I'm studying geodesics in my differential geometry course. We derived and proved Clairaut's relation (see also Shifrin P73.), which states that for a surface of revolution, the geodesics satisfy the equation $$r\cos\phi=\text{const}$$ where $r$ is the distance from the axis of revolution and $\phi$ is the angle between the geodesic and the parallel.

We have not done any spherical trig. so far, in particular, we haven't been introduced to the law of cosines.

However, following the section on Clairaut, an exercise asks for a proof of the spherical law of sines $$\frac{\sin\alpha}{\sin a}=\frac{\sin\beta}{\sin b}=\frac{\sin\gamma}{\sin c}.$$ The hint is to put one corner of the triangle on a pole, then use Clairaut's relation. If we put one vertex on the pole, then two edges will be longitudinal, and thus parallels will intersect those edges at right angles. Right triangles are much easier to handle, but the parallels "cut off" the angle from its opposing side. Is something here perhaps?

As most sine-law proofs I've seen use the spherical law of cosines, I thought maybe I should prove that using the hint, then derive the sine-law from that. But then I might as well go through the standard derivation of the law of cosines, which is a.s. not the point of the exercise.

Is it possible to use Clairaut's relation to show the spherical law of sines more directly? If not, is it possible to use it to derive the cosine-law?

  • $\begingroup$ Interesting exercise (not mine, although you referenced my book). Clairaut refers to the angle between the geodesic and the parallel, and the angle you're looking at will be the complementary angle (with the vertical geodesic), so it looks quite promising. $\endgroup$ – Ted Shifrin Jan 21 at 20:04
  • $\begingroup$ On the other hand, perhaps relevant is my Exercise 19 on p. 77 to use Clairaut to show that all great circles are in fact geodesics. $\endgroup$ – Ted Shifrin Jan 21 at 20:45
  • $\begingroup$ @TedShifrin Actually we don't have a textbook, but I keep yours around because I find it very helpful! Could you perhaps suggest a line of inquiry? I'm starting to get really stuck (I'm confusing myself more than progressing). $\endgroup$ – Ruben Kruepper Jan 21 at 20:57
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    $\begingroup$ Put angle $\alpha$ at the north pole, and use Clairaut on the opposite side to show $\dfrac{\sin\beta}{\sin b}=\dfrac{\sin\gamma}{\sin c}$. Clairaut on this opposite side tells $r_1\cos\phi_1 = r_2\cos\phi_2$ at the two vertices. Relate $r_1$ and $r_2$ to $b$ and $c$ and relate $\phi_1$ and $\phi_2$ to $\beta$ and $\gamma$, as I already suggested. ... Then permutation of variables gets the remaining equality (using Clairaut on the longitudes isn't very helpful). $\endgroup$ – Ted Shifrin Jan 21 at 21:56
  • $\begingroup$ Did you figure it out? :) $\endgroup$ – Ted Shifrin Jan 22 at 19:28

Following the hint and @TedShifrin's comment, place $\alpha$ on the north pole, then we may apply Clairaut on the side $a$, which is part of a geodesic. In particular, we may consider the intersections of $a$ with the parallels at the corners $\beta,\gamma$. Let $\theta_\beta, \theta_\gamma$ denote these angles and $r_\beta, r_\gamma$ denote the distance from the axis, we know $$r_\beta\cos\theta_\beta=r_\gamma\cos\theta_\gamma$$

Wlog we may assume that $\beta\le\gamma$, so because parallels intersect longitudinal geodesics at right angles, we have $$\frac\pi2=\theta_\beta+\beta=\theta_\gamma-\gamma$$ so $\cos\theta_\beta=\sin\beta, \cos\theta_\gamma=-\sin\gamma$. Furthermore, using some elementary properties of circle-sections with the lengths $\lvert b\rvert, \lvert c\rvert$, we must have $r_\beta=\sin\lvert c\rvert, r_\gamma=\sin\lvert b\rvert$. The claim then follows when we release the orientation of angles and plug these results into the first equation.


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