How optical axis is related to dielectric tensor?

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1. Dec 16, 2015

AAS

I want to know the relationship between the optical axis direction of a crystal and the dielectric constants in different directions in an anisotropic material.

2. Dec 16, 2015

DrDu

3. Dec 16, 2015

AAS

4. Dec 23, 2015

AAS

Can you please explain how the first expression formed?

5. Dec 23, 2015

blue_leaf77

From the curl relations for E and H and the assumption that the propagating fields are of plane waves, one can obtain
\begin{aligned} \mathbf{k}\times \mathbf{E} = \omega \mu \mathbf{H} \\ \mathbf{k}\times \mathbf{H} = -\omega \mathbf{D} \end{aligned}
Combining these will yield $\mathbf{k} (\mathbf{k}\cdot\mathbf{E})-k^2\mathbf{E} = -\omega^2 \mu \mathbf{D}$. Then use the relations like $\mu = 1/(c^2\epsilon_0)$, $\mathbf{D} = \epsilon \mathbf{E}$, and $\epsilon = \epsilon_r\epsilon_0$ to transform the RHS into $-\frac{\omega^2}{c^2}\epsilon_r\mathbf{E}$. So now,
$$\mathbf{k} (\mathbf{k}\cdot\mathbf{E})-k^2\mathbf{E} = -\frac{\omega^2}{c^2}\epsilon_r\mathbf{E}$$
Then substitute $\mathbf{k} = \frac{\omega}{c}n\mathbf{s}$ to eliminate $\omega$ in both sides. By taking element-by-element comparison between right and left sides you should see this expression leads to what is written in that slide.
NOTE: I don't think there should be $\mu$ in the LHS of the equation in the slide as its presence will violate the requirement that the LHS and RHS should have the same dimensions.