Past exam of the condensed matter physics course of the University of Oxford / 2012 / 2

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.

The FCC conventional unit cell has four atoms, so the cube containsatoms, neglecting surface corrections. A phonon is one excitation quantum of a vibrational normal mode; there can be many quanta per mode, and zero-point motion does not add half a phonon to the thermal population.

For the low-temperature thermal phonon number, integrate the Bose-Einstein distribution against the Debye model phonon density of states:At $10\mathrm{K}$, ${\Theta }_D/T=21.5$, so extending the upper limit to infinity has negligible effect at the precision required. The supplied integral yieldsThis is the thermally excited phonon number, not the number $3N$ of available modes.

The continuum approximation fails when the thermally occupied wavelengths become comparable with the cube size $L$. From the Debye temperature,A lowest nontrivial acoustic frequency is of order $v/L$; a standing-wave estimate ${\omega }_{\min }\sim \pi v/L$ corresponds to $\hslash {\omega }_{\min }/k_B\sim 0.25\mathrm{K}$. Periodic boundary conditions instead give $2\pi v/L$ and $0.50\mathrm{K}$. The precise spectrum depends on mechanical boundary conditions and branch speeds, neither of which is specified.

This finite-crystal phonon freeze-out requires summing discrete normal modes. For a finite positive lowest frequency, $N_{\mathrm{ph}}\sim g_{\min }{e}^{-\hslash {\omega }_{\min }/(k_{B}T)}$, rather than extrapolating the bulk $T^3$ law indefinitely. Rigid translations and rotations of an isolated crystallite are not internal thermal phonons. Surface acoustic modes can modify the intermediate crossover, but do not justify inventing a unique boundary spectrum.

The harmonic phonon contribution to the internal energy isFor $T\gg {\Theta }_D$, every retained mode has $x=\hslash \omega /(k_{B}T)\ll 1$. Expanding $1/(e^x-1)=1/x-1/2+O(x)$ cancels the explicit half quantum and gives $k_{B}T$ per mode to leading order. ThereforeEquivalently, differentiating before integrating givesThe integrand tends to $x^2$, and its integral is $({\Theta }_D/T{)}^3/3$, producing the same Dulong-Petit law. In classical language, the equipartition theorem assigns $k_{B}T/2$ each to the kinetic and elastic parts of every harmonic normal mode. Fixed zero-point energy does not contribute to the heat capacity.