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9 Geodesics Geodesics are the curves in a surface that a bug living in the surface would perceive to be straight For example, the shortest path between two points in a surface is always a geodesic We shall actually begin by giving a quite different definition of geodesics, since this definition is easier to work with We give various methods of finding geodesics on surfaces, before finally making contact with the idea of shortest paths towards the end of the chapter 9.1 Definition and basic properties If we drive along a ‘straight’ road, we not have to turn the wheel of our car to the right or left (this is what we mean by ‘straight’ !) However, the road is not, in fact, a straight line as the surface of the earth is, to a good approximation, a sphere and there can be no straight line on the surface of a sphere If the road ă will be non-zero, but we perceive is represented by a curve γ, its acceleration ă is zero, in the curve as being straight because the tangential component of ă is perpendicular to the surface This suggests other words because γ Definition 9.1.1 ă (t) is zero or perpendicular A curve on a surface S is called a geodesic if γ to the tangent plane of the surface at the point γ(t), i.e., parallel to its unit normal, for all values of the parameter t Andrew Pressley, Elementary Differential Geometry: Second Edition, Springer Undergraduate Mathematics Series, DOI 10.1007/978-1-84882-891-9 9, c Springer-Verlag London Limited 2010 215 216 Geodesics Equivalently, γ is a geodesic if and only if its tangent vector γ˙ is parallel along γ (see Section 7.4) Note that this definition makes sense for any surface, orientable or not There is an interesting mechanical interpretation of geodesics: a particle moving on the surface, and subject to no forces except a force acting perpendicular to the surface that keeps the particle on the surface, would move along a geodesic This is because Newtons second law of motion states that ă , which would therefore the force on the particle is parallel to its acceleration γ be perpendicular to the surface We begin our study of geodesics by noting that there is essentially no choice in their parametrization Proposition 9.1.2 Any geodesic has constant speed Proof Let γ(t) be a geodesic on a surface S Then, denoting d/dt by a dot, d d = 2ă = ( Ã ) Ã dt dt ă is perpendicular to the tangent plane and is therefore Since γ is a geodesic, So ă Ã = and the last equation shows perpendicular to the tangent vector γ that γ˙ is constant It follows from this proposition that a unit-speed reparametrization of a ˜ (t) = γ(t/λ) is a unitgeodesic γ is still a geodesic For, if = , then 2 ă , and hence is speed reparametrization of γ and ddtγ2˜ = λ12 ddtγ2 is parallel to γ also perpendicular to the surface Thus, we can always restrict to unit-speed geodesics if we wish In general, however, a reparametrization of a geodesic will not be a geodesic (see Exercise 9.1.2) We observe next that there is an equivalent definition of a geodesic expressed in terms of the geodesic curvature κg (see Section 7.3) Of course, this is why κg is called the geodesic curvature ! Proposition 9.1.3 A unit-speed curve on a surface is a geodesic if and only if its geodesic curvature is zero everywhere 9.1 Definition and basic properties 217 Proof Let γ be a unit-speed curve on the surface S, and let p ∈ S Let σ be a surface patch of S with p in its image, and let N be the standard unit normal of σ, so that ă Ã (N ì ) (9.1) g = γ (changing σ may change the sign of N, and hence that of g , but that is ă is parallel to N, it is obviously not relevant to the present discussion) If γ ˙ so by Eq 9.1, g = perpendicular to N ì , ă is perpendicular to N × γ ˙ But Conversely, suppose that κg = Then, γ ˙ N and N × γ˙ are perpendicular unit vectors in R3 (see the then, since , ă