electromagnetism - Why does a dielectric have a frequency dependent resistivity?

This question has come about because of my discussion with Steve B in the link below.

Related: Why is glass much more transparent than water?

For conductors, I can clearly see how resistivity $\rho\,\,(=1/\sigma)$ can depend on frequency from Ohm’s law, $\mathbf{J}=\sigma\mathbf{E}$. So if the E-field is an electromagnetic wave impinging on a conductor, clearly the resistivity is frequency dependent. In a similar fashion, the frequency dependence of the electric permittivity $\epsilon=\epsilon_0n^2(\omega)$ can be derived through the frequency dependence of the electric polarization and impinging electromagnetic wave (see How Does $\epsilon$ Relate to the Dampened Harmonic Motion of Electrons?).

1. 
   

   What does it mean physically for a dielectric to have a frequency dependent resistivity from (i) classical and (ii) quantum viewpoints? I am especially interested in the optical frequency range.

   
   


2. 
   

   Can a simple mathematical relationship be derived similar to the frequency dependent resistivity (for conductors) and electric permittivity (for dielectrics)?

   
   

Thank you in advance for any help on this question

Answer

A simple model that explains the frequency dependency of the resistivity of metals reasonably well is the Drude model (http://en.wikipedia.org/wiki/Drude_model). There we have frequency dependency because the electrons in a plasma are not moving arbitrarily fast, which is consistent with Xurtio's explanation. The cutoff frequencies are usually in the optical domain. For dielectrics similar models exist, which are often a sum of Lorentzian resonances. These have their origin in resonant absorption which is a quantum physical effect.

The imaginary part of the permittivity is related to the conductivity. This can be seen as follows: Amperes law is

$\nabla \times \mathbf{H} = \mathbf{J} +i \omega \epsilon_r \epsilon_0 \mathbf E$

and insert Ohms law in differential form

$\mathbf{J} = \sigma \mathbf{E}$

then you get

$\nabla \times \mathbf{H} = i \omega (\epsilon_r \epsilon_0 -i \sigma/\omega) \mathbf E$

which is just of the same form of as original form of amperes law but without the explicit $\mathbf{J}$ term. In conclusion Ohms law can be integrated in free space Maxwells equations (without the source terms) when the relative permittivity $\epsilon_r$ is taken as a complex value ($\widetilde\epsilon_r = \epsilon_r - i \sigma/(\omega \epsilon_0)$), where an imaginary part is added related to the conductivity. This essentially models the effect of moving charges under the influence of an oscillating field (light).

So the relation between polarization ($\mathbf D = \widetilde{\epsilon}_r \epsilon_0 \mathbf E = \mathbf P + \epsilon_0 \mathbf E$) and conductivity $\sigma$ is given as

$\mathbf{P} = \epsilon_0 (\epsilon_r - i \sigma/\omega - 1) \mathbf E$.

Since the real part of the permittivity is frequency dependent, so is the conductivity. This is because of the Kramers-Kronig relations which follow from a causality relation.