1
$\begingroup$

Consider the spherical triangle $\mathcal{P}$ with vertices $P_1 = (1,0,0)$, $P_2 = (0,1,0)$ and $P_3 = (1/\sqrt{3}, 1/\sqrt{3},1/\sqrt{3})$. Find the angles $\phi_1, \phi_2, \phi_3$ of $\mathcal{P}$ at $P_1, P_2, P_3$ respectively.

My first question: what's the relation between cartesian to spherical coordinate change $(x,y, z)$ $= (r\sin \theta \cos \phi, r \sin\theta \sin \phi, r \cos \theta)$ and angles of the spherical triangle?

How do I find $\phi_1, \phi_2, \phi_3$? I've calculated $\theta$ from $r\cos{\theta} = 0$ which resulted in $\theta = \pi/2$.

But I suspect my $\theta$ and $\phi$ may not be related to $\phi_1, \phi_2, \phi_3$. Is this the case? Thank you.

I think I know how to find the area after I've found these angles (regarding the title).

$\endgroup$
4
  • $\begingroup$ If you aren't required to use spherical coords, you could simply compute $$\cos\angle P_1OP_2 = P_1\cdot P_2 \qquad \cos\angle P_2OP_3 = P_2\cdot P_3 \qquad \cos\angle P_3OP_1 = P_3\cdot P_1$$ (The fact that all the vectors have unit length simplifies things!) That effectively gives $\angle P_1OP_2$, etc, the sides of the triangle. From there, use the spherical law of cosines to get the angles. (Some values here are "obvious", such as $\angle P_1OP_2=\pi/2$. However, the approach I describe works even with not-so-nice coordinates.) $\endgroup$
    – Blue
    Aug 5 '18 at 6:55
  • $\begingroup$ You can find the lengths of the three sides and then the angles from the Cosine rule. $\endgroup$
    – farruhota
    Aug 5 '18 at 6:58
  • $\begingroup$ If Q = (0,0,1) then P1P2Q is an equilateral right-angled triangle with P3 as its midpoint. $\endgroup$ Aug 5 '18 at 7:56
  • $\begingroup$ @blue Thank you. Indeed I'm not required to use spherical coordinates. I'm confused by what's what in the Wikipedia article, though, because there are sines in that formula. Do you mind showing me how to find one of $\phi_1, \phi_2, \phi_3$ as an example? $\endgroup$
    – Notsredt
    Aug 5 '18 at 7:59
1
$\begingroup$

As mentioned in my comment, the specific points used in this problem make various values "obvious". However, I'll crunch through a computational solution without leveraging that obvious-ness (but see below), because it's good to have ways to proceed when obvious things don't seem so obvious.


To recap a bit of spherical geometry ... Side $P_2P_3$ of the spherical triangle $\triangle P_1P_2P_3$ is the great circle arc subtending central angle $\angle P_2OP_3$. Since the sphere (and the great circle) has unit radius, arc-length and central-angle-measure are interchangeable; therefore, we can calculate the cosines of the sides of the triangle with dot products. (In what follows, I define $p_1 := |P_2P_3|$, etc.)

$$\begin{align} \cos p_1 = \cos\angle P_2OP_3 = P_2\cdot P_3 &= 1/\sqrt{3} = \cos p_2 \\[4pt] \cos p_3 = \cos\angle P_1OP_2 = P_1\cdot P_2 &= 0 \end{align}$$ Since $\sin^2\theta + \cos^2\theta = 1$ (and "knowing" that sine is non-negative for the angles in question), we may calculate $$\sin p_1 = \sin p_2 = \sqrt{1-\left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\sqrt{2}}{\sqrt{3}} \qquad\qquad \sin p_3 = 1$$

From here, we leverage the spherical law of cosines. For instance, $$\cos\phi_1 = \frac{\cos p_1 - \cos p_2\cos p_3}{\sin p_2 \sin p_3} = \frac{\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}\cdot 0}{\frac{\sqrt{2}}{\sqrt{3}}\cdot 1} = \frac{1}{\sqrt2}$$ $$\cos\phi_2 = \frac{\cos p_2 - \cos p_3\cos p_1}{\sin p_3 \sin p_1} = \frac{\frac{1}{\sqrt{3}}-0\cdot\frac{1}{\sqrt{3}}}{1\cdot\frac{\sqrt{2}}{\sqrt{3}}} = \frac{1}{\sqrt2}$$ $$\cos\phi_3 = \frac{\cos p_3 - \cos p_1\cos p_2}{\sin p_1 \sin p_2} = \frac{0-\frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{3}}}{\frac{\sqrt{2}}{\sqrt{3}}\cdot \frac{\sqrt{2}}{\sqrt{3}}} = -\frac12$$

From here, we see that $\phi_1$, $\phi_2$, $\phi_3$ are the "special" angles $\frac14\pi$, $\frac14\pi$, $\frac23\pi$. (The reader can compute the triangle's area from these.) It's no coincidence that two of these are half-right angles, while the last is one-third of a full turn; this is, in fact, a consequence of the "obvious-ness" I ignored from the beginning.

As hinted in @MichealBehrend's comment, if we introduce a fourth point ---I'll call it $P_4 = (0,0,1)$--- then $P_1$, $P_2$, $P_4$ are the vertices of an equilateral spherical triangle that fills a full octant of the sphere; each angle of this spherical triangle has measure $\frac12 \pi$ (a fact we'll use shortly). The point $P_3$, as with any point of the form $(k,k,k)$, lies on the line through the origin and the "center" of that octant, so that $P_3$ is the "center" of the triangle.

  • By three-fold rotational symmetry, we expect the angles formed by joining $P_3$ to the vertices of the triangle to be congruent; thus, each of them (including $\phi_3$) is one-third of a full turn.

  • Reflective symmetry implies that center $P_3$ should lie on the angle bisectors at $P_1$, $P_2$, $P_4$. Thus, for instance, $\phi_1 := \angle P_3P_1P_2$ should be half of $\angle P_4P_1P_2$, but that latter angle is a right angle; consequently, $\phi_1 = \frac14\pi$. (Likewise for $\phi_2$.)

This gives the various angles, from which the area of $\triangle P_1P_2P_3$ can be computed. However, there, too, is an "obvious" solution: that triangle is one-third of the full-octant triangle $\triangle P_1P_2P_4$, and the full-octant triangle has measure $4\pi/8 = \pi/2$. Therefore, $|\triangle P_1P_2P_3|= \pi/6$, which agrees with the "spherical excess" calculation $\phi_1+\phi_2+\phi_3-\pi$.

$\endgroup$
1
  • $\begingroup$ It's amazing how much I've learned from this post! Thank you very much! $\endgroup$
    – Notsredt
    Aug 5 '18 at 11:01

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.