Modes in 1D cavity with loss

Question

Let consider the simple problem of :

1D electromagnetic cavity with perfect conducting walls filled with a conducting medium

Let the length of the cavity be L, the conductivity be $\sigma$ and dielectric be $\epsilon$.

General solution

From Maxwell equations the solution of the wave equation inside this medium is :

$E_z=Ae^{-jk_{m}y}+Be^{+jk_{m}y} $

With $k_{m}$ given by the complex dispersion relation : $k_{m}=\frac{\omega}{c}\bar{n_{m}}=\beta-j\alpha $

Whose complex refractive index(which can easily be expanded) is : $\bar{n_{m}}=\sqrt{1+j\frac{\sigma}{\epsilon\omega}}$

Continuity at the PEC boundary $E_z(0,L)=0$

We consequently have the 2 sets of equations :

$A+B=0$

$Ae^{-j\bar{k_{m}L}}+Be^{+j\bar{k_{m}L}}=0 $

Since $k_m$ is complex how do you manage to find the resonant frequencies along with their corresponding mode shapes???

I have tried manipulation but when you look at the amplitude of the complex sinh on a graph, it only passes by 0 once. What am I missing??

Answer 1

As you say, I think that your system of equations does not have a non trivial solution of the form $$ E_z(y,t) = ( A e^{ik_m y} + B e^{-ik_m y} ) e^{i\omega t}, $$ with $\omega\in\mathcal{R}$. These solutions would be "everlasting", in the sense that they are not decaying with time. Maybe the fact of having a complex refractive index (and therefore dissipation) forbids their existence.

If, on the other hand, you look for solutions decaying in time, you might find modes. I mean, suppose that you look for solutions of the form $$ E_z(y,t) = ( A e^{ik_m y} + B e^{-ik_m y} ) e^{i(\omega+i\eta) t}, $$ with $\omega,\eta\in\mathcal{R}$ and $\eta>0$. The condition to find solutions now (that is, the determinant of your system of equations for $A$ and $B$ being equal to 0) is then $$2\frac{\omega+i\eta}{c}\sqrt{1+i\frac{\sigma}{\epsilon(\omega+i\eta)}}=m\pi,$$ with $m\in\mathcal{Z}$. You can calculate $\omega$ and $\eta$ with this equation, finding then solutions damped with time, which may be the only ones existing.

Does this make sense to you?