The <Drude response>
$$
G_R(\omega)=\frac{\sigma_0}{1-i\omega t_0}
$$
has a pole at $\omega=-i/t_0$. Analyticity in the upper half plane therefore requires
$$
\boxed{t_0>0}.
$$
For real frequency,
$$
\operatorname{Im}G_R(\omega)
=\frac{\sigma_0\omega t_0}{1+\omega^2t_0^2}.
$$
The sign condition from part ii then requires
$$
\boxed{\sigma_0t_0\leq0},
$$
so with a nonzero causal relaxation time the convention used in this question has $\sigma_0<0$. In conventions where the physical conductivity is defined with an additional minus sign, its static value is positive. The parameter $t_0$ is the relaxation time: after forcing is removed, the corresponding current or response decays as $e^{-t/t_0}$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 337 / 1 / vii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-337/1/vii/solution