# Minimum energy magnetic field has zero current?

Following section 9.2 of Bellan's "Fundamentals of Plasma Physics" suppose we have some domain (I assume simply connected) $D$ with a magnetic field $\mathbf{B}$ inside, but no electrostatic field. There is a boundary condition satisfied by this field but the book is vague about what exactly this is and seems to imply the result holds whatever boundary condition is chosen.

The claim (if I understand it correctly) is that of all fields satisfying these boundary conditions, the one with the lowest energy, given by

$E=\int_D dV\frac{1}{2}\mathbf{B}^2$

is the one with no current, ie. $\nabla\times\mathbf{B}=0$. The variational argument (slightly rearranged by me) is:

$\begin{array}{} \delta E &=& \int_D dV\mathbf{B}.\delta\mathbf{B} \\ &=& \int_D dV\mathbf{B}.(\nabla\times\delta\mathbf{A}) \\ &=& \int_{\partial D} \delta\mathbf{A}\cdot(\mathbf{B}\times d\mathbf{S})+\int_D dV \delta\mathbf{A}\cdot(\nabla\times\mathbf{B})\\ \end{array}$

(Writing $\delta\mathbf{B}$ as $\nabla\times\delta\mathbf{A}$ for some $\delta\mathbf{A}$.)

The book now claims the surface integral is zero and hence (because we're considering arbitrary $\delta\mathbf{A}$) $\nabla\times\mathbf{B}=0$. I can't see why the surface integral should be zero. The book says something about $\delta\mathbf{A}$ vanishing on the surface because of the boundary conditions on the variation of the magnetic field but I can't see why that would be true. Can anyone explain?

Additionally: does this theorem have a name and who first derived it?

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