Electromagnetic waves in a conductor

Question

I am trying to understand the solution for electromagnetic plane waves in a conductor. I understand the derivation of the wave equation:

$$\nabla^2 \boldsymbol{\mathrm{E}} = \mu\epsilon \frac{\partial^2 \boldsymbol{\mathrm{E}}}{\partial t^2} + \mu\sigma\frac{\partial \boldsymbol{\mathrm{E}}}{\partial t} $$

Then what most authors immediately do is say it admits plane wave solutions with a complex wave vector $\boldsymbol{\tilde{k}}=\boldsymbol{k}+i\boldsymbol{\kappa}$. What I don't understand is why both the real and imaginary parts of this complex wave vector both have to be in the same direction. Why can't the wave be, for example, travelling in the $x$ direction, but decaying in the $z$ direction, like an evanescent wave?

Answer 1

What other direction could the imaginary part point?

It’s not a facetious question.

With an (assumed) isotopic medium, there’s only one vector direction in the problem, which becomes $\hat{k}$.

To define another direction for the imaginary part, there has to be something physical to define that. An an-isotropy can do that. But absent that, the only available direction is along or against the only available vector: $\hat{k}$.

Answer 2

Try the general solution of the plane wave form $$ \textbf{E}=\textbf{E}_0 e^{i(\vec \lambda \cdot \vec r -\omega t)} \tag{1} $$ with \begin{align} \vec \lambda \cdot \vec r = \lambda_x x+\lambda_y y+\lambda_z z \end{align} and you will find that $\lambda$ must have a real and imaginary part. This follows because you have first and second order derivatives in $t$ so that one $t$-derivative of your ansatz (1) will return an imaginary multiple of $\textbf{E}$ while two derivatives in $t$ will return $\textbf{E}$ multiplied by a real number.

After all, the form of (1) is dictated by the assumption of a plane wave and by the general method of undetermined coefficients.

Answer 3

The reason is that you have assumed isotropic dielectric response. If you replace $\epsilon$ and $\sigma $ by tensors you can have the more general wave.