There is a connection between the linear response equations for a dielectric medium:
## P(x,t)=\int \chi(x-x',t-t') E(x',t') \, d^3 x' \, dt' ## and ## J_p(x,t)=\int \sigma(x-x',t-t') E(x', t') \, d^3x' dt' ##. Taking Fourier transforms these become: ## \tilde{P}(k,\omega)=\tilde{\chi}(k,\omega) \tilde{E}(k,\omega) ## and ## \tilde{J}_p(k,\omega)=\tilde{\sigma}(k,\omega) \tilde{E}(k,\omega) ##. The equation ## J_p=\dot{P} ## (for the polarization current=it follows also from the continuity equation) and its Fourier transform ## \tilde{J}_p(k,\omega)=-i \omega \tilde{P}(k,\omega) ## tie these together, along with ## D(x,t)=\int \epsilon(x-x',t-t') E(x',t') \, d^3 x' \, dt' ## and its Fourier transform, ## \tilde{D}(k, \omega)=\tilde{\epsilon}(k,\omega) \tilde{E}(k,\omega) ## so that ## \tilde{\epsilon}(k,\omega)=1+4 \pi \tilde{\chi}(k,\omega) ##. ## \tilde{\sigma}(k,\omega) ## is the conductivity, and ## \tilde{\epsilon}(k,\omega) ## is the dielectric constant. I used cgs units so that ## D=E+4 \pi P ##, but conversion to any other units can be readily done. Hopefully this was helpful. (With a little algebra, you can solve for ## \tilde{\sigma}(k,\omega) ## in terms of ## \tilde{\epsilon}(k,\omega) ##). For a reference, Ichimaru's Plasma Physics book has much of this in the first couple of chapters.