# Help understanding proof involving Maxwell equation

by U.Renko
Tags: equation, involving, maxwell, proof
 P: 57 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.
 Homework Sci Advisor HW Helper Thanks P: 12,463 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: $$\nabla\cdot\vec B = 0\\ \nabla\times \vec B = \mu_0\epsilon_0\frac{\partial}{\partial t}\vec E\\ \nabla\times \vec E = -\frac{\partial}{\partial t}\vec B$$ https://www.google.co.nz/search?q=ma...hrome&ie=UTF-8