Say there is a sphere on which there is an ant and the ant wants to go to another point. The ant can't definitely travel through the sphere. So it has to travel along a curve. My question is what is the least distance between the two points i.e. distance between 2 points on a sphere.
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These curves are called geodesics, and found using the Christoffel equations.Those are part of great circles, and you can compute the length using integrals. – mich95 May 29 '15 at 16:07
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Can you tell me where to find these equations. – Abhijith S. Raj May 29 '15 at 16:10
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Or can you tell me where I can learn about them. – Abhijith S. Raj May 29 '15 at 16:12
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Learning them alone is, in my opinion, pretty hard if you have not taken any differential geometry class. There are plenty of books. The one I used is Differential Geometry for Curves and Surfaces by Manfredo Do Carmo. – mich95 May 29 '15 at 16:14
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I'll check it out. Thankyou. – Abhijith S. Raj May 29 '15 at 16:16
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4Calculating arc lengths of geodesics would be crushing a peanut with a steamroller. :) – Andrew D. Hwang May 29 '15 at 16:44
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If $a = (a_{1}, a_{2}, a_{3})$ and $b = (b_{1}, b_{2}, b_{3})$ are points on a sphere of radius $r > 0$ centered at the origin of Euclidean $3$-space, the distance from $a$ to $b$ along the surface of the sphere is $$ d(a, b) = r \arccos\left(\frac{a \cdot b}{r^{2}}\right) = r \arccos\left(\frac{a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}}{r^{2}}\right). $$ To see this, consider the plane through $a$, $b$, and the origin. If $\theta$ is the angle between the vectors $a$ and $b$, then $a \cdot b = r^{2} \cos\theta$, and the short arc joining $a$ and $b$ has length $r\theta$.
Andrew D. Hwang
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I am sorry if my question is stupid, but if we consider the exact same scenario in general dimension n >= 2, does this formula also apply? Also, if we change the norm with respect to which we consider the sphere, for example we consider the spheres associated with the L_p-norms, can we still obtain an explicit formula? – Akurishen Jul 08 '24 at 09:24
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1@Akurishen (i) Yes, the same formula works in arbitrary dimension (because the center of the sphere and two points determine a plane). (ii) This does not give a distance formula on a ball in the $p$-norm. Offhand, except in special cases ($p = 1$, $2$, and $\infty$), I expect the distance between points on a ball in the $p$-norm is not an elementary function (in the technical sense) of the points' coordinates. – Andrew D. Hwang Jul 09 '24 at 12:09
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After practicing with Sine/Cosine Rules of spherical Trig and getting their flavor I find ready-made
useful, especially to check special cases.
EDIT1:
The Clairauts Law of geodesics says
$ r \sin \beta = a \sin \lambda $
where r is radius, $a$ sphere radius, $\beta$ is path's angle to meridian, and $ \lambda $ is co-latitude.
Narasimham
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Distances can be calculated using the metric tensor from differential geometry, see:
Metric Tensor on Sphere: $ g_{ij} = \text{diag}(R^2, R^2 \sin^2 \theta) $
Line Element on Sphere: $ ds^2 = R^2 d\theta^2 + R^2 \sin^2 \theta d\phi^2 $
Where ds represents infinitessimal displacements. R, theta, psi are the spherical coordinates, and the formula for the line element is the sum of the coordinate differentials squared times the diagonal entries of the metric.
Hope this helps!
Ishan Jha
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