How Does Each Element in the Permittivity Tensor Matrix Represent an Anisotropic Material?

[lholmes135]

TL;DR Summary

Please explain meaning of each element of permittivity as a tensor.

If I have an anisotropic material with permittivity:
$$\epsilon=
\begin{pmatrix}
\epsilon_{ii} & \epsilon_{ij} & \epsilon_{ik} \\
\epsilon_{ji} & \epsilon_{jj} & \epsilon_{jk} \\
\epsilon_{ki} & \epsilon_{kj} & \epsilon_{kk} \\
\end{pmatrix}
$$
What exactly does each element represent in this matrix? For example, if I have an electric field polarized in the x direction, and moving in the y direction, how do I determine the permittivity in this case?

 

[jasonRF]

I don't know what you mean by "moving" in the y direction. In any case, the permittivity tensor relates the electric flux density to the electric field
$D_i = \epsilon_{ij} E_j$
Or, in matrix form,
$\left( \begin{array}{c} D_x \\ D_y \\ D_z \end{array} \right) = \left( \begin{array}{ccc} \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} \end{array}\right) \left( \begin{array}{c} E_x \\ E_y \\ E_z \end{array} \right)$
And this electric flux density shows up in Ampere's law
$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \frac{\partial}{\partial t} \mathbf{D}$
where $\mathbf{J}$ includes all currents that weren't lumped into $\mathbf{D}$.
jason

 

[lholmes135]

I should have specified, I am talking about in an electromagnetic wave.

 

[kuruman]

I don't know what you mean by "moving" in the y direction.

I am guessing that if one applies an external field $\vec E=E_x~\hat x+ E_y~\hat y+E_z~\hat z$ to the anisotropic sample, and then "moves" in the y-direction, the electric displacement vector is $D_y=\epsilon_{yx}E_x+\epsilon_{yy}E_y+\epsilon_{yz}E_z.~$ By contrast, "moving" in any direction $\hat r$ in an isotropic sample ($\epsilon_{ij}=\epsilon$), one has $D_r=\epsilon E_r.$

 

[lholmes135]

For calculating D it makes sense, but there are many other equations where it seems strange to use a tensor. How about for calculating the index of refraction? $n=\sqrt{\epsilon_r\mu_r}$. Since $\epsilon$ and $\mu$ depend on the direction of the electromagnetic wave and its polarization, this tells me that the index of refraction is also directional. How can you determine the index of refraction in this case? Is there not a way to determine a specific scalar value of permittivity, given the direction the wave is traveling and the direction it is polarized?

 

[Vanadium 50]

this tells me that the index of refraction is also directional

What's wrong with that? Calcite does this. Many perovskites do this.

 

[vanhees71]

If the material is not isotropic the linear response of the charged particles making up this medium to a (sufficiently weak) external electromagnetic field is not any longer described by scalar permittivities (and permeabilities if the medium is also "magnetic") but by a permittivity tensor. It's usually symmetric (if no additional magnetic field is present). Physically it describes birefringence.

 

[jasonRF]

For calculating D it makes sense, but there are many other equations where it seems strange to use a tensor. How about for calculating the index of refraction? $n=\sqrt{\epsilon_r\mu_r}$. Since $\epsilon$ and $\mu$ depend on the direction of the electromagnetic wave and its polarization, this tells me that the index of refraction is also directional. How can you determine the index of refraction in this case? Is there not a way to determine a specific scalar value of permittivity, given the direction the wave is traveling and the direction it is polarized?

You start with Maxwell's equations to answer these kinds of questions. If you want to understand how plane waves behave, then the standard approach is to represent the electric field as the real part of $\mathbf{E}\, e^{i(\omega t - \mathbf{k \cdot r})}$, where $\mathbf{E}$ is a constant complex-valued vector, and likewise for the magnetic field. $\mathbf{k}$ is of course the wave vector, which is orthogonal to the planes of constant phase. When you insert them into Faraday's law, for example, you should find $\mathbf{k\times E} = \omega \mathbf{B}$, which is algebraic. If you do the same for Ampere's law (and assume $\mathbf{J=0}$ and the media is non-magnetic), the equation becomes $\mathbf{-k\times B} = \omega\mu_0 \, \epsilon \mathbf{E}$. You should verify those for yourself. Finally, you combine these two equations to get
$\omega^2 \mu_0 \, \epsilon \mathbf{E} +\mathbf{k\times k \times E} = 0$
I like to then define a relative permittivity tensor $\epsilon_r = \epsilon / \epsilon_0$, and the unit vectors $\hat{\mathbf{k}} = \mathbf{k}/k$. We then have
$\epsilon_r \mathbf{E} + n^2 \mathbf{\hat{k}\times \hat{k} \times E} = 0$
where $n^2 = c^2 k^2/\omega^2$. Again, I recommend going through the algebra yourself.

What does this mean? For a given $\mathbf{\hat{k}}$, which simply specifies the direction of the wave, the equation above is a generalized eigenvalue problem. The eigenvalues are the squared indexes of refraction $n_\alpha^2$ ($\alpha =$ 1, 2 or 3), and the eigenvectors $\mathbf{E}_\alpha$ indicate the polarizations of the waves associated with each eigenvalue. In general, $\epsilon_r$ is a function of frequency and often other parameters such a temperature and/or perhaps an imposed magnetic field.

For some "simple" media and directions, the above procedure is much less daunting than it looks. For example, for a uniaxial crystal we have,
$\epsilon_r = \left( \begin{array}{ccc} \epsilon_1 & 0 & 0 \\ 0 & \epsilon_1 & 0 \\ 0 & 0 & \epsilon_2 \end{array}\right).$
If we consider waves propagating in the x direction so that $\mathbf{\hat{k}} = \mathbf{\hat{x}}$, then the above equation can be written
$\left( \begin{array}{ccc} \epsilon_1 & 0 & 0 \\ 0 & \epsilon_1 - n^2 & 0 \\ 0 & 0 & \epsilon_2-n^2 \end{array}\right)\mathbf{E} = 0.$
This has a non-trivial solution for $\mathbf{E}$ only if the determinant of the matrix is zero, which tells us $(\epsilon_1-n^2)(\epsilon_2-n^2) = 0$. This equation has only two solutions for $n^2$. One is $n_1^2 = \epsilon_1$ with corresponding $\mathbf{E}_1 = \mathbf{\hat{y}}E_1$, and the other is $n_2^2 = \epsilon_2$ with $\mathbf{E}_2 = \mathbf{\hat{z}}E_2$. This means that a linearly polarized wave traveling in the x direction will have index of refraction $n= \sqrt{\epsilon_1}$ if it is polarized along the y direction, and $n= \sqrt{\epsilon_2}$ if it is polarized along the z direction.

Jason

 

[vanhees71]

One should note that of course the diagonal form of the symmetric permittivity tensor is taken when choosing a particular coordinate system, with the principle axes as the Cartesian basis. For any symmetric tensor you always can find a Cartesian basis, where it is diagonal.

 

[jasonRF]

One should note that of course the diagonal form of the symmetric permittivity tensor is taken when choosing a particular coordinate system, with the principle axes as the Cartesian basis. For any symmetric tensor you always can find a Cartesian basis, where it is diagonal.

Thanks vanhees71 - I'm glad you caught that!

jason