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I am trying to solve the geodesic equations on a unit sphere:

\begin{align} \frac{d^2 \theta}{d \lambda^2} - \sin\theta\cos\theta \frac{d \phi}{d \lambda} \frac{d \phi}{d \lambda} = 0 \\ \frac{d^2 \phi}{d \lambda^2} + 2 \cot\theta \frac{d \theta}{d \lambda} \frac{d \phi}{d \lambda} = 0 \\ \end{align}

The tangent vector I wish to transport is given by $(0,0,1)$ in the $(r,\theta, \phi)$ coordinate system at the initial point $(1,\pi/4, 0)$.

I'm not sure how to go about the problem. Since a great circle isn't traversered in the $\theta = \pi/4$ plane, the circle isn't a geodesic, as can be easily be easily checked. Does this mean I have to solve the two coupled differential equations to find the most general solution of the geodesic? Or is there a simple way to go about the problem given the initial point and the tangent vector?

Junaid Aftab
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  • You haven't made a Web search before. In 15 seconds, I found this document: see equations (19)-(20) etc... of (http://www.physicspages.com/2013/03/31/geodesic-equation-geodesics-on-a-sphere/) – Jean Marie May 08 '17 at 08:27
  • @Junaid: If $\theta$ denotes longitude (as it seems to from context, since $(0, 0, 1)$ is tangent to the sphere at $(1, \pi/4, 0)$), then $\theta = \pi/4$ does define a great circle...? – Andrew D. Hwang May 08 '17 at 10:50
  • See Example 6.3 at the bottom of the document: https://drive.google.com/file/d/0B8SHtf9-swltWkstLTVJUWZZOHc/view – Andrew Tawfeek May 08 '17 at 14:08

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