Math Tutorials and More
by George

In a vacuum Maxwell's equations reduce to

\begin{equation} \text{div }{\mathbf E} = 0\tag{1}\\ \end{equation} \begin{equation} \text{div }{\mathbf B} = 0 \tag{2}\\ \end{equation} \begin{equation} \text{curl }{\mathbf E} = -\frac{1}{c}\frac{\partial\mathbf B}{\partial t}\tag{3}\\ \end{equation} \begin{equation} \text{curl }{\mathbf B} = \frac{1}{c}\frac{\partial\mathbf E}{\partial t}\tag{4} \end{equation} Any vector field $\mathbf{F}$ satisfies the identity where $\nabla$ is the gradient operator and $\pmb{\triangle}$ is the Laplacian.

Applying equation (5) to $\mathbf{E}$ and $\mathbf{B}$ and making use of equations (1) and (2), we get

\begin{equation} \text{curl }\text{curl }\mathbf{E} = -\pmb{\triangle}{\mathbf E}\tag{6} \end{equation} \begin{equation} \text{curl }\text{curl }\mathbf{B} = -\pmb{\triangle}{\mathbf B}.\tag{7} \end{equation} Taking the curl of equation (3) and applying (6), we get \begin{equation} \pmb{\triangle}\mathbf{E}=\frac{1}{c}\frac{\partial\text{ curl }\mathbf{B}}{\partial t}.\tag{8} \end{equation} Substituting equation (4) into equation (8), we obtain the wave equation \begin{equation} \pmb{\triangle}\mathbf{E}=\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2}.\tag{9} \end{equation} In a similar manner we can show that $\mathbf{B}$ satisfies the wave equation \begin{equation} \pmb{\triangle}\mathbf{B}=\frac{1}{c^2}\frac{\partial^2\mathbf{B}}{\partial t^2}.\tag{10} \end{equation}

It is well known that differential equations such as (9) and (10) possess wave-like solutions.

To see that there can be no purely electrical waves, suppose that $\mathbf{B}=\mathbf{0}$. Then it follows from equation (4) that the time derivative of $\mathbf{E}$ is zero, i.e., $\mathbf{E}$ doesn't change with time. In a similar way we can see that there can be no purely magnetic waves.

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