Download Geodesic Equations and Clairaut Relation - Lecture Notes | MATH 4080 and more Study notes Geometry in PDF only on Docsity! 5.2. GEODESIC EQUATIONS AND CLAIRAUT RELATION 55 5.2 Geodesic Equations and Clairaut Relation Now that we know what geodesics are, how can we do the calculations? Let’s consider an orthogonal patch x(u, v), (i.e., F = xu · xv = 0). Let σ be a geodesic in the patch x. Then, we know that we can write σ(t) as σ(t) = x(u(t), v(t)) and it follows that σ′ = xuu′ + xvv′ and σ′′ = xuuu′ 2 + xuvv′u′ + xuu′′ + xvuu′v′ + xvvv′ 2 + xvv′′ Now, from our previous work, we can replace xuu, xuv, and xvv by their representations in {xu,xv, U}. We find σ′′ = xu [ u′′ + Eu 2E u′ 2 + Ev E u′v′ − Gu 2E v′ 2 ] + xv [ v′′ − Ev 2E u′ 2 + Gu G u′v′ − Gv 2G v′ 2 ] + U [ lu′ 2 + 2mu′v′ + nv′2 ] So the first two terms give us the tangential part of σ′′. For σ to be a geodesic, these terms must be zero, so it is necessary and sufficient that the following geodesic equations are satisfied. u′′ + Eu 2E u′ 2 + Ev E u′v′ − Gu 2E v′ 2 = 0 v′′ − Ev 2E u′ 2 + Gu G u′v′ − Gv 2G v′ 2 = 0 (Geodesic Equations) The geodesic equations are a system of 2nd order differential equations. Theorem 5.1 Let p = x(u0, v0) be a point on a surface M with patch x(u, v) and let v ∈ TpM . Then there is a unique geodesic σ : (−r, r)→M with σ(0) = p and σ′(0) = v. The proof of this is based on the theory of differential equations, and the existence of a solution to a system with certain intial values. Our geodesic equations satisfy the hypothesis of the theorem, so such a solution exists. Let’s look at an example. Example 5.3 The Unit Sphere, S2. Take the standard patch for the unit sphere. x(u, v) = (cosu cos v, sinu cos v, sin v) We have calculated previously that E = cos2 v, F = 0, and G = 1. The geodesic equations for the unit sphere become u′′ − 2 tan v v′ u′ = 0 v′′ + sin v cos v u′2 = 0. How in the world will we ever be able to solve such a system of 2nd order non-linear differential equations?!? In fact, it is the case that we usually CANNOT solve the differential equations directly. In the case of the sphere, though, we have a few tricks up our collective sleeves to bring to bear on the problem. 56 1. Assume that σ(t) = x(u(t), v(t)) is a unit speed geodesic. Then since σ′ = u′xu+v′xv we have the unit speed relation Eu′ 2 +Gv′2 = 1 which on the sphere is the condition cos2 v u′2 + v′2 = 1. 2. Solve the first geodesic equation as follows:∫ u′′ u′ = ∫ 2 tan v v′ ln(u′) = −2 ln(cos v) + C u′ = c cos2 v 3. Now replace u′ in the unit speed equation by c cos2 v 1 = c2 cos4 v cos2 v + v′2 v′ 2 = 1− c 2 cos2 v v′ = √ cos2 v − c2 cos2 v 4. Now, divide u′ by v′ and we get a separable differential equation du dv = c cos v √ cos2 v − c2 . 5. Integrate this to find u = ∫ c cos v √ cos2 v − c2 dv = ∫ c sec2 v√ 1− c2 sec2 v dv = ∫ c sec2 v√ 1− c2 − c2 tan2 v dv = ∫ dw√ 1− w2 substitute w = c tan v√ 1− c2 = arcsin(w) + d = arcsin ( c tan v√ 1− c2 ) + d Thus, sin(u− d) = λ tan v where λ = c√ 1−c2 . Now, use the Angle Sum formula for Sines to get sin(u − d) = sinu cos d− sin d cosu.
