# Wiedemann-Franz law

The Wiedemann-Franz law states that the ratio of thermal conductivity $\kappa$ and electrical conductivity $\sigma$ for metals fairly accurately obeys $\kappa/\sigma = LT$, where $T$ is the temperature and $L$ is the Lorenz number, whose value is in the order of $2 \cdot 10^{-8}$ in SI units.

Assuming that $\kappa$ refers only to the electronic contribution to heat capacity, I understand how one can derive this law, or at least justify it, in a free electron model. In the classical Drude model, one then finds $L = 3k_B^2/(2e^2)$, and in a quantum mechanical free electron treatment, one finds $L = \pi^2k_B^2/(3e^2)$.

However, what about the lattice contribution to thermal conductivity? I understand that the electronic contribution might be dominant for most metals. The example of diamond (which is not a metal and hence the Wiedemann-Franz law is not expected to hold), however, shows that the lattice contribution can have the same order of magnitude as the electronic contribution, since diamond has comparable heat conductivity metals at room temperature. One important reason for the high thermal conductivity of diamond is the scarcity of defects. Does that mean that if we could produce a sample of metal with very few defects, the lattice contribution would be important? Is it important even for metals with defects?

• The high thermal conductivity of diamond is due to its high stiffness and low density which result in very high speed of phonons (speed of sound). Metals have contributions from both mechanisms but the electronic contribution is dominant (about two orderr of magnitude higher for some metals). – nasu May 1 '17 at 20:50
• Actually the wiedemann-franz law only applies at low temperatures where the lattice contribution becomes negligible. At room temperature it usually does not hold. – KF Gauss Jan 4 '19 at 18:07