 Solution

For degenerate carriers with approximately elastic scattering and a common transport relaxation time, $\sigma =n{e}^2\tau /m^{\ast }$ and ${\kappa }_e=(C_e/V)v_F^2\tau /3$. Using $C_e/V={\pi }^{2}n{k}_B^{2}T/(2E_F)$ and $v_F^2=2E_F/m^{\ast }$ givesThis is the Wiedemann–Franz law. The same mobile electrons carry charge and heat, but Pauli statistics set the electronic heat capacity and hence the Lorenz number. The relation applies to electronic thermal conductivity, not automatically to total electrical conductivity including phonons. Inelastic electron-phonon scattering can relax energy and charge currents differently; nondegeneracy and strong energy dependence of scattering also limit the simple form.

A meaningful test combines a four-terminal electrical measurement with a steady thermal-gradient measurement on the same sample and temperature range. Contact heat resistance, parasitic heat leaks, phonon conduction and sample geometry must be controlled before interpreting a measured Lorenz ratio as an electronic departure from the law.