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
\[\begin{equation} \begin{aligned} \mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{E}) &=-\mathbf{\nabla} \times (\mu \frac{\partial H}{\partial t}) \\ &=- \mu \frac{\partial}{\partial t} (\mathbf{\nabla} \times \mathbf{H}) \\ &= - \mu \epsilon \frac{\partial}{\partial t} (\frac{\partial \mathbf{E}}{\partial t}) \\ &=- \mu \epsilon \frac{\partial^2 \mathbf{E}}{\partial t^2} \end{aligned} \end{equation}\]
 
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  
\[ - \nabla^2 \mathbf{E}=- \mu \epsilon \frac{\partial^2 \mathbf{E}}{\partial t^2} \rightarrow \nabla^2 \mathbf{E}=\mu \epsilon \frac{\partial^2 \mathbf{E}}{\partial t^2} \]

Add comment

Security code
Refresh