 Solution

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.