Field with rot = 0 but non conservative.

[LCSphysicist]

Homework Statement

I want to know what is the problem with this field and why it is not conservative.

Relevant Equations

The curl.

x and y are different than 0
As you can check, rot (del x F) is equal zero. Immediately someone could imagine this is a conservative field. But it is not, it is not path independent in certain occasions.
I just know one way, involving polygonal, but there is not another way to check if F is conservative?

 

[etotheipi]

The curl is $\vec{0}$, which means that you can express $\vec{F} = -\nabla \phi$ wherever $\vec{F}$ is defined, however you have a singularity at the origin $(x,y) = (0,0)$. That means that any line integral around a closed curve that encloses the origin is not necessarily zero. If the closed curve does not enclose the singular point, then the line integral will be zero.

 

[LCSphysicist]

The curl is $\vec{0}$, which means that you can express $\vec{F} = -\nabla \phi$ wherever $\vec{F}$ is defined, however you have a singularity at the origin $(x,y) = (0,0)$. That means that any line integral around a closed curve that encloses the origin is not necessarily zero. If the closed curve does not enclose the singular point, then the line integral will be zero.

You unearthed a quote from Kleppner that was I don’t know where in my brain, involving a function similar to this with the same problem lol! The brain is a bizarre thing, and thank you ;) This is a more logic way to evaluate the functions

 

[vanhees71]

The original field is defined on $\mathbb{R}^2$ with the $3$-axis excluded. That's a multiply-connected region, i.e., you cannot shrink all closed curves continuously to a single point. Any closed curve with the $z$ axis intersecting all surfaces with this curve as a boundary cannot be contracted to a single point in a continuous way. Any other closed curve, however can.

You can easily show that in any open simply connected part of where a vector field is defined, which fulfills $\vec{\nabla} \times \vec{V}=0$ has a potential, $\vec{V}=-\vec{\nabla} \phi$, when restricted to this open simply-connected region.

In your case you can get a maximal such region, by just excluding not only the $3$-axis from the domain of the vector field but an entire half-plane with the $3$-axis as a boundary. You can find a unique potential in this restricted domain, but it will have a jump across the excluded half-plane by a constant $2 \pi$.

It's a good exercise to do the calculation to find this kind of "muti-valued potential".

Hint: Work in cylinder coordinates with some arbitrary domain for $\varphi \in [\alpha,\alpha+ 2 \pi[$ with $\alpha \in [-\pi,\pi]$ arbitrary. Note that cylinder coordinates are singular along the $3$-axis!