Talk:Crystal optics

Symmetry of Electric Susceptibility Tensor

The article currently claims that the symmetry of the electric susceptibility tensor can be derived from thermodynamic arguments. Although a citation is not given, I have seen this same claim elsewhere, but I am unable to find a description of the thermodynamic arguments. Could someone find out what they are and maybe incorporate them into the article?

Tobycrisford (talk) 22:18, 31 May 2011 (UTC)[reply]

I believe the argument is that the dielectric tensor is derivative of D w.r.t. E and D itself is a derivative of the free energy w.r.t. E, thus epsilon is the second derivative of F and therefore does not depend on the order of differentiation. Don't have time right now to check/edit, but e.g. Landau-Lifshitz electrodynamics of continuous media mentions it I believe. 12.104.156.31 (talk) 02:30, 9 March 2012 (UTC)[reply]

How does the case of nonzero absorption look like?

The article says that the basis in which epsilon can be diagonalized is orthogonal only if epsilon is real. What happens in the general case where one has both refraction and absorption? In isotropic media, n=Re[sqrt(epsilon)] and absorption k=Im[sqrt(epsilon)], but how would those expressions be generalized to the tensor case? Would be nice if somebody could add the corresponding expressions. 12.104.156.31 (talk) 02:36, 9 March 2012 (UTC)[reply]