- #1
AAS
- 3
- 0
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.
How optical axis is related to dielectric tensor?
- Context: Graduate
- Thread starter AAS
- Start date
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Discussion Overview
The discussion focuses on the relationship between the optical axis direction of a crystal and the dielectric constants in different directions within anisotropic materials. It involves theoretical considerations and mathematical derivations related to electromagnetic wave propagation in such materials.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant seeks to understand how the optical axis direction relates to the dielectric constants in anisotropic materials.
- Another participant provides a reference to a document that may contain relevant information, specifically pointing to pages 3 and 8.
- A later post requests clarification on how a specific expression was formed, indicating a need for deeper understanding of the derivation.
- One participant presents a mathematical derivation involving curl relations for electric and magnetic fields, leading to an expression that relates wave vector, electric field, and dielectric properties.
- There is a concern raised about the presence of the permeability term in the left-hand side of an equation, suggesting it may violate dimensional consistency.
Areas of Agreement / Disagreement
Participants do not appear to reach a consensus on the correctness of the mathematical expressions discussed, as one participant questions the dimensional consistency of a term in the equation. The discussion remains unresolved regarding the implications of this concern.
Contextual Notes
The discussion includes complex mathematical relationships and assumptions about wave propagation in anisotropic media, which may not be fully articulated or agreed upon by all participants.
- #2
DrDu
Science Advisor
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See here, especially pages 3 and 8:
http://home.strw.leidenuniv.nl/~keller/Teaching/China_2008/CUK_L03_handout.pdf
http://home.strw.leidenuniv.nl/~keller/Teaching/China_2008/CUK_L03_handout.pdf
- #3
AAS
- 3
- 0
Thank youDrDu said:
- #4
AAS
- 3
- 0
AAS said:
- #5
blue_leaf77
Science Advisor
- 2,637
- 786
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.
$$
\begin{aligned}
\mathbf{k}\times \mathbf{E} = \omega \mu \mathbf{H} \\
\mathbf{k}\times \mathbf{H} = -\omega \mathbf{D}
\end{aligned}
$$
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.
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