At temperatures much lower than the Debye temperature (T ≪ T_D), what does the Debye model predict for the heat capacity of an insulating solid?
AC_V = 3Nk (the classical Dulong-Petit value), since quantum effects are always small at any temperature
BC_V scales linearly with T, since phonons behave like free particles at low temperature
CC_V scales as T³ (cubic temperature dependence), approaching zero as T → 0
DC_V is temperature-independent below T_D, then rises sharply to 3Nk above it
At T ≪ T_D, most phonon modes have energies ℏω ≫ kT and are frozen out — they cannot be thermally excited. Only the lowest-frequency modes (long-wavelength acoustic phonons with ω ≪ ω_D) contribute meaningfully. The ω² density of states gives precisely the right weighting to produce C_V ∝ T³. This T³ law is one of the landmark successes of quantum statistical mechanics and is experimentally confirmed for virtually all insulators at low temperature. A linear T dependence would occur for a metal (from the electronic contribution), not from phonons.
Question 2 Multiple Choice
What is the key physical improvement the Debye model makes over the Einstein model?
ADebye includes anharmonic interactions between atoms; Einstein assumes perfectly harmonic potentials
BEinstein ignores quantum mechanics; Debye correctly includes zero-point energy of lattice vibrations
CDebye treats the crystal as having a continuous spectrum of phonon frequencies (ω² density of states up to a cutoff), rather than assuming all modes vibrate at the same frequency
DDebye uses a more realistic interatomic potential derived from first principles
The Einstein model assumed all 3N vibrational modes have the same frequency ω_E. This captured quantum suppression at low temperature but failed quantitatively — especially at low T where real crystals show T³ behavior but the Einstein model gives exponential suppression. Debye's insight was to model the crystal as an elastic continuum, giving a density of states g(ω) ∝ ω² (from the volume of a sphere in k-space) up to a cutoff ω_D. This continuous spectrum correctly captures the long-wavelength acoustic phonons that dominate at low temperature, producing the T³ law. Both models are quantum-mechanical; the key difference is the density of states.
Question 3 True / False
In the Debye model, the cutoff frequency ω_D is chosen so that the total number of phonon modes equals 3N — matching the number of degrees of freedom in a crystal of N atoms.
TTrue
FFalse
Answer: True
This is the defining constraint of the Debye model. Unlike photons in a blackbody cavity, which span an infinite spectrum, a crystal with N atoms has exactly 3N vibrational modes (3 degrees of freedom per atom). Debye imposed a hard cutoff ω_D such that ∫₀^{ω_D} g(ω) dω = 3N, where g(ω) ∝ ω². This cutoff defines the Debye temperature T_D = ℏω_D/k, a material-specific scale that separates quantum behavior (T ≪ T_D, T³ regime) from classical behavior (T ≫ T_D, Dulong-Petit regime).
Question 4 True / False
At temperatures much higher than the Debye temperature (T ≫ T_D), the Debye model predicts that heat capacity continues to increase without bound as temperature rises.
TTrue
FFalse
Answer: False
At T ≫ T_D, the Debye model predicts C_V → 3Nk, the classical Dulong-Petit value — a constant, not an increasing function of temperature. In this regime, kT ≫ ℏω for all modes, so the quantum Planck distribution reduces to the classical limit kT per mode, giving U = 3NkT and C_V = 3Nk. The heat capacity saturates because all 3N modes are fully excited and quantum effects are negligible. This classical limit is reached smoothly above T_D and represents the maximum heat capacity the model predicts.
Question 5 Short Answer
Why does the Debye model's ω² density of states produce a T³ temperature dependence of heat capacity at low temperatures, rather than some other power law?
Think about your answer, then reveal below.
Model answer: At T ≪ T_D, only modes with ℏω ≲ kT are thermally excited; the upper limit of the integral is effectively kT/ℏ rather than ω_D. The total energy is U ≈ ∫₀^{kT/ℏ} g(ω) · ℏω/(e^{ℏω/kT}−1) dω. With g(ω) ∝ ω² and the upper limit proportional to T, a change of variables x = ℏω/kT converts the integral to U ∝ T⁴ ∫₀^∞ x³/(e^x−1) dx — a pure number times T⁴. Therefore C_V = dU/dT ∝ T³. The T³ law emerges from the ω² density of states combined with the thermal cutoff at ω ~ kT/ℏ; a different power law in g(ω) would give a different power law in C_V.
This dimensional argument is the core insight. The T⁴ scaling of energy (and thus T³ scaling of C_V) at low temperature is not specific to phonons — it applies to any 3D system of bosonic excitations with a linear dispersion and g(ω) ∝ ω². The same calculation gives the Stefan-Boltzmann T⁴ law for blackbody radiation (photons). The Debye model's success is precisely that long-wavelength acoustic phonons do have approximately linear dispersion, so the elastic continuum approximation works well in the regime that matters most for the T³ law.