Solving the source free Maxwell equations for plane waves

Question

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.

Answer 1

I'm starting from the relations you derive.

The correct - or at least normal - way to phrase plane wave solutions is not with everything being a function of $\omega$, but with a dispersion relationship $\omega = \omega(\mathbf k)$ (and for very anisotropic relations you also include any polarization vectors), and with polarization vectors $\mathbf V_0$ lying in some space determined by $\mathbf k$.

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Notice that by taking a cross-product with $\mathbf k$, $$\mathbf k \times \mathbf V = \omega \mathbf B_0 \implies (\mathbf k \cdot \mathbf V)\mathbf k - |\mathbf k|^2 \mathbf V = -\omega^2 \mu_0 \mathbf D_0$$ and therefore $\mathbf D_0\equiv \mathbf d,\omega$ must solve two (sets of) linear equations: $$ (\hat{\mathbf k} \cdot A^{-1} \mathbf d) \hat{\mathbf k} - A^{-1}\mathbf d = -\frac{\epsilon_0 \mu_0 \omega^2}{|\mathbf k|^2} \mathbf d,\qquad \hat{\mathbf k} \cdot \mathbf d = 0$$

Take an inner product with $\mathbf d$ to find $$\omega^2 = \frac 1{\mu_0\epsilon_0}\frac{\mathbf d^T A^{-1} \mathbf d}{\mathbf d^T \mathbf d} |\mathbf k|^2 \quad \implies \quad \boxed{\omega = \pm \frac{|\mathbf k|}{\sqrt{\mu_0\epsilon_0}} \sqrt{\hat {\mathbf d}^T A^{-1} \hat {\mathbf d}}}$$ which gives the dispersion relationship in terms of the as yet unknown polarization $\mathbf d$.

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Let's use $\mathbf e = A^{-1}\mathbf d$ for simplicity. Then we have to solve $$ (\hat{\mathbf k} \cdot \mathbf e) \hat{\mathbf k} - \mathbf e = -\alpha^2 A\mathbf e,\qquad \hat{\mathbf k} \cdot A\mathbf e = 0$$ where $\alpha$ can be found from the above; but treating $\alpha$ as undetermined, the former is a linear equation via $(\hat{\mathbf k} \hat{\mathbf k}^T-I+\alpha^2 A)\mathbf e = 0$.

Edit: The matrix $\mathbf v \mathbf v^T$ is given by the usual matrix product of the $1\times n,n\times 1$ matrices given, so that $(\mathbf v \mathbf v^T)_{ij} = v_i v_j$. Notice that $(\mathbf v \mathbf v^T)\mathbf w = (\mathbf v \cdot \mathbf w) \mathbf v$.

I'll let you pick it up from here, since you fancied doing this yourself! Note that $\mathbf B_0$ is determined by $\mathbf e$, so all that remains to do is find what sort of $\mathbf e$ is possible. (You might find it useful to choose a basis in which $A$ is a diagonal matrix, so $D=I-\alpha^2 A$ is diagonal; then see what the different possibilities for the kernel of $\hat{\mathbf k} \hat{\mathbf k}^T-D$ are.)