Phonons and the Debye Model

A crystal of $N$ atoms has $3 N$ vibrational degrees of freedom. Classically each contributes $k_{B} T$ to the energy by equipartition, giving the constant heat capacity $3 N k_{B}$ that Dulong and Petit measured at room temperature, but the heat capacity of every solid falls to zero at low temperature, and it falls as $T^{3}$ . The Einstein solid gets the freeze-out qualitatively but predicts an exponential falloff, too fast. The defect is the assumption that every atom vibrates at one frequency. The lattice actually supports a continuum of collective modes — sound waves of every wavelength down to the atomic spacing — and quantizing those modes as bosons gives the observed $T^{3}$ law.

Phonons as a boson gas

The normal modes of a harmonic lattice are collective oscillations, each an independent harmonic oscillator of some frequency $ω$ . Quantizing a mode gives evenly spaced levels $(n + \frac{1}{2}) ℏ ω$ ; the excitation quantum is a phonon, a boson carrying energy $ℏ ω$ and crystal momentum $ℏ k$ . Phonon number is not conserved — raising a mode's excitation creates phonons, thermal contact destroys them — so, as for photons, the chemical potential is zero and the mean occupation of a mode is

⟨ n (ω)⟩ = \frac{1}{e ^{ℏ ω / k_{B} T} - 1} .

The construction parallels the photon gas exactly, with three differences: the wave speed is the speed of sound $v_{s}$ rather than $c$ ; there are three polarizations per wavevector (one longitudinal, two transverse) rather than two; and the number of modes is not infinite but exactly $3 N$ , because a lattice of $N$ atoms cannot vibrate at a wavelength shorter than roughly the interatomic spacing.

Dispersion and the Debye approximation

The true dispersion $ω (k)$ of a monatomic lattice is linear at long wavelength, $ω \approx v_{s} k$ , and bends over to a maximum at the edge of the first Brillouin zone, where the wavelength reaches twice the lattice spacing. The exact relation is complicated and material-specific. The Debye model makes two simplifications: it takes the dispersion to be linear, $ω = v_{s} k$ , for every mode, and it replaces the polyhedral Brillouin zone by a sphere in $k$ -space chosen to enclose exactly $3 N$ modes. The sphere radius sets a maximum frequency $ω_{D}$ , the Debye frequency.

The true acoustic branch rises linearly then flattens at the Brillouin-zone edge; the Debye model keeps the linear slope and truncates it at a cutoff frequency chosen to preserve the total mode count.

The density of modes follows the photon counting with the three replacements. For one polarization the number of modes with wavevector up to $k$ is $V k^{3} /6 π^{2}$ , so with three polarizations and $ω = v_{s} k$ ,

g (ω) = \frac{3 V}{2 π ^{2} v _{s}^{3}} ω^{2}, 0 \leq ω \leq ω_{D},

and $g (ω) = 0$ above the cutoff. The cutoff is fixed by requiring the total mode count to equal $3 N$ :

0 \int ω_{D} g (ω) d ω = \frac{V ω _{D}^{3}}{2 π ^{2} v _{s}^{3}} = 3 N ⟹ ω_{D} = v_{s} (6 π^{2} \frac{N}{V})^{1/3} .

The cutoff wavelength $2 π v_{s} / ω_{D}$ is of order the interatomic spacing, the physical shortest wave the lattice supports. The Debye temperature packages the cutoff as an energy scale.

The Debye density of states rises as frequency squared and stops abruptly at the cutoff; the Einstein model concentrates all modes at a single frequency instead, a spike.

Energy and heat capacity

Dropping the temperature-independent zero-point term, the thermal energy sums the mode energy over the density of states,

U = 0 \int ω_{D} g (ω) \frac{ℏ ω}{e ^{ℏ ω / k_{B} T} - 1} d ω .

Substituting $x = ℏ ω / k_{B} T$ and $x_{D} = Θ_{D} / T$ , and using $ω_{D}^{3} = 6 π^{2} N v_{s}^{3} / V$ to eliminate the constants,

U = 9 N k_{B} T (\frac{T}{Θ _{D}})^{3} 0 \int x_{D} \frac{x ^{3} d x}{e ^{x} - 1} .

