# Help understanding proof involving Maxwell equation

1. Apr 2, 2014

### U.Renko

help understanding "proof" involving Maxwell equation

I'm currently taking a course on mathematical methods for physics.
(Like always I'm a bit confused about where exactly I should post these questions, should it go to the homework forum? )

anyway as I was reading the lecture notes I found this demonstration that if $\vec{J} = 0$ in Maxwell-Ampere's law satisfies the wave equation: $\left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 \right)\vec B = \vec 0$ :

There is one small step I'm not sure why/how he did it.

he took the curl of $\nabla \times\vec B$, that is, $\nabla\times\left(\nabla\times\vec B \right) = \nabla\left(\nabla\cdot\vec B\right) - \nabla^2\vec B$

then since $\nabla\cdot\vec B = 0$ and $\nabla\times\vec B = \mu_0\epsilon_0\frac{\partial\vec E}{\partial t}$ he says $\nabla\times\left(\nabla\times\vec B \right) = \nabla\left(\mu_0\epsilon_0\frac{\partial\vec E}{\partial t}\right) = \mu_0\epsilon_0\left(\nabla\times\frac{\partial\vec E}{\partial t}\right)$

so far I understand everything, but now he says:
$\mu_0\epsilon_0\left(\nabla\times\frac{\partial\vec E}{\partial t}\right) = \mu_0\epsilon_0\frac{\partial\left(\nabla\times\vec E\right)}{\partial t}$ and it is this very last step that I'm not sure if it's allowed. And if is, why it is?

the rest of the demonstration follows easily, supposing this last step is correct.

2. Apr 3, 2014

### Matterwave

That step just says that time derivatives commute with the curl. You should try this out for yourself and see that it is so. (Basically it's true simply because partial derivatives commute and the basis vectors are not time-dependent.)

Just write out the curl of E, take the time derivative. Then write out the curl of the time derivative of E and they are identical.

3. Apr 3, 2014

### Simon Bridge

You missed one:
$$\left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2 \right)\vec B = 0\\ \implies \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec B =\nabla^2\vec B\\ \implies \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec B = \nabla(\nabla\cdot\vec B)-\nabla\times (\nabla\times \vec B)$$

Note:
$$\nabla\cdot\vec B = 0\\ \nabla\times \vec B = \mu_0\epsilon_0\frac{\partial}{\partial t}\vec E\\ \nabla\times \vec E = -\frac{\partial}{\partial t}\vec B$$