My question is exactly the same as this question, but I cannot do the variation of integral referred in the Addendum, i.e., \begin{align} S_B\propto\int dx\; B^2=\int \left(\nabla\times A\right)^2\; dx, \end{align} where $S_B$ is some magnetic action, $B$ the magnetic field and $A$ the vector potential. I have problems when I vary wrt $A$. I don't obtain the last equality: \begin{align} \delta S_B\propto 2\int dx\;\left[\left(\nabla\times A\right)\cdot \left(\nabla\times\delta A\right)\right]=2\left[\delta A\times (\nabla\times A)\right]+2\int dx\; \left[(\nabla\times\nabla\times A)\cdot \delta A\right]. \end{align}
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$\begingroup$ Did you already derived the Euler-Lagrange equations ? It's the same thing here. See e.g. en.wikipedia.org/wiki/Euler–Lagrange_equation and references therein. $\endgroup$– FraSchelleCommented Oct 14, 2017 at 8:26
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$\begingroup$ I think my question was not well defined. I know to derive E-L equations, and I understand I have to vary with respect to $A$, but I cannot obtain the last equality. Maybe is some easy calculation, but I don't see it. $\endgroup$– hyriusenCommented Oct 14, 2017 at 12:31
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Use the vector calculus identity $$ \nabla\cdot(\delta {\bf A}\times(\nabla \times {\bf A}))= (\nabla\times \delta{\bf A})\cdot(\nabla\times {\bf A})- \delta {\bf A} \cdot (\nabla\times(\nabla \times {\bf A})), $$ and Stokes' theorem. Note that you need an extra pair of parenthesis in your last volume integral ---vector products are not associative so what you have written is ambiguous--- and the first term on the RHS should be a surface integral.
answered Oct 14, 2017 at 15:25