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The Debye model treats a three-dimensional solid as an isotropic elastic continuum with three independent acoustic phonon polarizations. It assumes harmonic normal modes, a linear dispersion relation $\omega =vq$, and equilibrium Bose-Einstein statistics with zero phonon chemical potential. Real anisotropy and nonlinear dispersion are replaced by one effective Debye sound speed. A cutoff restores the finite number of atomic vibrational degrees of freedom; the continuum itself would otherwise have infinitely many modes.

For sample volume $V$, the wavevector state density is $V/(2\pi {)}^3$ per polarization. Counting a sphere givesThere are $3N$ modes for $N$ atoms. HenceThe associated phonon density of states isIf longitudinal and transverse sound speeds differ, use $3/v^3=1/v_L^3+2/v_T^3$ for the low-frequency count. The Debye frequency is an effective mode-counting cutoff, not a claim that every actual branch terminates at exactly that frequency.

In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

The Pauli exclusion principle confines thermally excited conduction electrons to an energy interval of order $k_{B}T$ around the Fermi surface. Their excess energy is therefore proportional to $T^2$, giving an electronic heat capacity $\gamma T$. More precisely, $\gamma =({\pi }^2/3)k_B^2{g}_{\mathrm{mol}}({ϵ}_F)$. In a three-dimensional Debye model, acoustic phonons have density of states proportional to ${\omega }^2$; their thermal energy scales as $T^4$, givingper mole of atoms. This assumes a normal metal without a lower-temperature superconducting or magnetic anomaly.

For ${\theta }_D\ll T\ll T_F$, the lattice approaches the Dulong-Petit value, so $C_V\simeq 3R+\gamma T$, usually dominated by $3R$. Each classical vibrational mode contributes $k_B$ through kinetic and potential equipartition. This is a high-temperature solid-state regime, not a claim that the solid persists to arbitrarily high temperature; anharmonicity and $C_P-C_V$ can matter.

The body-centred cubic cell has two atoms, so $n_{\mathrm{at}}=2/a^3$. Counting three acoustic branches gives $k_D^3=6{\pi }^2{n}_{\mathrm{at}}$. By the Debye-temperature definition,This is an effective Debye sound speed; separate longitudinal and transverse speeds are not determined. For isotropic branches, $3/v_D^3=1/v_L^3+2/v_T^3$.