is perpendicular to , it follows that ă discussion in Section 7.3), and since is parallel to N The following result gives the simplest examples of geodesics Proposition 9.1.4 Any (part of a) straight line on a surface is a geodesic By this, we mean that every straight line can be parametrized so that it is a geodesic A similar remark applies to other geodesics we consider and whose parametrization is not specified (see Exercise 9.1.2) Proof This is obvious, for any straight line has a (constant speed) parametrization of the form γ(t) = a + bt, ă = where a and b are constant vectors, and clearly γ Example 9.1.5 All straight lines in the plane are geodesics, as are the rulings of any ruled surface, such as those of a (generalized) cylinder or a (generalized) cone, or the straight lines on a hyperboloid of one sheet The next result is almost as simple: 218 Geodesics Proposition 9.1.6 Any normal section of a surface is a geodesic Proof Recall from Section 7.3 that a normal section of a surface S is the intersection C of S with a plane Π, such that Π is perpendicular to the surface at each point of C We showed in Corollary 7.3.4 that κg = for a normal section, and so the result follows from Proposition 9.1.3 Example 9.1.7 All great circles on a sphere are geodesics For a great circle is the intersection of P O Π the sphere with a plane Π passing through the centre O of the sphere, and so if P is a point of the great circle, the straight line through O and P lies in Π and is perpendicular to the tangent plane of the sphere at P Hence, Π is perpendicular to the tangent plane at P Example 9.1.8 The intersection of a generalized cylinder with a plane Π perpendicular to the rulings of the cylinder is a geodesic For it is clear that the unit normal N is perpendicular to the rulings It follows that N is parallel to Π, and hence that Π is perpendicular to the tangent plane 9.1 Definition and basic properties 219 N Π EXERCISES 9.1.1 Describe four different geodesics on the hyperboloid of one sheet x2 + y − z = passing through the point (1, 0, 0) 9.1.2 A (regular) curve γ with nowhere vanishing curvature on a surface S is called a pre-geodesic on S if some reparametrization of γ is a geodesic on S (recall that a reparametrization of a geodesic is not usually a geodesic) Show that: ă Ã (N × γ) ˙ = every(i) A curve γ is a pre-geodesic if and only if γ where on γ (in the notation of the proof of Proposition 9.1.3) (ii) Any reparametrization of a pre-geodesic is a pre-geodesic (iii) Any constant speed reparametrization of a pre-geodesic is a geodesic (iv) A pre-geodesic is a geodesic if and only if it has constant speed 9.1.3 Consider the tube of radius a > around a unit-speed curve γ in R3 defined in Exercise 4.2.7: σ(s, θ) = γ(s) + a(cos θ n(s) + sin θ b(s)) Show that the parameter curves on the tube obtained by fixing the value of s are circular geodesics on σ 220 Geodesics 9.1.4 Let γ(t) be a geodesic on an ellipsoid S (see Theorem 5.2.2(i)) Let ˙ 2R(t) be the length of the diameter of S parallel to γ(t), and let S(t) be the distance from the centre of S to the tangent plane Tγ(t) S Show that the curvature of γ is S(t)/R(t)2 , and that the product R(t)S(t) is independent of t 9.1.5 Show that a geodesic with nowhere vanishing curvature is a plane curve if and only if it is a line of curvature 9.1.6 Let S1 and S2 be two surfaces that intersect in a curve C, and let γ be a unit-speed parametrization of C (i) Show that if γ is a geodesic on both S1 and S2 and if the curvature of γ is nowhere zero, then S1 ad S2 touch along γ (i.e., they have the same tangent plane at each point of C) Give an example of this situation (ii) Show that if S1 and S2 intersect orthogonally at each point of C, ˙ is parallel to N1 at then γ is a geodesic on S1 if and only if N each point of C (where N1 and N2 are unit normals of S1 and S2 ) Show also