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  "body": "\n## Proof\n\nWe compute the curl in coordinates and use the symmetry of second derivatives.\n\n1. Let $\\phi \\colon U \\to \\R$ be of [[class $C^2$|0TCBZMEYGN]] on the open set $U \\subseteq \\R^3$. Its first partial derivatives are continuous, so $\\phi$ is differentiable by [[continuous partial derivatives imply differentiability|0KXSK8TEEB]].\n2. By the coordinate form of the [[gradient|0NS2QYVC4Y]], $\\nabla \\phi = (\\phi_x, \\phi_y, \\phi_z)$, where $\\phi_x = \\partial \\phi / \\partial x$ and so on.\n3. The components of $\\nabla \\phi$ have first partial derivatives on $U$, since $\\phi$ is $C^2$. For variables $u, v \\in \\set{x, y, z}$, write $\\partial_u \\phi_v$ for the [[second partial derivative|02RSXVM1ZN]] $\\partial^2 \\phi / \\partial u \\, \\partial v$.\n4. By the definition of the [[curl|0GDNHS9J2X]] with $(P, Q, R) = (\\phi_x, \\phi_y, \\phi_z)$, $\\nabla \\times (\\nabla \\phi) = \\big( \\partial_y \\phi_z - \\partial_z \\phi_y, \\ \\partial_z \\phi_x - \\partial_x \\phi_z, \\ \\partial_x \\phi_y - \\partial_y \\phi_x \\big)$.\n5. By [[Schwarz's theorem|0NNQY8WCW6]], $\\partial_y \\phi_z = \\partial_z \\phi_y$, $\\partial_z \\phi_x = \\partial_x \\phi_z$ and $\\partial_x \\phi_y = \\partial_y \\phi_x$ at each point of $U$.\n6. So each component in step 4 is $0$, and $\\nabla \\times (\\nabla \\phi) = 0$ on $U$.\n\n## Idea\n\nEach component of the curl of $\\nabla \\phi$ is a difference of two mixed second partial derivatives of $\\phi$, taken in the two possible orders. For a $C^2$ function these agree, so every component vanishes.\n",
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