Electromagnetic Wave Penetration into Conductors

Question

Right now, I'm looking at two straightforward derivations of the dispersion relations of electromagnetic waves in conductors, and they seem to disagree on the low frequency attenuation. Could someone supply some insight into the regimes of validity of these very basic methods?

In my electromagnetism class, we derived the dispersion relation of waves in homogeneous highly conductive media as follows:

We have Ampere's Law and Faraday's Law

$\nabla\times H=\partial_t D+ J$

$\nabla\times E=\partial_t B$

Use the constitutive relations $D=\varepsilon E ; B=\mu H$, and Ohm's law $J=\sigma E$

$\nabla \times H = (\varepsilon\partial_t + \sigma)E$

$\nabla\times E=\mu\partial_t H$

For purposes, let's just assume exponential solutions $E=\hat{x}E_0e^{i(\omega t-kz)}$

$\nabla \times H = (i\varepsilon \omega + \sigma)E$

$\nabla\times E=i\mu \omega H$

Curl of Ampere's Law, plug in Faraday's, use vector identity, as usual, and arrive at

$k^2 = (\epsilon \mu \omega^2 -i\mu\omega\sigma)= \varepsilon\mu\omega^2 (1-i\frac{\sigma}{\epsilon\omega}) $

Suppose we have a really good conductor ($\sigma\gg\epsilon\omega$), then the right-hand side is almost pure imaginary, and we can write $k\approx \sqrt{\mu\sigma \omega/2} (1-i)$. One can easily see that increasing $\omega$ increases (the imaginary part of) $k$, which increases the attenuation of the wave. Higher frequencies attenuate more sharply in good conductors.

In my solids class, we derived the dispersion relation of waves in highly conductive media in the free electron gas model.

See Introduction to Solid-State Physics by Kittel (pg 396-398)

Starting from the free electron in an E-field

$m\partial_t^2x=-eE$

Solve for the magnitude of the oscillations:

$x=eE/m\omega^2$

Relate to the polarization

$P=-nex=-\frac{ne^2}{m\omega^2}E$

Then get the relative permittivity from that relation

$\varepsilon_r = 1 -\frac{ne^2}{m\omega^2} = 1 -\frac{\omega_p^2}{\omega^2}$

Which leads to the dispersion

$k^2 = \mu\epsilon (\omega^2-\omega_P^2) $

For, say copper, $\omega_P$ should be something like 6 PHz. For frequencies above that (like UV), $k$ is real and the light propagates. For frequencies below that, $k$ is imaginary and the waves must be attenuated/reflected.

I want to compare with the result from above. I assume for my E&M class, we're dealing with frequencies much lower than PHz, since these are electrical engineers. So, looking at this result in the $\omega \ll \omega_P$ limit, we see that increasing $\omega$ decreases the magnitude of $k$, which means that waves will be attenuated less sharply at higher frequencies.

These two models seem to disagree on the $\omega \ll\omega_P$ behaviour. Why is that/ how does the latter model break down these frequencies? Thanks!

Answer 1

One key difference between the two treatments is an assumption noted by Kittel: "In the absence of collisions...". He's deriving an expression for the dielectric response in the lossless limit. Conversely, the low-frequency EEs include dissipation ($J= \sigma E$), but leave the dielectric constant constant.

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Update: I found a unified treatment in "Fields and Waves in Communication Electronics" by Ramo, Whinnery, and Van Duzer, section 13.3.

The free electron equation of motion (velocity $\boldsymbol{v}$) is:

$$m \frac{d\boldsymbol{v}}{dt} = - e \boldsymbol{E} - m \, \nu \, \boldsymbol{v} $$

The first term models polarization (displacement current) while the second accounts for collisions, with $\nu$ the collision rate (ohmic losses). Assuming periodic steady-state solutions with angular frequency $\omega$, one finds:

$$ \boldsymbol{J} = - n_e e \boldsymbol{v} = \frac{n_e e^2}{m(\nu + i \omega)} \boldsymbol{E} $$

Note that this term includes elements both in-phase (ohmic current) and in quadrature (polarization / displacement current) components.

Substituting into Maxwell: $$ \boldsymbol{\nabla \times H} = i \omega \epsilon_c \boldsymbol{E} + \boldsymbol{J} = i \omega \epsilon_c \boldsymbol{E} + \frac{n_e e^2}{m(\nu + i \omega)} \boldsymbol{E} $$

$$ = i \omega \left[\left(\epsilon_c - \frac{n_e e^2}{m(\nu^2 + \omega^2)} \right) -i \frac{n_e e^2 \nu}{\omega m(\nu^2 + \omega^2)} \right] \boldsymbol{E} $$

Here $\epsilon_c$ is added by hand to represent the polarization response of lattice atom core electrons (different from the free electrons modeled by the above equation).

Now we can look at two extreme cases that correspond to the two analyses in the question:

1) low frequency ($\omega^2<<\nu^2$) and the core electron dielectric response dominates. $$ \epsilon = \epsilon_c - i \frac{\sigma}{\omega}$$

where $\sigma=n_e e^2/(m\nu)$ is the low-frequency conductivity.

2) ionized gas (so $\epsilon_c=\epsilon_0$) with $\nu << \omega$ $$ \epsilon = \epsilon_0 \left[1 - \left(\frac{\omega_p}{\omega} \right)^2 \right]$$

where $\omega_p = \sqrt{\frac{n_e e^2}{\epsilon_0 m}}$

It's nice to see these two cases emerge from a common framework.

Answer 2

You have treated the high-$\omega$ limit inconsistently in the first model.

Assuming a conductor is good requires you to ask that $\sigma\gg\epsilon\omega$ since that is the only characteristic scale of conductance. However, this requires you to stay within the bounds of $\omega\ll\sigma/\epsilon$ and the limit of high $\omega$ is meaningless.

If you want to take that limit, you should return to the original, general expression: $$k^2=\epsilon\mu\omega^2(1+i\frac\sigma{\epsilon\omega})\approx \epsilon\mu\omega^2.$$ Thus at high frequency the material is not even dispersive. (This needs, of course, $\epsilon$ and $\mu$ to be independent of $\omega$, which they are not, in the same way that your previous argument requires it of $\sigma$.)

The overall lesson, I think, is that whether a material is a good conductor or not is completely dependent on the timescales involved. Purcell has a good discussion on this and drives home very effectively the comparison with whether a material is solid or liquid: that too depends completely on the timescales involved.