 Solution

Choose the $e^{-i\omega t}$ convention. The steady displacement in the Lorentz oscillator model has complex amplitudeThe induced electric dipole moment is $p_{\omega }=q{u}_{\omega }$, so the microscopic electric polarizability is $\alpha (\omega )=q^2/[m({\omega }_T^2-{\omega }^2-i\gamma \omega )]$. For $N$ identical oscillators per unit volume, $P_{\omega }=N{p}_{\omega }={ϵ}_0{\chi }_{\omega }{E}_{\omega }$. ConsequentlyHere the dimensionless quantity called polarizability in the paper is the electric susceptibility; the single-electron polarizability $\alpha$ has different units. The sign of the Electron charge disappears because both the driving force and dipole moment contain $q$.

Multiplying by the complex conjugate of the denominator givesFor positive damping and positive frequency, ${ϵ}_r^{\prime \prime }\ge 0$, as required for passive wave absorption in this Fourier convention. At zero frequency, ${ϵ}_r^{\prime }=1+{\omega }_p^2/{\omega }_T^2$ and ${ϵ}_r^{\prime \prime }=0$. Near a weakly damped resonance the imaginary part has a narrow positive peak, while the real part rises on the low-frequency side and then falls steeply through ${ϵ}_r^{\prime }=1$ at ${\omega }_T$. It can become negative above resonance for sufficiently strong oscillator strength and weak damping. At large frequency, ${ϵ}_r^{\prime }=1-{\omega }_p^2/{\omega }^2+O({\omega }^{-4})$ and ${ϵ}_r^{\prime \prime }\sim {\omega }_p^2\gamma /{\omega }^3$: both approach the vacuum values. The absorption maximum is close to, but need not be exactly at, ${\omega }_T$ for finite damping.