Geodesics of the 2-sphere in terms of the arc length

In summary: So our final conclusion is that the straight-line distance between two points on the sphere is equal to the length of the shortest path between those points that passes through the center of the sphere.In summary, the arc length parameterization of the 2-sphere geodesics is given by:$$\cos(\theta)=\sqrt{1-K^2}\cdot \sin(\frac{s}{R})$$$$\tan(\phi+\frac{\pi}{3}-n \cdot \pi)=\sqrt{\frac{3}{7}} \cdot \tan(\frac{s}{R})$$What respects to the limits of integration:$$P_{1}=(\frac{\pi
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Adrian555
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Parametrize the geodesics of a 2-sphere in terms of the arc lenght.
I'm trying to evaluate the arc length between two points on a 2-sphere.

The geodesic equation of a 2-sphere is:

$$\cot(\theta)=\sqrt{\frac{1-K^2}{K^2}}\cdot \sin(\phi-\phi_{0})$$

According to this article:http://vixra.org/pdf/1404.0016v1.pdfthe arc length parameterization of the 2-sphere geodesics is given by:

$$\cos(\theta)=\sqrt{1-K^2}\cdot \sin(\frac{s}{R})$$

$$\tan(\phi-\phi_{0})=K \cdot \tan(\frac{s}{R})$$However, when I evaluate the integral:

$$s=\int_{s_{1}}^{s_{2}}{ds}=s_{2}-s_{1}$$I don't obtain the right solution. Is there another way of parametrize the geodesics of a 2-sphere in terms of the arc lenght?
 
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I haven’t looked at the linked article yet, but you should be aware that vixra is generally an unreliable source - in fact, the forum rules specifically disallow using it.

It’s likely that one of our other members will be able to point you towards a better starting point.
 
  • #3
The rotation group ##SO(3)## acts by isometries on ##S^2##. Given two points on the sphere, you could find a rotation that takes them both to the equator, and measure distance there, where the computation is much easier.
 
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Nugatory said:
I haven’t looked at the linked article yet, but you should be aware that vixra is generally an unreliable source - in fact, the forum rules specifically disallow using it.

It’s likely that one of our other members will be able to point you towards a better starting point.

Thanks for your answer. The problem is that I have two points in spherical coordinates:

$$P_{1}=(\frac{\pi}{4},0)$$

$$P_{2}=(\frac{\pi}{3},\frac{\pi}{2})$$

The great circle which passes through these two points is:

$$\cot(\theta)=\frac{2}{\sqrt{3}}\cdot \sin(\phi+\frac{\pi}{3}-n \cdot \pi)$$

The parametric equations are:

$$\cos(\theta)=\frac{2}{\sqrt{7}}\cdot \sin(\frac{s}{R})$$

$$\tan(\phi+\frac{\pi}{3}-n \cdot \pi)=\sqrt{\frac{3}{7}} \cdot \tan(\frac{s}{R})$$

What respects to the limits of integration:

$$P_{1}=(\frac{\pi}{4},0) --> s=1,209$$
$$P_{2}=(\frac{\pi}{3},\frac{\pi}{2}) --> s=0,7227$$

And the integral I evaluate is:

$$s=\int_{1,209}^{0,7227}{ds}=0,7227-1,209$$
 
  • #5
I'll assume that your 2-sphere is the round 2-sphere (the locus of all points in 3-space whose distance from the origin is equal to 1). Now suppose you have calculated that the straight-line distance between two points of the sphere (as points of 3-space) — the usual square-root of the sum of the squared differences of corresponding coordinates — is equal to D. Using basic trigonometry, you can figure that D = 2 sin(θ/2), where θ is the angle between the radii of the sphere that end at each of the two points. (It's a good exercise to prove this.) From that, we conclude that θ = 2 arcsin(D/2). (Note that θ will be in the interval [0, π].) Now recall that the geodesic distance along a unit sphere is exactly that angle θ.
 
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