{
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  "body": "\n## Counterexample\n\nOne could guess that each $C^1$ vector field with zero [[curl|0GDNHS9J2X]] on an open set is [[conservative|05R9NF2J49]].\nThis is false.\nLet $U = \\{(x, y, z) \\in \\mathbb{R}^3 : (x, y) \\ne (0, 0)\\}$, which is $\\mathbb{R}^3$ minus the $z$-axis, and let\n$$F(x, y, z) = \\left( \\frac{-y}{x^2 + y^2}, \\ \\frac{x}{x^2 + y^2}, \\ 0 \\right).$$\nThen $\\nabla \\times F = 0$ on $U$, but $F$ is not conservative on $U$.\n\n## Remarks\n\nThe third component of $\\nabla \\times F$ is $\\partial_x \\big( x / (x^2 + y^2) \\big) - \\partial_y \\big( -y / (x^2 + y^2) \\big) = \\frac{y^2 - x^2}{(x^2 + y^2)^2} - \\frac{y^2 - x^2}{(x^2 + y^2)^2} = 0$, and the other two components are zero.\nOn the unit circle $\\gamma(t) = (\\cos t, \\sin t, 0)$, $t \\in [0, 2\\pi]$, we have $F(\\gamma(t)) \\cdot \\gamma'(t) = 1$. So the circulation is $2\\pi \\ne 0$, and [[path independence|0BFN21TF5F]] fails.\nThe domain $U$ is not [[simply connected|075C629XTH]], so this does not contradict the theorem for [[convex|0B22P010D6]] or simply connected sets.\nLocally, $F = \\nabla \\theta$, where $\\theta$ is the polar angle of $(x, y)$. But no continuous choice of $\\theta$ exists on all of $U$.\n\n## Explanations\n\n### L1 human\n\nThe field circles around the $z$-axis, and its strength falls off with distance in just the right way to have no local swirl.\nBut a loop around the axis picks up work $2\\pi$, so no potential can exist. The hole in the domain causes this.\n",
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        "locator": "Irrotational vector fields",
        "url": "https://en.wikipedia.org/wiki/Conservative_vector_field"
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  "name": "Curl-free field that is not conservative",
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