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?

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    $\begingroup$ That is NOT a potential for your field across its entire domain. For one, $\arctan(y/x)$ prohibits $x=0$, but plenty of good points in your domain have $x=0$. $\endgroup$ – Randall Oct 20 '17 at 15:21
  • $\begingroup$ @Randall would appreciate if you could elaborate. $\endgroup$ – Rax Adaam Oct 20 '17 at 15:22
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    $\begingroup$ Matthew Leingang below nailed it. $\endgroup$ – Randall Oct 20 '17 at 15:22
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    $\begingroup$ You are dipping your toes into de Rham cohomology here! $\endgroup$ – Lord Shark the Unknown Oct 20 '17 at 15:49
  • $\begingroup$ @LordSharktheUnknown & here I thought one couldn't drown with only toes in the water! Hopefully I'll get there... $\endgroup$ – Rax Adaam Oct 20 '17 at 20:47

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 $U$.

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    $\begingroup$ I've wasted so much time on this. Thank you for ending my suffering. Wish they took the time to underscore this when presenting either the definition or the example... $\endgroup$ – Rax Adaam Oct 20 '17 at 15:26
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    $\begingroup$ @RaxAdaam: Congratulations on having the question in the first place. It's a good one. I think domains are always mentioned in the definitions, or in hypotheses to theorems, but this is one where it makes a big difference. There are more, like why the Intermediate Value Theorem requires a closed and bounded interval for a domain. $\endgroup$ – Matthew Leingang Oct 20 '17 at 15:37
  • $\begingroup$ Maybe you could mention homotopy and winding number so as to give a higher topological perspective of the problem involved. $\endgroup$ – Hans Oct 20 '17 at 16:03
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    $\begingroup$ +1 to all this. Certainly domains are mentioned, but instructors really need to point out how it matters and can greatly influence both your methodology and answers. $\endgroup$ – Randall Oct 20 '17 at 18:43
  • $\begingroup$ @Matthew, thanks for the kind words of encouragement. I have avoided most questions of domain and continuity for multivariate functions because I haven't found any clear presentation of how to deal with them (i.e. one that doesn't just seem to pull out one unrelated trick after another; vs. single-var where there are a handful of related strategies). As taught at my school, we are basically equipped to disprove multivariate continuity at a point, but not prove it. Would appreciate any pointers towards good sources. I'll definitely look into the IVT conditions - thanks for the heads up! $\endgroup$ – Rax Adaam Oct 20 '17 at 20:45

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