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  "body": "\n## Definition\n\nLet $n \\ge 2$ and let $\\mu$ be a finite positive [[Borel measure|DYR1QFNGNJ]] on $\\R^n$ with compact [[support|FQTXV6BQZG]].\nLet $k$ be the [[Newtonian kernel|6TJTWQ5HCR]] $k(x) = \\abs{x}^{2-n}$ when $n \\ge 3$, and the [[logarithmic kernel|VDZ5FKYC2T]] $k(x) = \\log(1/\\abs{x})$ when $n = 2$, with $k(0) = +\\infty$.\nThe *energy* of $\\mu$ is\n$$I(\\mu) = \\iint k(x - y)\\, d\\mu(x)\\, d\\mu(y).$$\nFor $n \\ge 3$ it lies in $[0, +\\infty]$; for $n = 2$ it lies in $(-\\infty, +\\infty]$, because $k(x - y)$ is bounded below on the compact set $\\operatorname{supp}\\mu \\times \\operatorname{supp}\\mu$.\n\n## Remarks\n\nEquivalently, $I(\\mu) = \\int U^\\mu \\, d\\mu$, where $U^\\mu(x) = \\int k(x - y)\\, d\\mu(y)$ is the [[potential|JHKSF3XFY2]] of $\\mu$.\nSome authors include the normalising constant of the fundamental solution in $k$, which multiplies the energy by a constant.\nThe infimum of $I(\\mu)$ over probability measures on a compact set $K$ determines the [[capacity|4YZVFGJVSZ]] of $K$.\nFor instance, when $n \\ge 3$ and with this kernel, the Newtonian capacity of $K$ is the reciprocal of that infimum.\n\nExample: let $\\sigma$ be the uniform probability measure on the unit sphere in $\\R^3$.\nIts potential is $U^\\sigma(x) = 1/\\max(\\abs{x}, 1)$, which equals $1$ on the sphere, so $I(\\sigma) = 1$.\nIn the plane, the uniform probability measure on the unit circle has potential $\\log(1/\\max(\\abs{z}, 1))$, which is $0$ on the circle, so its energy is $0$.\n\nNon-example: the Dirac measure $\\delta_0$ is a positive Borel measure with compact support, but $I(\\delta_0) = k(0) = +\\infty$, so it does not have finite energy.\n",
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