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Conservative vector field conditions 
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#1
Dec509, 07:59 PM

P: 13

My calculus book states that a vector field is conservative if and only if the curl of the vector field is the zero vector. And, as far as I can tell a conservative vector field is the same as a pathindependent vector field.
The thing is, I came across this: http://www.math.umn.edu/~nykamp/m2374/readings/pathindex/ The site shows a vector field where the curl is equal to the zero vector, yet the vector field is not conservative. As far as I can tell, saying "F is conservative iff Curl(F) = 0" contradicts the claims of the site I posted. What conditions must be met for a vector field to be conservative? 


#2
Dec609, 01:12 AM

P: 986

Like the site says, the curl should be zero and the domain should be simply connected. Your calculus book probably implies that you work in R^n or some other simply connected space.
A conservative field is usually defined as one that is a gradient of some scalar field. Curl of gradient is automatically zero. On the other hand, if you have a field whose curl is zero and its domain is simply connected, you can smoothly deform any closed path into a point, and then it's possible to prove that the field is pathindependent, therefore you can construct the scalar field out of the vector field. 


#3
Dec609, 06:24 AM

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P: 39,565

"Simply connected" means that any closed curve can be "shrunk" to a single point while staying inside the region. Since (0,0) is not in the region (F is not defined at (0,0) so Curl(F)= 0 is not true there), any closed curve that includes (0,0) in its interior cannot be "shrunk" to a single point without passing through (0,0) which is not in the region. 


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