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!