How optical axis is related to dielectric tensor?

[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.

 

[DrDu]

See here, especially pages 3 and 8:
http://home.strw.leidenuniv.nl/~keller/Teaching/China_2008/CUK_L03_handout.pdf

 

[AAS]

See here, especially pages 3 and 8:
http://home.strw.leidenuniv.nl/~keller/Teaching/China_2008/CUK_L03_handout.pdf

Thank you

 

[AAS]

Thank you

Can you please explain how the first expression formed?

 

[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.