 Solution

For a cold, unmagnetized, collisionless plasma, take an equilibrium Electron density $n_e$, Electrons of mass $m_e$ and charge $-e$, and approximately stationary ions. Use fields proportional to $e^{-i\omega t}$. The Electron displacement satisfiesIts electric polarization is $\mathbf{P}=-n_{e}e\mathbf{x}$, hence the cold collisionless plasma dielectric responseThe negative susceptibility arises from the inertial Electron response, rather than a restoring force binding an Electron to a particular atom.

A transverse electromagnetic wave then obeysFor $\omega >{\omega }_p$, the plasma supports propagation, with refractive index $n_r=\sqrt{1-{\omega }_p^2/{\omega }^2}<1$. The phase velocity is greater than $c$, while the group velocity is $c\sqrt{1-{\omega }_p^2/{\omega }^2}<c$; their product is $c^2$, and no superluminal signal is implied. Below the plasma frequency, the bulk wavenumber is imaginary and the field is evanescent. A sufficiently thick plasma largely reflects such radiation in the lossless model.

Including an Electron momentum-relaxation rate ${\nu }_c$ replaces the denominator by $\omega (\omega +i{\nu }_c)$:Its imaginary part describes absorption with the stated time convention. Finite Electron temperature also produces plasma screening of slowly varying electrostatic fields, with ${\lambda }_D=\sqrt{{ϵ}_0{k}_B{T}_e/(n_e{e}^2)}$. That spatial, longitudinal screening problem is distinct from simply taking $\omega =0$ in the cold transverse-wave formula. Magnetic bias, ion motion and kinetic resonances require more general dielectric responses.