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There are two questions here at MSE that are very similar to mine, namely Does zero curl imply a conservative field? and Why does $\vec\nabla\times\vec A=\vec0$ imply $\vec A=\vec\nabla B$? But they don't fully answer my question.

The accepted answer for the first linked question claims that $\operatorname{curl}(\mathbf a)=\mathbf0$ does not imply that $\mathbf a$ is conservative but the point that confuses me is that once you extend from the field defined over a simply connected subspace

to the "non conservative" version over a space that is not simply connected

then there is no continuous scalar field whose gradient can produce such a vector field. There would be a discontinuity at the line where they join.

With this, can we say that if $\mathbf a\colon U\subset\mathbb R^m\to\mathbf{R}^m$ is a vector field (with $m=2, 3$) then $\operatorname{curl}(\mathbf a)=\mathbf 0\nRightarrow\mathbf{a}=\boldsymbol{\nabla}b$ unless $U$ is simply connected?

Also, if $\mathbf{a}=\boldsymbol{\nabla}b$ can we say that $\mathbf{a}$ is conservative even if $U$ is not simply connected?

  • The answer to both your questions is yes. (What's your definition of a conservative force field? Some people use being a gradient field as the definition. Equivalent is path independence of the work line integral.) – Ted Shifrin Nov 27 '19 at 22:18
  • I was referring to path independence of line integrals. –  Nov 27 '19 at 22:39
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    Yes, the equivalence should be in your textbook. If not, you can check out my YouTube lectures. See, in particular, MATH 3510 day 31, starting approximately around minute 10. – Ted Shifrin Nov 27 '19 at 22:45
  • Thank you, that is illustrative (although I am not very versed with the language of differential forms I was able to understand the outline of the proof). And about the first question, do you know how it can be proved? –  Nov 27 '19 at 23:11
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    If I understand you correctly, it's a counterexample — rather than a proof — that's required. The usual example is $\vec F = (-y,x)/(x^2+y^2)$ on $\Bbb R^2-{0}$. – Ted Shifrin Nov 27 '19 at 23:14
  • But how would you go about showing that there exists no scalar field $b$ such that $\vec\nabla b = \vec F$? That's my main question –  Nov 27 '19 at 23:42
  • Ahh, do you use the fact that it's non-conservative and then the equivalence shown in the video? –  Nov 27 '19 at 23:43
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    Think about your pictures above. What's a closed path around which you get positive work? – Ted Shifrin Nov 27 '19 at 23:43
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    Right. By the Fundamental Theorem for Line Integrals, gradient fields do zero work around a closed curve. – Ted Shifrin Nov 27 '19 at 23:45
  • Thank you very much, professor. –  Nov 27 '19 at 23:46

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