Consider the picture below. I have a sphere of radius $r$, centered at $C$. The angle $\varphi$ is the dihedral angle between the plane defined by the shaded area and a plane through the indicated diameter with a normal coinciding with the bisector of $\theta$ when the planes are orthogonal (I think it is intuitively clear what this angle defines, even though the precise statement is a little long).

The lune on which the arc $AB$ is located has an area of $2r^2\theta_{\varphi=\pi/2}$, where $\theta_{\varphi=\pi/2}$ is the angle $\theta$ at $\varphi=\pi/2$. The angle $\theta$ is related to $AB$ by length$(AB) = r\theta$, but I would like to know the relation between $\theta$ and $\varphi$, once $\theta_{\varphi=\pi/2}$ has been fixed.


This is essentially so I can integrate some function from $\varphi=0$ to $\varphi=\pi$ to get the area of the lune. I need that to get an alternative definition, by integration, of the surface area of the $n$-sphere.

Edit: If it is not clear, the arcs from the north pole (topmost emphasized point) to $A$ and to $B$ are the same length, and $AB$ is a segment of a great circle (i.e. a geodesic).

  • $\begingroup$ So $\theta_{\varphi=\pi/2}$ is the angle between the great circles containing $A$ and $B$? $\varphi$ is the angle between the pole and $A$ and $B$? $\endgroup$ – robjohn May 9 '13 at 18:22
  • $\begingroup$ It might be simpler to define $\varphi$ as the angle $NC\frac{A+B}2$ where $N$ is the pole. $\endgroup$ – Peter Taylor May 9 '13 at 18:24
  • $\begingroup$ @robjohn Yes, that's a better description $\endgroup$ – Jānis Lazovskis May 9 '13 at 20:07
  • $\begingroup$ @robjohn Regarding your answer below, the side $\varphi$ should be $\varphi r$, and $\theta/2$ should be $\theta r/2$. I am assuming (though not 100% sure) that will change your relation accordingly. Is that correct? $\endgroup$ – Jānis Lazovskis May 9 '13 at 20:20
  • $\begingroup$ @JimboBimbo: In spherical trigonometry, sides are measured by the angle they subtend at the center of the sphere. While it is true that the linear length of the side is $\varphi r$ where $r$ is the radius of the sphere, for use in spherical trig formulas, the side has angular length $\varphi$. $\endgroup$ – robjohn May 9 '13 at 20:24

Assuming that $\theta_{\varphi=\pi/2}$ is the angle between the great circle containing $N$ and $A$ and the great circle containing $N$ and $B$, and that $\varphi$ is the angle between the north pole and the great circle containing $A$ and $B$, then $\triangle N(\frac{A+B}{2})B$ is a right spherical triangle with the right angle at $\frac{A+B}{2}$. The basic relations for right spherical triangles yield $$ \tan\left(\frac{\theta_{\varphi=\pi/2}}{2}\right)\sin(\varphi)=\tan\left(\frac{\theta}{2}\right) $$ $\hspace{3.2cm}$enter image description here


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