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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.

The Debye model approximates a three-dimensional crystal as an elastic continuum supporting quantized normal modes, or phonons. At long wavelengths, the acoustic phonon branches have approximately linear dispersion relations ${\omega }_s(k)=v_{s}k$: one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total $3N$ vibrational degrees of freedom for $N$ atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.

For periodic boundary conditions in volume $V$, each allowed wavevector occupies $(2\pi {)}^3/V$ in the space of wavevectors. Counting a spherical shell for each branch gives the density of states per unit angular frequencyAn effective speed of sound $v_D$ can be defined by $3/v_D^3=1/v_l^3+2/v_t^3$. In the single-speed Debye model, choose ${\omega }_D=v_D{k}_D$ so thatThus the Debye wavevector, Debye frequency and Debye temperature satisfyHere $g$ has units of seconds when angular frequency is measured in ${\mathrm{s}}^{-1}$; its integral is the dimensionless number of normal modes. The spherical cutoff replaces the real Brillouin zone by an equal-count continuum construction.

Each phonon mode is a quantum harmonic oscillator. Since phonon number is not conserved at thermal equilibrium, the chemical potential is zero and its Bose-Einstein distribution gives mean thermal energy $\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]$. The zero-point energy is independent of temperature in the harmonic, fixed-volume approximation and does not contribute to the heat capacity at constant volume. Excluding that constant term,Differentiating the first integral at fixed volume avoids losing the temperature-dependent upper limit in the second one. The resulting Debye heat capacity isFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\zeta (4)={\pi }^4/15$. ThereforeThis $T^3$ law follows from the three-dimensional ${\omega }^2$ density of states and the availability only of low-frequency modes at low temperature. It is a central success of the Debye model.

For $T\gg {\Theta }_D$, the Bose-Einstein distribution reduces to the classical equipartition theorem: every mode contributes thermal energy $k_{B}T$. Consequentlythe Dulong-Petit law, or $3R$ per mole of atoms. The simple model omits detailed phonon dispersion relation, distinct optical phonon branches, and anharmonic effects such as thermal expansion. In a metal, the low-temperature total heat capacity additionally contains a term approximately proportional to $T$ from electrons, so the lattice $T^3$ term should not be mistaken for the whole measurement.

Only states within an energy window of order $k_{B}T$ around the Fermi energy can change occupation in a degenerate metal. A Sommerfeld expansion gives electronic energy above its zero-temperature value proportional to $g(E_F)(k_{B}T{)}^2$, hence $C_e=\gamma T$. Three-dimensional linear-dispersion acoustic phonons have density of states proportional to ${\omega }^2$; integrating their Bose occupations gives thermal energy proportional to $T^4$ and the Debye heat capacity $C_{mph}=\beta T^3$. Thus $C/T=\gamma +\beta T^2$ at sufficiently low temperature in the normal state.

Reading the straight low-temperature fit gives approximately $\gamma \simeq 33\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$ and $\beta \simeq 3.8\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-4}$. These estimates agree with the reported fit to this material in the original experiment.^[ref] Treating the graph values as approximate, the molar density of states isThe stipulated two electrons per formula unit give $n=2N_A/V_m\simeq 2.62\times 10^{27}{\mathrm{m}}^{-3}$. For a single spin-degenerate parabolic band, the molar density of states is $g(E_F)=3(2N_A)/(2E_F)=3N_A/E_F$. HenceThese are model-dependent effective parameters inferred from the measured heat capacity and assumed carrier count, rather than an independent determination of a unique band structure.

Using the Debye prefactor stated in the question givesThis prefactor counts three effective modes per formula unit. Consistently choosing the same count for the Debye wavevector gives $k_D=(6{\pi }^2{N}_A/V_m{)}^{1/3}\simeq 4.3\times 10^9{\mathrm{m}}^{-1}$. The corresponding average sound speed isThere is a normalization choice here: a formula unit has 31 atoms. With the conventional full atomic-mode count, the molar Debye coefficient contains an additional factor 31, and both ${\Theta }_D$ and $k_D$ increase by $31^{1/3}$. That convention gives ${\Theta }_D\simeq 251\mathrm{K}$ and $k_D\simeq 1.34\times 10^{10}{\mathrm{m}}^{-1}$, while leaving the inferred sound speed unchanged. The mode count in the Debye model must be used consistently; mixing the printed temperature with an atomic-count wavevector gives an erroneous sound speed.

