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A field that is conservative must have a curl of zero everywhere. However, I was wondering whether the opposite holds for functions continuous everywhere: if the curl is zero, is the field conservative? Can someone please give me an intuitive explanation and insight into this, and if it is true, why? Also, please try to not be too rigorous (only in grade 9).

John A.
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    Do you know Stokes theorem? The important fact that leads to the existence of a potential is that the line integral just depends on the start and end point and not curve connecting these. – Fabian Feb 04 '18 at 18:15
  • @Math_QED, I mean that path independence holds, so the field can be written as the gradient. – John A. Feb 04 '18 at 18:18
  • @Fabian, yes I know Stokes theorem, but can you elaborate on your comment? – John A. Feb 04 '18 at 18:19
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    You were a bit hasty in accepting an answer. The one you’ve chosen tacitly makes some important assumptions about the domain of the vector field, without which your proposition is false. – amd Feb 04 '18 at 19:24
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    There's a factually incorrect answer rated at -2 and a correct answer rated at 7, and the former is the one that's accepted? – anomaly Feb 04 '18 at 21:05
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    @anomaly - in fairness to the OP, the argument in the accepted answer becomes correct if one assumes the OP is simply-connected; meanwhile, the other answer is correct in greater generality but doesn't provide an argument. – Ben Blum-Smith Feb 04 '18 at 21:12
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    @BenBlum-Smith: That's not a reasonable assumption, though, even in the physics world. The other answer links to a counterexample and states the assumption necessary to make the conclusion valid; actually proving it requires algebraic topology that's beyond the scope of the original question. – anomaly Feb 05 '18 at 04:11
  • Thanks all for the help! – John A. Feb 05 '18 at 21:17

3 Answers3

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Any conservative vector field $F :U \to \mathbb{R}^3$ is irrotational, i.e. $\mathbf{curl} (F)=0$, but the converse is true only if the domain $U$ is simply connected (see here for a classical example).

Emilio Novati
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Not necessarily. Look at the following potential $A%$ defined on some region:

The associated vector field $F=\mathrm{grad}(A)$ looks like this:

Since it is a gradient, it has $\mathrm{curl}(F)=0$. But we can complete it into the following still curl-free vector field:

This vector field is curl-free, but not conservative because going around the center once (with an integral) does not yield zero.

This happens because the region on which $F$ is defined is not simply connected (i.e. it has a hole). If you are only interested in vector fields on all of $\Bbb R^3$, then you are safe: $\Bbb R^3$ is simply connected and every curl-free vector field is conservative.

M. Winter
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  • Thanks for the help! – John A. Feb 05 '18 at 21:18
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    Great illustrations! How did you make them? – Hee Jin May 06 '18 at 16:53
  • @Emily Thanks :). These are made with Mathematica and some post-processing in some standard image editing software (Paint.NET in my case). – M. Winter May 06 '18 at 16:54
  • I don't think every R3 curl-free field is conservative. We could take that 2D non-conservative field example you gave and simply assign 0 to the z-component to lift it into 3d space. It would have a continuous singularity along the z-axis. It is curl free and undefined for z-axis but not conservative for similar reason. – Shuheng Zheng Dec 04 '20 at 16:56
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    @ShuhengZheng If you have a singularity at the $z$-axis, then the vector field is not defined on $\Bbb R^3$, but on $\Bbb R^3$ minus the $z$-axis, which is not a simply connected space. The statement of the answer holds and is a well-known theorem (see also here). – M. Winter Dec 04 '20 at 18:33
  • @M.Winter What if the vector field $\vec{F}$ is defined on a simply connected region, and it is curl-free. However, it is defined such that $\vec{F}=\vec{F}(\dot{\vec{r}})$? Does that theorem still holds true? – Jack Oct 01 '23 at 16:31
  • @Jack I am not exactly sure what you are asking. If the domain is simply connected and the field has zero curl, then the theorem applies, no matter how your field is defined. I am also not sure what you mean by $\dot{\vec r}$. – M. Winter Oct 02 '23 at 11:17
  • @M.Winter $\dot{\vec{r}}=\frac{d\vec{r}}{dt}$ – Jack Oct 02 '23 at 13:50
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    @Jack The fields in question for this theorem depend on the point $\vec r$ only. Time derivatives make no real sense in this context. – M. Winter Oct 02 '23 at 19:57
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You have to keep in mind that a vector field is not just a set of functions, but also a domain. For instance, the vector field $\mathbf{F} = \left<-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right>$ on the set $U = \left\{(x,y) \neq (0,0)\right\}$ has a curl of zero. But it's not conservative, because integrating it around the unit circle results in $2\pi$, not zero as predicted by path-independence.

On the other hand, the same vector field restricted to $U' = \left\{x>0\right\}$ is conservative. A potential function is $f(x,y) = \arctan\left(\frac{y}{x}\right)$.

The difference is that $U'$ is simply connected, while $U$ is not. In fact, this is an symbolic version of M. Winter's graphical example.

Matthew Leingang
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  • Here if we take $M$ and $N$ as component function of $F$ then $M_y =N_x$ On $\mathbb R^2 \setminus {(0,0)}$. So Mdx+Ndy is exact and by necessary and sufficient condition we have potential function on $\mathbb R^2 \setminus {(0,0)}.$ which is not the case . So does it mean that while proving sufficient part of exact ness we need domain D simply connected? – Meet Patel Mar 01 '24 at 16:33
  • @MeetPatel: You got it! – Matthew Leingang Mar 01 '24 at 20:44