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The Lorenz number is the ratio $L=K_e/(\sigma T)$. The Sommerfeld value $L_0=({\pi }^2/3)(k_B/e{)}^2\simeq 2.44\times 10^{-8}\mathrm{W}\mathrm{\Omega }{\mathrm{K}}^{-2}$ applies under the assumptions of the Wiedemann-Franz law. The ratio must use electronic thermal conductivity if lattice heat flow is appreciable.

In a metal, mobile Electrons transport heat as well as electric charge. A temperature gradient creates an imbalance between carriers arriving from warmer and colder regions. The resulting heat current obeys Fourier's law ${\mathbf{j}}_Q=-\kappa \nabla T$ for an isotropic material, where thermal conductivity $\kappa$ has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a bar of area $A$ carrying steady heat flow $P$, $P=-\kappa AdT/dx$ when parasitic losses are negligible.

A kinetic estimate for the electronic contribution iswhere $C_e$ is the electronic heat capacity per volume, $v_F$ the Fermi velocity, and ${\ell }_{\kappa }=v_F{\tau }_{\kappa }$ the heat-transport mean free path. Fermi-Dirac statistics and the Pauli exclusion principle restrict thermal rearrangements to states near the Fermi surface, rather than allowing every Electron to contribute a classical heat capacity. A degenerate parabolic band givesUsing the Drude model form $\sigma =n{e}^2{\tau }_{\sigma }/m$ for electrical transport, the relation between thermal and electrical relaxation times in metals becomesFor scattering that relaxes heat and charge currents similarly, ${\tau }_{\kappa }\simeq {\tau }_{\sigma }$ and the Wiedemann-Franz law follows:Thus good electrical conductors are usually good thermal conductors. The correct quantum Lorenz number differs from a naive classical electron-gas estimate.

Scattering from impurities, boundaries and phonons limits the heat-transport path. In an impurity-dominated low-temperature regime with approximately constant relaxation time, ${\kappa }_e$ is approximately proportional to $T$. Where phonon-limited electrical resistivity is proportional to $T$ and the Wiedemann-Franz approximation is applicable, ${\kappa }_e$ can be roughly temperature independent. Inelastic processes can relax heat and charge currents differently, so the Lorenz ratio is not universal at all temperatures. Lattice phonons add a separate positive thermal contribution; the measured total is $\kappa ={\kappa }_e+{\kappa }_{\mathrm{ph}}$, and the electronic Wiedemann-Franz expression should not be applied blindly to that total.

In a metal, a partially filled energy band or overlapping energy bands supplies Electrons with nearby unoccupied states. At thermal equilibrium, oppositely directed group velocities are equally populated and the net electric current vanishes. An applied electric field distorts the occupation distribution, producing a small mean drift velocity. A filled energy band alone has no such low-field intraband response because its velocity contributions cancel over the entire Brillouin zone.

In the Drude model, let $n$ be the mobile Electron number density, let $e>0$ be the elementary charge, and let $\tau$ be the momentum relaxation time. The mean motion obeysThe steady state is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m$. The current density is opposite to the Electron drift velocity:This gives microscopic Ohm's law, with electrical conductivity $\sigma$ in $\mathrm{S}{\mathrm{m}}^{-1}$ and electrical resistivity ${\rho }_{\mathrm{el}}$ in $\Omega \mathrm{m}$. For a uniform wire of length $\ell$ and cross-sectional area $A$, its electrical resistance is $R={\rho }_{\mathrm{el}}\ell /A$. The dissipated power per volume is $\mathbf{J}\cdot \mathbf{E}=\sigma E^2$, the Joule heating supplied by the external source.

The free electron model with Fermi-Dirac statistics improves the microscopic picture. The Pauli exclusion principle fills states up to the Fermi energy at low temperature. For a spherical, parabolic energy band with two spin states,Only states near the Fermi surface change their occupation in a weak-field response; the large equilibrium Fermi velocity is not the small drift velocity. The same $n{e}^2\tau /m$ formula results for this isotropic parabolic model when $\tau$ is evaluated for transport near the Fermi surface. In a crystal, the energy band group velocity, effective mass and possibly direction-dependent scattering replace the simplest bare-particle picture. The mean free path is approximately ${\ell }_{\mathrm{mfp}}=v_F\tau$.

