3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Matthew Leingang below nailed it. $\endgroup$ – Randall Oct 20 '17 at 15:22
  • 1
    $\begingroup$ 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
3
$\begingroup$

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 $U$.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ +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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.