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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^3$ be open, and let $\\phi \\colon U \\to \\mathbb{R}$ be of class $C^2$.\nThen\n$$\\nabla \\times (\\nabla \\phi) = 0 \\quad \\text{on } U.$$\n\n## Remarks\n\nProof: the first component of the [[curl|0GDNHS9J2X]] of $\\nabla \\phi$ is $\\partial_y \\partial_z \\phi - \\partial_z \\partial_y \\phi$, and this is zero by [[Schwarz's theorem|0NNQY8WCW6]]. The other components are similar.\nSo a [[conservative|05R9NF2J49]] $C^1$ vector field is irrotational. This gives a quick test: if $\\operatorname{curl} F \\ne 0$ at some point, then $F$ has no scalar potential.\nIn $\\mathbb{R}^n$, the same argument shows $\\partial_i F_j = \\partial_j F_i$ for all $i, j$ when $F = \\nabla \\phi$ with $\\phi$ of class $C^2$.\nThe converse needs a condition on $U$. It holds on [[convex sets|0B22P010D6]], but [[not on every open set|0GRM3BS0DD]].\n\n## Explanations\n\n### L1 human\n\nA gradient field has no swirl.\nGoing around a small loop in a landscape, you return to the same height. So the field $\\nabla \\phi$ has no circulation, and its curl is zero.\n",
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