# Speed of light in lossy dielectric medium

1. Jan 31, 2013

### WhiteHaired

It is usually written that the speed of light in a dielectric medium is $v=\frac{c}{\sqrt{\epsilon_r}}$, where $c$ is the speed of light in vacuum and $\epsilon_r$ is the relative permittivity. But, how can it be calculated for lossy and not necessarily low-loss dielectrics, i.e. those with a complex permittivity $\epsilon_r=\epsilon'-j\epsilon"$?

a) $v=\frac{c}{\sqrt{\epsilon'}}$?
b) $v=\frac{c}{\sqrt{|\epsilon_r|}}$?
c) None of above?

Related to previous question: If the natural resonance frequency of a resonant mode in an empty (vacuum) microwave cavity is $f_0$, which is the natural resonance frequency for the same mode, but with the cavity completely-filled with the previous lossy dielectric material?

a) $f=\frac{f_o}{\sqrt{\epsilon'}}$?
b) $f=\frac{f_o}{\Re{(\sqrt{\epsilon'-j\epsilon"})}}$?
b) $f=\frac{f_o}{\sqrt{|\epsilon_r|}}$?
c) None of above?

All of the textbooks and webs (that I've found) about this topic, consider only lossless dielectrics or approximations for low-loss dielectrics, but no general expressions.

2. Feb 1, 2013

### K^2

It helps to understand the equation, not just know it. Ignoring, for the moment, frequency dependence of permittivity and permeability, electromagnetic wave is described by the following equation.

$$\nabla^2 E = \mu \epsilon \frac{\partial^2 E}{\partial t^2}$$

In your case, $\small \mu = \mu_0$ and $\small \epsilon = \epsilon_r \epsilon_0$. And to make things easier, lets take solution of the form $E = E_0 e^{i(k\cdot x - \omega t)}$.

$$- k^2 E = - \mu_0 \epsilon_r \epsilon_0 \omega^2 E$$

Since $\small c^2 = 1/\mu_0\epsilon_0$, the above gives you $k^2 = \frac{\epsilon_r \omega^2}{c^2}$.

Now we can look at velocity. Since you seem to be interested in the phase velocity, we are looking at points in space and time where $\small E$ has same phase as $\small E_0$. In other words, $\small k \cdot x - \omega t$ is purely imaginary. Realizing that $\small x$ and $\small t$ are always real, that gives you a simple enough formula.

$$Re(k)\cdot x = Re(\omega)t$$

Differentiating both sides, we get the equation for phase velocity.

$$v = \frac{dx}{dt} = Re\left(\frac{\omega}{k}\right)$$

From earlier, we have $\frac{\omega^2}{k^2} = \frac{c^2}{\epsilon_r}$. So we have to take the square root here, keeping in mind that we are working with complex numbers.

$$v = Re\left(\frac{c}{\sqrt{\epsilon_r}}\right)$$

If $\small \epsilon_r$ is real, that trivially simplifies to your formula. It is, however, not. We need to take square root of a complex number. In other words, we are looking for some $\small z = a + ib$ such that $\small z^2 = \epsilon_r$.

$$a^2 - b^2 + 2iab = \epsilon' + i\epsilon''$$

This is a system of equations with four possible solutions, but only one is physical. It's a bit messy, so I only give the result for a, which is all we need for the real part above.

$$a = \frac{\sqrt{\epsilon' + \sqrt{\epsilon'^2 + \epsilon''^2}}}{\sqrt{2}}$$

So the correct formula for phase velocity is as follows.

$$v = \frac{\sqrt{2}c}{\sqrt{\epsilon' + \sqrt{\epsilon'^2 + \epsilon''^2}}}$$

Which again simplifies to formula you have whenever $\small \epsilon'' = 0$.

3. Feb 4, 2013

### WhiteHaired

Thank you very much.