# An electrodinamic identity: starting point [closed]

With this request, I would like to ask you kindly how you can prove this identity. I thank you for those who can help me.

$$$$\overline{\nabla} \times (\overline{\nabla} \times \overline{E})=-\frac{\partial}{\partial t}(\overline{\nabla} \times \overline{B})\tag{1}$$$$

## closed as off-topic by AccidentalFourierTransform, tpg2114♦Jul 19 at 12:31

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Maxwell's version of Faraday's law of induction is $$\nabla\times {\bf E}= -\frac{\partial {\bf B}}{\partial t}.$$ Now apply $$\nabla\times$$ to both sides. Note that $$\frac{\partial }{\partial t}(\nabla\times {\bf X})= \nabla\times\left( \frac{\partial {\bf X}}{\partial t}\right).$$
• For the same reason that $\partial^2/\partial x\partial y=\partial^2/\partial y\partial x$. It doesn’t matter in which order you take partial derivatives. – G. Smith Jul 12 at 21:32
• You can develop both sides and you'll see that the only thing that will change is the order of partial derivatives. For example, according to the second equation you have, for the $x$ axis: $\dfrac{\partial}{\partial t}\left(\dfrac{\partial X_z}{\partial y}-\dfrac{\partial X_y}{\partial z}\right)=\dfrac{\partial}{\partial y}(\dfrac{\partial X_z}{\partial t})-\dfrac{\partial}{\partial z}(\dfrac{\partial X_y}{\partial t})$ Which is true because of the Schwarz theorem en.wikipedia.org/wiki/Symmetry_of_second_derivatives – Syrocco Jul 12 at 21:46