A perfectly periodic static lattice is already included in Bloch states and does not by itself give a finite electrical resistivity. Momentum relaxation comes from defects, impurities, boundaries, and electron-phonon scattering. Independent scattering mechanisms approximately add their rates, giving Matthiessen's rule,The approximately temperature-independent ${\rho }_0$ is the residual resistivity. In ordinary simple metals, the phonon contribution is approximately proportional to temperature at high temperature; a three-dimensional acoustic-phonon transport model gives a $T^5$ contribution well below the relevant Debye temperature. Other mechanisms can dominate in particular materials or regimes, so neither power law is universal.

The classical Drude model incorrectly treats the equilibrium carrier statistics. Fermi-Dirac statistics explains the small electronic heat capacity, while electrons carrying both charge and heat give the approximate Wiedemann-Franz law ${\kappa }_e/(\sigma T)={\pi }^2{k}_B^2/(3e^2)$ for degenerate carriers with appropriate nearly elastic scattering. Both the simple electrical conductivity formula and this thermal relation depend on the assumptions about energy bands and scattering, rather than being exact for all metals.

Let $n=N/\mathcal{V}$ and $c_e=C_e/\mathcal{V}$ be the carrier density and electronic heat capacity per unit volume. In isotropic kinetic theory, the angular average $⟨v_x^2⟩=v_F^2/3$ gives the electronic thermal conductivitywhere ${\ell }_Q=v_F{\tau }_Q$ is the mean free path associated with the heat-current relaxation time. With $v_F^2=2E_F/m^{\ast }$,The Drude model gives the electrical conductivity $\sigma =n{e}^2{\tau }_{\sigma }/m^{\ast }$, where ${\tau }_{\sigma }$ is the momentum relaxation time. ThereforeWhen the same approximately elastic collisions relax heat and charge currents, ${\tau }_Q={\tau }_{\sigma }$. Using $A={\pi }^2/2$ gives the Wiedemann-Franz law$L_0$ is the Lorenz number. Degenerate electrons move at the Fermi velocity, not the classical thermal velocity; using classical equipartition would give the wrong coefficient.

The Wiedemann-Franz law applies to a normal degenerate metal when the same itinerant Electrons carry heat and charge, scattering is approximately elastic over the thermal shell, and the transport probability varies slowly across the energy interval $k_{B}T$ around the Fermi level. The comparison uses the electronic thermal conductivity $K_e$, with any lattice contribution removed.

It works particularly well as $T\to 0$ when elastic impurity or defect scattering dominates. It also often holds approximately for ${\Theta }_D\ll T\ll T_F$, where thermally populated phonons act as a nearly elastic classical scattering field. Room temperature need not lie above the Debye temperature for every metal.

At intermediate temperatures, inelastic small-angle electron-phonon scattering can relax heat much more efficiently than charge. Then ${\tau }_Q\ne {\tau }_{\sigma }$ and $K_e/(\sigma T)$ can be smaller than $L_0$. Strong energy dependence of scattering, additional neutral heat carriers, or a transition out of the normal metallic state also invalidates the simple derivation. An observed deviation alone does not identify which assumption failed.

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.

For a degenerate isotropic parabolic electron band, electrical conductivity is $\sigma =n{e}^2{\tau }_{\sigma }/m$ and electronic thermal conductivity is ${\kappa }_e={\pi }^{2}n{k}_B^{2}T{\tau }_{\kappa }/(3m)$. Their ratio is ${\kappa }_e/(\sigma T)=L_0{\tau }_{\kappa }/{\tau }_{\sigma }$, where $L_0=({\pi }^2/3)(k_B/e{)}^2$ is the Lorenz number. Nearly elastic scattering that relaxes charge and heat currents similarly gives ${\tau }_{\kappa }\simeq {\tau }_{\sigma }$ and the Wiedemann-Franz law. Inelastic scattering can relax heat current and directed momentum differently, invalidating the simplest identification even in a good metal.

The Wiedemann-Franz law relates the electronic thermal conductivity of a normal degenerate metal to its electrical conductivity:An isotropic kinetic theory estimate gives $K_e=c_e{v}_F^2{\tau }_Q/3$. With electronic heat capacity density $c_e={\pi }^{2}n{k}_B^{2}T/(2E_F)$, $v_F^2=2E_F/m^{\ast }$ and $\sigma =n{e}^2{\tau }_{\sigma }/m^{\ast }$, the ratio is $L_0{\tau }_Q/{\tau }_{\sigma }$. Approximately elastic scattering with a smooth energy dependence gives equal heat and charge relaxation times. Inelastic electron-phonon scattering can make them different; the law also concerns the electronic contribution, not additional lattice heat conduction.^[ref]