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Length of Curve in $2D$ is $l_{\gamma}(\mathbb{R}^2)=\int_{0}^{1}\sqrt{(dr/dt)^2+r^2(d\theta/dt)^2}$

Length of a curve in $3D$ is $l_{\gamma}(\mathbb{R}^3)=\int_{0}^{1}\sqrt{(dr/dt)^2+r^2(d\phi/dt)^2+r^2\sin^2\phi(d\theta/dt)^2}$ so when the curve lie on $S^2$ the second expression becomes $$l_{\gamma}(S^2)=\int_{0}^{1}\sqrt{(d\phi/dt)^2+\sin^2\phi(d\theta/dt)^2}$$ I myself calculated that shortest distance between any two points must be straight line by analyzing the formula $l_{\gamma}(\mathbb{R}^2)$, could any one tell me how to analyze and find the shortest distance between any two points on $S^2$ by analyzing the formula $2D$ is $l_{\gamma}(\mathbb{R}^2)$ and $l_{\gamma}(S^2)$?

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  • $\begingroup$ could you please give a detail answer? $\endgroup$ – Ding Dong Mar 12 '13 at 15:03
  • $\begingroup$ @Tomás: what you write is not true for points on the same parallel. $\endgroup$ – Georges Elencwajg Mar 12 '13 at 15:38
  • $\begingroup$ @GeorgesElencwajg Yes you are right, I am gonna fix it in the answer. $\endgroup$ – Tomás Mar 12 '13 at 15:40
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    $\begingroup$ Your title should be «shortest distance between two points in $S^2$»: there is no shortest distance between any two points on the sphere, as the inf of the set of distances is zero. The word any is best avoided, I'd say... $\endgroup$ – Mariano Suárez-Álvarez Mar 12 '13 at 15:58
  • $\begingroup$ @CityOfGod You can directly use Euler-Lagrange formula to minimize the functional $\endgroup$ – AnilB Mar 12 '13 at 16:36
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The point here is that the sphere admits a distance-nonincreasing map to a fixed longitudal curve (say the Greenwich one), by sending each point of the sphere to the corresponding point with the same latitude on the greenwich curve.

To convince oneself that this map is distance-nonincreasing, just think of an accordion folding ball decoration that collapses the said toy to a half-circle.

More formally, the metric on the sphere is $d\phi^2+\sin^2\phi\, d\theta^2$ (for the usual spherical coordinates where $\phi$ is the angle between the point and the north pole). Since $d\phi^2+\sin^2\phi\, d\theta^2\geq d\phi^2$, the projection is distance non-increasing.

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Take the great circle that contains the two points. By changing the coordinates, you can suppose that this great circle is the parallel (an great circle too) given by the equation $\phi$ varying and $\theta$ constant (the interval where $\phi$ vary and the constant depends on the parametrization, for example, we can suppose that $\phi\in (0,2\pi)$ and $\theta \in (-\frac{\pi}{2},\frac{\pi}{2}$), which implies that $\theta=0$). Then, for any curve joining these points, we have that \begin{eqnarray} l_\gamma (S^2) &=& \int_I\sqrt{\Big(\frac{d\phi}{dt}\Big)^2+\sin^2(\phi)\Big(\frac{d\theta}{dt}\Big)^2} \nonumber \\ &\geq& \int_I \Big|\frac{d\phi}{dt}\big| \nonumber \end{eqnarray}

From the last inequality, you can conclude.

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  • $\begingroup$ from the second line I dont understand, please be rigorious or give some picture or add some more sentenses to make me understand :( $\endgroup$ – Ding Dong Mar 12 '13 at 16:33
  • $\begingroup$ The sphere is all symmetric. Take two points on it. Now you can rotate the sphere in such a way that these two points, lies in the circle contained in the $xy$ plane. $\endgroup$ – Tomás Mar 12 '13 at 16:36
  • $\begingroup$ Oh okay that I agree but it is still not clear to me that what is the shortest distance between two points on a sphere. $\endgroup$ – Ding Dong Mar 12 '13 at 16:43
  • $\begingroup$ The last term in inequality show that the arc of that circle in $xy$ plane is the shortest path. When you fix $\theta$ and vary $\phi$ you are "walking" on the parallels. In our case we are "walking" on the bigest parallel, to wit, the great circle in the $xy$ plane. Now note that the last integral is the length of the arc (on the parallel) joining these points. $\endgroup$ – Tomás Mar 12 '13 at 16:48
  • $\begingroup$ so what is the analogy between $\theta,\phi$ and $r,\theta$ in Polar in plane? $\endgroup$ – Ding Dong Mar 13 '13 at 4:33

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