The upward curvature above about $3\mathrm{K}$ indicates low-energy vibrations missing from a pure acoustic $T^3$ model. In this cage compound, the original analysis fits an additional localized mode with $\hslash \Omega /k_B\simeq 32\mathrm{K}$, assigned to a rattling mode of a barium guest atom.^[ref] At temperatures high compared with all relevant phonon energies, the full lattice approaches the Dulong-Petit law, $C_{mlattice}\to 3(31)R=93R\simeq 773\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The reduced three-mode model alone approaches $3R$ and omits the other branches. An approximately linear electronic term can still be added while the carriers remain degenerate, but cannot be extrapolated beyond their Fermi temperature.

In the three-dimensional Debye model, the lattice has $3N$ normal modes for $N$ atoms. Replacing their dispersion by a linear acoustic dispersion up to a cutoff gives the Debye heat capacityAt low temperature the upper limit can be replaced by infinity. Since the integral then equals $4{\pi }^4/15$,At high temperature each harmonic mode contributes $k_B$ by the equipartition theorem, giving the Dulong-Petit lawFor one mole replace $N{k}_B$ by $R$. The curve starts with zero slope, rises as $T^3$, and levels off toward the classical limit. This is the lattice contribution; a metal's electronic heat capacity is a separate low-temperature term.

The Debye wavevector follows by counting three polarizations inside the cutoff sphere:With the supplied speed of sound treated as the effective Debye speed, ${\omega }_D=v{k}_D$ andHere $k_D\simeq 1.52\times 10^{10}{\mathrm{m}}^{-1}$. Real crystals have different longitudinal and transverse sound speeds; the one-speed estimate uses the stated value, rather than inferring those unprovided branches.

For one conduction electron per atom, let $n_a$ be the atom density. State counting in a Bloch band gives the Fermi wavevector $k_F=(3{\pi }^2{n}_a{)}^{1/3}$. The Debye model preserves the three vibrational modes per atom by a spherical Debye wavevector cutoff $q_D=(6{\pi }^2{n}_a{)}^{1/3}$, soBoth wavevectors are of order the inverse atomic spacing. The numerical factor uses the spherical Debye approximation; a real zone boundary and the largest-frequency mode depend on direction and branch. Because $q_D<2k_F$, short-wavelength acoustic phonons can scatter Fermi surface electrons through substantial angles. For nearly elastic scattering within a spherical Fermi surface,Near room temperature, when a substantial fraction of the phonon spectrum is occupied, electron-phonon scattering is often the main source of electrical resistance. For $T\gg {\Theta }_D$, the phonon population and scattering rate are approximately proportional to $T$, giving $\rho \propto T$, $\sigma \propto T^{-1}$ and roughly constant $K_e$ when the Wiedemann-Franz law holds.

For small-angle electron-phonon transport at low temperature, thermally active acoustic modes instead haveThey predominantly scatter forward. The factor $1-\cos \theta$ makes such collisions inefficient at relaxing an electrical current, while a phonon energy of order $k_{B}T$ can still substantially change an electron's thermal excitation. Heat-current relaxation therefore need not share the charge-current relaxation time. In the clean three-dimensional acoustic-phonon model this gives ${\tau }_Q^{-1}\propto T^3$ but ${\tau }_{\sigma }^{-1}\propto T^5$, so the phonon-limited electrical resistivity scales as $T^5$ and $K_e\propto T^{-2}$. These laws describe the regime before residual elastic scattering takes over; at sufficiently low temperature impurities give nearly constant residual resistivity and $K_e\propto T$, restoring the usual Lorenz number.^[ref]

Phonons also transport heat themselves, with $K_{\mathrm{ph}}\simeq c_{\mathrm{ph}}{c}_s{\ell }_{\mathrm{ph}}/3$. At low temperature their population decreases and their phonon mean free path can become boundary- or defect-limited. At higher temperature Umklapp scattering can relax the phonon heat current. Such lattice conduction is an additional contribution, not the quantity constrained by the electronic Wiedemann-Franz law.