## Proof That the E and H Fields Satisfy the Wave Equation

Maxwell's Laws are:
$\mathbf{\nabla} \cdot \mathbf{E}=0$
(1)
$\mathbf{\nabla} \cdot \mathbf{H}=0$
(2)
$\mathbf{\nabla} \times \mathbf{E}=- \mu \frac{\partial H}{\partial t}$
(3)
$\mathbf{\nabla} \times \mathbf{H}=- \epsilon \frac{\partial E}{\partial t}$
(4)
Take the curl of both sides of (3) to obtain
\begin{aligned} \mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{E}) &=-\mathbf{\nabla} \times (\mu \frac{\partial H}{\partial t}) \\ &=- \mu \frac{\partial}{\partial t} (\mathbf{\nabla} \times \mathbf{H}) \\ &= - \mu \epsilon \frac{\partial}{\partial t} (\frac{\partial \mathbf{E}}{\partial t}) \\ &=- \mu \epsilon \frac{\partial^2 \mathbf{E}}{\partial t^2} \end{aligned}

Now use the identity
$\mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{V})=\mathbf{\nabla} ( \mathbf{\nabla} \cdot \mathbf{V}) - \nabla^2 \mathbf{V}$
and remeber that
$\mathbf{\nabla} ( \mathbf{\nabla} \cdot \mathbf{E}) = \mathbf{\nabla} \mathbf{0}=\mathbf{0}$
Hence
$\mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{E})=- \nabla^2 \mathbf{E}$

Hence
$- \nabla^2 \mathbf{E}=- \mu \epsilon \frac{\partial^2 \mathbf{E}}{\partial t^2} \rightarrow \nabla^2 \mathbf{E}=\mu \epsilon \frac{\partial^2 \mathbf{E}}{\partial t^2}$