that, in this case, γ is a geodesic on both S1 and S2 if and only if C is part of a straight line 9.2 Geodesic equations Unfortunately, Propositions 9.1.4 and 9.1.6 are not usually sufficient to determine all the geodesics on a given surface For that, we need the following result: Theorem 9.2.1 A curve γ on a surface S is a geodesic if and only if, for any part γ(t) = σ(u(t), v(t)) of γ contained in a surface patch σ of S, the following two equations are satisfied: d (E u˙ + F v) ˙ = (Eu u˙ + 2Fu u˙ v˙ + Gu v˙ ), dt d (F u˙ + Gv) ˙ = (Ev u˙ + 2Fv u˙ v˙ + Gv v˙ ), dt where Edu2 + 2F dudv + Gdv is the first fundamental form of σ The differential equations (9.2) are called the geodesic equations (9.2) 9.2 Geodesic equations 221 Proof Since {σu , σv } is a basis of the tangent plane of σ, γ is a geodesic if and only ă is perpendicular to u and v Since γ˙ = uσ ˙ u + vσ ˙ v , this is equivalent to if γ d d (uσ ˙ u + vσ (uσ ˙ u + vσ ˙ v ) · σ u = and ˙ v ) · σ v = (9.3) dt dt We show that these two equations are equivalent to the two geodesic equations The left-hand side of the first equation in (9.3) is equal to d dσ u ˙ v ) · σ u ) − (uσ ˙ u + vσ ˙ v) · ((uσ ˙ u + vσ dt dt d ˙ v ) · (uσ ˙ uu + vσ ˙ uv ) ( 9.4) = (E u˙ + F v) ˙ − (uσ ˙ u + vσ dt d = (E u˙ + F v) ˙ u · σ uv + σ v · σ uu ) + v˙ (σ v · σ uv )) ˙ − (u˙ (σ u · σ uu ) + u˙ v(σ dt Now, Eu = (σ u · σ u )u = σ uu · σ u + σ u · σ uu = 2σu · σ uu , so σ u · σ uu = 12 Eu Similarly, σ v · σ uv = 12 Gu Finally, σ u · σ uv + σ v · σ uu = (σ u · σ v )u = Fu Substituting these values into (9.4) gives d d (uσ ˙ u + vσ ˙ − (Eu u˙ + 2Fu u˙ v˙ + Gu v˙ ) ˙ v ) · σ u = (E u˙ + F v) dt dt This shows that the first equation in (9.3) is equivalent to the first geodesic equation in (9.2) Similarly for the other equations The geodesic equations are non-linear differential equations, and are usually difficult or impossible to solve explicitly The following example is one case in which this can be done Another is given in Exercise 9.2.3 Example 9.2.2 We determine the geodesics on the unit sphere S by solving the geodesic equations For the usual parametrization by latitude θ and longitude ϕ, σ(θ, ϕ) = (cos θ cos ϕ, cos θ sin ϕ, sin θ), we found in Example 6.1.3 that the first fundamental form is dθ2 + cos2 θ dϕ2 222 Geodesics We might as well restrict ourselves to unit-speed curves γ(t) = σ(θ(t), ϕ(t)), so that θ˙2 + ϕ˙ cos2 θ = 1, and if γ is a geodesic the second equation in (9.2) gives d (ϕ˙ cos2 θ) = 0, dt so that ϕ˙ cos2 θ = Ω, where Ω is a constant If Ω = 0, then ϕ˙ = and so ϕ is constant and γ is part of a meridian We assume that ϕ˙ = from now on The unit-speed condition gives Ω2 , θ˙2 = − cos2 θ so along the geodesic we have 2 θ˙2 dθ = = cos2 θ(Ω−2 cos2 θ − 1), dϕ ϕ˙ and hence dθ √ , −2 cos θ Ω cos2 θ − where ϕ0 is a constant The integral can be evaluated by making the substitution u = tan θ This gives u du √ = sin−1 √ , ±(ϕ − ϕ0 ) = Ω−2 − − u2 Ω−2 − ±(ϕ − ϕ0 ) = and hence tan θ = ± Ω−2 − sin(ϕ − ϕ0 ) This implies that the coordinates x = cos θ cos ϕ, y = cos θ sin ϕ and z = sin θ of γ(t) satisfy the equation z = ax + by, √ √ where a = ∓ Ω−2 − sin ϕ0 , and b = ± Ω−2 − cos ϕ0 This shows that γ is contained in the intersection of S with a plane passing through the origin Hence, in all cases, γ is part of a great circle The geodesic equations can be expressed in a different, but equivalent, form which is sometimes more useful than that in Theorem 9.2.1 9.2 Geodesic equations 223 Proposition 9.2.3 A curve γ on a surface S is a geodesic if and only if, for any part γ(t) = σ(u(t), v(t)) of γ contained in a surface patch σ of S, the following two equations are satisfied: u ¨ + Γ111 u˙ + 2Γ112 u˙ v˙ + 122 v = vă + 211 u + 2Γ212 u˙ v˙ + Γ222 v˙ = Proof As we noted after Definition 9.1.1, γ is a geodesic if and only if γ˙ is parallel ˙ v , the equations in the statement of the proposition