+ dxk. dt 2. dt Γi km dxm. . Its equations of motion are second order differential equations. with intitial conditions

Transcription

1 Homework 7. Solutions 1 Show that great circles are geodesics on sphere. Do it a) using the fact that for geodesic, acceleration is orthogonal to the surface. b ) using straightforwardl equations for geodesics c) using the fact that geodesic is shortest. a) See Lecture notes, subsection.. b) ) The straightforward equations for geodesic: d x i + dxk Γi km dxm = 0 are just equation of motion for free Lagrangian on the Riemannian surface. Hence in the case of sphere the are equations of motion of the Lagrangian of free particle on the sphere is L = R θ +R sin θ ϕ. Its equations of motion are second order differential equations θ sin θ cos θ ϕ = 0 ϕ + cotan θ θ ϕ = 0 with intitial conditions θt) t=0 = θ 0, θt) t=0 = a ϕt) t=0 = ϕ 0, ϕt) t=0 = b 1) for geodesics θt), ϕt) starting at the initial point p = θ 0, ϕ 0 ) with initial velocit v 0 = a θ + b ϕ. All Christoffel smbols vanish except Γ θ ϕϕ = sin θ cos θ, and Γ ϕ ϕθ = Γϕ θϕ = cotan θ.) This differential equation is not ver eas to solve in general case. On the other hand use the fact that rotations are isometries of the sphere. Rotate the sphere in a wa such that the initial point transforms to the point the point θ 0 = π, ϕ 0 = 0 and then rotate the sphere with respect to the axis 0X such that θ-component of velocit becomes zero. We come to the same differential equation but with changed initial conditions: θ sin θ cos θ ϕ = 0 ϕ + cotan θ θ ϕ = 0 with intitial conditions θt) t=0 = π, θt) t=0 = 0 ϕt) t=0 = 0, ϕt) t=0 = Ω 0 ) where we denote b Ω 0 the magnitude of initial velocit. One can eas check that the functions θt) = π ϕt) = Ω 0 t are the solution of the differential equations for geodesic with initial conditions ). Hence this is geodesic passing through the point θ 0 = π, ϕ = 0) with initial velocit Ω 0 ϕ. We see that this geodesic is the equator of the sphere. We proved that an arbitrar geodesic after appling the suitable rotation is the great-circle equator. On the other hand an equator is the great circle the intersection of the sphere x + + z = R with the plane z = 0) and the rotation transforms the equator to the another great circle. Hence all arcs of great circles are geodesics and all geodesics are arcs of great circles. c) See the lecture notes the subsection.4.1 Again on geodesics on sphere and on Lobachevsk plane.) Consider in E a vector X = attached at the point p: x = R, = 0, z = ) R. Consider also a sphere x + + z = R in E and the following two curves: a curve C 1 which is the intersections of this sphere with plane = 0 and a curve C 1 which is the intersections of this sphere with the plane z = R. Both curves C 1, C pass through the point p. Show that the vector X is tangent to the sphere and express this vector in spherical coordinates. Describe the parallel transport of the vector X along these closed curves. What will be the result of parallel transport of the vector X along these closed curves? 1

