# prove that minor arc of a great circle is the shortest distance

How to prove that the minor arc of a great circle is the shortest path connecting two points on the surface of a sphere?

The key to solving the problems is that we must take all curves connecting the two points , not just great-circle or parametrized differentiable curve, into account!

A similar problem is the straight line distance(shortest distance between two points): we must take all curves connecting the two points into account!

• What's wrong with the proof on the wiki page you linked to? – nbubis Jul 13 '13 at 20:59
• @nbubis Assumption of differentiability, which the OP wants to avoid? – 40 votes Jul 13 '13 at 21:03
• we can approximate all continuous curves by diffenrentiable ones s.t. the length of them goes to the length of original curve, so we can prove the theorem by proving it for differentiable curves. – jouge Jul 13 '13 at 21:04
• @nbubis parametrized differentiable curve . $\theta(t), \phi(t)$ are differentiable. – ziang chen Jul 13 '13 at 21:08

Here is a low-tech approach which does not use derivatives much. Given distinct points $a,b$ on the sphere, let $\gamma$ be the shorter arc of great circle between them. Let $V$ be the surface obtained by rotating $\gamma$ about the line $ab$. Note that $V$ is convex (i.e., bounds a convex body) and lies inside the sphere. The nearest-point projection onto a convex set does not increase Euclidean distance and therefore does not increase length of curves (defined by means of sums of Euclidean distances over partitions). Therefore, it suffices to prove that a curve connecting $a$ to $b$ on surface $V$ cannot be shorter than $\gamma$.
Let $\Gamma$ be any such curve. Pick partition points $p_k$ on $\Gamma$ to approximate its length. For each $k$, draw the plane through $p_k$ that is orthogonal to line $ab$. This plane crosses $\gamma$ at a point $q_k$. It is easy to see that $|p_k-p_{k+1}|\ge |q_k-q_{k+1}|$ for each $k$. (Just write the Euclidean distance in cylindrical coordinates with $ab$ as the axis.) Since $\sum_k |q_k-q_{k+1}|$ is greater than or equal to the length of $\gamma$, the conclusion follows.
Assume there existed a continues but non-differentiable path, which was non-differentiable only at a finite number of points. for each such point which has slope $\phi'(x_-)$ on one side, and slope $\phi'(x_+)$ on the other, we can round the corner with a circle of radius $\epsilon$ who's contribution to the total length goes to zero.