Calculating n^2 for Free Electrons with Friction Interactions

[Skwishm]

Homework Statement

The index of refraction is given by

n^2 = (N q / ε m) (1 / (ω0^2 - ω^2 + iγω))

Where N is the electron density, q is the charge of an electron, ε is the permittivity of free space, m is the mass of an electron, ω0 is the resonant frequency, ω is the incident frequency, and γ is the friction coefficient.

Consider the case of a free electron with a friction coefficient given by interactions with positive ions. What is n^2 for this case?

2. The attempt at a solution
I'm not super sure how to approach this problem. I figure that ω0 has to be 0 for free electrons, but I'm not sure what to do with the friction term. A gentle nudge in the right direction would be greatly appreciated.

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?