Metals — Question 3

Question 3

The ac Drude response separates dissipation from electron inertia

Use phasors proportional to $e^{-i\omega t}$ and $\dot J+J/\tau=\sigma_0E/\tau$.

1. Derive complex conductivity and its real and imaginary parts.

2. At $\omega\tau=1$, find current amplitude and phase relative to a real cosine field, and mean absorbed power density.

Question 3 - Solution

The ac Drude response separates dissipation from electron inertia

(a) Method and calculation

$$(1-i\omega\tau)\widetilde J=\sigma_0\widetilde E\Rightarrow\boxed{\widetilde\sigma=\sigma_0/(1-i\omega\tau)}.$$$$\boxed{\sigma_1=\frac{\sigma_0}{1+\omega^2\tau^2},\quad\sigma_2=\frac{\sigma_0\omega\tau}{1+\omega^2\tau^2}}.$$

(b) Interpretation and further calculation

$$\boxed{J(t)=\frac{\sigma_0E_0}{\sqrt2}\cos(\omega t-\pi/4)}\quad(\omega\tau=1).$$$$\boxed{\langle JE\rangle=\tfrac12\sigma_1E_0^2=\sigma_0E_0^2/4}.$$

Physical check

With this phasor convention a positive imaginary conductivity corresponds to current lag. Only the in-phase component produces cycle-averaged heating.