# Dispersion relation for TE and TM waves in general anisotropic medium

I want to calculate the dispersion relation (the relation between $\bf k$ and permittivity and permeability tensors and $\omega$) for a TE and a TM wave with wave vector $\mathbf k=k_x\mathbf {\hat x}+k_z\mathbf {\hat z}$ propagating in an anisotropic medium with general diagonal permeability and permittivity tensors $\mu=\mu_0 \tilde \mu$ and $\epsilon=\epsilon_0 \tilde \epsilon$ with $\tilde \epsilon$ and $\tilde \mu$ given below.

I write the Maxwell's curl equations for the fields with a space-time dependence of the form $\exp(-ik_0\bf k\cdot r+i\omega t)$ (with $k_0=\omega\sqrt{\epsilon_0\mu_0})$ in the $k$ domain as usual:

$$\bf k\times \bf E=\bf {\tilde \mu} \bf B$$ $$\bf k\times \bf H=-\bf{ \tilde \epsilon} \bf E$$

where $\bf{ \tilde \epsilon}$ and $\bf {\tilde \mu}$ are dimensionless matrices: $$\begin{pmatrix} \epsilon_x & 0 &0 \\ 0 & \epsilon_y & 0\\ 0 &0 &\epsilon_z \end{pmatrix},\begin{pmatrix} \mu_x & 0 &0 \\ 0 & \mu_y & 0\\ 0 &0 &\mu_z \end{pmatrix}$$ writing the vector products $\bf k\times \bf H$ and $\bf k\times \bf E$ as matrix multiplication $\bar k \bf H$ with $$\bar k=\begin{pmatrix} 0 & -k_z &k_y \\ k_z & 0 & -k_x\\ -k_y &k_x &0 \end{pmatrix}$$

I rewrite the curl equations as: $$\cases{ \bar k \bf H=-\tilde \epsilon\bf E\\\bar k \bf E=\tilde \mu \bf H}\to\cases{(\bar k \tilde \mu^{-1}\bar k+\tilde \epsilon)\bf E=0 \\(\bar k \tilde \epsilon^{-1}\bar k+\tilde \mu)\bf H =0 }$$ Now, the determinant of the coefficient matrices $\bar k \tilde \epsilon^{-1}\bar k+\tilde \mu$ and $\bar k \tilde \mu^{-1}\bar k+\tilde \epsilon$ must be zero for the systems to have non-trivial solutions.

My question is, where in this process I should consider the assumption of the waves being TM or TE? I know that the final dispersion relation depends on this.(being TE or TM)

• Of course the most general medium doesn't have a diagonal permittivty/permeability tensor. – user21433 Feb 24 '14 at 17:44
• @SeanD It's doesn't have anything to do with my question, but we can diagonalize every non-diagonal permeability/permittivity tensor, I presume? – user215721 Feb 24 '14 at 17:59
• @I've never seen anyone try, probably because transforming away from the cartesian basis would be pretty awkward experiment-wise. I can't think of any good reason either way though. Maybe someone more knowledgeable can chime in. – user21433 Feb 24 '14 at 18:14
• @user215721: Generally the tensors are symmetric and thus diagonalizable, but note that implicit in your formulation is the notion that $\mu$ and $\epsilon$ are simultaneously diagonalizable, ie, there is a choice of principal axes in which both $\mu$ and $\epsilon$ become diagonal. I'd guess that's usually possible for most materials, but I don't know for sure. – DumpsterDoofus Feb 24 '14 at 18:15
• @SeanD: Actually, most commercial crystal optics data is given in the principal axis coordinate system of the crystal material, for that exact reason. The notion of "fast axes" and "slow axes" is just a rephrasing of the principal axis condition. – DumpsterDoofus Feb 24 '14 at 18:18

TE and TM waves only interact with those elements of permittivity and permeability tensors that the wave has a nonzero electric and magnetic field along them, respectively. so with a wave vector $\mathbf k=k_x\mathbf {\hat x}+k_z\mathbf {\hat z}$, for a TE wave only $\epsilon_y$ interacts with the E field and only $\mu_z$ and $\mu_x$ interact with the H field. For a TM wave, only $\epsilon_z$ and $\epsilon_x$ interact with E the field and only $\mu_y$ interacts with the H field.
After calculating the determinant that you have obtained, and incorporating the above facts (setting non-interacting elements of material tensors to zero for each case, plus $k_y=0$) we will find :
$$\frac{k_x^2}{\epsilon_z}+\frac{k_z^2}{\epsilon_x}=\omega ^2\mu_y\tag {TM}$$
$$\frac{k_x^2}{\mu_z}+\frac{k_z^2}{\mu_x}=\omega ^2\epsilon_y\tag{TE}$$