Electrical Conductivity from the Boltzmann Equation

Statement

Starting from the semiclassical Boltzmann transport equation for the electronic distribution \(f(\vec{r},\vec{k},t)\), we linearize about local equilibrium in the relaxation-time approximation (RTA) and solve the steady, spatially homogeneous problem in a uniform electric field \(\vec{E}\). This yields the linear-response current \(j_i=\sigma_{ij}E_j\) with the conductivity tensor \(\sigma_{ij}=e^2\!\int\!\frac{2\,d^3k}{(2\pi)^3}\,\tau(\vec{k})\,v_i v_j\!\left(-\frac{\partial f_0}{\partial\varepsilon}\right)\), which reduces to a Fermi-surface average and, for an isotropic free-electron gas, recovers the Drude result \(\sigma=\dfrac{ne^2\tau}{m}\).

Why it matters

The Boltzmann equation is the master equation of semiclassical transport: essentially every DC and low-frequency transport coefficient in a metal or semiconductor — electrical conductivity, thermal conductivity, thermopower, magnetoresistance, the Hall coefficient — is a moment of the same linearized distribution. Deriving conductivity carefully shows exactly which approximations turn a rigorous kinetic theory into the one-line Drude formula that undergraduates memorize, and where each approximation earns its keep.

It also reframes the phenomenological Drude model as a controlled limit rather than a guess. The relaxation time \(\tau\) acquires a microscopic meaning as a lifetime of the perturbed distribution, transport becomes an integral over the Fermi surface rather than over all electrons, and the same machinery, with \(-\partial f_0/\partial\varepsilon\) replaced by an energy-weighted kernel, immediately gives the Wiedemann–Franz law and Mott's formula for the thermopower.

Assumptions

Derivation

Result

Limiting cases

• High-frequency (AC) field \(\vec E=\vec E_0e^{-i\omega t}\): the steady-state assumption is dropped and \(\tau\to\tau/(1-i\omega\tau)\), giving the Drude dielectric response \(\sigma(\omega)=\dfrac{ne^2\tau/m}{1-i\omega\tau}\); reduces to the DC result as \(\omega\tau\to0\).
• Strong scattering / dirty limit \(\tau\to0\): \(\sigma\to0\) linearly, resistivity \(\rho=1/\sigma\to\infty\); the residual resistivity of a metal (Matthiessen's rule, \(\tau^{-1}=\tau_{\rm imp}^{-1}+\tau_{\rm ph}^{-1}\)).
• Ballistic limit \(\tau\to\infty\) (or \(\ell\gtrsim L\)): the RTA still gives \(\sigma\to\infty\), but the physical conductance saturates at the Sharvin/quantum limit — Boltzmann breaks (see below).
• Isotropic band \(v_iv_j\to\tfrac13v^2\delta_{ij}\): tensor collapses to scalar Drude \(ne^2\tau/m\).
• Nondegenerate (classical) gas, \(k_BT\gg\varepsilon_F-\mu\): the sharp window \(-\partial f_0/\partial\varepsilon\) is replaced by a broad Maxwell–Boltzmann weight; recovers the semiconductor mobility \(\mu=e\langle\tau\rangle/m^*\) with an energy-averaged \(\tau\).

Breaks when

Failure modes

• Confusing lifetime with transport time. Using the single-particle scattering rate \(1/\tau\) where the transport rate \(1/\tau_{\rm tr}=\int(1-\cos\theta)\,W(\theta)\,d\theta\) is required; small-angle scattering barely degrades current but shortens the lifetime, so \(\tau_{\rm tr}\gg\tau\) is possible.
• Thinking all \(n\) electrons conduct. The Drude \(n\) is an accident of cancellation; forgetting that only the Fermi-surface shell responds leads to wrong intuition for thermopower and thermal transport, where the cancellation is incomplete.
• Sign errors in \(g\). Dropping the minus sign in \(\partial f_0/\partial\varepsilon<0\), or mishandling the electron charge \(-e\), flips the direction of current relative to \(\vec E\).
• Keeping \(f_0\) in the current integral. Failing to note that \(\int\vec v\,f_0=0\) by parity and double-counting the equilibrium (zero) current.
• Using \(-\partial f_0/\partial\varepsilon\to\delta(\varepsilon-\varepsilon_F)\) for thermal transport. The leading \(\delta\)-function gives zero thermopower; one must Sommerfeld-expand to next order in \((k_BT/\varepsilon_F)\).
• Forgetting the spin factor 2 or the \((2\pi)^{-3}\) density, giving \(\sigma\) off by \(2\) or by \((2\pi)^3\).
• Treating \(\tau\) as \(T\)-independent to explain \(\rho(T)\); the temperature dependence of resistivity lives entirely in \(\tau(T)\), not in the prefactor \(ne^2/m\).

