Consider the metric of ##S^{2}##: $$ds^{2}=d\theta^{2}+\sin^{2}(\theta)d\phi^{2}$$ Then in order to determine the geodesics on this surface one can minimise the integral $$s=\int_{l_{1}}^{l_{2}}\sqrt{\left(\frac{d\theta}{dl}\right)^{2}+\sin^{2}(\theta)\left(\frac{d\phi}{dl}\right)^{2}}dl$$ where ##l## parametrises a path connecting two points on the surface. We can identify the Lagrangian as ##L=\sqrt{\left(\frac{d\theta}{dl}\right)^{2}+\sin^{2}(\theta)\left(\frac{d\phi}{dl}\right)^{2}}##.(adsbygoogle = window.adsbygoogle || []).push({});

If one parametrises the path by its arc-length ##s## then ##L=1##. Now, in the case of space-time the arc-length of a time-like path is equal to the proper time between the two endpoints of the path, and hence an arc-length parametrisation corresponds in this case to choosing proper time to parametrise the path.

My question is, is the analogue of this for a curved surface (such as ##S^{2}##) theproperdistancebetween two points on the surface, defined as ##dl=\sqrt{ds^{2}}##, such that, in the case of ##S^{2}##, the equations of motion are: $$\frac{d^{2}\theta}{dl^{2}}-\sin(\theta)\cos(\theta)\left(\frac{d\phi}{dl}\right)^{2}=0\\ \frac{d^{2}\phi}{dl^{2}}+2\cot(\theta)\frac{d\theta}{dl}\frac{d\phi}{dl}=0$$

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# I Geodesics on surfaces

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