Suppose $V\subset \mathbb{R}^3$ be non-empty and at least twice differentiable (Smooth) and let $S$ be the surface that encloses $V$ (for example a sphere). Let $\textbf{F}\in \mathbb{R}^3$ be a smooth vector field for all space. Let $\textbf{n}$ denote the normal to the surface $S$. Does the surface integral over $S$ preserve the curl operation with respect to the vector field $\textbf{F}$. In other words, Does the surface integral of $\textbf{n}\times\textbf{F}$ commute with the curl operation $$\textbf{curl}\biggl(\oint_{S}^{}{\textbf{n}\times\dfrac{\textbf{F}(\textbf{r}^\prime)}{|\textbf{r}-\textbf{r}^\prime|} ~ds}\biggr) = \oint_{S}^{}{\textbf{n}\times \dfrac{ \textbf{curl} (\textbf{F}) (\textbf{r}^\prime)}{|\textbf{r}-\textbf{r}^\prime|}~ds}?$$ Here the surface intergrals are evaluated with respect to the position $\textbf{r}^\prime$ and produce vector fields.
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$\begingroup$ on the left-hand-side, the integral no longer depends on spatial variables, so how can you take the curl? $\endgroup$– Carlo BeenakkerCommented Feb 16, 2023 at 16:05
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$\begingroup$ @CarloBeenakker The notation here is the same notation used in the wiki page en.wikipedia.org/wiki/…. So that these integrals produce vector fields. $\endgroup$– MrPieCommented Feb 16, 2023 at 16:17
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$\begingroup$ there is no factor $|r-r'|$ in your integrals, as in the wiki page you cite; the function $F$ depends on position, you integrate over position, how can some position dependence remain? $\endgroup$– Carlo BeenakkerCommented Feb 16, 2023 at 16:56
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$\begingroup$ @CarloBeenakker I just put it into $\textbf{F}$ I can rewrite it differently with that term if I need to. I guess i should have put that. I fixed it $\endgroup$– MrPieCommented Feb 16, 2023 at 16:58
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$\begingroup$ @MrPie : So, both surface integrals are functions of $\mathbf r$, right? $\endgroup$– Iosif PinelisCommented Feb 16, 2023 at 18:16
2 Answers
$\renewcommand\r{\mathbf r}\newcommand\n{\mathbf n}\newcommand\F{\mathbf F}\newcommand\0{\mathbf 0}\newcommand\curl{\operatorname{\mathbf{curl}}}$No. E.g., let $S$ be the unit sphere and let $\F:=(1,0,0)$, so that $\curl\F=\0$ and hence the right-hand side of the identity in question is $\0$.
On the other hand, the left-hand side of the identity in question is \begin{equation} \begin{aligned} \curl\oint_{S}ds(\r')\,\n(\r')\times\dfrac{\F(\r')}{|\r-\r'|} &=\oint_{S}ds(\r')\,\curl\Big(\n(\r')\times\dfrac{\F(\r')}{|\r-\r'|}\Big), \end{aligned} \end{equation} where the latter $\curl$ is of course with respect to $\r$.
For $\r=(1,0,0)$, the first coordinate of the latter (vector) integral (rewritten in the spherical coordinates) is \begin{equation} \int_0^\pi d\phi\int_0^{2\pi}d\theta\, g(\phi,\theta), \end{equation} where \begin{equation} g(\phi,\theta):=-\frac{\cos^2\phi+\sin^2\phi\;\sin^2\theta} {(2-2 \cos \theta \sin\phi)^{3/2}} \, \sin\phi, \end{equation} which is manifestly $<0$ for $(\phi,\theta)\in(0,\pi)\times(0,2\pi)$.
So, the left-hand side of the identity in question is not $\0$, and thus
the identity does not hold in general. $\quad\Box$
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$\begingroup$ @losifPinelis what if $\textbf{F}$ is curl free? It seems that has an impact on this question. $\endgroup$– MrPieCommented Feb 16, 2023 at 21:43
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$\begingroup$ @MrPie : I don't understand the point of your question. In my example, $\mathbf F$ is curl free. $\endgroup$ Commented Feb 16, 2023 at 22:17
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$\begingroup$ @losifPinelis sorry I meant if $\textbf{F}$ is not curl free. The contradiction you found was easy because the right hand side vanished quickly but the left doesn't. Can you find such a field with $\textbf{Curl}(\textbf{F})\neq 0$ that still makes this a contradiction? is it true otherwise? $\endgroup$– MrPieCommented Feb 16, 2023 at 22:22
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1$\begingroup$ @MrPie : This is easy. Take your favorite non-curl-free field $\mathbf G$, and replace my curl-free $\mathbf F$ by $\mathbf F_\delta:=\mathbf F+\delta\,\mathbf G$ with a nonzero real $\delta$. Then $\mathbf F_\delta$ is not curl free, but, by continuity, your identity will still fail to hold with $\mathbf F_\delta$ in place of $\mathbf F$ if $|\delta|$ is small enough. $\endgroup$ Commented Feb 16, 2023 at 22:31
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$\begingroup$ @losifPinelis hmmm interesting thank you!! I posted another question related to this topic if you're interested. It is here mathoverflow.net/questions/441022/… $\endgroup$– MrPieCommented Feb 16, 2023 at 22:54
These are two different integrals. To see they are different, you could for example take $\textbf{F}(\textbf{r})=\textbf{r}$. Then the curl of $\textbf{F}$ vanishes, so the integral on the right-hand-side is zero, but the left-hand-side integral does not vanish for arbitrary $V$ (without any spherical symmetry).
