Under which circumstances would a rotation free vector field conservative ?

Question

It's obvious that every conservative vector field is rotation free (the curl of the vector field is equal to zero) in an open space, but is there any theorem that proves other way around so these two theorems form a criterium ?

Thanks

Answer 1

Let $\Omega\subset{\mathbb R}^n$, $n\geq2$, be a connected open set, and let ${\bf F}$ be a vector field on $\Omega$ with ${\rm curl}({\bf F})\equiv0$. Then ${\bf F}$ is locally conservative on $\Omega$. This means that each point $p\in\Omega$ has a neighborhood $U$ such that ${\bf F}\restriction U$ is conservative. We then have a local potential $f:\>U\to{\mathbb R}$ such that $\nabla f(x)={\bf F}(x)$ for $x\in U$.

The question remains whether we can concatenate all these local potentials (determined up to an additive constant) to a single potential $f:\>\Omega\to{\mathbb R}$. This depends on the global topological properties of $\Omega$. A sufficient condition is that $\Omega$ is simply connected. Another example is the following: If $\Omega\subset{\mathbb R}^2$ is the punctured plane (hence not simply connected) and $\int_\gamma {\bf F}(z)\cdot dz=0$ for a single loop around the origin then the rotation free ${\bf F}$ is conservative.