Given the field $P(x,y)=\frac{y}{y^2-(x-1)^2}$, $Q(x,y)=\frac{1-x}{y^2-(x-1)^2}$ determine whether the field $(P(x,y,Q(x,y))$ is conservative within its defined range.


This question follows two questions from the same problem:

  1. Prove that $\int_cP\,dx+Q\,dy=0$ given curve $(x-3)^2+(y-3)^2=1$
  2. Calculate $\int_cP\,dx+Q\,dy$ if the curve is $c(t)=\langle 1-\cos t, \sin t\rangle$, $0\le\theta\le 2\pi$.

First we see that $P_y=Q_x$ so it's a conservative field in the range except the point $(1,0)$ where the field is undefined. So in question 1) the integral automatically is $0$.

In the 2) question the integral is $2\pi$.

The correct answer is that the field $(P(x,y,Q(x,y))$ is not conservative because in the 2) question the integral was not equal to $0$.

This is extremely confusing to me. We clearly see that $P_y=Q_x$ which means the field is conservative! Now there's the point $(1,0)$ indeed I guess if the range includes that point then the field is not defined hence not conservative? But the answer depends on the range in that case. But the question doesn't explicitly mention the range.

  • $\begingroup$ The domain $\Bbb R^2-\{(1,0)\}$ has nontrivial cohomology. $\endgroup$ – Angina Seng Aug 8 '17 at 18:32
  • $\begingroup$ The error is in the following sentence: "We clearly see that $P_y=Q_x$ which means the field is conservative!" It is not true that if $P_y=Q_x$, then the field is conservative. There are topological obstructions (depending on the "shape" of the domain) that prevent the vanishing of the 2D "curl" $P_y-Q_x$ from entailing conservativity. $\endgroup$ – symplectomorphic Aug 8 '17 at 18:42

The field is locally conservative, meaning that around any point in its domain there is a small neighborhood on which the field is indeed conservative. This is also known as curl-free, for the simple reason that the curl of the field is zero. However, once there is a hole in the domain you open the door for larger-structure non-conservativeness, just like you discovered here.

  • $\begingroup$ So when asked whether a field is conservative we should be thinking of the domain as the whole of the axes and disregard instances of "local conservativeness"? $\endgroup$ – Yos Aug 8 '17 at 18:45
  • 1
    $\begingroup$ If it's not locally conservative, then it can't be conservative, so it still might be worthwhile to check. However, if it is locally conservative it might not be conservative. It's just that you now only have to check around the holes in the domain because that's the only way a locally conservative field may fail to be conservative. $\endgroup$ – Arthur Aug 8 '17 at 18:51

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