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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 $\mathbb R^3$?2011-05-17
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    Why do you want to do this? It would be helpful if you could explain your motivation for trying to do this calculation.2011-05-17
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    @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 me2011-05-17

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