 Solution

Introduce the electron momentum-relaxation rate $\nu$ through $m_e(\dot{\mathbf{v}}+\nu \mathbf{v})=-e\mathbf{E}$. For a harmonic field $\mathbf{E}=\mathrm{Re}({\mathbf{E}}_0{e}^{-i\omega t})$, the real part of the electrical conductivity is $\mathrm{Re}\sigma =n_e{e}^2\nu /[m_e({\omega }^2+{\nu }^2)]$. At weak damping, the mean heating is $Q=\frac{1}{2}\mathrm{Re}\sigma |E_0{|}^2$ and the beam intensity is $I=\frac{1}{2}{ϵ}_{0}cn|E_0{|}^2$, where $n=\sqrt{1-{\omega }_p^2/{\omega }^2}$. ThereforeWith the relaxation convention $\nu =1/(2{\tau }_{ei})$, this becomes the printed resultConvention qualification: a $90^{\circ }$ deflection time does not by itself specify a momentum-relaxation rate without a collision model. If instead ${\tau }_{ei}$ is defined by $\nu =1/{\tau }_{ei}$, the intensity absorption coefficient is twice the printed one. Amplitude and intensity damping also differ by two. The result above states the extra convention needed rather than silently identifying these quantities. It assumes $\omega >{\omega }_p$, $\nu \ll \omega$ and sufficiently weak damping away from the cutoff; the apparent divergence near cutoff is outside that approximation.