DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

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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)
§4. Covariant Differentiation, Parallel Translation, and Geodesics 69for some constant c. Now, using (∗∗), and switching to Leibniz notation for obvious reasons,we obtain√dudt = ± 1 −dtc 2sin 2 u(t) ,dvdvdu = dt c csc 2 u= ± √du 1 − c 2 csc 2 u ;and sothus, separating variables givesdv = ±c csc2 udu√1 − c 2 csc 2 u = ± c csc 2 udu√(1 − c 2 ) − c 2 cot 2 u .Now we make the substitution c cot u = √ 1 − c 2 sin w; then we havec csc 2 ududv = ± √(1 − c 2 ) − c 2 cot 2 u = ∓dw,and so, at long last, we have w = ±v + a for some constant a. Thus,c cot u = √ 1 − c 2 sin w = √ 1 − c 2 sin(±v + a) = √ 1 − c 2 (sin a cos v ± cos a sin v),and so, finally, we have the equationc cos u + √ 1 − c 2 sin u(A cos v + B sin v) =0,which we should recognize as the equation of a great circle! (This curve lies on the plane√1 − c 2 (Ax + By)+cz = 0.) ▽We can now give a beautiful geometric description of the geodesics on a surface of revolution.Proposition 4.4 (Clairaut’s relation). The geodesics on a surface of revolution satisfy theequation(♦)r cos φ = const,where r is the distance from the axis of revolution and φ is the angle between the geodesic and theparallel. Conversely, any curve satisfying (♦) that is not a parallel is a geodesic.Proof. For the surface of revolution parametrized as in Example 7 of Section 2, we have E =1,F =0,G = f(u) 2 ,Γuv v =Γvu v = f ′ (u)/f(u), Γvv u = −f(u)f ′ (u), and all other Christoffel symbolsare 0 (see Exercise 2.3.5d.). Then the system (♣♣) ofdifferential equations becomes(† 1 )(† 2 )u ′′ − ff ′ (v ′ ) 2 =0v ′′ + 2f ′f u′ v ′ =0.Rewriting the equation († 2 ) and integrating, we obtainv ′′ (t)′v ′ (t) = −2f (u(t))u ′ (t)f(u(t))ln v ′ (t) =−2lnf(u(t)) + constv ′ c(t) =f(u(t)) 2 ,
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DIFFERENTIALGEOMETRY:A First Course
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118 INDEXisometry, 107K, 48k-point
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