1
$\begingroup$

Consider the differential form:

$$\omega=-\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy$$

I believe that this form is not exact. i.e: There does not exist a $\psi$ such that $d\psi=\omega$.

In particular, this is supposed to be true since the line integral along a loop, $$\int_{C}\omega = 2\pi \neq 0$$

But, what about $$\psi=-\arctan(\frac x{y})$$

According to Maple,

As you can see, Maple says that $d\psi=\omega$. So what's the deal here? Is $\omega$ exact or not?

Thank You!

$\endgroup$
  • 2
    $\begingroup$ The relevant concept is "exact on a given domain". Note that $\arctan(x/y)$ is defined in two half planes that don't contain the circle $C$. $\endgroup$ – Gribouillis Nov 20 '17 at 8:17
  • $\begingroup$ arctan is not a "single-valued" function. With the usual definitions it will have discontinuities on the punctured plane. $\endgroup$ – Lord Shark the Unknown Nov 20 '17 at 8:18
  • $\begingroup$ We see $$\omega=-\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy=\dfrac{xdy-ydx}{x^2 + y^2}dy=d\arctan\dfrac{y}{x}$$ this relation holds where $(x,y)\neq(0,0)$. $\endgroup$ – Nosrati Nov 20 '17 at 8:28
2
$\begingroup$

Your $\psi$ is just defined for $y\neq 0$. Therefore $d\psi=\omega$ holds on $A=\{(x,y)\in\mathbb R^2~:~ y>0\}$ and on $B=\{(x,y)\in\mathbb R^2~:~y<0\}$. You can conclude that for any closed curve $\gamma$ in $A$ or in $B$ you get $\int_\gamma \omega =0$. But your loop doesn't belong to $A$ or $B$. Moreover, you can conclude from $\int_C\omega\neq 0$ for a loop $C$ that there isn't any $\psi$ such that $d\psi=\omega$ on $\mathbb R^2\setminus\{(0,0)\}$.

$\endgroup$
1
$\begingroup$

The differential form $\omega$ is exact in some region where $\arctan(y/x)$, or more generally the phase $\operatorname{atan}(x,y)$, makes sense. For example, it is exact in the domain you get when you throw away the non-positive half of the $x$-axes.

However, it is not exact in the punctured plane $\Bbb{R}^2\setminus\{0,0\}$. That non-vanishing path integral around the closed loop proves that.

$\endgroup$
  • $\begingroup$ This is rather likely to be a duplicate, but I don't have the time to look for one. Switching this to CW while we wait to close. $\endgroup$ – Jyrki Lahtonen Nov 20 '17 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.