If $\vec{\nabla} × \vec{A}=\vec{0}$, does this qualify as the necessary and sufficient condition for being able to writ $\vec{A}=\vec{\nabla}$ f where f is any scalar function?
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$\begingroup$ $f$ cannot be any scalar function obvsly. $\endgroup$– DigitallisCommented Sep 25, 2023 at 18:04
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$\begingroup$ Integrate $A$ along a path like $\int \vec{dl}\cdot \vec{A}$ (a path with a fixed starting point and end point equal to the independent variable in question, but otherwise an arbitrary path) to get a good candidate for $f$. $\endgroup$– 5th decileCommented Sep 25, 2023 at 18:21
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Necessary, yes, which is quite easy to prove.
If you're working with two or three dimensional vector fields, the condition is also sufficient, but not in any other dimension (but that begs the question how the cross product is defined in other dimensions).
The keyword you're looking for is Conservative Vector field.
PS: If you're just working on the subset of $\mathbb{R}^3$, make sure that it is simply connected!
