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I seem to have this seemingly trivial Problem, but can't figure it out.

Situation: I have my Unit Sphere, $S^2$ defined as a Riemannian Manifold. Parameterizing Geodesics (Great circles) on this sphere is absolutely no Problem, IF I embed it in $\mathbb{R}^3$. BUT, I don't want to do this.

What I really want to do, is use riemannian polar coordinates. Now genereally, this isn't a problem, as I could just take any point on the geodesic and call it $(0,0)$ (with $(r,\varphi)$ being my riemannian polar coordinates). And all the lines going through the origin would be my great circles.

However, The origin of the exponential map I'm using does not lie on the geodesic of interest. I do however know 2 points (say $P_1, P_2$) and their polar coordinates.

So basically, I have $\text{exp}_p^{-1} : S^2\to \mathbb{R}^2$ and my two points in $\mathbb{R}^2$ defined by $(r_1,\varphi_1)$ and $(r_2,\varphi_2)$.

I'm looking for some parameterisation to connect these two points via a great circle. And I just can't think of anything sensible.

Any help would be highly appreciated.

Edit: To clarify, I have a working method using an $\mathbb{R}^3$ embedding, but it's very bulky and "unpretty".

Further, I'm using this geodesic to "bound" one of my integrals and shifting the Origin is therefore not an option

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I'm not sure I understand exactly what you mean by not wanting to embed $S^2$ in $\mathbb R^3$. You can transform your two points to coordinates in $\mathbb R^3$, find an appropriate rotation to rotate, say, the $x$-$y$-plane into the plane of these points, apply that rotation to the unit circle in the $x$-$y$-plane, and then transform the resulting coordinates back into your polar coordinates. The result would contain no traces of $\mathbb R^3$ and would give you the correct intrinsic coordinates. Are you saying that you want to avoid such an intermediate detour via $\mathbb R^3$? –  joriki May 17 '11 at 11:07
    
Why do you want to do this? It would be helpful if you could explain your motivation for trying to do this calculation. –  Zhen Lin May 17 '11 at 11:23
    
@joriki, yes, I'm currently trying to avoid the $\mathbb{R}^3$ detour. Currently I'm using that method, but it's far from elegant. @Zhen, my motivation is, that I'm not able to simplify my current approach. It's rather "bulky", and I simply can't work with it. The Formula I'm using takes up roughly an A3-Sheet of paper. I'm hoping a different approach will help me –  user11008 May 17 '11 at 13:27

2 Answers 2

Why don't you just use spherical trigonometry? You have a "base triangle" with vertices $O$, $A_1$ and $A_2$, sides $O A_i$ of length $r_i$ and an angle $\alpha:=|\phi_2-\phi_1|$ between them. Now consider an arbitrary point $P$ on the third side $A_1 A_2$. The "line" $OP\ $ has length $r$ and encloses angles $\alpha_i$ with the sides $OA_i$. The formulas of spherical trigonometry will give you an equation connecting $r$ and the $\alpha_i$.

Let $s_i$ be the lengths of the sides $A_i P$. Then $$\cos s_i=\cos r_i \cos r +\sin r_i \sin r \cos\alpha_i\qquad(i=1,2)$$ and $$\cos(s_1+s_2)=\cos r_1\cos r_2+\sin r_1\sin r_2 \cos(\alpha_1+\alpha_2)\ .$$ Eliminate $s_1$ and $s_2$ from these three equations to get the desired result. Hint: Square the identity $\sin s_1 \sin s_2=\cos s_1 \cos s_2-\cos(s_1+s_2)$ and replace $\sin^2 s_i$ by $1-\cos^2 s_i$.

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Thanks! While it does give me some ideas, I still have to find a function that tells me $r$ in dependence of $\alpha_i$. Which brings me to my original problem. I've been looking at the Trigonometry Formulas, but couldn't figure it out, do you know how to do this? –  user11008 May 17 '11 at 17:46

Neither approach I propose here are pleasant, but it demonstrates that it can be done in principle.

You can write down an explicit expression for the exponential map in coordinates $(r, \theta) \mapsto (x, y, z)$. If we put the origin of Riemannian polar coordinates at $O = (0, 0, 1)$, then we have $(x, y, z) = (\cos \theta \sin r, \sin \theta \sin r, \cos r)$. You can then try to invert this to obtain expressions for great circles in $(r, \theta)$.

A slightly more clever approach would be to use stereographic projection $\pi : S^2 \to \mathbb{R}^2 \cup \{ \infty \}$: $$\displaystyle (u, v) = \left( \frac{x}{1-z}, \frac{y}{1-z} \right)$$ So changing to Riemannian polar coordinates, $$\displaystyle (u, v) = \left( \frac{\sin r \cos \theta}{1 - \cos r}, \frac{\sin r \sin \theta}{1 - \cos r} \right)$$ we see that $$\displaystyle \sqrt{u^2 + v^2} = \frac{\sin r}{1 - \cos r}$$ and the polar angle in the $(u, v)$ plane is equal to $\theta$. We know what the action of $\text{SO}(3)$ on the sphere corresponds to in the $(u, v)$ plane (namely certain Möbius transformations), so we can write down expressions for rotations of the sphere in terms of $(r, \theta)$. (These will probably be quite ugly.) This then allows us to reduce to the case where $P_1 = O$, so we can apply the inverse rotation to get the expression for a great circle through $P_1$ and $P_2$ in Riemannian polar coordinates centred at $O$.

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Thanks for the Input. I've been using a similar approach, and as you said, it isn't very pretty. However, there is a reason I can't simply transform my map so that $P_1 = 0$. I'm using such a geodesic as a boundary for an integral (Integrating over a geodesic triangle). I'm already sitting on one Point ($P_3$ in this case) of this triangle, transforming it would sadly only shift my problem. –  user11008 May 17 '11 at 13:32

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