along γ Since γ˙ = uσ ˙ u + vσ follow from Proposition 7.4.5 It can of course be verified directly that the differential equations in Proposition 9.2.3 are equivalent to those in Theorem 9.2.1 (see Exercise 9.2.6) Proposition 9.2.3 makes it obvious that the geodesic equations are secondorder ordinary differential equations for the functions u(t) and v(t) Even though we may be unable in many situations to solve these equations explicitly, the general theory of ordinary differential equations provides valuable information about their solutions This leads to the following result, which tells us exactly ‘how many’ geodesics there are Proposition 9.2.4 Let p be a point of a surface S, and let t be a unit tangent vector to S at p Then, there exists a unique unit-speed geodesic γ on S which passes through p and has tangent vector t there In short, there is a unique geodesic through any given point of a surface in any given tangent direction Proof The geodesic equations in Proposition 9.2.3 are of the form u ă = f (u, v, u, v), vă = g(u, v, u, v), (9.5) where f and g are smooth functions of the four variables u, v, u˙ and v ˙ It is proved in the theory of ordinary differential equations that, for any given constants a, b, c, and d, and any value t0 of t, there is a solution of Eqs 9.5 such that ˙ ) = c, v(t ˙ ) = d, (9.6) u(t0 ) = a, v(t0 ) = b, u(t 224 Geodesics and such that u(t) and v(t) are defined and smooth for all t satisfying |t − t0 | < , where is some positive number Moreover, any two solutions of Eqs 9.5 satisfying (9.6) agree for all values of t such that |t − t0 | < , where is some positive number ≤ We now apply these facts to the geodesic equations Suppose that p lies in a patch σ(u, v) of S, say p = σ(a, b), and that t = cσ u + dσ v , where a, b, c, and d are scalars and the derivatives are evaluated at u = a, v = b A unit-speed curve γ(t) = σ(u(t), v(t)) passes through p at t = t0 if and only if u(t0 ) = a, v(t0 ) = b, and has tangent vector t there if and only if ˙ ) = u(t ˙ )σ u + v(t ˙ )σ v , cσ u + dσ v = t = γ(t i.e., u(t ˙ ) = c, v(t ˙ ) = d Thus, finding a (unit-speed) geodesic γ passing through p at t = t0 and having tangent vector t is equivalent to solving the geodesic equations subject to the initial conditions (9.6) But we have said above that this problem has a unique solution Example 9.2.5 We already know that all straight lines in a plane are geodesics Since there is a straight line in the plane through any given point of the plane in any given direction parallel to the plane, it follows from Proposition 9.2.4 that there are no other geodesics Example 9.2.6 Similarly, on a sphere, the great circles are the only geodesics, for there is clearly a great circle passing through any given point of the sphere in any given direction tangent to the sphere (If p is the point and t the tangent direction, let Π be the plane passing through the origin parallel to p and t (i.e., with normal p × t); then take the intersection of the sphere with Π.) The following consequence of Theorem 9.2.1 can also be used in some cases to find geodesics without solving the geodesic equations Corollary 9.2.7 Any local isometry between two surfaces takes the geodesics of one surface to the geodesics of the other ... surface patch σ of S, the following two equations are satisfied: u ă + 111 u + 21 12 u v + 122 v = vă + 21 1 u + 2? ?21 2 u˙ v˙ + ? ?22 2 v˙ = Proof As we noted after Definition 9.1.1, γ is a geodesic if and... The unit-speed condition gives ? ?2 , θ? ?2 = − cos2 θ so along the geodesic we have 2 θ? ?2 dθ = = cos2 θ(Ω? ?2 cos2 θ − 1), dϕ ϕ˙ and hence dθ √ , ? ?2 cos θ Ω cos2 θ − where ϕ0 is a constant The... d? ?2 + cos2 θ d? ?2 22 2 Geodesics We might as well restrict ourselves to unit-speed curves γ(t) = σ(θ(t), ϕ(t)), so that θ? ?2 + ϕ˙ cos2 θ = 1, and if γ is a geodesic the second equation in (9 .2)