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I have heard that the following is true:

Let $\vec{F}$ be a vector field such that $\nabla \times \vec{F} = \vec{0}$ everywhere on its domain, $D$. Further require that $D$ is simply connected. Then there exists a scalar field $f(x,y,z)$ such that $\vec{F} = \nabla f$.

It's that second portion of my definition, that $D$ needs to be simply connected, that concerns me. Here is my issue:

let's go backwards. Let $f(x,y,z)$ be defined on a domain that is not simply connected. Can I not still compute $\nabla f$? This would be a vector field, and I presume it exists (if I'm wrong here then the next part is also not correct). Assuming the gradient exists on this domain, then wouldn't it have zero curl on that domain? In short, why is it important that my domain be simply connected in order to go back to my scalar field?

Michael Stachowsky
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1 Answers1

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The idea is that for a given vector field $\vec F:X\subseteq\Bbb R^n\to\Bbb R^n$ of class $C^1$ one cannot find any function $f:X\subseteq\Bbb R^n\to\Bbb R$ of class $C^2$ such that $\nabla f=\vec F$, if $\vec F$ is not conservative. If the domain $X$ is simply connected, then $\vec F$ is conservative (i.e. $\vec F=\nabla f$ for some $f$) if and only if $\vec\nabla\times\vec F=\vec 0$ on $X$.

When the domain is not simply connected the above equivalence does not hold. In fact we only have that: if $\vec F$ is conservative then $\nabla\times\vec F=\vec 0$ on $X$. An example to show that $\vec\nabla\times\vec F=\vec 0\implies\text{$\vec F$ conservative}$ does not need to hold if $X$ is not simply connected is $$F:\Bbb R^2/\{\vec 0\}\to\Bbb R^2~~~~~F(x,y)=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$$ Note that to compute the curl one would just put a zero in the $z$-component of $\vec F$. The above function has $\vec 0$ curl (scalar curl), however, there exists no $C^2$ function $f$ such that $\vec F=\nabla f$. Also, $X$ is not simply connected here because the origin is removed from the $xy$-plane (i.e. $\Bbb R^2/\{\vec 0\}$ is not simply connected).

Dave
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  • I see. So then it's not that the gradient doesn't exist, it's that it is no longer sufficient that $\nabla \times \vec{F} = \vec{0}$ to ensure that we are looking at a gradient? – Michael Stachowsky Jul 10 '17 at 17:18
  • @MichaelStachowsky when the domain is not simply connected, $\vec\nabla\times\vec F=\vec 0$ is not sufficient (although it is necessary) to say that $\vec F=\nabla f$ for some $f$. So if $\vec\nabla\times\vec F=\vec 0$, then it may or may not be the case that $\vec F=\nabla f$, but we need further work to determine this (when the domain is not simply connected). – Dave Jul 10 '17 at 17:25
  • I see. Is there a similarly clear test, or would it depend on the properties of the domain itself? – Michael Stachowsky Jul 10 '17 at 17:28
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    Another way to see if a vector field is conservative is to simply differentiate the component functions of $\vec F$ to obtain mixed partial derivatives of the yet determined $f$. If the mixed partial derivatives are equal then $f$ exists since $f$ is $C^2$. – Dave Jul 10 '17 at 18:04