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Green's theorem is

$$\oint_{\partial D} (P\, dx+Q\, dy) = \iint_D dx\,dy \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$$

where one can see that the RHS is asymmetric in $x$ and $y$. Why is this, and what is the physical significance?

I suspect the answer has to do with using a right-handed coordinate system (e.g. one can use Stokes' theorem to show the above, which involves a vector product, which is right-handed), or possibly that the boundary is traversed anti-clockwise, but I'm unable to make any deeper or more accurate statement than that...

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    $\begingroup$ Yes, that's basically it. You can take the integral of a differential form only on an oriented curve, and to orient the boundary of $D$ you need to use the orientation of the ambient space, which gives you the anti-clockwise orientation. If you had taken the clockwise orientation you would get the opposite formula. (Hopefully someone can write an answer more precise than my comment.) $\endgroup$ Apr 8, 2015 at 13:40
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    $\begingroup$ Talking out of my a** here, but I think the reason is quite deep and has to do with wedge products. You'll have to explore far and yonder into differential geometry to make a "deep and accurate statement" I fear. I'm still at the beginning and this is all I can say. $\endgroup$
    – GPerez
    Apr 8, 2015 at 13:43
  • $\begingroup$ The left side is also asymmetric in $x$ and $y$. See my answer below. $\endgroup$ Apr 8, 2015 at 14:12
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    $\begingroup$ Not "asymmetric" but "antisymmetric" I would say. Switching $x$ and $y$ produces a sign change. (This is ok since this transformation changes orientation of space, thus changing sign in the left hand side as well. All of this is explained very clearly in other answers). $\endgroup$ Apr 8, 2015 at 14:28

3 Answers 3

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Both the left side and the right side are asymmetric in $x$ and $y$: The boundary $\partial D$ goes around $D$ in a particular direction and not in the other direction.

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The lack of symmetry is only apparent. Green's formula is a special case of what's known as the (general) Stokes's Theorem, which you should look up in Wikipedia for starters: but think of it like this, if you exchange the coordinates $x$ with $y$ you must also exchange the vector fields components $P$ with $Q$ to be consistent.

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    $\begingroup$ (+1) It's worth pointing out that that exchanging $x$ and $y$ also reverses the orientation $\partial D$, which reverses the sign of the l.h.s. of the formula given in the theorem. This is necessary, as exchanging $P \leftrightarrow Q$ and $x \leftrightarrow y$ reverses the sign of the r.h.s. $\endgroup$ Apr 8, 2015 at 13:55
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It is linked with the orientation of space. Denoting $\omega$ the differential form $\,P\mathrm d\mkern1.5mu x+Q\mathrm d\mkern1.5mu y$, Green-Riemann's formula can be written: $$\int_{\partial D} \omega= \int_D \mathrm d\mkern1.5mu\omega $$ where $\,\mathrm d\mkern1.5mu\omega$ is the exterior differential of $\omega$, defined by: \begin{align*}\mathrm d\mkern1.5mu (P\,\mathrm d\mkern1.5mu x+Q\,\mathrm d\mkern1.5mu y)&= \mathrm d\mkern1.5mu P\wedge \mathrm d\mkern1.5mu x+\mathrm d\mkern1.5mu Q\wedge\mathrm d\mkern1.5mu y\\ &=\Bigl(\frac{\partial P}{\partial x}\mathrm d\mkern1.5mu x+ \frac{\partial P}{\partial y}\mathrm d\mkern1.5mu y\Bigr)\wedge\mathrm d\mkern1.5mu x +\Bigl(\frac{\partial Q}{\partial x}\mathrm d\mkern1.5mu x+ \frac{\partial Q}{\partial y}\mathrm d\mkern1.5mu y\Bigr)\wedge\mathrm d\mkern1.5mu y \\ & = \frac{\partial P}{\partial y}\mathrm d\mkern1.5mu y\wedge\mathrm d\mkern1.5mu x+ \frac{\partial Q}{\partial x}\mathrm d\mkern1.5mu x\wedge\mathrm d\mkern1.5mu y = \Bigl(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Bigr)\mathrm d\mkern1.5mu x\wedge\mathrm d\mkern1.5mu y \end{align*}

since $\,\mathrm d\mkern1.5mu x\wedge\mathrm d\mkern1.5mu x=\mathrm d\mkern1.5mu y\wedge\mathrm d\mkern1.5mu y=0\,$ and $\,\mathrm d\mkern1.5mu y\wedge\mathrm d\mkern1.5mu x=-\mathrm d\mkern1.5mu x\wedge\mathrm d\mkern1.5mu y$.

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