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I've been trying to solve the maxwell equations: $$\nabla\cdot\vec{D}=0,\quad \nabla\cdot\vec{B}=0$$ $$\nabla\times\vec{E}=-\frac{\partial \vec{B}}{\partial t},\quad \nabla\times\vec{H}=\frac{\partial \vec{D}}{\partial t}$$ Where $\vec{D}=\epsilon \vec{E}$, $\vec{B}=\mu_0 \vec{H}$ Here $\epsilon$ is the three by three matrix: $$\epsilon=\epsilon_0A$$ Where $A$ is symmetric and $\mu_0$ and $\epsilon_0$ are the usual constants. I've been looking for plane wave solutions, real part implied, of the form $$\vec{D}=\vec{D_0}e^{i(\vec{k}\cdot\vec{x}-\omega t)},\quad \vec{B}=\vec{B_0}e^{i(\vec{k}\cdot\vec{x}-\omega t)}$$ By solve here I mean find conditions on $\vec{D_0},\vec{B_0},\vec{k}$ only in terms of $\omega$. By substitution I have found these relations: $$\vec{k}\cdot\vec{D_0}=0,\quad\vec{k}\cdot\vec{B_0}=0$$ $$\vec{k}\times\frac1\epsilon_0A^{-1}\vec{D_0}=\omega\vec{B_0},\quad-\vec{k}\times\vec{B_0}=\mu_0\omega\vec{D_0}$$ I want to figure this out on my own, but I've been playing with these things for a long time now and I believe to get this in the form $\vec{D_0}=\vec{D_0}(\omega),\vec{B_0}=\vec{B_0}(\omega),\vec{k}=\vec{k}(\omega)$, requires some knowledge outside of my skill set, I would much appreciate it if anyone could point me in the right direction here.

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Try picking orthogonal directions (i.e., $\hat{k}\perp\vec{B}\perp\vec{D}$) –  Kyle Kanos Aug 28 '13 at 18:44
    
@KyleKanos: Ok, so $\vec{k}$ is perpendicular to $\vec{D}$ and $\vec{B}$, however by the 4th equation $\vec{D}$ is perpendicular to $\vec{k}$ and $\vec{B}$. So they must all be perpendicular to each other? But the 3rd equation states that $\vec{B}$ is perpendicular to $\vec{k}$ and a transformed $\vec{D}$, does this imply my calculations were wrong? –  Freeman Aug 28 '13 at 18:53
    
Yes, take each to be perpendicular to each other. Recall that $\hat{z}\times\hat{x}=\hat{y}$. –  Kyle Kanos Aug 28 '13 at 18:56
    
@KyleKanos: So they are wrong, ok. However I feel that I would get a similar set of equations anyway, what would the general approach be there to achieve my desired result? –  Freeman Aug 28 '13 at 18:57
    
How are they wrong? –  Kyle Kanos Aug 28 '13 at 18:59

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