Electromagnetic Wave Equation Derived

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In a vacuum Maxwell's equations reduce to

\begin{align} &\text{div }{\mathbf E} = 0\\[1em] &\text{div }{\mathbf B} = 0\\[1em] &\text{curl }{\mathbf E} = -\frac{1}{c}\frac{\partial\mathbf B}{\partial t}\\[1em] &\text{curl }{\mathbf B} = \frac{1}{c}\frac{\partial\mathbf E}{\partial t}. \end{align}

Any vector field $\mathbf{F}$ satisfies the identity

\begin{equation} \text{curl }\text{curl }\mathbf{F}=\nabla\text{div }\mathbf{F}-\pmb{\triangle}{\mathbf F} \end{equation}

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}\end{equation} \begin{equation}\text{curl }\text{curl }\mathbf{B} = -\pmb{\triangle}{\mathbf B}. \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}. \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}. \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}. \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.