# Beyond the frequency cutoff in Debye model

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?

• 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.] – Philip Wood Feb 28 '18 at 16:03
• 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? – Shing Mar 2 '18 at 15:43
• 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. – Philip Wood Mar 2 '18 at 15:53