# Is this equation for the density of states of an elastic isotropic material an approximation?

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?

$$\omega(q)=cq$$