Maxwell's Equations Divergence Question

Question

$$ \left\{ \begin{align} \text{div } \textbf{E} & =0, \\ \text{div } \textbf{H} & =0, \\ \text{curl } \textbf{E} & = \frac{-1}{c} \frac{\partial \textbf{H}}{\partial t}, \\ \text{curl } \textbf{H} & = \frac{1}{c} \frac{\partial \textbf{E}}{\partial t}. \end{align} \right.$$ Using the equations above, and the assumption that $\textbf{E}$ and $\textbf{H}$ are infinitely differentiable, show that the following identity holds true.

$$\nabla^2\textbf{E}=\dfrac{1}{c^2}\dfrac{\partial ^2\textbf{E}}{\partial t^2}$$

If $\text{div }\textbf{E}=0$, how could this be true?

Thank you for any suggestions.

Answer 1

You have $$ \frac{1}{c}\frac{\partial}{\partial t}\left(\frac{1}{c}\frac{\partial E}{\partial t} \right) = \frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}. $$ On the other hand, $$ \frac{1}{c}\frac{\partial}{\partial t}\left(\frac{1}{c}\frac{\partial E}{\partial t} \right)=\frac{1}{c}\frac{\partial}{\partial t}\left(\mbox{curl}\left(H\right) \right) = \mbox{curl}\left(\frac{1}{c}\frac{\partial H}{\partial t}\right)=-\mbox{curl}\left(\mbox{curl}E\right). $$ Now, a vector calculus identity given is that for any vector valued function $\phi\in C^2(\mathbb{R}^3;\mathbb{R}^3)$ $$ \left(\mbox{curl}(\mbox{curl}\phi)\right)_j= \partial_j\left(\mbox{div}\phi\right) - \sum_{i=1}^{3}\partial_{ii}\phi_j. $$ So if you use the unfortunately misleading notation that $$ (\nabla^2\phi)_j = (\sum_{i=1}^{3}\partial_{ii} \phi)_j, $$ you have obtained $$ \frac{1}{c^2}\frac{\partial^2 E}{\partial t^2} = \nabla^2E -\nabla(\mbox{div}E), $$ and since div$E=0$, you are done.

The reason why you were confused is the motivation for not using the $\nabla^2$ notation, of dubious meaning: many prefer to write $\Delta$ instead. Indeed, $\nabla$ goes from 1 dimension to $3$, and then $\nabla^2$ arguably should go from $3$ to $9$ dimension (and many people use to for that purpose to denote the Hessian matrix). Whereas $\Delta$ does not change dimension, $1$ to $1$ and $3$ to $3$. It just acts on each component of the vector the way you would expect it to.

Answer 2

Hint: the vector Laplacian of a vector field $A$ satisfies the identity,

$$\nabla^2A=\nabla (\nabla\cdot A)-\nabla \times(\nabla\times A)$$

Answer 3

The notation $\nabla^2\mathbf E$ can be misleading. It looks as if it should mean $\nabla(\nabla\mathbf E)$, and indeed it does mean that provided you understand the inner $\nabla$ as the gradient operator, applied to each component of $\mathbf E$ individually, not the divergence operator applied to the whole vector $\mathbf E$. So the intended $\nabla\mathbf E$ is not a scalar but rather 3 separate vectors, one for each component of $\mathbf E$, namely the gradient of that component. The outer $\nabla$ then produces the divergence of each of these three gradients. The final result is that $\nabla^2\mathbf E$ is obtained by applying "div grad" to each component of $\mathbf E$ separately.

Of course, the right side of your equation is also obtained by applying the operator "second derivative with respect to time" to each component separately. So the whole equation says that each component of $\mathbf E$ satisfies the wave equation.