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

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SUMMARY

The discussion focuses on evaluating the arc length between two points on a 2-sphere using geodesic equations. The geodesic equation is defined as $$\cot(\theta)=\sqrt{\frac{1-K^2}{K^2}}\cdot \sin(\phi-\phi_{0})$$, and the arc length parameterization is given by $$\cos(\theta)=\sqrt{1-K^2}\cdot \sin(\frac{s}{R})$$ and $$\tan(\phi-\phi_{0})=K \cdot \tan(\frac{s}{R})$$. The user encounters difficulties in integrating the arc length between the points $$P_{1}=(\frac{\pi}{4},0)$$ and $$P_{2}=(\frac{\pi}{3},\frac{\pi}{2})$$, leading to a request for alternative parametrization methods. The discussion suggests using the rotation group ##SO(3)## to simplify calculations by transforming points to the equator.

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  • Knowledge of integral calculus for evaluating arc lengths.
  • Basic trigonometry, particularly involving sine and cosine functions.
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  • Research the properties of the rotation group ##SO(3)## and its application to spherical geometry.
  • Study the derivation and application of geodesic equations on a 2-sphere.
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TL;DR
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.
 
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:

$$s=\int_{1,209}^{0,7227}{ds}=0,7227-1,209$$
 
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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