Discussion

The deep content of this derivation is that transport is a Fermi-surface phenomenon. The exact result \(\sigma_{ij}=\tfrac{e^2}{4\pi^3}\int_{\rm FS}(dS/\hbar v)\,\tau\,v_iv_j\) makes no reference to electrons deep in the sea; they are Pauli-blocked and cannot be accelerated. That the isotropic average nonetheless returns \(ne^2\tau/m\), a formula containing the total density \(n\), is a genuine coincidence of the free-electron band: the density of states \(g(\varepsilon_F)=3n/mv_F^2\) carries the same \(v_F^{-2}\) that the two velocity factors supply, and they cancel. In any anisotropic metal this cancellation is imperfect and the tensor structure of \(\sigma_{ij}\) directly images the shape of the Fermi surface — the origin of the enormous transport anisotropies in layered and quasi-1D conductors.

The relaxation-time approximation is where physics is quietly buried. Replacing the collision integral \(\left(\partial f/\partial t\right)_{\rm coll}=\sum_{\vec k'}[W_{\vec k'\vec k}f_{\vec k'}(1-f_{\vec k})-W_{\vec k\vec k'}f_{\vec k}(1-f_{\vec k'})]\) by \(-g/\tau\) presumes that scattering always relaxes the distribution toward equilibrium at a rate independent of the shape of the perturbation. This is exact only for elastic, isotropic scattering; more generally the true eigenvalue problem for the collision operator has many relaxation times, and the RTA keeps one. The virtue is that it reduces an integral equation to algebra while preserving the correct structure of linear response — which is why it remains the workhorse of transport theory.

A more careful statement connects the RTA to the memory-function and Kubo formalisms. The Kubo formula \(\sigma_{ij}(\omega)=\tfrac{1}{\hbar\omega}\int_0^\infty dt\,e^{i\omega t}\langle[j_i(t),j_j(0)]\rangle\) gives the same DC conductivity when the current-current correlator decays exponentially with the transport rate, establishing that \(\tau\) in the Boltzmann result is the transport lifetime of the current, not merely a single-particle quantity. Vertex corrections in the diagrammatic (Kubo) approach are precisely the \((1-\cos\theta)\) weighting that distinguishes \(\tau_{\rm tr}\) from \(\tau\); the Boltzmann equation resums the ladder diagrams automatically. Moreover, the same linearized equation with the driving term \(\vec v\cdot\nabla_r T\,(\partial f_0/\partial T)\) instead of \(-e\vec E\cdot\vec v\,\partial_\varepsilon f_0\) yields the thermal conductivity and, through the \(\varepsilon-\mu\) weighting, the Wiedemann–Franz law \(\kappa/\sigma T=L_0=\tfrac{\pi^2}{3}(k_B/e)^2\) — with the crucial caveat that this Lorenz number is exact only for elastic scattering, and inelastic phonon scattering at intermediate \(T\) drives \(L\) below \(L_0\).

Common misconceptions. (i) The drift velocity is tiny (mm/s) even though the Fermi velocity is \(\sim10^6\) m/s — conduction is a slight asymmetry of a fast, nearly balanced flux, not a slow bulk drift. (ii) \(\tau\) is not the time between collisions of a single electron in the classical sense; it is the decay time of the macroscopic non-equilibrium distribution. (iii) A larger \(n\) does not straightforwardly mean a better conductor, because \(\tau\) and \(m^*\) typically vary with \(n\) and band filling; the "good metal" copper owes its conductivity as much to a long \(\tau\) as to its carrier density.

Worked examples

Problems

1. (A) Estimate the DC conductivity of sodium, treating it as a free-electron metal with \(n=2.65\times10^{28}\ \text{m}^{-3}\) and relaxation time \(\tau=3.1\times10^{-14}\ \text{s}\). Compare to the measured \(\rho_{\rm Na}\approx4.7\times10^{-8}\ \Omega\,\text{m}\).
   Solution

   \(\sigma=ne^2\tau/m=(2.65\times10^{28})(2.56\times10^{-38})(3.1\times10^{-14})/(9.11\times10^{-31})\). Numerator: \(2.65\times2.56=6.784\Rightarrow6.784\times10^{-10}\); \(\times3.1\times10^{-14}=2.103\times10^{-23}\). Divide: \(2.103\times10^{-23}/9.11\times10^{-31}=2.31\times10^{7}\ \text{S/m}\), i.e. \(\rho\approx4.3\times10^{-8}\ \Omega\,\text{m}\), within \(\sim10\%\) of the measured value — good for a single-\(\tau\) model.
2. (B) In the RTA, resistivity is \(\rho=m/(ne^2\tau)\). At high temperature phonon scattering gives \(1/\tau\propto T\). Show how \(\rho\) scales with \(T\), and explain physically why \(n,m\) do not carry the temperature dependence.
   Solution

