= Solution
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 obeys
$$
m\frac{d\mathbf v_d}{dt}=-e\mathbf E-\frac{m\mathbf v_d}{\tau}.
$$
The steady state is $\mathbf v_d=-e\tau\mathbf E/m$. The <current density> is opposite to the <Electron> <drift velocity>:
$$
\boxed{\mathbf J=-ne\mathbf v_d=\sigma\mathbf E,\qquad
\sigma=\frac{ne^2\tau}{m},\qquad \rho_{\rm el}=\frac{m}{ne^2\tau}.}
$$
This gives microscopic <Ohm's law>, with <electrical conductivity> $\sigma$ in $\mathrm{S\,m^{-1}}$ and <electrical resistivity> $\rho_{\rm el}$ in $\Omega\,\mathrm m$. For a uniform wire of length $\ell$ and cross-sectional <area> $A$, its <electrical resistance> is $R=\rho_{\rm el}\ell/A$. 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,
$$
k_F=(3\pi^2n)^{1/3},\qquad E_F=\frac{\hbar^2k_F^2}{2m},\qquad v_F=\frac{\hbar k_F}{m}.
$$
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 $ne^2\tau/m$ 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 $\ell_{\rm mfp}=v_F\tau$.
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>,
$$
\frac1\tau\simeq\frac1{\tau_{\rm imp}}+\frac1{\tau_{\rm ph}}+\cdots,\qquad
\rho_{\rm el}(T)\simeq\rho_0+\rho_{\rm ph}(T).
$$
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> $\kappa_e/(\sigma T)=\pi^2k_B^2/(3e^2)$ 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>.
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