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Why is it true that$$\displaystyle curl (\vec{F})=0 \iff \vec{F} \text{ is conservative}$$ i.e. $$\displaystyle \exists f \text{ s.t. }\nabla f=\vec{F}$$

Jivan Pal
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Alex
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    Because of stokes theorem. If you want to proove that: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/4.-triple-integrals-and-surface-integrals-in-3-space/part-c-line-integrals-and-stokes-theorem/session-92-proof-of-stokes-theorem/MIT18_02SC_MNotes_v13.3.pdf – Gappy Hilmore Jun 13 '15 at 12:16
  • I am very new to these topics and I can't understand that pdf to be honest. Could you offer some intuition as to why it might be true? I will come back to that pdf in a few weeks when I get more understanding. Thanks. – Alex Jun 13 '15 at 12:18
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    Roughly - choose a base point $O$ and two paths $P_1$ and $P_2$ from $O$ to $A$. Use zero curl to show that integrating forward along $P_1$ and backward along $P_2$ i.e. a loop gives zero. So you can define a scalar field by the integral along a path from $O$ to $A$. And this is what you need. – Mark Bennet Jun 13 '15 at 12:20
  • I don't know its proof myself. I though maybe you wanted to. https://i.imgur.com/PkGvC4L.png This is stokes theorem. Insert curlF as 0. – Gappy Hilmore Jun 13 '15 at 12:22
  • @grdgfgr Perhaps what is missing is the equivalence $$\exists f \text{ s.t. }\nabla f = F \iff \int_C F = \int_{C'} F$$ for any pair of regular curves $C, C'$. – GPerez Jun 13 '15 at 12:31
  • Hi @MarkBennet, does the region where F is curl-free need to be simply connected? Or can we relax this assumption? I saw that it was necessary, per an MIT OCW lecture, but perhaps there is an improvement in the result. Thanks, – User001 Aug 03 '15 at 01:20

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Assume for simplicity that the domain $\Omega\subset{\mathbb R}^3$ of the vector field ${\bf F}$ is convex and contains the origin ${\bf 0}$, and consider a tiny parallelogram $P$ centered at some point ${\bf p}\in\Omega$. Its boundary $\partial P$ consists of four vectorial segments ${\bf U}$, ${\bf V}$, $-{\bf U}$, and $-{\bf V}$. Doing the calculation one finds that $$\int_{\partial P}{\bf F}\cdot d{\bf r}={\rm curl}\,{\bf F}({\bf p})\cdot({\bf U}\times{\bf V})\ +o\bigl({\rm area}(P)\bigr)\ .$$ In particular, if ${\rm curl}\,{\bf F}\equiv{\bf 0}$, for any given $\epsilon>0$ one has $$\left|\int_{\partial P}{\bf F}\cdot d{\bf r}\right|\leq\epsilon\>{\rm area}(P)\tag{1}$$ for all sufficiently small parallelograms $P\subset\Omega$. It follows that in fact $$ \int_{\partial R}{\bf F}\cdot d{\bf r}=0\tag{2}$$ even for "large" rectangles $R$ aligned to the coordinate axes, because such an $R$ can be subdivided into a million small $P$'s for which $(1)$ holds, and the contributions along all inner boundary edges cancel.

We now define a potential $f:\>\Omega\to{\mathbb R}$ as follows: For any point ${\bf x}\in\Omega$ let $\gamma({\bf x})$ be a path connecting ${\bf 0}$ with ${\bf x}$, and consisting of three line segments parallel to the coordinate axes. Then put $$f({\bf x}):=\int_{\gamma({\bf x})}{\bf F}\cdot d{\bf r}\ .$$ From $(2)$ it follows that $f({\bf x})$ does not depend on the chosen $\gamma({\bf x})$ , and using a proper choice for each partial derivative of $f$ it is then easy to show that $$\nabla f({\bf x})={\bf F}({\bf x})\qquad({\bf x}\in\Omega)\ .$$

Christian Blatter
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