I'm studying Waves in Cold plasmas right now, but I guess my question is generalizable. It's about the 4th Maxwell Equation in polarizable / conductive media:

$\nabla \times H = \frac{1}{c} \frac{\partial D}{\partial t} + \frac{4\pi}{c}J $

Now from our electrodynamics lectures we know that $D = \epsilon E$ with $\epsilon$ being the dielectric tensor, but also the crucial modellization of Ohm's law for conductive media $J = \sigma E$ is well within our memory. So my first reflex would now be to just plug in everything and obtain an equation for $E$ in Fourier-space and see where I get to. However this seems never to happen in the literature, and ppl as Stix (Waves in Plasmas, Chap1) or Jackson establish the relationship $\epsilon = 1 - \frac{4 \pi \sigma}{i \omega}$.

This is now what confuses me severly:
$D$ arises from the argumentation that if we have bound charges in a medium, they will linearly answer with a polarisation $P$ so that $D = E + 4\pi P$. The conductivity involves a (to be specified) steady-state modell of free charges in response to a electric field.

Although fundamentally different views, the bound and free modell seem to be connected by $\epsilon = 1 - \frac{4 \pi \sigma}{i \omega}$, but I'd be glad for some physical motivation. In the end (for cold plasmas at least), we always use $J = n m v$ and $v$ from the equation of motion to obtain a relationship between $J$ and $E$, thus obtaining $\sigma$. But there are also sources which do stuff with the dielectric tensor which I dont really understand (like Padmanabhan, Theoretical Astrophysics Vol I., Chap 9.5).

So concluding, my questions would now be the following:

  • Naively plugging in $D = \epsilon E$ and $J = \sigma E$ into Maxwell 4 seems to be wrong, also due to the physical reasonings behind those modells. We can't have both in the equation. True/False?
  • Why can such a relation as $\epsilon = 1 - \frac{4 \pi \sigma}{i \omega}$ exist between a static and a dynamic modell? Or am I mislead here and the $i\omega$ hints at a dynamic origin for $\epsilon$ too?

Thanks in advance, for anyone taking the time!


2 Answers 2


First Question Well, the first thing is that $J$ = $\sigma \ E$ happens to be an approximation (fun side note: there is no general derivation of Ohm's law from first principles [at least that's what J.D. Jackson argues]). The conductivity is actually a tensor, which contains parts similar to your expression for epsilon.

In and wave plasma theory, one wants to find a relationship between $\mathbf{D}$ and $\mathbf{E}$. If we define $\overleftrightarrow{\boldsymbol{\sigma}}$ as the conductivity tensor, then we can define: $$ \mathbf{j} = \overleftrightarrow{\boldsymbol{\sigma}} \cdot \mathbf{E} \\ \mathbf{D} = \varepsilon_{o} \ \overleftrightarrow{\mathbf{K}} \cdot \mathbf{E} $$ where $\vec{\mathbf{K}}$ is called the dielectric tensor. Then we can use Ampere's law to show: $$ \overleftrightarrow{\mathbf{K}} = \overleftrightarrow{\mathbf{I}} - \frac{ \overleftrightarrow{\boldsymbol{\sigma}} }{ i \ \varepsilon_{o} \ \omega } $$ where $\overleftrightarrow{\mathbf{I}}$ is the unit tensor. We don't really care too much about this, we want the dispersion relation, which is a function that defines the relationship between $\omega$ and $\mathbf{k}$ (or wavenumber). To do this, we linearize (i.e., after Fourier transforming, $\nabla$ $\rightarrow$ $i \ \mathbf{k}$ and $\partial_{t}$ $\rightarrow$ -$i \ \omega$) both Ampere's and Faraday's law to get: $$ i \mathbf{k} \times \mathbf{E} = -i \ \omega \mathbf{B} \\ i \mathbf{k} \times \mathbf{B} = \frac{ -i \ \omega }{ c^{2} } \overleftrightarrow{\mathbf{K}} \cdot \mathbf{E} $$ which gives us an equation only dependent upon $\mathbf{E}$, seen as: $$ \mathbf{k} \times \left( \mathbf{k} \times \mathbf{E} \right) + \frac{ \omega^{2} }{ c^{2} } \overleftrightarrow{\mathbf{K}} \cdot \mathbf{E} = 0 \\ \mathbf{n} \times \left( \mathbf{n} \times \mathbf{E} \right) + \overleftrightarrow{\mathbf{K}} \cdot \mathbf{E} = 0 $$ Either of these equations can be written in tensor form as $\overleftrightarrow{\mathbf{D}} \cdot \mathbf{E}$ = 0. If $\overleftrightarrow{\mathbf{D}}$ has a determinant that goes to zero, then there is a non-trivial solution for $\mathbf{E}$. This solution is the dispersion relation.

Second Question In both a plasma and metal, electrons and ions can move and oscillate. Simple approximations for the theory I briefly described above take advantage of the disparity in the electron and ion masses and assume $M_{i}$ $\rightarrow$ $\infty$ or $m_{e}$ $\rightarrow$ 0. They make this assumption by arguing that in the limit of high frequencies, the ions do not really move fast enough to matter. In a conducting metal, the idea is similar except that the ions are far more rigidly constrained to a lattice, thus $M_{i}$ $\rightarrow$ $\infty$ is a better approximation. However, even so both the ions and electrons can still oscillate. Typically in solids, one only considers the electrons and your expression for $\epsilon \left( \omega \right)$ $\rightarrow$ $\epsilon \left( \omega_{pe} \right)$, where $\omega_{pe}$ is the electron plasma frequency.

Physical Motivation Think of how a system responds to external forces. In both a solid and plasma, the idea is to try and figure out how the system will respond under ideal conditions. When looking at cold plasma theory, you are trying to find the normal modes of a system similar to what is done in trying to describe the motion of a mass attached to a complex series of springs.

Let me know if you need further clarification.


Maybe I can partially answer your question. In your equation on the r.h.s. you write the term: $\frac{4\pi}{c} J$. Stix doesn't include this term in his treatment. All plasma currents which are responsible for the plasma waves are inside $D$. For example here is the standard way to derive $D(\omega)$:

$\nabla \times B = \mu_0 \vec j + \epsilon_0 \mu_0 \frac{\partial E} {\partial t}$. Then expressing $\vec j$ using Ohm's law and Fourier transforming we find something like that $D(\omega) = (\frac{i\bar \sigma} {\omega \epsilon_0} + \bar I) \vec E$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.