2 The vector X 0 = attached at the point p = x = R, = 0, z = ) R of the sphere x + +z = R is tangent to this sphere since X0 x + + z R ) = 0: X x + + z R ) p = 0 = x + + z R ) p = p = 0. The vector X 0 = attached at the point p is proportional to the vector ϕ : X 0 = b 0 ϕ. The spherical coordinates of the point p are θ 0 = π 6 cos θ 0 = ). Since the metric on the sphere is G = R dθ + sin θdϕ ) then calculating the length of the vector X 0 in ambient space in Cartesian coordinates and on the sphere in spherical coordinates we come to X 0 = = 1 = R sin θ 0 dϕ b ) = Rb ϕ b = 1 = R sin θ 0 R, X 0 = ϕ = R sin θ 0 R ϕ. Using the formula X = dxt) + ẋ k Γ i km Xm = 0 for parallel transport we see that for an arbitrar curve θt), ϕt) on the sphere starting at the initial point θt 0 ) = θ 0, ϕt 0 ) = ϕ 0 the parallel transport Xt) = at) θ + bt) ϕ of the initial vector X t0 = a 0 θ + b 0 ϕ over the curve C is defined b the first order differential equation: ȧ + ϕγ θ ϕϕ b = ȧ ϕ sin θ cos θb = 0 ḃ + ϕγ ϕ ϕθ a + θγ ϕ θϕ b = ḃ + ϕa + θb)cotan θ = 0 with initial conditions at0 ) = a 0 bt 0 ) = b 0 ) We use here the information about Christoffel smbols of sphere.) Appl this equation for both curves For the first curve there is another much more clear solution: see below.) First consider the curve C 1. For the curve C 1 in spherical coordinates: θt) = θ 0 + t, ϕt) = 0, θ = 1, ϕ = 0. The differential equation ) becomes: ȧ ϕ sin θ cos θb = ȧ = 0 ḃ + ϕa + θb)cotan θ = ḃ + bcotan π 6 + t) = 0 with initial conditions at0 ) = 0 bt 0 ) = b 0 = R a) 1 The solution of this equation is at) = 0, bt) = R sin θt), X = 1 R sin θt) ϕ = 1 R sint+ π 6 ) ϕ. This answer is evident without an calculation: Namel the vector X = during transport along the curve C 1 in the ambient space E, remains tangent to the sphere. Hence Xt) = at an point of this circle. The vector 1 at point θt) equals to. We come to the answer. R sin θt) ϕ Note that during parallel transport along the curve C 1 the final vector coincides with initial one. For the curve C. This curve in spherical coordinates is θt) = θ 0 = π 6, ϕt) = t, θ = 0, ϕ = 1. Hence the equation ) for parallel transport becomes: ȧ ϕ sin θ cos θb = ȧt) sin θ0 cos θ 0 bt) = 0 ḃt) + at)cotan θ 0 = 0 with initial conditions at0 ) = 0 bt 0 ) = b 0 ) Here we have two first order differential equations. Introduce b t) = sin θ 0 bt). We have dat) cos θ 0 b ) ) ) t) = 0 db t) + cos θ 0 at) = 0, i.e. d at) 0 1 at) b = cos θ t) b t) or in other wa d a + ib ) = i cos θ 0 a + ib )

3 The solution is at) + ib t) = e it cos θ 0 a 0 + ib 0) = cost cos θ 0 ) i sint cos θ 0 ))a 0 + ib 0) In particular as a result of parallel transport along the closed latitude that is the vector rotates on the angle Φ = π cos θ 0. at) + ib t) t=π = e πi cos θ 0 a 0 + ib 0) We see that both answers for curves C 1 and C are in coincidence with the formula of rotation of the vector during parallel transport along the closed curve: Φ = Kdσ. Indeed for the sphere K = 1 R. For the first curve C 1 the first case Φ = Kdσ = 1 R πr = π and for the curve C Φ = Kdσ = 1 R πr 1 cos θ 0 ) = π1 cos θ 0 ) Show that vertical lines x = a are geodesics non-parameterised) on Lobachevsk plane. We consider here the realisation of Lobachevsk plane hperbolic plane) as upper half of Euclidean plane x, ): > 0} with the metric G = dx +d. Consider second order differential equations defining geodesics with initial conditions such that horisontal velocit equals to zero: we uses the information from Homework 6 or from Lecture notes about Christoffels for Lobachevsk plane: Γ x xx = 0, Γ x x = Γ x x = 1, Γx = 0, Γ xx = 1, Γ x = Γ x = 0, Γ = 1.) x ẋẏ = 0 + ẋ ẏ = 0 xt) t=t0 = x 0, ẋt) t=t0 = 0 t) t=t0 = 0, ẏt) t=t0 = ẏ 0 This equation has a solution and it is unique. One can see that if we put xt) 0, i.e. curve is vertical then we come to the equation ẏ = 0. Solution of these equation gives curve x = x 0, = t): ẏ = 0. The image of this curve clearl is vertical ra x = x 0, > 0. 4 Show that the following transformations are isometries of Lobachevsk plane: a) horizontal translation r r + a where a = a, 0), b) homothet: r λr λ > 0), c) inversion with the centre at the points of the line x = 0 : r a + r a where a = a, 0): r a x = a + x a = x a) +. x a) + We have to show that Riemannian metric G = dx +d remains invariant under these transformations. a) horizontal translation. If x x + a, then dx and d do not change. Hence G is invariant under horisontal translations. b) Homothet. If x λx, λ where λ > 0 is a constant then dx +d λ dx +λ d λ = dx +d does not change too. c) inversion with the centre at the points of the line = 0. Since we proved that horisontal translation is isometr it suffices to consider inversion with centre at the point x = = 0: r r x r : = x x + = x + This line is called absolute.