Differentiating gives the Debye heat capacity,

C_{V} = 9 N k_{B} (\frac{T}{Θ _{D}})^{3} 0 \int x_{D} \frac{x ^{4} e ^{x}}{( e ^{x} - 1 ) ^{2}} d x,

a universal function of the single ratio $T / Θ_{D}$ . This scaling — that every insulating solid has the same $C_{V} /3 N k_{B}$ once temperature is measured in units of its own Debye temperature — is the model's sharpest prediction, and it holds across materials whose $Θ_{D}$ span an order of magnitude.¹

The two limits

High temperature ( $T ≫ Θ_{D}$ ). Then $x_{D} \to 0$ and the integrand $x^{4} e^{x} / (e^{x} - 1)^{2} \to x^{2}$ over the whole range, so $\int_{0}^{x_{D}} x^{2} d x = x_{D}^{3} /3$ and $C_{V} \to 9 N k_{B} (\frac{T}{Θ _{D}})^{3} \frac{1}{3} (\frac{Θ _{D}}{T})^{3} = 3 N k_{B} .$ The Dulong-Petit value is recovered: all $3 N$ modes carry $k_{B} T$ each.
Low temperature ( $T ≪ Θ_{D}$ ). Then $x_{D} \to \infty$ and the integral becomes the constant $\int_{0}^{\infty} x^{3} / (e^{x} - 1) d x = π^{4} /15$ , so $U \to \frac{3 π ^{4}}{5} N k_{B} \frac{T ^{4}}{Θ _{D}^{3}}, C_{V} \to \frac{12 π ^{4}}{5} N k_{B} (\frac{T}{Θ _{D}})^{3} .$

The cubic law has the same origin as the Stefan-Boltzmann $T^{4}$ energy: at low temperature only the long-wavelength modes with $ℏ ω ≲ k_{B} T$ are excited, and their number grows as $T^{3}$ from the $ω^{2}$ density of states. The finite cutoff is irrelevant at low $T$ because those high modes are frozen anyway, which is why the Debye and exact dispersions agree there.

The Debye heat capacity rises from zero as temperature cubed at low temperature and saturates at the classical Dulong-Petit value once all modes are active above the Debye temperature.

Debye against Einstein and experiment

The two models share the high-temperature limit $3 N k_{B}$ but split at low temperature. Einstein's single frequency $ω_{E}$ gives every mode the same Boltzmann suppression $e^{- ℏ ω_{E} / k_{B} T}$ , so $C_{V} \sim e^{- Θ_{E} / T}$ falls exponentially — faster than any measured solid. Debye's spectrum of arbitrarily low frequencies keeps a population of easily excited modes down to $T = 0$ , giving the gentler $T^{3}$ falloff that experiment shows.

	Einstein model	Debye model
Spectrum	single frequency $ω_{E}$	continuum up to $ω_{D}$
Density of states	delta spike	$g \propto ω^{2}$
Low- $T$ heat capacity	$\propto e^{- Θ_{E} / T}$	$\propto T^{3}$
High- $T$ limit	$3 N k_{B}$	$3 N k_{B}$
Agreement with data	qualitative	quantitative below $Θ_{D}$

Neither model is exact — the true density of states has van Hove peaks the Debye $ω^{2}$ curve smooths over — but the Debye interpolation is accurate to a few percent across the whole temperature range with one fitted parameter, $Θ_{D}$ . The Einstein model survives as the description of the optical phonon branches of a polyatomic crystal, whose flat dispersion really does concentrate modes near a single frequency.

Measured heat capacity against temperature with both models fitted to the same high-temperature value. The Debye curve tracks the data down to low temperature; the Einstein curve drops below it, falling off exponentially.

Summary

Lattice vibrations quantize into phonons: bosons of energy $ℏ ω$ , crystal momentum $ℏ k$ , and $μ = 0$ , with occupation $1/ (e^{ℏ ω / k_{B} T} - 1)$ .
The Debye model linearizes the dispersion $ω = v_{s} k$ and cuts it at $ω_{D} = v_{s} (6 π^{2} N / V)^{1/3}$ , chosen so the $g (ω) = 3 V ω^{2} /2 π^{2} v_{s}^{3}$ spectrum holds exactly $3 N$ modes. The Debye temperature is $Θ_{D} = ℏ ω_{D} / k_{B}$ .
The heat capacity is a universal function of $T / Θ_{D}$ : it reaches $3 N k_{B}$ (Dulong-Petit) for $T ≫ Θ_{D}$ and falls as $C_{V} = \frac{12 π ^{4}}{5} N k_{B} (T / Θ_{D})^{3}$ for $T ≪ Θ_{D}$ .
The $T^{3}$ law matches experiment; the Einstein model's single frequency gives an exponential falloff that is too fast, though it describes flat optical branches well.

Schroeder, §7.5, derives the $9 N k_{B} (T / Θ_{D})^{3}$ prefactor and the $T^{3}$ law; Pathria & Beale, §7.4, tabulate Debye temperatures and the universal-curve collapse. Ashcroft & Mermin, Ch. 23, treat the true lattice dynamics the Debye sphere approximates. ↩