I am experiencing a slight dilemma here. Starting from the Divergence theorem, I am trying to show how a divergence-free field $\vec F$ implies that $\vec F$ can be written as the curl of another vector, say $\vec G$. Below I have written some steps which outline my current chain of logic.

Note that I'm assuming we don't already know that $\vec F =\vec\nabla\times \vec G$, since this is what I am trying to prove. Also, I am not using anything like the Helmholtz theorem either.

Here is goes:

Step $0$: Start with the divergence theorem in $\mathbb R^3$:

$$\iint_S\vec F\cdot d\vec S=\iiint _V\text{div}(\vec F)\;dV,$$

where $S$ is of course a closed surface.

Step $1$: If $\vec F$ is a $C^1$ vector field on an open region containing the volume V and $\text{div}(\vec F)=0$ everywhere, then

$$\iint_S\vec F\cdot d\vec S=0,$$

which says the flux through any closed surface must be zero.

Step $2$: Suppose we break the closed surface $S$ into two orientable surfaces $S_1$ and $S_2$ (note that $S_1$ and $S_2$ share the same boundary curve). Let one surface assume the orientation of the other. E.g, if $S_2$ assumes the orientation of $S_1$, then $S_2$ starts off with "negative" orientation, which we will denote by $-S_2$. Since $\iint_{-S_2}\vec F\cdot d\vec S =-\iint_{S_2}\vec F\cdot d\vec S$, hence

$$0=\iint_S\vec F\cdot d\vec S=\iint_{S_1} \vec F\cdot d\vec S + \iint_{-S_2} \vec F\cdot d\vec S=\iint_{S_1} \vec F\cdot d\vec S - \iint_{S_2} \vec F\cdot d\vec S.$$


$$\iint_{S_1}\vec F \cdot d\vec S=\iint_{S_2}\vec F \cdot d\vec S.$$

This says that the flux of a divergence-free vector field through any two open surfaces is surface independent and only depends on the surface boundary. In other words, for any fixed non-empty boundary, we may deform the surface $S$.

You may view this thread for more details regarding this step: Let $F$ be a vector field in $\mathbb{R}^3$. If $F$ is divergence free, we may deform the surface. Why?

Step $3$: This is where I am stuck. I've seen several articles somehow relate this surface independence property back to Stokes' theorem. In particular, I've seen: If $\text{div}(\vec F)=0$ everywhere, then

$$\tag{$\star$}\iint_{\text{any open surface}}\!\!\!\!\vec F\cdot d\vec S=\text{constant}=\oint_{\partial S}\vec G\cdot d\vec r=\iint_S\text{curl}(\vec G)\cdot d\vec S,$$ which implies $\vec F=\vec\nabla\times\vec G$. Here's the link to the article where that's from (on page 3): https://physics56.files.wordpress.com/2016/01/tutorial-7-scalar-and-vector-potential3.pdf

My confusion lies with equation $(\star)$. I just don't see how one simply relates the LHS to Stokes' theorem. I know this is analogous to the case when $\text{curl}(\vec F)=0$, which by Stokes' theorem implies path independence and moreover, the existence of a scalar function $\psi$, such that $\vec F=\vec\nabla \psi$. We can show that this is indeed the case for curl-free fields by using the fundament theorem of calculus. Again, I'm trying to come to this conclusion using vector calculus but not Helmholtz theorem or calculus on manifolds. Just following my outlined chain of logic starting from the Divergence theorem.

I appreciate any input I can get. :)

  • $\begingroup$ Helmholtz theorem is what allows you to conclude the existence of an appropriate $G$ such that the equation is satisfied; the Stokes Theorem does not contain any such construction to my knowledge $\endgroup$ Dec 28, 2022 at 7:51
  • $\begingroup$ I wasn't aware of that. I have only recently discovered Helmholtz theorem. That particular theorem also referred to as the fundamental theorem of vector calculus correct? $\endgroup$
    – whitenoise
    Dec 28, 2022 at 7:54
  • 3
    $\begingroup$ A link says more than a thousand words. $\endgroup$
    – Kurt G.
    Dec 28, 2022 at 7:57
  • $\begingroup$ Happily digging down the rabbit hole! $\endgroup$
    – whitenoise
    Dec 28, 2022 at 7:58
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    $\begingroup$ In differential form language, this is the Poincaré lemma (assuming your hypothesis holds on all of $\Bbb R^3$ or on a subset $S$ with $H^2_{\text{deRham}}(S) = 0$. It does not follow from the divergence theorem, as you actually have to integrate some partial differential equations. $\endgroup$ Jan 3, 2023 at 19:29

1 Answer 1


After revisiting this topic, I realized that there was a simpler explanation to the original question I had.

My question was: How can we how show that $$ \text{div}(\vec F)=0 \quad\implies\quad \vec F=\vec \nabla\times \vec G, $$ using the Divergence theorem (sometimes called the Gauss-Green theorem).

It turns out that the theorem itself does not directly imply this property unless we know that fact that

the divergence of the curl always vanishes.

This is analogous to

the curl of the gradient always vanishes.

Knowing these properties, once can make the classic loop of implications we often find in most introductory vector calculus texts:

$$\text{div}(\vec F)=0 \quad (\text{i.e. } \vec F=\vec\nabla\times \vec G) \quad\implies\quad \oint_{\partial V}\vec F\cdot \hat n\,dS=0 \quad\implies\quad \iint_{S_1}\vec F\cdot \hat n\,dS = \iint_{S_2}\vec F\cdot \hat n\,dS\;\; (\text{i.e. surface independance}) $$ Important: We must note that this not a "one-way" chain of consequences. That is, any single one of these properties implies the other two. Again, the same analogy can be made for gradient fields. That is, curl-free fields which are in fact the gradient of some scalar function.

This is all I have.

Thank to everyone who commented on my OP! :))

  • $\begingroup$ You still need the hypothesis that the vector field is $C^1$ on all of $\Bbb R^3$, or on a subset with the appropriate topological condition. $\endgroup$ Jun 8, 2023 at 22:18
  • $\begingroup$ You are correct. I had assumed those conditions were already implied. Thanks. $\endgroup$
    – whitenoise
    Jun 8, 2023 at 22:48
  • $\begingroup$ And I need to emphasize that the first of your chain of implications is non-obvious and non-trivial; the others are immediate consequences of the Divergence Theorem. Indeed, the third and fourth follow from the first; you do not need the second for that, provided the force is smooth and divergence-free everywhere inside $V$ (or inside the union of $S_1$ and $S_2$). So this is still sort of misleading. $\endgroup$ Jun 8, 2023 at 23:05
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    $\begingroup$ I see what you mean @TedShifrin. I went ahead and changed the presentation to account for that. Hopefully it's a it more logic. :) $\endgroup$
    – whitenoise
    Jun 9, 2023 at 5:54

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