DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

70 Chapter 2. Surfaces: Local Theoryso along a geodesic the quantity f(u) 2 v ′ = Gv ′ is constant. We recognize this as the dot productof the tangent vector of our geodesic with the vector x v , and so we infer that ‖x v ‖ cos φ = r cos φis constant. (Recall that the tangent vector of the geodesic has constant length.)To this point we have seen that the equation († 2 )isequivalent to the condition r cos φ = const,provided we assume ‖α ′ ‖ 2 = u ′2 + Gv ′2 is constant as well. But ifwe differentiate and obtainu ′ (t) 2 + Gv ′ (t) 2 = u ′ (t) 2 + f(u(t)) 2 v ′ (t) 2 = const,u ′ (t)u ′′ (t)+f(u(t)) 2 v ′ (t)v ′′ (t)+f(u(t))f ′ (u(t))u ′ (t)v ′ (t) 2 =0;substituting for v ′′ (t) using († 2 ), we findu ′ (t) ( u ′′ (t) − f(u(t))f ′ (u(t))v ′ (t) 2) =0.In other words, provided u ′ (t) ≠0,aconstant-speed curve satisfying († 2 ) satisfies († 1 )aswell. (SeeExercise 5 for the case of the parallels.) □Remark. We can give a simple physical interpretation of Clairaut’s relation. Imagine a particleconstrained to move along a surface. If no external forces are acting, then angular momentum isconserved as the particle moves along a geodesic. In the case of our surface of revolution, the verticalcomponent of the angular momentum L = α × α ′ is—surprise, surprise!—f 2 v ′ , which we’ve shownis constant. Perhaps some forces normal to the surface are required to keep the particle on thesurface; then the resulting torques have no vertical component, inasmuch as (α × n) · (0, 0, 1) = 0.Returning to our original motivation for geodesics, we now prove the followingTheorem 4.5. Geodesics are locally distance-minimizing.Proof. Choose P ∈ M arbitrary and a geodesic γ through P . Start by drawing a curve C 0through P orthogonal to γ. Wenow choose a parametrization x(u, v) sothat x(0, 0) = P , theQv-curvesγPu-curvesC 0Figure 4.4u-curves are geodesics orthogonal to C 0 , and the v-curves are the orthogonal trajectories of theu-curves. (It follows from Theorem 3.3 of the Appendix that we can do this on some neighborhoodof P .) We wish to show that for any point Q = x(u 0 , 0) on γ, any path from P to Q is at least aslong as the length of the geodesic segment.In this parametrization we have F =0and E = E(u) (see Exercise 10). Now, if α(t) =x(u(t),v(t)), a ≤ t ≤ b, isany path from P = x(0, 0) to Q = x(u 0 , 0), we have∫ b √∫ b √length(α) = E(u(t))u ′ (t) 2 + G(u(t),v(t))v ′ (t) 2 dt ≥ E(u(t))|u ′ (t)|dtaa