Trying to evaluate a solid angle of a spherical cap with Stokes theorem:

$$ \begin{gathered} \int_\Omega \frac{\hat r}{r^2}\cdot d\vec\omega\\ A = - \frac{\cot\theta}{r} \hat\phi\\ \nabla \times A = \frac{\hat r}{r^2} \end{gathered} $$

Since in this case $A$ is not continuous at $\theta=0$, I've considered a second reversed inner loop taken to the limit of 0, and then obtained: $$ \int_\Omega\frac{\hat r}{r^2}\cdot d\vec\omega = \int_\gamma -\frac{\cot\theta}{r} \cdot r\sin\theta\space d\phi - \int_{\gamma_0} A\cdot d\vec\gamma = -2\pi\cos\theta + 2\pi $$ This is the normal solid angle formula for the cap of a sphere.

Then I tried with a different $A$: $$ \begin{gathered} A = -\phi\frac{\sin\theta}{r}\space\hat\theta\\ \nabla\times A = \frac{\hat r}{r^2} \end{gathered} $$

But the tangent vector of the loop $\gamma$ has only $\hat\phi$ component, which means: $$ A\cdot d\vec\gamma = 0 $$ So the whole integral evaluates to $0$.

Am I missing something simple here?


Resolved by achille hui in the comments. $A$ is not continuous in the second case too because it jumps a value when $\phi$ turns from $2\pi$ to $0$. The cap is a multivalued point for $A$ as well. Both issues are solved simultaneously by making a small loop near the cap, then breaking both loops at $\phi'$ near 0 and then stitching them together instead along $\hat\theta$. So the integral is now non-zero and evaluates to the correct formula (in the limit of $\phi'\rightarrow0$, the angle at which the loops were broken).

  • $\begingroup$ Your second $A$ is not single valued on the cap $\theta \le \theta_0$. In particular, it has a jump type discontinuity at the cut $\phi = 0$ / $\phi = 2\pi$. $\endgroup$ – achille hui Oct 29 '17 at 15:13
  • $\begingroup$ Indeed, I missed that cut! Now that I've split the loop even more (in order to not have the cut), it works quite nicely! Thanks a lot! Can comments be upvoted? $\endgroup$ – user2941240 Oct 29 '17 at 15:37

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