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.

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

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@John: how is the index of refraction or permittivity (which I relate to polarization), related to the frequency dependent resistivity? I don't know the connection. –  Carlos Jul 27 '13 at 16:43
Carlos: sorry, I completely misread your question. Please disregard my comment. I've removed my duplicate flag. –  John Rennie Jul 28 '13 at 7:01

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.

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Another great answer! I want to interpret this since I’ve never seen “ac conductivity.” If I have a wave impinging on a conducting medium, I expect that this term will be added to the complex wave number and therefore, contribute to the skin depth. Would this be a correct assumption? If so, is there an “easy interpretation” on the role that the ac conductivity plays? If glass has a larger conductivity value then I expect a larger skin depth. I am most interested in the frequency dependence and the role that ac conductivity plays, especially at optical frequencies. –  Carlos Jul 28 '13 at 12:54
Yes, the skin depth is zero if the conductivity is infinite. At frequency f AC conductivity is a complex number. The absolute value of this number describes the ratio of the amplitudes of E and J, and the angle of this complex number describes the phase shift (for a sinusoidal wave of frequency f). At DC you dont have a phase shift, so conductivity is a positive real number. The phase-shift (and amplitude) variation with frequency is due to the mentioned dynamic effects (i.e. that polarization does not follow instantaneously the E field). –  Andreas H. Jul 28 '13 at 16:41
Wow – this is awesome. Up to this point, I have only been exposed to the phase shift due to “ac permittivity,” which of course, leads to larger than one index of refraction and its connection to the apparent speed of light in a medium. Now you are showing me that ac conductivity contributes to these effects as well. In some ways, I allowed myself to be misled by Griffiths because he assumes dc conductivity. Without ac conductivity (as Steve B and you has shown me), I was making lots of incorrect assumptions on the skin depth for dielectrics like its independence of frequency. –  Carlos Jul 28 '13 at 18:23
If I understand correctly, ac conductivity plays the dominant role in determining why glass has a larger skin depth than water at optical frequencies. Thank you very much! –  Carlos Jul 28 '13 at 18:24
Well I am not sure about the last point, don't forget the real part of the permittivity. This is also different for glass and water. But it is true that real and imaginary part (ac conductivity) both influence the refractive index. –  Andreas H. Jul 28 '13 at 19:26

A di-electric experiences polarization int he presence of an electric field. The magnitude of polarization will present an effective resistance (more polarization against the field = more apparent resistance).

But the polarization takes time (it's not instantaneous). So think about polarization delay vs. the change of the source electric field (i.e. the "frequency"). The faster the source field changes, the less time the dielectric has to polarize. For very slow frequencies, the polarization will be able to keep up with the changes in electric field.

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That is a really nice simple explanation. Do you know how polarization and resistivity are mathematically connected? –  Carlos Jul 27 '13 at 16:52
keeping with a classical treatment, I see it more as a voltage source reversed than a resistor. Either way, you do experience a voltage drop across the dielectric and its proportional to the polarization (so basically compute the potential generated by the polarization.) If you're ok with differential equations, you could model the capacitor and solve for V (t). –  Xurtio Jul 28 '13 at 16:57

Well I hesitate to even venture into this question, because technical terms are being misused creating the problem.

Resistance, and resistivity, is something that arises from Ohm's Law.

Namely, for a certain class of materials (mostly metallic conductors) , if all other physical parameters are held constant (difficult to do), the ratio of the current flowing to the applied Voltage, is constant.

So Ohm's law simply says: R is CONSTANT .

And R does not vary with frequency either. with varying currents, Ohm's law applies to all instants of time, so with AC Voltages and Currents the two are ALWAYS in phase.

The practical problem arises, in that when you have a current flowing in a resistive medium; say a wire, there is a magnetic field set up, that surrounds the current flow, and that magnetic field is also inside the wire, and the magnitude of the field depends on the ENCLOSED current. So the center of the wire has a lower current, so it generates a smaller magnetic field.

If the current varies, then the magnetic field is restricted in its movement, or change, by the velocity of EM wave propagation (c).

As a consequence of this time lag, the current carrying conductor now exhibits, an Inductance effect, so the equivalent circuit is no longer a simple resistor with constant Ohmic resistance; it is a series circuit of a resistor in series with an inductor; approximately 3 nano-Henries per centimeter of a straight wire.

So you now have an AC impedance that is Z = R + j.2.pi.f.L

So now the current will be less, and as you raise the frequency, the inductive reactance will increase linearly with frequency, so the current will drop.

The resistance has not changed one iota; the impedance has. Eventually, you will end up with the current in the center of the wire going completely backwards, compared to the outer layers. That backwards current further diminishes the current for a given Voltage, so the wire center is now more of a nuisance than a useful conductor. So you might as well get rid of it, and use a hollow tube.

This is the essence of "Skin effect", it has nothing whatsoever to do with the resistance or the resistivity of the conductor, which remains completely frequency independent. It is the AC impedance that IS INCREASING, not THE RESISTIVITY.

If it is frequency dependent, it is NOT a RESISTOR, complying with Ohm's Law; it is a complex AC circuit involving Inductance, and also Capacitance, when you get into it.

Words, have meaning, and when scientists use the wrong words; specially ones that also have colloquial common meanings; it creates havoc for all; this question for example.

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The OP did not claim that a dielectric is a resistor. Resistivity is the ratio of electric field to current density and that is a frequency dependent and material dependent quantity (see e.g. the Drude model). I do not see where he did use wrong words. This is in addition to the effect that there might be an complex impedance. –  Andreas H. Jul 27 '13 at 23:20