Does zero curl imply a conservative field?

Solution 1:

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).

Solution 2:

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.

Solution 3:

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.