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In https://arxiv.org/abs/hep-ph/0010057 the following vector calculus equality is claimed without proof although in note [4] the cryptic comment is made that "The relation is essentially the momentum space identity $(\mathbf{k}\times\mathbf{A})^2=\mathbf{k}^2\mathbf{A}^2-(\mathbf{k}\mathbf{A})^2$ in position space":


Indeed there is the vector relation: \begin{align} \int \mathbf{A}^2(x)d^3 x & = \frac{1}{4\pi} \int d^3 x d^3 x' \frac{[\nabla \times \mathbf{A}(x)] \cdot [\nabla \times\mathbf{A}(x')]}{\vert \mathbf{x}-\mathbf{x'} \vert} \\ & \qquad + \frac{1}{4\pi} \int d^3 x d^3 x' \frac{[\nabla \cdot \mathbf{A}(x)] [\nabla \cdot \mathbf{A}(x')]}{\vert \mathbf{x}-\mathbf{x'} \vert} \tag{6}\label{6} \\ & \qquad + \rm{surface\ terms} \end{align} Each of the two terms is positive; hence (up to the surface term question) we can minimize the integral of $\mathbf{A}^2$ by choosing $\nabla \cdot \mathbf{A} = 0.$ With this choice the integral of $\mathbf{A}^2$ is minimal in accord with our above remarks and is expressed only in terms of the magnetic field $\nabla \times \mathbf{A}$


This $\eqref{6}$ is indeed a very interesting identity and Gubarev, et al, go on to show it also in relativistically invariant form. When $\mathbf{A}$ is the vector potential, $\mathbf{B}=\nabla\times\mathbf{A}$, then in the Coulomb gauge $\nabla\cdot\mathbf{A}=0$ and $$\int \mathbf{A}^2(x)d^3 x = \frac{1}{4\pi} \int d^3 x d^3 x' \frac{\mathbf{B}(x) \cdot \mathbf{B}(x')}{\vert \mathbf{x}-\mathbf{x'} \vert} + \rm{surface\ terms}$$ Ignoring the "surface terms" in the infinity and assuming that the integrals of $\eqref{6}$ are positive indeed then we have the gauge independent minimum on the right side dependent only on the $\mathbf{B}$ field: $$\int \mathbf{A}^2(x)d^3 x \ge \frac{1}{4\pi} \int d^3 x d^3 x' \frac{\mathbf{B}(x) \cdot \mathbf{B}(x')}{\vert \mathbf{x}-\mathbf{x'} \vert}.$$ I have two questions:

  1. I would like to see a more detailed explanation of the proof based on the momentum space - position space equality
  2. Why is it obvious that on the right side of $\eqref{6}$ each of the two integrals is positive?
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  • $\begingroup$ I'm not so sure about your question 2, and I'm only directing you elsewhere for question 1, so I think a comment is best. The relation mentioned in note [4] is a easy to prove for any two vectors by simply brute forcing the expansion. That being said, it is not apparent to me that that relation is actually relevant to deriving (6); that instead looks like work similar to derive classic Helmholtz-type decompositions. That math seems helpful: en.wikipedia.org/wiki/Helmholtz_decomposition. $\endgroup$ – flevinBombastus Sep 6 '20 at 15:42
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As @flevinBombastus has suggested here is a sketch of the proof of the equality in Equation $(6)$ based on [1]. Start with $$\nabla^2\frac{1}{|\mathbf x - \mathbf x'|}=-\delta(\mathbf x - \mathbf x')$$ and $$\nabla \times (\nabla \times \mathbf v)=\nabla (\nabla\cdot \mathbf v) - \nabla^2 \mathbf v$$ Then $$\mathbf{A}(\mathbf x) = \int d^3x' \mathbf{A}(\mathbf x')\delta(\mathbf x - \mathbf x') =-\int d^3x' \mathbf{A}(\mathbf x')\nabla'^2\frac{1}{|\mathbf x - \mathbf x'|},$$ therefore $$\int d^3x \mathbf{A}(\mathbf x)\cdot\mathbf{A}(\mathbf x)=-\int\int d^3x d^3x' \mathbf{A}(\mathbf x)\cdot \mathbf{A}(\mathbf x')\nabla'^2\frac{1}{|\mathbf x - \mathbf x'|}$$

Now integrate RHS by parts over $\mathbf x'$. If $\mathbf A$ vanishes at infinity then the surface term will vanish, and after some more rearrangements and partial integration we get the required identity Eq. (6). Interestingly, the same procedure also works for the scalar product of two vector fields. This takes care of the 1st question.

[1] Durand: "On an identity for the volume integral of the square of a vector field,” Am.J.Phys. 75 (6), June 2007

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