4 Now b straightforward calculations one can show that dx +d = dx +d. To avoid the straightforward calculations consider coordinates r, ϕ: x = r cos ϕ, = r sin ϕ then dx + d = dr + r dϕ r sin ϕ = 1 dr sin ϕ r + dϕ sin ϕ = 1 sin ϕ d log r) + dϕ sin ϕ. Under transformation of inversion in these coordinates ϕ does not change, r 1 r, log r log 1 r. It is evident that in coordinates u = log r, ϕ metric does not change. Hence inversion with centre at the point x = = 0) is isometr. Hence an inversion with a centre at the arbitrar point a, 0) and with an arbitrar radius is isometr, since horisontal translation and homothet are isometries. In complex cooridinates this is so called Mobius transformation transformation z 1 z learned in the course Hperbolic Geometr.) which ou 5 Show that upper arcs of semicircles x a) + = R, > 0 are non-parametersied) geodesics. You ma do this exercise solving explicitl differential equations for geodesics, but it is much more nice to use inversion Mobius ) transformation studied in the previous exercise: Consider the inversion of the Lobachevsk plane with the centre at the point x = a R, = 0 see the exercise above). This inversion does not change Riemannian metric, it is isometr. Isometr transforms geodesics to geodesics. On the other hand it transforms the semicircle x a) + = R, > 0 to the vertical ra x = a R + 1 R, > 0. This can be checked directl. On the other hand the vertical ra is geodesic. Hence the initial curve was the geodesic too. 6 Let ABC be triangle formed b geodesics on the sphere of the radius R. Express the area of this triangle via its angles. Do the previous exercise for the triangle on the Lobachevsk plane. We use 1) the Theorem from Lecture notes about the rotation of the vector during parallel transport along closed curve C = D: Φ = D Kdσ a ) the fact that velocit vector is covariantl constant along geodesic ) during parallel transport scalar product and hence the angle between two vectors does not change. Let ABC be the triangle formed b arcs of geodesics. We denote b α, β, γ angles at the vertices A, B, C. Let vector A 0 be velocit vector at the vertex A along the arc AB. Under parallel transport it becomes the vector A 1 which is the velocit vector of the arc AB attached at the vertex B. Take the vector B 0 which is velocit vector at the vertex B along the arc BC. The vector B 0 has the angle β with the vector A 1. Do the parallel transport of two vectors A 1 and B 0 along the arc of geodesic BC. We arrive at the vertex C with the vector B 1 which is the velocit vector along the curve BC at the vertex C and the vector A which has the angle β with the vector B 1 since angles do not change during parallel transport. Now take the vector C 0 which is velocit vector at the vertex C along the arc CA. The vector C 0 has the angle γ with the vector B 1 and it has the angle β + γ with the vector A. Now make the final parallel transport along the arc of geodesic CA of three vectors: vector C 0, the vector B 1 and the vector A. We come to three vectors: vectors C 1, vector B and the vector A at the vertex A. The vector C 0 will transform to the vector C 1 which is the velocit vector of the curve CA at the vertex A. The vector B has the angle β with the vector C and the vector A has the angle α with the vector C 1. Hence the vector A has the angle β + γ with the vector C 1. The vector C 1 has the angle π α with initial velocit vector A 0. We see that after parallel transport along the geodesics the vector A 0 becomes the vector A which has the angle α + β + γ π with the initial vector A 0. Now using the Theorem we come to α + β + γ π = Kdσ 4 ABC

