Drift velocity, mobility and conductivity in metals

Drift velocity

Without a field, conduction electrons in a metal move randomly at high speed (about $10^6$ m/s) and their average velocity is zero, so there is no net current. When a field $\mathcal{E}$ is applied, each electron is accelerated opposite to the field between collisions with vibrating ions and defects. The small average velocity gained in the direction of the force is the drift velocity $v_d$.

  E field  -------->
  e- path:  \/\/\/\_/\/\/\_/\  (random + slow drift)
            <--- net drift v_d

Between collisions the acceleration is $a = e\mathcal{E}/m_e$. If $\tau$ is the mean free time (relaxation time) between collisions, and each collision randomises the velocity, the average gained velocity is

$$v_d = \frac{e\tau}{m_e}\mathcal{E}$$

Typical $v_d$ is only mm/s to cm/s, far less than the thermal speed.

Drift mobility

The drift mobility $\mu_d$ is the drift velocity per unit electric field:

$$\mu_d = \frac{v_d}{\mathcal{E}} = \frac{e\tau}{m_e} \qquad (\text{m}^2\text{V}^{-1}\text{s}^{-1})$$

It measures how easily electrons move through the lattice. It is large when $\tau$ is long (few collisions). For copper, $\mu_d \approx 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$.

Conductivity

With $n$ free electrons per unit volume moving at $v_d$, the current density is

$$J = e n v_d = e n \mu_d \mathcal{E}$$

Comparing with Ohm's law $J = \sigma\mathcal{E}$:

$$\sigma = e n \mu_d = \frac{n e^2 \tau}{m_e}, \qquad \rho = \frac{1}{\sigma}$$

Example (copper): $n = 8.5 \times 10^{28}\ \text{m}^{-3}$, $\mu_d = 43 \times 10^{-4}\ \text{m}^2/\text{V s}$ gives $\sigma = 8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 43 \times 10^{-4} \approx 5.9 \times 10^{7}\ \text{S/m}$.

Factors

In metals $n$ does not change with temperature, so the fall in $\sigma$ (rise in resistivity, $\rho \propto T$) comes entirely from the fall in mobility due to increased lattice scattering.