Why is this vector field not conservative, even though it has a potential? (what is the actual definition of a conservative vector field?)

Question

My question is really ''what is the definition of a conservative vector field''?

I've consulted 3 textbooks that all say a vector field $\vec{F}$ is conservative by definition if there exists a scalar potential $\phi$ such that $\nabla \phi = \vec{F}$. Then, they go on to talk about connected domains, path independence and the equality of mixed partials and how they are all related.

In particular, they emphasize that e.g. in $\mathbb{R}^2$ given $\vec{F} = \bigl<F_1,\,F_2\bigr>$, if $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$ on a simply-connected domain, then $\vec{F}$ is conservative on that domain.

However, without fail, all of them then offer the example of $\vec{F} = \bigl< \frac{-y}{x^2+y^2},\, \frac{x}{x^2+y^2} \bigr>$, pointing out:

1. it's line integral is not path independent, even though
2. $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$

and this is explained by pointing out that the domain is not simply connected (if the path contains the origin).

That much makes sense to me: we cannot conclude $\vec{F}$ is conservative based on the partial derivatives, because the domain is not simply connected - totally consistent with what has been presented.

What none of them address is why $\vec{F} = \bigl< \frac{-y}{x^2+y^2},\, \frac{x}{x^2+y^2} \bigr>$ is not conservative when there exists a potential $\phi = \arctan(y/x)$ such that $\nabla \phi = \vec{F}$. None of the texts mention any necessary conditions on the scalar potential. So is the existence of a scalar potential the definition of a vector field being conservative or not?

Answer 1

Any mapping, be it a vector field or a scalar function or something else, requires a domain.

It is true that where $\phi(x,y)$ is defined, $\nabla \phi = \vec F$. But $\vec F$'s domain is the plane minus the origin, while $\phi$'s domain is the plane minus a line (the $y$-axis).

Since there's no function with the same domain as $\vec F$ whose gradient is $\vec F$, $\vec F$ is not conservative.

Notice that the right half of the plane is simply connected, and as you've shown, $\vec F$ restricted to that domain is conservative. $\phi$ works as a potential on that domain.

The upshot is that the question of whether $\vec F$ is conservative on $U$ is a question not just about the component functions of $\vec F$ but the “shape” (we say topology) of $U$.