On a two-sphere of radius with metric

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Differential geometry of curved spacetime · Differential geometry of curved spacetime

The Geometry of Gravity, Ch Symmetry: Lie derivatives, Killing vectors and conserved quantities

Problem

On a two-sphere of radius aa with metric ds2=a2(dθ2+sin⁡2θ dϕ2)ds^2=a^2(d\theta^2+\sin^2\theta\,d\phi^2), θ\theta the colatitude, a bead constrained to the surface and subject to no tangential force moves on a geodesic (a great circle), parametrised by arc length ss so that a2θ˙2+a2sin⁡2θ ϕ˙2=1a^2\dot\theta^2+a^2\sin^2\theta\,\dot\phi^2=1, a dot denoting d/dsd/ds. Use the chapter's result that ξμx˙μ\xi_\mu\dot x^\mu is constant along any geodesic when ξ\xi is a Killing vector. ((a)) For ξ=∂ϕ\xi=\partial_\phi, write the conserved quantity hh in terms of θ\theta and ϕ˙\dot\phi. Eliminate ϕ˙\dot\phi from the unit-speed condition to show that a2θ˙2=1−h2a2sin⁡2θ,a^2\dot\theta^2=1-\frac{h^2}{a^2\sin^2\theta}, and deduce that the bead can never reach a colatitude with asin⁡θ<∣h∣a\sin\theta<|h|. ((b)) State the maximum number of independent Killing vectors of a two-dimensional space and confirm the sphere attains it. ((c)) A great circle crosses the equator (θ=π/2\theta=\pi/2) making an angle α=30∘\alpha=30^\circ with the equator itself, so the component of its unit tangent along the local ϕ\phi-direction there is cos⁡α\cos\alpha. Use (a) to find the smallest colatitude it reaches, and hence the greatest latitude.

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