Lecture 3 21
Figure 1.15. Decomposing tangent vectors to show that a
straight line is the shortest smooth curve between two points.
are great circles. While the first fact is an article of faith in elementary
geometry, it requires a proof using a certain amount of calculus. We
will sketch the proof, but for a reader not familiar with calculations
involving arbitrary curves, we recommend carrying out the argument
in detail as an exercise.
Consider an arbitrary parametrised curve with endpoints p and
q, and project it to the straight line pq. As a parametrised curve, the
projection is no longer than the original curve—in fact, it is strictly
shorter if the original curve does not lie entirely on the line.
If the curve is smooth, this follows from the formula for the length
of the curve as the integral of the length of its tangent vector, which
decomposes into two components, one parallel to the line pq, and
one perpendicular (Figure 1.15). For an arbitrary curve, one can
use an approximation by a polygonal curve—in either case, having
established that the length of the original curve is greater than or
equal to the length of the projected curve, one uses integration to
show that the length of the projected curve is greater than or equal to
the length of the interval pq, with equality if and only if the parameter
is monotone (so that the curve is a reparametrised interval).
A very similar argument can be carried out on the sphere, using
geographic coordinates around the point p and projection along par-
allels to the meridian (great circle) passing through p and q. In fact,
once it is understood just what is needed for this argument, it can be
adapted in many cases to find geodesics.
It is sometimes the case that one can find geodesics on other
surfaces by reducing the question to a known situation. For example,
the following exercise can be solved by reducing the question to the
case of the Euclidean plane.