Density of states: Debye phonons vs free electons 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.)
 A: 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.
