shortest distance between two points on $S^2$ 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  $l_{\gamma}(S^2)$?
 A: 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.
A: 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.
