{
  "aliases": [
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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^3$ be open, and let $A \\colon U \\to \\mathbb{R}^3$ be a vector field of class $C^2$.\nThen\n$$\\nabla \\cdot (\\nabla \\times A) = 0 \\quad \\text{on } U.$$\n\n## Remarks\n\nProof: with $A = (A_1, A_2, A_3)$, the [[divergence|02NSMQVQWK]] of the [[curl|0GDNHS9J2X]] is $\\partial_x (\\partial_y A_3 - \\partial_z A_2) + \\partial_y (\\partial_z A_1 - \\partial_x A_3) + \\partial_z (\\partial_x A_2 - \\partial_y A_1)$. The six terms cancel in pairs by [[Schwarz's theorem|0NNQY8WCW6]].\nSo a field that has a vector potential is divergence-free. On an open ball, the converse is also true: a $C^1$ divergence-free field is the curl of some field.\nThe identity is a special case of the rule [[d d = 0|0X308R2DBE]] for the exterior derivative.\n\n## Explanations\n\n### L1 human\n\nThe curl of a field describes rotation, and pure rotation does not create or destroy anything.\nSo the curl of any smooth field has no sources or sinks, which means its divergence is zero.\n",
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  "msc": [
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  "name": "Divergence of a curl is zero",
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