5 For the sphere we have K = 1 R, hence: α + β + γ π = For the Lobachevsk plane K = 1 and α + β + γ π = ABC ABC Kdσ = Area of spheric triangle ABC R Kdσ = Area of spheric triangle ABC) 7 Let Xt) be parallel transport of the vector X along the curve on the surface M embedded in E, i.e. v X = 0, where v is a velocit vector of the curve C and Levi-Civita connection induced connection) on the surface. Compare this condition v X = 0 for internal observer) with the condition for external observe that for the vector Xt) dxt) is orthogonal to the surface ). We know that Levi-Civita connection equals to the connection : X Y = can.flat X Y) tangent. Show that the definitions of External and Internal Observers are equivalent. Denote b v the velocit vector of curve C on the surface M. For external observer parallel transport means that dx = v X is colinear to the normal vector to the surface. This is equivalent to the fact that tangential component equals to zero: v X tangent = 0. According to definition of induced Levi-Civita) connection this means that v X = 0. Thus we prove that both observers have equivalent definitions. 8 Let r = rt) be an arbitrar geodesic on the Riemannian manifold. Show that magnitudes I = g ik ẋ i ẋ k is preserved along geodesic. Let r = rt) be an arbitrar geodesic on the sphere. Show that magnitudes I = sin θ ϕ and E = sin θ ϕ + θ are preserved along geodesics. Let r = rt) be an arbitrar geodesic on Lobachevsk plane. Show that magnitudes I = v x are preserved along geodesics. There are two fundamental facts: 1. For Lagrangian L = Lx i, ẋ i ) the magnitude E = ) ẋ k ẋk L and E = v x +v is preserved on equations of motions. This magnitude is called energ. More in details this means that for an x i = x i t) which is the solution of Euler-Lagrange differential equations d ) ẋ = the magnitude k x k ) Et) = E x i t) = ẋ k ẋk L x i t) is a constant. The proof is immediate: using Euler-Lagrange equations of motions we come de = d ) ẋ k ẋk L = d ) ẋ k ẋ k + ẋ k x k dl = x k ẋk + ẋ k x k dl = dl dl = 0. if x i = x i t) are solutions of Euler-Lagrange equations. The second fundamental fact is that if Lagrangian L = Lx i, ẋ i ) does no depend on variable x m x m = 0) then the magnitude p m = x m ) We defined parallel transport in Geometr course using the second condition 5

6 is preserved on equations of motions, i.e. for an x i = x i t) which is the solution of Euler-Lagrange differential equations d ) ẋ = the magnitude k x k p m t) = p m x i t) = ẋ m x i t) is a constant. The proof is even simpler than for the first fact: using Euler-Lagrange equations of motions we come dp m t) = d ) ẋ m x i t) = x m = 0. The magnitude p m is called generalised momentum corresponding to the coordinate x m ) Now it is ver eas to solve the problems for sphere and Lobachevsk plane. Geodesics are solutions of Euler-Lagrange equations of motion for Lagrangian L = gpqẋp ẋ q. The energ E = ) ẋ k ẋk L = L L = L is preserved. Hence the Lagrangian itself preserves on geodesics. We see that for sphere the Lagrangian L = g pqẋ p ẋ q = 1 θ + sin θ ϕ ) is preserved and for Lobachevsk plane the Lagrangian L = g pqẋ p ẋ q = ẋ +ẏ is preserved. The Lagrangian of the sphere L = 1 θ + sin θ ϕ ) does not depend explicitl on ϕ, hence I = ϕ = sin θ ϕ is preserved. The Lagrangian of the Lobachevsk plane L = ẋ +ẏ does not depend explicitl on x, hence I = ẋ is preserved. 6