The density of states of phonons can be calculated with

$$ Z(\omega)=\frac{V}{(2\pi)^3}\int_{\omega=\text{const}}\frac{d\vec{f}_\omega}{|\vec\nabla\omega|} $$

where $\omega$ is the phonon frequency and $d\vec{f}_\omega$ is the part of the infinitesimal wave vector $d\vec q$ that is perpendicular to the plane $\omega(\vec q)=\text{const}$.

Now, for each branch $i$ (longitudinal or one of the two degenerate transversal states) in an elastic isotropic material, and its speed of sound $c_i$, my textbook states that:

$$ |\vec\nabla\omega|=c_i\\ \int_{\omega=\text{const}}d\vec{f}_\omega=4\pi q^2 $$

This finally leads to the equation

$$ Z(\omega)=\frac{V}{2\pi^2}\left(\frac{1}{c_L^3}+\frac{2}{c_T^3}\right)\omega^2 $$

I understand the result of the areal integral, since it's just the area of the sphere with radius $q$. However, I do not understand the result of $|\vec\nabla\omega|$. I learned that the speed of sound is the phase velocity $c=\omega/q$. In fact, this relation was used to get to the final equation for the DOS. Now, since the derivative of $\omega$ is also equal to the speed of sound, that means a linear dispersion relation $\omega(\vec q)\propto q$ must have been assumed.

However, this seems to be an approximation to me, and I don't understand why it can be used in this case. Am I right with my assessment that this is an approximation? If yes, under what circumstances can it be used?


1 Answer 1


It seems that this derivation was using the Debye model, where

$$ \omega(q)=cq $$

It gets introduced later in the book, I don't know why it wasn't mentioned there.


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