# Does zero curl imply a conservative field?

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).

• 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 4 '18 at 18:15
• 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 4 '18 at 19:24
• 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 4 '18 at 21:05
• @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 4 '18 at 21:12
• @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 5 '18 at 4:11

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.

• Thanks for the help! – John A. Feb 5 '18 at 21:18
• Great illustrations! How did you make them? – Hee Jin May 6 '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 6 '18 at 16:54

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).

• Thanks for your help! – John A. Feb 5 '18 at 21:18
• you are welcome! :) – Emilio Novati Feb 5 '18 at 21:28

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.