Why can some gradient fields not be simply connected?

[Leo Liu]

Homework Statement

Reference: https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-c-greens-theorem/session-72-simply-connected-regions-and-conservative-fields/MIT18_02SC_MNotes_v5.pdf
(Last 2 pages)

Relevant Equations

As I understand it, being "simply connected" means that the closed curves in the domain region contain some area(s) that are not in the domain. In other words, the region has got some hole(s) in it.

For example,
$$\left\langle
\frac x {r^3},
\frac y {r^3}
\right\rangle
= \nabla \left(
-\frac 1 r
\right)$$
where $r=\sqrt{x^2+y^2}$, is a gradient field even though it is undefined at the origion. I get that it is physically possible since it is similar to the equation of the electric field of a positive charge place at the origion, and electric field is the gradient of the gradient of the potential function. But what is the mathematical explanation here? Thanks.

 

[WWGD]

Homework Statement:: Reference: https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-c-greens-theorem/session-72-simply-connected-regions-and-conservative-fields/MIT18_02SC_MNotes_v5.pdf
(Last 2 pages)
Relevant Equations:: As I understand it, being "simply connected" means that the closed curves in the domain region contain some area(s) that are not in the domain. In other words, the region has got some hole(s) in it.

For example,
$$\left\langle
\frac x {r^3},
\frac y {r^3}
\right\rangle
= \nabla \left(
-\frac 1 r
\right)$$
where $r=\sqrt{x^2+y^2}$, is a gradient field even though it is undefined at the origion. I get that it is physically possible since it is similar to the equation of the electric field of a positive charge place at the origion, and electric field is the gradient of the gradient of the potential function. But what is the mathematical explanation here? Thanks.

In the notation I am familiar with, it is regions ( open, connected sets) that may be simply/not simply -connected; not vector fields. I think a standard definition is that in a non-simply-connected region, curves may wind around points not in the set, as you said. But a Vector Field is a function ( that assigns n-ples of vectors) and not a region. It may be defined on a region, but it is not itself a region. I believe the result you are considering here is that every vector field in a simply-connected region is conservative/ independent of path. Edit: Vector Field on a simply-connected region is Conservative if it is curl-free

 

[Leo Liu]

In the notation I am familiar with, it is regions ( open, connected sets) that may be simply/not simply -connected; not vector fields. I think a standard definition is that in a non-simply-connected region, curves may wind around points not in the set, as you said. But a Vector Field is a function ( that assigns n-ples of vectors) and not a region. It may be defined on a region, but it is not itself a region. I believe the result you are considering here is that every vector field in a simply-connected region is conservative/ independent of path. Edit: Vector Field on a simply-connected region is Conservative if it is curl-free

Thank you. But if the domain of the vector field given is not simply connected, and for any curve C (including those that enclose the undefined origin), $$\oint_C \vec F \cdot d \vec r$$ can we still conclude that it is a gradient field? Why?

 

[FactChecker]

It is not the hole a the domain that necessarily causes a problem. You can take any gradient in a simply connected domain and arbitrarily say that it is not defined in a hole in the domain. It would still be a gradient on the rest of the domain.
The issue is whether an integral in a closed curve that winds around the hole is non-zero. If the vector field has a zero integral on every closed curve that does not wind around the hole, it can be considered to be a gradient on a Riemannian manifold. The integrals on curves that wind around the hole gives a multiple-valued function which is single-valued on each branch of the manifold.

 

[Leo Liu]

It is not the hole a the domain that necessarily causes a problem. You can take any gradient in a simply connected domain and arbitrarily say that it is not defined in a hole in the domain. It would still be a gradient on the rest of the domain.
The issue is whether an integral in a closed curve that winds around the hole is non-zero. If the vector field has a zero integral on every closed curve that does not wind around the hole, it can be considered to be a gradient on a Riemannian manifold. The integrals on curves that wind around the hole gives a multiple-valued function which is single-valued on each branch of the manifold.

What do you call the case in which the path integral of a closed curve C enclosing an undefined region is 0 (eg the gravitational field of a neutron star)?

 

[Infrared]

Thank you. But if the domain of the vector field given is not simply connected, and for any curve C (including those that enclose the undefined origin), $$\oint_C \vec F \cdot d \vec r=0$$ can we still conclude that it is a gradient field? Why?

(Added an $=0$ at the end which I think you meant to put).
Yes, this is true. Pick any point $p_0$ in your domain. For any other point, $p$, let $C(p)$ by any path from $p_0$ to $p$ and define $f(p)=\int_{C(p)}\vec{F}\cdot d\vec{r}.$ The answer you get is independent of which path you pick by the condition on $\vec{F}$ and you can check that $\vec{F}$ is the gradient of $f.$

 

[FactChecker]

What do you call the case in which the path integral of a closed curve C enclosing an undefined region is 0 (eg the gravitational field of a neutron star)?

That makes the line integral between two points path independent. Therefore, it can be used to define a well-defined potential for which it is the gradient.