Geodesics as Lines in Differential Geometry, Papers of Geometry

Download Geodesics as Lines in Differential Geometry and more Papers Geometry in PDF only on Docsity! Chapter 5 Geodesics, Metrics, and Isometries 5.1 Geodesics and their Definition In an axiomatic approach to geometry we study the properties of points and lines (and planes in 3-dimensional geometry). Most of the theorems in axiomatic geometry deal with the relationships between points and lines. If we are to see how the differential geometry we have been studying is to relate to axiomatic geometry, we need some method for developing an abstract definition of a line. This is different from our axiomatic technique of taking a line as an undefined term. Once we have defined a line we need to make certain that it has the appropriate properties in the typical models of Euclidean and non-Euclidean geometries. For example, if we have the right definition, then the lines on the sphere will be great circles. What will be lines in hyperbolic geometry? The differential geometric concept of a line should involve derivatives, possibly curvature, and what else? We previously proved that a straight line in the plane is the shortest distance between two points. Will this distance minimization criterion serve as our definition? If we do, it would be a hard condition to check and thus make the use of lines in our surfaces difficulty, if not impossible. We need an easier condition, if we are to be able to use it. If we look back at our theory of curves, we saw a different condition for a curve to be a line. We showed that a curve is a line if its second derivative vanishes. This looks like a good candidate, because it fits into our framework of differential geometry and it is an easy condition to check. There is a small problem with this condition. If we have a curve, α, on a surface M and place the condition that α′′ = 0 on the curve, then α is a straight line in R3. That does not insure that the curve remains on M . Let’s not give up on this notion. Maybe all we need to do is to look at the situation from the frame of reference of the surface M . Let us not think about the third dimension given by the unit normal U . Recall that the acceleration along a curve can be broken up into the tangential and normal components (Lemma 1.4.2) α′′ = ν ′T+κν2N, but because we cannot see anything outside the tangent plane we do not see the normal component of acceleration. Suppose that α has unit speed. Then we have two perpendicular unit vectors T = α′ and U , the unit normal of M . We get a third unit vector orthogonal to T and U by taking T × U . These three vectors have to form a basis for R3. Thus, α′′ = aT + bT × U + cU where a = α′′ ·T b = α′′ ·T× U c = α′′ · U. 52 5.1. GEODESICS AND THEIR DEFINITION 53 Rewrite α′′ = (α′′ ·T)T + (α′′ ·T× U)T× U + (α · U)U. Now, recall that α′ ·α′ = 1 so differentiating we again find α ·α′′ = 0 or α′′ ·T = α′′ ·α′ = 0. Thus, there is no T-component for α′′, α′′ = (α′′ ·T× U)T× U + (α′′ · U)U. Now, we know that U ·(T×U) = 0. Yet another vector identity is u ·(v×w) = −v ·(u×w). Therefore, we have α′′ · (T× U) = −α′′ · (U ×T) = −(−U · (α′′ ×T) = U · (α′′ × α′) Our first condition above (U · (T× U) = 0) tells us that T× U lies in TpM for all p ∈M . The second condition tells us α′′ · (T× U) = U · (α′′ × α′) = ‖U‖‖α′′ × α′‖ cos θ = ‖α′′ × α′‖ cos θ = κα cos θ where κα is the curvature of α and θ is the angle between α′′ × α′ and U . This quantity is called the geodesic curvature of α and is denoted κg = κα cos θ. Thus, we can decompose the acceleration into two components α′′tan = κg T× U and α′′normal(α′′ · U)U. Now, if we consider only the frame of reference of the surface M , then we cannot see the normal component of acceleration and the acceleration of a curve will be zero exactly when α′′tan = 0. We define a curve α in M to be a geodesic if α ′′ tan = 0. Geodesics are going to be the lines of our geometry. Lemma 5.1 A geodesic has constant speed. Proof: The speed of α is v = ‖α′‖, so v2 = α′ · α′. Differentiate both sides and we get 2v dv dt = 2α′ · α′′ = 0, since α′′ = α′′normal and α ′′ normal · α′ = 0. Thus, either v = 0 (and is hence constant) or dv dt = 0 which makes v a constant.