DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

68 Chapter 2. Surfaces: Local TheoryAs we saw before, κ n gives the normal component of the curvature vector; κ g gives the tangentialcomponent of the curvature vector and is called the geodesic curvature. This terminology arisesfrom the fact that α is a geodesic if and only if its geodesic curvature vanishes.Example 5. We saw in Example 1 that every great circle on a sphere is a geodesic. Are thereothers? Let α be a geodesic on a sphere centered at the origin. Since κ g =0,the accelerationvector α ′′ (s) must be a multiple of α(s) for every s, and so α ′′ × α = 0. Therefore α ′ × α = A isa constant vector, so α lies in the plane passing through the origin with normal vector A. That is,α is a great circle. ▽Using the equations (♣), let’s now give the equations for the curve α(t) =x(u(t),v(t)) to bea geodesic. Since X = α ′ (t) =u ′ (t)x u + v ′ (t)x v ,wehave a(t) =u ′ (t) and b(t) =v ′ (t), and theresulting equations are(♣♣)u ′′ (t)+Γ u uuu ′ (t) 2 +2Γ u uvu ′ (t)v ′ (t)+Γ uvvv ′ (t) 2 =0v ′′ (t)+Γ v uuu ′ (t) 2 +2Γ v uvu ′ (t)v ′ (t)+Γ vvvv ′ (t) 2 =0.The following result is a consequence of basic results on differential equations (see Theorem 3.1of the Appendix).Proposition 4.3. Given a point P ∈ M and V ∈ T P M, V ≠ 0, there exist ε>0 and a uniquegeodesic α: (−ε, ε) → M with α(0) = P and α ′ (0) = V.Example 6. We now use the equations (♣♣) tosolve for geodesics analytically in a few examples.(a) Let x(u, v) =(u, v) bethe obvious parametrization of the plane. Then all the Christoffelsymbols vanish and the geodesics are the solutions ofu ′′ (t) =v ′′ (t) =0,so we get the lines α(t) =(u(t),v(t)) = (a 1 t + b 1 ,a 2 t + b 2 ), as expected. Note that α doesin fact have constant speed.(b) Using the standard spherical coordinate parametrization of the sphere, we obtain (see Example1 or 2 of Section 3) the equations(∗)u ′′ (t) − sin u(t) cos u(t)v ′ (t) 2 =0=v ′′ (t)+2cotu(t)u ′ (t)v ′ (t).Well, one obvious set of solutions is to take u(t) =t, v(t) =v 0 (and these, indeed, givethe great circles through the north pole). More generally, let’s now impose the additionalcondition that the curve be arclength-parametrized:(∗∗)u ′ (t) 2 + sin 2 u(t)v ′ (t) 2 =1.Integrating the equation in (∗) weobtain ln v ′ (t) =−2lnsinu(t)+const, sov ′ c(t) =sin 2 u(t)