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.

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


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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 $$ \frac{\partial }{\partial t}(\nabla\times {\bf X})= \nabla\times\left( \frac{\partial {\bf X}}{\partial t}\right). $$

  • $\begingroup$ I premise that my policy that I adopt in all sites including TeX.SE always and however positively rate the effort of a user. I think I have understood how to get there. But why is your second identity true? $\endgroup$ – Sebastiano Jul 12 at 21:14
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    $\begingroup$ 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. $\endgroup$ – G. Smith Jul 12 at 21:32
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    $\begingroup$ 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 $\endgroup$ – Syrocco Jul 12 at 21:46
  • $\begingroup$ @G.Smith I vote positively for the comment and an your answer. $\endgroup$ – Sebastiano Jul 12 at 22:00
  • $\begingroup$ @Syrocco See, please, the previous comment. $\endgroup$ – Sebastiano Jul 12 at 22:01

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