Sufficient condition for a vector field to be conservative

[DottZakapa]

Homework Statement:: F is not conservative because D is not simply connected
Relevant Equations:: Theory

Having a set which is not simply connected is a sufficient conditiond for a vector field to be not conservative?

 

[FactChecker]

No. Not if D is the domain. You can take any conservative vector field and arbitrarily say that a hole in it is not part of the domain. That does not change that it is conservative in the remaining domain. Path independence of the path integral between two points is key.

 

[PeroK]

Homework Statement:: F is not conservative because D is not simply connected
Relevant Equations:: Theory

Having a set which is not simply connected is a sufficient conditiond for a vector field to be not conservative?

I guess you are referring to the theorem that a vector field defined on a simply connected set is conservative iff it is irrotational.

Note that a conservative vector field is always irrotational, but you can have an irrotational, non-conservative field defined on a set that is not simply connected. The holes are the problem.

This is very different from saying that a field is non-conservative just because it's defined on a set that is not simply connected. That's a logical misreading of what the theorem says.

This is discussed here:

https://mathinsight.org/conservative_vector_field_determine

 

[zinq]

Think of a conservative vector field as the gradient of some function — which should be thought of as the derivative of the function. So if you integrate this vector field (= grad(f)) along a curve, just as with 1-dimensional integration of a derivative, you get the difference of the values of f at the endpoints of the curve. So if you integrate the vector field all around a closed curve, you must get 0.

In a small neighborhood, you can detect a local gradient vector field by the fact that its curl is everywhere 0. For instance in the plane (without the origin), consider the vector field V(x,y) = (-y/(x² + y²), x/(x² + y²) ). Near enough to any point (x,y) ≠ (0,0), this is the gradient of the angle function. But the angle function is not defined continuously in any open set that goes around the origin. Sure enough, V(x,y) is not conservative on such an open set. (To see this, take the line integral of V(x,y) around a circle enclosing the origin, and sure enough you don't get 0. If you haven't done this, it's worth doing this calculation.)