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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)

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  1. Ngô Quốc Anh
    May 5, 2010 at 9:35 pm

    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.

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