= Solution
The <free electron model> replaces the microscopic ionic potential and interacting electronic motion by independent mobile <electrons> in a uniform neutralizing background, with approximately parabolic dispersion $\epsilon=\hbar^2k^2/(2m^*)$. Its quantum version fills states according to <Fermi-Dirac statistics> and the <Pauli exclusion principle>. Scattering is represented by a <momentum relaxation time> $\tau$, rather than derived from the model's ideal free trajectories. This model explains a <Fermi energy> much larger than room-temperature thermal <energy>, an <electronic heat capacity> proportional to <temperature> at low <temperature>, and a Drude <electrical conductivity>. With appropriate scattering assumptions it also yields the <Wiedemann-Franz law>. It cannot by itself explain <band gaps>, detailed band anisotropy, strong correlations, or the microscopic <temperature> dependence of $\tau$.
The inertial term is <mass> times the rate of change of mean <drift velocity>. The electric <force> is $eE$, taking the printed $e$ as the signed <electron> charge, and $-\gamma\langle v\rangle=-m^*\langle v\rangle/\tau$ describes loss of directed momentum to scattering. <electrons> retain large microscopic <velocities>; this <damping> is of their mean directed motion. In a steady DC field,
$$
\langle v\rangle=\frac{e\tau}{m^*}E,\qquad
j=ne\langle v\rangle,\qquad
\boxed{\sigma_0=\frac{ne^2\tau}{m^*}.}
$$
Both charge factors are negative for <electrons>, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the <force> and current equations each require a minus sign and give the same result.
For AC response choose physical fields as <real parts> of <phasors> proportional to $e^{-i\omega t}$. Substitution gives
$$
(-i\omega+1/\tau)\widehat v=\frac{e}{m^*}\widehat E,\qquad
\boxed{\sigma(\omega)=\frac{\sigma_0}{1-i\omega\tau}.}
$$
Thus
$$
\boxed{\operatorname{Re}\sigma=\frac{\sigma_0}{1+(\omega\tau)^2},\qquad
\operatorname{Im}\sigma=\frac{\sigma_0\omega\tau}{1+(\omega\tau)^2}.}
$$
The <real part> decreases from $\sigma_0$ to zero. The <imaginary part> starts at zero, reaches $\sigma_0/2$ at $\omega\tau=1$, and falls as $1/\omega$. The <real part> falls as $1/\omega^2$ at high <frequency>. With an $e^{+i\omega t}$ convention, only the sign of the <imaginary part> changes. Positive $\operatorname{Re}\sigma$ gives positive cycle-averaged dissipation $\langle jE\rangle=\operatorname{Re}\sigma\,|\widehat E|^2/2$.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-2-drude-chain.png]
{title=Drude conductivity and the dispersion and density of states of a monatomic harmonic chain}
The supplied DC conductivity determines the transport time:
$$
\tau=\frac{m_e\sigma_0}{ne^2}=2.49\times10^{-14}\,\mathrm s.
$$
Inertia matters when $\omega\tau\sim1$, giving \b[$\omega_c\simeq4.02\times10^{13}\,\mathrm{rad\,s^{-1}}$], or \b[$\nu_c=\omega_c/(2\pi)\simeq6.40\times10^{12}\,\mathrm{Hz}$]. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.
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