Home > Curves & Surfaces, Geometry, Problem Solving > Shortest path on a sphere

## Shortest path on a sphere

Show that the shortest path between two points on a sphere can be acheved by walking on a great circle of the sphere passing through those two points.
Solution: Pick coordinates such that the sphere is centered in $(0,0,0,)$, $P$ has coordinates $(0,0,1)$ and $Q$ has coordinates $(\sin u_0,0,\cos u_0)$. We can pick $u_0 \in [0,\pi]$, because if it’s the other way around, then by a symmetry, the problem can be transformed to the case where $u_0 \in [0,\pi]$. A curve which joins $P$ and $Q$ and is contained in the surface of our sphere $\mathbb{S}$ can be parametrized like this:
$\alpha(t)=(\sin u(t) \cos v(t),\sin u(t)\sin u(t),\cos u(t))$
where $u : [0,1] \to [0,\pi],\ u(0)=0,\ u(1)=u_0$ and $v: [0,1] \to [0,2\pi],\ v(0)=v(1)=0$. Denote by $L$ the length from $P$ to $Q$ on $\alpha$ and get
$L=\int_0^1 \| \alpha^\prime(t)\|\rm{d}t$.
A few calculations yeld
$\| \alpha^\prime(t)\|=\|(\cos u(t)\cos v(t) u^\prime (t)-\sin u(t)\sin v(t)v^\prime (t),\cos u(t)\sin v(t) u^\prime (t)+ \sin u(t)\cos v(t)v^\prime (t),-\sin u(t)u^\prime (t))\|=$
$\sqrt{(\cos u(t)\cos v(t) u^\prime (t)-\sin u(t)\sin v(t)v^\prime (t))^2+(\cos u(t)\sin v(t) u^\prime (t)+ \sin u(t)\cos v(t)v^\prime (t))^2+(-\sin u(t)u^\prime (t))^2}=$ $\sqrt{(u^\prime(t))^2+\sin^2 u(t) (v^\prime(t))^2}$
Therefore
$L=\int_0^1 \| \alpha^\prime(t)\|{\rm d}t=\int_0^1 \sqrt{(u^\prime(t))^2+\sin^2 u(t) (v^\prime(t))^2}{\rm d}t \geq \int_0^1 \sqrt{(u^\prime(t))^2} {\rm d}t=\int_0^1 |u^\prime (t)| {\rm d} t \geq \int_0^1 u^\prime (t) {\rm d} t =u(1)-u(0)=u_0$. (1)

Considering the great circle which passes through $P(0,0,1)$ and $Q(\sin u_0,0,\cos u_0)$, we see that $u_0$ is exactly the length of the small arc with ends $P$ and $Q$, because $u_0 \in [0,\pi]$.

If $P,Q$ are not antipodal, then this distance can only be achieved on a unique path, namely, the one described above. We should have equality in (1), which is equivalent to
$\sin u(t)v^\prime (t)=0$ and $u^\prime (t)\geq 0$. This would imply $v \equiv 0$ and $u$ increasing, which is equivalent to walking on a small arc of a great circle of the sphere.

See the following pdf for another approach. Geodesics on sphere (source: http://education.uncc.edu/droyster/courses/fall98/math4080/classnotes/geodesiceqn.pdf)

This shortest path is usually called geodesic. As a complete manifold, the sphere $\mathbb S^n$ itself satisfies the Hopf–Rinow theorem, that proves the existence of geodesic.