Magnetic helicity

Remarks

The boundary condition makes $\mathcal{H}$ independent of the choice of $A$ up to gauge. If $\chi$ is a smooth function on the closure of $V$ and $A' = A + \nabla \chi$, then $\nabla \cdot (\chi B) = \nabla \chi \cdot B$, so by the divergence theorem $$\int_V A' \cdot B \, dV - \int_V A \cdot B \, dV = \int_{\partial V} \chi \, B \cdot n \, dS = 0.$$ On a region that is not simply connected, adding to $A$ a curl-free field that is not a gradient can change the integral, and one fixes extra conditions on $A$.

Example: for two thin untwisted closed flux tubes with fluxes $\Phi_1$ and $\Phi_2$, the helicity is $2 \, \mathrm{Lk} \, \Phi_1 \Phi_2$, where $\mathrm{Lk}$ is the linking number of their core curves. So linked tubes have nonzero helicity and unlinked untwisted tubes have helicity $0$.

Non-example: the uniform field $B = (0, 0, 1)$ on the open unit cube does not satisfy $B \cdot n = 0$ on the top and bottom faces. There the integral depends on the potential: $A = (-y, 0, 0)$ gives $\int A \cdot B \, dV = 0$, while $A = (-y, 0, 1)$, which has the same curl, gives $1$.

The same integral, written for the vorticity of a fluid, is the kinetic helicity of the flow.