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We say a vector field $\mathbf{F}$ is conservative if there exists some $f$ such that $\nabla f=\mathbf{F}$. I know that this implies that the field is path independent, and this makes sense to me: the work done moving something along a path on a conservative vector field is independent of the path. But I can't figure out any intuitive way to think about the definition itself. What does it look like? My intuition is that most vector fields ought to be a gradient of some function if they are at least moderately well-behaved, but this is clearly not the case. Is there a way to eyeball a vector field and know that there can be no such $f$?

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    $\begingroup$ In $\mathbb{R}^3$ you can compute the curl: a field is conservative iff its curl is zero. One of the Stokes' theorem variants shows that this is because the curl is the limit of small line integrals around circles centered at the point, as the radius of the circle goes to zero. $\endgroup$ – Ian Jul 20 '14 at 4:41
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    $\begingroup$ In order to see the usefulness of conservative field theory, you should read both Fundamental Theorems of Line Integrals. About eyeballing vector fields, you need to look for the concept of Curl (rot in some literature). $\endgroup$ – Daniel Charry Jul 20 '14 at 4:45
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Due to Stokes' theorem (either the specific case, or the very general statement on differential forms), and the Poincare Lemma we know that at least on a contractible (think, could be squashed to a point) subset of $\mathbb{R}^3$ these should be precisely the "curl-less", or irrotational vector fields. A classical example is the vector field

$(x,y,z) \to (y,-x,0)$ which basically gives a big curling vector field around the $z$-axis. The idea is that there are obvious paths (the curls) on which the field is always doing strictly positive work.

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