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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.

The Wiedemann-Franz law relates the electronic part of thermal conductivity to electrical conductivity:For degenerate electrons with elastic scattering and a transport function smooth over $k_{B}T$, a Sommerfeld expansion makes the same Fermi-level transport factor appear in charge conduction and in heat conduction. Their ratio cancels that factor and leaves the displayed Lorenz number. Heat conductivity is defined at zero electric current, so thermoelectric subtraction is part of the precise transport expression.

The law does not relate total solid heat conductivity to charge conductivity when phonons contribute substantially. Strong inelastic scattering, sharp energy dependence, nondegenerate carriers or nonstandard electronic regimes can change the ratio. Low-temperature residual elastic-scattering transport is a common setting where the standard value is recovered. The unit W·ohm/K² follows because ${\kappa }_e/(\sigma T)$ contains one extra inverse temperature beyond thermal conductivity itself.

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.

The free electron model replaces the detailed periodic potential by a constant background, treating conduction electrons as mobile particles with approximately parabolic dispersion $E={\hslash }^2{k}^2/(2m)$. It is useful to distinguish the classical Drude model from the quantum free electron gas. They share a simple transport picture, but differ fundamentally in occupation statistics.

With a momentum-relaxation time $\tau$, both lead in a simple isotropic one-band approximation toThis explains approximately Ohmic longitudinal conduction and a carrier-density scale from the Hall effect. The relaxation time is phenomenological: introducing it does not itself predict scattering from defects or phonons. A four-probe resistance measurement separates the sample's longitudinal response from current-contact voltage drops; field and current reversals isolate the transverse signal.

The classical model incorrectly assigns roughly $3N{k}_B/2$ of electronic heat capacity, far too large in a degenerate metal. Fermi-Dirac statistics fixes this: most occupied states cannot change occupation after a small thermal perturbation because nearby low-energy states are already filled. Only a fraction of order $k_{B}T/E_F$ near the Fermi energy participates. For $T\ll T_F$,The quantum model also gives a finite Fermi velocity even at zero temperature and Pauli paramagnetism. Under predominantly elastic scattering, its electronic heat and charge conductivities obey the Wiedemann-Franz law,This is a quantum success; classical equipartition does not give the correct small heat capacity and Lorenz number together.

Its limitations are equally instructive. Without the periodic potential it has no band gaps, so it cannot distinguish a filled-band insulator from a metal or explain semiconductor doping. The simplest single-electron Hall formula cannot explain positive Hall coefficients; band curvature and electron holes, or a multiband transport treatment, are needed. It neglects much Fermi-surface anisotropy and detailed optical interband absorption. Noninteracting electrons alone do not explain ferromagnetism or superconductivity, and lattice thermal behavior needs phonons. The nearly-free electron model, which allows reciprocal-lattice Fourier components of the potential to mix plane waves, provides a controlled first correction near Bragg planes.

Its main success is a simple quantitative baseline for degenerate conduction electrons; its main failure is treating the crystal and interactions as too featureless. A fit of conductivity alone is not a test of the whole model: compare heat capacity, Hall sign, optical response and temperature dependence using independently constrained parameters.

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.

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.

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.