In the Debeye approximation the density of states goes with phonon-energy^2, while the density of states for free electrons goes with sqrt(energy of the electrons), why is that?
(I use Introduction to Solid State Physics, Charles Kittel as learning book.)
The density-of-states is given by $$D(E) \propto \int dk_x dk_y dk_z \delta\left[E-\epsilon(\mathbf{k})\right],$$ where I omitted the prefactor for simplicity. The result of this integration depends on the form of the dispersion law $\epsilon(\mathbf{k})$. Assuming that the dispersion is isotropic (which is really rarely the case in real materials) we have for electrons $$\epsilon(\mathbf{k}) = \frac{\hbar^2k^2}{2m},$$ whereas for phonons $$\epsilon(\mathbf{k}) = vk.$$ The integration in the density of states is reduced to the integration over the magnitude of the momentum (after transition to the spherical coordinates): $$D(E)\propto 4\pi \int_0^{+\infty}dk k^2\delta\left[E-\epsilon(k)\right]= \frac{4\pi (k^*)^2}{\left|\frac{d\epsilon(k)}{dk}|_{k=k^*}\right|},$$ where $k*$ is determined from the equation $E=\epsilon(k^*)$, which for electrons gives $$k^*=\frac{1}{\hbar}\sqrt{2mE}, \frac{d\epsilon(k)}{dk}|_{k=k^*} = \frac{\hbar^2k^*}{m},$$ whereas for phonons $$k^* = v, \frac{d\epsilon(k)}{dk}|_{k=k^*} = v.$$ Thus for electrons the density-of-states behaves as $D(E)\propto k* \propto \sqrt{E}$, whereas for phonons it is $D(E)\propto (k^*)^2\propto E^2$.
Finally, it is worth noting that the results are different, if this calculation is done in two or one dimensions, which is quite relevant for nanostructure physics.
