Stationary Maxwell (Wave) Equations

We seek to solve the sationary ($\partial / \partial t$ = $i \omega$) Maxwell equations in the absence of free charges. These equations are:
\begin{equation} \label{eq:E} \nabla \times E = \frac{-i\omega}{c}\mu \cdot H \end{equation}

\begin{equation} \label{eq:H} \nabla \times H = \frac{i\omega}{c}\epsilon \cdot E \end{equation}
These combine to form two vector Helmholtz equations:
\begin{equation} \label{eq:vE} \nabla \times \mu^{-1} \cdot ( \nabla \times E) = \frac{\omega^{2}}{c}\epsilon \cdot E \end{equation}

\begin{equation} \label{eq:vH} \nabla \times \epsilon^{-1} \cdot ( \nabla \times H) = \frac{\omega^{2}}{c}\mu \cdot H \end{equation}
Notice the duality between $ (H,\mu) $ and $ (E,\epsilon) $ ,both E and H are solutions of a vector Helmholtz equation we can solve for either by solving the "same" equation. Note however, that E and H satisfy different boundary conditions at the interface of a perfect conductor; for E the tangential component of the field vanishes at the boundary whereas for H the normal component is zero at the conductor.
Due to the periodicty of the crystal we seek Bloch wave solutions $ \tilde E e^{i \hat k \cdot \hat x} $ where $ \tilde E $ is periodic. Solving for $ \tilde E $ (or $ \tilde H $ ) we get
\begin{equation} \label{eq:Efinite} (\nabla + i \hat k) \times [\mu^{-1} \cdot (\nabla + i \hat k) \times \tilde E] = \frac{\omega^{2}}{c^{2}}\epsilon \cdot \tilde E \end{equation}
Which we solve in the unit cell volume. This equation is obtained from (3) by replacing $ \nabla $ with $ \nabla + i k $ . Henceforth, we shall write $ \tilde E $ as E for simplicity.
Ryan McClarren
Last modified: Wed Aug 14 13:12:59 EDT 2002