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How optical axis is related to dielectric tensor?

  1. Dec 16, 2015 #1

    AAS

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    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. jcsd
  3. Dec 16, 2015 #2

    DrDu

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  4. Dec 16, 2015 #3

    AAS

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  5. Dec 23, 2015 #4

    AAS

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    Capture.PNG Can you please explain how the first expression formed?
     
  6. Dec 23, 2015 #5

    blue_leaf77

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    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.
     
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