   Since \(n,m,e\) are (to leading order) \(T\)-independent, \(\rho=m/(ne^2\tau)\propto1/\tau\propto T\). Hence the linear-in-\(T\) resistivity of metals above the Debye temperature. Physically, raising \(T\) increases the phonon population \(\bar n_{\rm ph}\propto k_BT/\hbar\omega\), so the electron–phonon scattering rate — and thus \(1/\tau\) — grows linearly; the carrier density (set by band filling) and effective mass (set by band curvature) are essentially frozen. At low \(T\) the Bloch–Grüneisen law gives \(\rho\propto T^5\) instead, from the phase-space restriction on small-angle phonon scattering.
3. (B) A tetragonal metal has an anisotropic effective-mass tensor with \(m_x^*=m_y^*=0.30\,m\) and \(m_z^*=1.2\,m\), and a common relaxation time \(\tau=4.0\times10^{-14}\ \text{s}\) and carrier density \(n=1.0\times10^{28}\ \text{m}^{-3}\). Compute the diagonal conductivities \(\sigma_{xx}\) and \(\sigma_{zz}\) using \(\sigma_{ii}=ne^2\tau/m_i^*\), and the anisotropy ratio.
   Solution

   \(ne^2\tau=(1.0\times10^{28})(2.56\times10^{-38})(4.0\times10^{-14})=1.024\times10^{-23}\). Then \(\sigma_{xx}=1.024\times10^{-23}/(0.30\cdot9.11\times10^{-31})=1.024\times10^{-23}/(2.733\times10^{-31})=3.75\times10^{7}\ \text{S/m}\). \(\sigma_{zz}=1.024\times10^{-23}/(1.2\cdot9.11\times10^{-31})=1.024\times10^{-23}/(1.093\times10^{-30})=9.37\times10^{6}\ \text{S/m}\). Ratio \(\sigma_{xx}/\sigma_{zz}=m_z^*/m_x^*=1.2/0.30=4.0\). The tensor structure mirrors the mass anisotropy directly.
4. (B) The mobility is \(\mu=e\tau/m^*\). For electrons in silicon with \(\mu=0.14\ \text{m}^2\text{V}^{-1}\text{s}^{-1}\) and conductivity effective mass \(m^*=0.26\,m\), find \(\tau\). Is the semiclassical picture (\(\ell\gg a\), lattice constant \(a=5.4\) Å) self-consistent if the thermal velocity is \(v_{\rm th}\approx2.3\times10^5\ \text{m/s}\)?
   Solution

   \(\tau=\mu m^*/e=(0.14)(0.26\cdot9.11\times10^{-31})/(1.6\times10^{-19})\). Compute \(m^*=2.369\times10^{-31}\ \text{kg}\); \(\mu m^*=0.14\cdot2.369\times10^{-31}=3.317\times10^{-32}\); \(/1.6\times10^{-19}=2.07\times10^{-13}\ \text{s}\). Mean free path (nondegenerate, use thermal velocity) \(\ell=v_{\rm th}\tau=(2.3\times10^5)(2.07\times10^{-13})=4.8\times10^{-8}\ \text{m}=48\ \text{nm}\gg a\). So \(\ell/a\sim90\gg1\) and the semiclassical Boltzmann treatment is self-consistent.
5. (C) Starting from \(\sigma_{ij}=e^2\int\frac{2d^3k}{(2\pi)^3}\tau v_iv_j(-\partial f_0/\partial\varepsilon)\), (a) prove \(\sigma_{ij}=\sigma_{ji}\) and that \(\sigma\) is positive semidefinite; (b) for a spherical band show it reduces to \(\sigma=ne^2\tau/m\) using \(g(\varepsilon_F)=3n/(mv_F^2)\).
   Solution

   (a) Symmetry: the integrand contains \(v_iv_j=v_jv_i\), and \(\tau,\,(-\partial f_0/\partial\varepsilon)\ge0\) are scalars, so \(\sigma_{ij}=\sigma_{ji}\) (Onsager at \(\vec B=0\)). Positive semidefinite: for any vector \(a_i\), \(a_i\sigma_{ij}a_j=e^2\int\frac{2d^3k}{(2\pi)^3}\tau(-\partial f_0/\partial\varepsilon)(\vec a\cdot\vec v)^2\ge0\) since every factor is nonnegative. (b) Isotropy lets \(v_iv_j\to\tfrac13 v^2\delta_{ij}\); on the Fermi surface \(v=v_F\), and \(\int\frac{2d^3k}{(2\pi)^3}(-\partial f_0/\partial\varepsilon)=g(\varepsilon_F)\). Thus \(\sigma_{ij}=\tfrac13 e^2\tau v_F^2 g(\varepsilon_F)\delta_{ij}\). Insert \(g(\varepsilon_F)=3n/(mv_F^2)\): \(\sigma=\tfrac13 e^2\tau v_F^2\cdot3n/(mv_F^2)=ne^2\tau/m\), with the \(v_F^2\) cancelling exactly.