## Phonon density of states of the Debye model

In the Debye model, the dispersion relation is linear, ω = c|k|, and the density of states is quadratic as it is in the long wavelength limit.

$$D(\omega )= \frac{3\omega^2}{2\pi^2 c^3}\quad \text{[s rad}^{-1}\text{ m}^{-3}\text{]}.$$

Here c is the speed of sound. This holds up to a maximum frequency called the Debye frequency ωD. In three dimensions there are 3 degrees of freedom per atom so the total number of phonon modes is 3n.

$$3n=\int\limits_0^{\omega_D}D(\omega)d\omega.$$

Here n is the atomic density. There are no phonon modes with a frequency above the Debye frequency. The Debye freqency is $\omega_D^3 = 6\pi^2nc^3$.

The form below generates a table of where the first column is the angular frequency ω in rad/s and the second column is the density of states D(ω) in units of s/(rad m³).

 Speed of sound: c = [m/s] Atomic density: n = [1/m³]

 Material Speed of sound [m/s] Atom density [m-3] Debye frequency [rad/sec] aluminum 6320 6.03×1028 9.66×1013 copper 4660 8.47×1028 7.98×1013 diamond 18000 17.70×1028 39.39×1013 gold 3240 5.91×1028 4.92×1013 lead 2160 3.30×1028 2.70×1013 ice(-4C) 3280 3.07×1028 4.00×1013 zinc 4170 6.57×1028 6.56×1013 titanium 6100 5.66×1028 9.13×1013 tin 3320 3.69×1028 4.31×1013 silicon 9620 4.99×1028 13.81×1013 nickel 5630 9.14×1028 9.88×1013 iron oxide(magnetite) 5890 1.34×1028 5.45×1013 tungsten 5180 6.42×1028 8.08×1013 zirconium 4650 4.34×1028 6.37×1013 molybdenum 6250 6.39×1028 9.74×1013 beryllium 12900 12.29×1028 25.01×1013

The density of states can be used to calculate the temperature dependence of thermodynamic quantities.

### The high temperature limit $k_BT > > \hbar\omega_D$

The energy spectral density is,

$$u(\omega) = \frac{3\omega^2}{2\pi^2 c^3}\frac{\hbar\omega}{\exp\left(\frac{\hbar\omega}{k_BT}\right) -1}.$$

In the high temperature limit, the exponential factor can be expanded as $\exp\left(\frac{\hbar\omega}{k_BT}\right)\approx 1 + \frac{\hbar\omega}{k_BT}$. The energy spectral density then becomes,

$$u(\omega) = \frac{3\omega^2}{2\pi^2 c^3}k_BT.$$

This can be integrated to yield the internal energy density,

$$u = \frac{\omega_D^3}{2\pi^2 c^3}k_BT= 3nk_BT.$$

The specific heat has the Dulong-Petit form,

$$c_v = 3nk_B.$$