Question

In Euclidean (flat) space, a circle’s circumference C is related to its radius r by C = 2πr. Formally, this can be shown by integrating the infinitesimal distance element ds around the circumference:

C = ! ds = ! 2π 0 rdθ = 2πr , (1)

where the second equality holds because the 2-D Euclidean metric is ds2 = dr2 + r2dθ2, and dr = 0 for a circle of constant radius r.

(a) In a 2-D space of positive curvature (κ = 1), the metric is ds2 = dr2 + R2 sin2 (r/R)dθ2, where R is the space’s radius of curvature. Set up an integral similar to (1) and calculate the circumference C of a circle of radius r.

(b) Using the result for part (a), show that the ratio C/r approaches 2π for r → 0, but C/r < 2π for finite r.

(c) Using the result for part (a), show that the circumference C has a maximum for a particular (and finite!) value of r. Find this value of r in terms of the curvature radius R, and make a sketch showing why this r gives the maximum C.

Answer 1