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Define a function $f : \mathbb{S}^2\times \mathbb{S}^2\times \mathbb{S}^2\rightarrow \mathbb{R}$ by $f(p,q,r)$ to be a area of the geodesic triangle $pqr$ in the unit sphere $\mathbb{S}^2$.

(Here the area of the triangle $pqr$ is $\angle p + \angle q +\angle r -\pi$).

Question 1 : Here how can we find $$ \int_{\mathbb{S}^2\times \mathbb{S}^2\times \mathbb{S}^2 } \ f(p,q,r) d{\rm Vol}(p,q,r) $$

Question 2 : Furthermore, how can we find the following $$\int_{\mathbb{S}^2 } \ f(p_0,q_0,r) d{\rm Vol}(r) $$

Question 3 : Consider the triangle $\Delta pqr$ where the order of $p,\ q,\ r$ is positively oriented. And define $$A = \frac{ q\times p }{| q\times p| },\ B = \frac{r\times q}{|r\times q|},\ C = \frac{p\times r}{|p\times r|} $$

Note that $\angle \ (A,B) =\pi-\angle p $ so that $$ \angle (A,B) + \angle (B,C)+\angle (A,C) = 2\pi - {\rm Area}\ \Delta pqr $$

Hence ${\rm perim}\ \Delta ABC$, the sum of all side lengths in $\Delta ABC$, is equal to $2\pi - {\rm Area}\ \Delta pqr$.

Hence we want to calculate $$ \int_{(A,B,C)\in \mathbb{S}^2 \times\mathbb{S}^2\times \mathbb{S}^2}\ {\rm perim}\ \Delta ABC \ d{\rm Vol}_{ (A,B,C) }$$

HK Lee
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  • Choose $p$ arbitrarily, then the probability $q$ is distance $t$ from $p$ is $\frac{1}{2}\sin(t)$ (i.e. the volume of a ball is $\int_0^R 2\pi \sin(t), dt$). Then the probability that $r$ is distance $s$ from the geodesic through $p$ and $q$ is $\frac{1}{2}\sin(\frac{\pi}{2}-s)$. Does that help? – 1Rock Jan 18 '21 at 03:14
  • I can't shake the feeling that Gauss-Bonnet might be useful somehow... – Ivo Terek Jan 18 '21 at 05:12
  • @1Rock : I can not understand "the probability $q$ is distance $t$ from $p$ is $\frac{1}{2}\sin\ (t)$" Is it $0$ ? – HK Lee Jan 19 '21 at 12:03
  • @Ivo Terek : How can we use the Gauss-Bonnet theorem ? – HK Lee Jan 19 '21 at 12:04
  • @HK Lee: you're right, I mean the PDF is $\frac{1}{2} \sin(t)$. – 1Rock Jan 21 '21 at 02:33

1 Answers1

1

The area of a spherical triangle $ABC$ on a unit sphere with respective angles $\alpha,\beta,\gamma$ in radians has an extremely nice formula:

$$\text{Area}(\triangle ABC) = \alpha+\beta+\gamma-\pi$$

There is a very cute proof of this formula using some clever geometric arguments given here among other places.

Because this formula is linear in the angles, we can just use linearity of expectation: compute the average value $x$ of an angle in a spherical triangle, and our answer will be $x+x+x-\pi$.

Without loss of generality, let $A$ be at the north pole, and fix $B$ to be on the Prime Meridian. Then the longitude of $C$ will be uniformly distributed on $[-180,180]$, so the angle $\alpha$ is uniformly distributed between $0$ and $\pi$; in particular, its expectation is $\pi/2$. So the expected area is simply

$$3\cdot\frac{\pi}{2}-\pi = \frac{\pi}2$$

or $1/8$ of the sphere.

RavenclawPrefect
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  • Thank you for your answer But I can not understand the part 'Because this formula is linear in the angles, we can just use linearity of expectation' Is there other way ? or Could you elaborate your answer ? – HK Lee Jan 18 '21 at 11:28
  • Linearity of expectation means that the expected value of the sum of two or more random variables is equal to the sum of their expected values. So, the expected value of the area of the triangle, which is the expected value of (sum of the angles minus $\pi$), is equal to the expected value of the first angle plus the expected value of the second angle plus the expected value of the third angle minus $\pi$. Does that make sense? – RavenclawPrefect Jan 18 '21 at 19:53
  • Your argument may be right. But I can not check whether your approach to this problem is right or not. When we put three points having $order$ on a sphere, the average angle at the second point is $\frac{\pi}{2}$ ? – HK Lee Jan 18 '21 at 22:26
  • If we put three points on a sphere independently and at random, the expected angle at any of the vertices is the same by symmetry. Each of those expected angles is $\pi/2$. – RavenclawPrefect Jan 19 '21 at 14:13
  • Yeah. In fact, expectation for one angle for three point is $\frac{\pi}{2}$. I can show this through simple calculation. But we have three points and three angles. These are related. If we move one point, then all three angles are changed. So how do you guarantee the independence ? – HK Lee Jan 19 '21 at 14:22
  • The angles aren’t independent! But what we care about is the expected sum of the angles (minus $\pi$), and the expectation of the sum of random variables is equal to the sum of their expected values, even if they are not independent. – RavenclawPrefect Jan 19 '21 at 14:31
  • All I’m using here is linearity of expectation; see e.g. https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2005/readings/ln14.pdf for a detailed explanation with examples. Is this helpful? I’m still not sure if I’m addressing your concern. – RavenclawPrefect Jan 20 '21 at 15:17
  • I know that there is a material about linearity of expectation. But why can we use the linearity of expectation on the problem in OP ? – HK Lee Jan 20 '21 at 23:17
  • Because we're adding random variables (namely, the measures of each angle). Linearity of expectation applies to any situation in which one is taking the sum of random variables; it's just the statement that $\mathbb{E}[X+Y]=\mathbb{E}[X]+\mathbb{E}[Y]$. So we have $\mathbb{E}[\text{Area}] = \mathbb{E}[\alpha+\beta+\gamma-\pi] = \mathbb{E}[\alpha]+\mathbb{E}[\beta]+\mathbb{E}[\gamma]-\mathbb{E}[\pi]=\pi/2+\pi/2+\pi/2-\pi=\pi/2$. – RavenclawPrefect Jan 20 '21 at 23:53
  • I don't understand why $E [\alpha ] = \frac{\pi}{2}$ – HK Lee Jan 21 '21 at 00:21
  • This is the paragraph starting with "Without loss of generality" in my answer. If we orient our perspective so that $A$ is at the north pole, then the other two points are at a random "longitude" and have an expected difference in angle (relative to some fixed direction from $A$) of $90^\circ$. – RavenclawPrefect Jan 21 '21 at 00:41
  • As I already commented, when we find one angle, then we can show that the expectation angle is $\pi/2$ But we have to find three angles. In this case, why $E( \alpha+\beta+\gamma)=E(\alpha)+E(\beta)+E(\gamma)$ ? – HK Lee Jan 21 '21 at 01:04
  • Let us continue this discussion in chat. – RavenclawPrefect Jan 21 '21 at 01:05