# Maxwell's Equations Curl Question

\left\{ \begin{align} \text{div } \textbf{E} &=0, \\ \text{div } \textbf{H} & =0, \\ \text{curl } \textbf{E} & = \dfrac{-1}{c} \dfrac{\partial\textbf{H}}{\partial t}, \\\text{curl } \textbf{H} & =\dfrac{1}{c} \dfrac{\partial\textbf{E}}{\partial t}, \end{align} \right. where $c$ is the speed of light.

Using the equations above, and the assumption that $\textbf{E}$ and $\textbf{H}$ are infinitely differentiable, show that the following identity holds true.

$$\nabla \text{x}(\nabla \text{x} \textbf{E})= \dfrac{-1}{c}\dfrac{\partial^2\textbf{E}}{\partial t^2}$$

$$\nabla \text{x} \textbf{E} =\dfrac{-1}{c}\dfrac{\partial\textbf{H}}{\partial t}$$

$$\nabla \text{x} \left(\dfrac{-1}{c}\dfrac{\partial\textbf{H}}{\partial t}\right) = \dfrac{-1}{c} \left(\nabla \text{x} \dfrac{\partial \textbf{H}}{\partial t}\right)$$

$$=\dfrac{-1}{c} \left( \dfrac{1}{c} \dfrac{\partial^2\textbf{E}}{\partial t^2} \right)$$

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

Could someone please correct this if it's wrong? I don't know if taking the curl of a derivative equals the derivative of the curl...if not, I don't know how else to prove this.

Thank you.

Yes, the space and time derivatives commute so you can exchange curl and $\partial/\partial t$. This follows from Schwarz' theorem which tells you that if a function has continuous second partial derivatives at any point (which $\mathbf{E}$ and $\mathbf{H}$ do since they are infinitely differentiable) then you can exchange the order of partial differentiation.
• @Fantini take the first component, we have $$\left( \frac{\partial}{\partial y} \frac{\partial H_z}{\partial t} - \frac{\partial}{\partial z} \frac{\partial H_y}{\partial t} \right)$$ How do we know that we can "factor" the time derivative out here? Jul 20, 2014 at 14:15
• @DanZimm I guess I got carried away because he wrote $d/dt$. The answer to your question is, of course, Schwarz's theorem. Jul 20, 2014 at 14:26