Examples where rot F = 0 don't imply conservative field

[Metaleer]

Hey, all.

Anyway, I've been looking at books and sources online, and the only counterexample to the wrongly stated theorem

$$\nabla \times \mathbf{F} = 0 \Leftrightarrow \text{conservative vector field}$$

seems to be $$\mathbf{F} = \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right),$$
or other vector fields based on this one. In one other example, a third component is added, leaving the original two. The reason that the "theorem" is wrongly stated is that it requires the additional hypothesis of the vector field's domain being simply-connected, which it isn't in this case.

Does anyone know any other vector fields where the domain isn't simply-connected, its curl vanishes, and it ends up being not conservative, and isn't based on the counterexample I gave?

Thanks in advance.

PS: To mods, I was in a hurry and the thread's title is wrong, could you please change "don't" to "doesn't"? Thank you. :)

 

[micromass]

Hello Metaleer!

You won't be finding any other vector field on $\mathbb{R}^2\setminus\{(0,0)\}$ with that property, this can be proven.

Basically, the first De Rham cohomology group of $\mathbb{R}^2\setminus\{(0,0)\}$ is $\mathbb{R}$, which means in les high-brow terms that every vector field on $\mathbb{R}^2\setminus\{(0,0)\}$ has the form

$$r\phi+\psi$$

for r a real number, $\phi$ the vector field you describe in your OP and $\psi$ conservative. This means that all counterexamples to your theorem must be based on the vector field $\phi$ in your OP.

Of course, if you check out other domains than $\mathbb{R}^2\setminus\{(0,0)\}$, then you might find more counterexamples, depending on what the first De Rham cohomology group is...

(If you want a good book on the topic, read Differential Forms by Weintraub)

 

[Metaleer]

Hey, micromass.

Thanks a lot for the info, I sort of had a sneaky suspicion this was related to de Rham cohomology. I'll give that book a read, thanks a lot!