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May 25th, 2015, 09:18 PM  #1 
Newbie Joined: Jun 2014 From: Hong Kong Posts: 7 Thanks: 0  Question about conditions for conservative field
In common textbooks' discussions about conservative vector field. There is always two assumptions about the region concerned, namely the region is simply connected and open. Usually in textbooks there is not much explanations on why these assumptions are necessary, no proof is given on why conservative field is not possible if the region is not simply connected or not open. I wonder whether these two assumptions are just for computational convenience or it is really logically not possible to have a conservative field in region that is not simply connected or is not open? 
May 26th, 2015, 12:38 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
No, it is not just for "computational convenience". It may not be true for a nonsimply connected region. For example, the field $\displaystyle \frac{\vec{r}}{r}$ satisfies the conditions everywhere except at the origin and so the conclusion, the existence of a "potential field", does not follow. If we were to remove the origin, the conditions would be satisfied everywhere in that nonsimply connected region but the potential field still does not exist. 

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