 Solution

The Debye model treats a solid's low-frequency lattice vibrations as acoustic phonons in an elastic continuum. There are three polarizations per wavevector, one longitudinal and two transverse, with approximately linear dispersion relations. A cutoff is chosen to preserve the total $3N$ vibrational modes of $N$ atoms. The phonon number itself is not fixed; mode occupations follow the Bose-Einstein distribution with zero chemical potential.

For one polarization of speed $v_s$, counting wavevector states in a spherical shell gives $g_s(\omega )=V{\omega }^2/(2{\pi }^2{v}_s^3)$. Define an average Debye speed by $3/v_D^3=1/v_L^3+2/v_T^3$. The total density of states can then be writtenThe Debye wavevector is $k_D=(6{\pi }^{2}N/V{)}^{1/3}$, the Debye frequency is ${\omega }_D=v_D{k}_D$, and the Debye temperature is ${\Theta }_D=\hslash {\omega }_D/k_B$.

Omitting the temperature-independent zero-point energy, the vibrational energy and resulting Debye heat capacity areAt low temperature the upper limit can be extended to infinity. The integral is $4{\pi }^4/15$, givingOnly the long-wavelength acoustic modes are appreciably populated, explaining the Debye T-cubed test. At high temperature every mode contributes $k_B$ by equipartition theorem, so $\boxed{C_V\to 3N{k}_B}$, the Dulong-Petit law.

The model succeeds especially at low temperature, where long-wavelength elasticity is reliable. It replaces real anisotropic branches by a simple continuum and cannot reproduce detailed optical phonon branches or every intermediate-temperature feature. Metals also have an electronic heat capacity approximately proportional to $T$; defects, magnetic excitations or a phase transition can add other terms. Experiments often measure heat capacity at constant pressure, which need not equal the harmonic model's $C_V$ exactly.