Tutorial Exercises: Geometric Connections

Transcription

1 Tutorial Exercises: Geometric Connections 1. Geodesics in the Isotropic Mercator Projection When the surface of the globe is projected onto a flat map some aspects of the map are inevitably distorted. Mapmakers can choose which specific properties to preserve. For navigation at sea the angles between directions are critical. Mercator wanted angles on the map to correspond to angles on the sphere. This will only occur if the metric for the map is isotropic: g ij is a multiple of the identity matrix, or equivalently ds 2 has the factorised form ds 2 = A(x, y(dx (a Starting from stard spherical polar coordinates show that the distance on a sphere of unit radius is given by the metric ds 2 = dλ 2 + cos 2 λdφ 2 where φ is the longitude λ = π 2 θ is the latitude. (b Now consider a transformation of coordinates to the 2D plane to draw maps. Let φ = x let λ be some unknown function of y. Find the relationship between λ y which makes the metric for the map isotropic identify the common coefficient A. It is convenient to make y = 0 the equator. (c Rearrange the result from above to show that y λ are related by the following interesting relations tan λ = sinh y sin λ = tanh y cos λ = sechy When these relationships hold y is said to be the Gudermannian of λ. You should now be able to write the coefficient A as a function of y thus have the metric in terms of the new variables. (d Write down the Lagrangian that is equivalent to the metric you have found. Using first integrals determine the equation for the shape of the geodesics. Show that 1. Solution: sinh y = α sin(x + β where α β are constants is the general solution (by substitution. (a Start from Since λ = π 2 ds 2 = dθ 2 + sin 2 θdφ 2. θ then dλ = dθ cos λ = sin θ thus (b If we change variables the metric becomes ds 2 = dλ 2 + cos 2 λdφ 2. ds 2 = ( dλ cos 2 λdx 2. 40

2 which is isotropic if the two coefficients are equal A = ( dλ 2 = cos 2 λ thus dλ = ± cos λ We choose the positive sign so that the map will have north in the positive y direction (unless we want to make one of those upside-down Australia at the top-of-the-world maps. This first order ODE can be solved by separation to give y = sec λdλ = ln(sec λ + tan λ + C We choose the constant C to be zero, so that the equator λ = 0 corresponds to the horizontal axis y = 0. We also have A(x, y = cos 2 λ but we need to write this is terms of x y which is the next part of the question. (c Re-arrange to give exp(y = sec λ + tan λ 1 exp( y = sec λ + tan λ = sec λ tan λ sec 2 λ tan 2 = sec λ tan λ λ By adding subtracting one can immediately get The given relationships follow from these. We also thus have A(x, y = sech 2 y or (d The equivalent Lagrangian is the equivalent Hamiltonian is tan λ = sinh y sec λ = cosh y ds 2 = sech 2 (y(dx L = 1 2 sech2 (y(ẋ 2 + ẏ 2. H = 1 2 cosh2 (y(p 2 x + p 2 y. Working with the Lagrangian we can identify two first integrals (one for t one for x E = 1 2 sech2 (y(ẋ 2 + ẏ 2 P = sech 2 (yẋ Since we are interested in the shape of the geodesic we will also use for the first derivative dx = ẏ ẋ. Combining these three results gives us the equation 2E P 2 = cosh2 (y[1 + ( dx 2 ] 41

3 This equation can be solved by separation... however since I have told you the solution you can start with that differentiate it implicitly to find Substituting this into the DE gives cosh y = α cos(x + β dx 2E P 2 = cosh2 (y + α 2 cos 2 (x + β = 1 + sinh 2 (y + α 2 cos 2 (x + β then substituting in the expression for sinh(y gives 2E P 2 = 1 + α2 sin 2 (x + β + α 2 cos 2 (x + β = 1 + α 2 Thus the equation does satisfy the DE. One of the two arbitrary constants α is related to E P as given above. The other arbitrary constant arises because you can choose the zero for longitude anyway you like: for example, passing through Greenwich in Engl. By the way, converting the curve back to longitude latitude yields or in spherical-polars tan λ = α sin(φ + β cot θ = α sin(φ + β the SAME equation for a great circle that was obtained in an earlier tute. 2. An Area-preserving map A small element of area on the surface of a unit sphere is given by da sphere = sin θdθdφ, whereas on a flat map an element of area is da flat = dx. (a Write the surface area element da sphere in terms of the latitude λ. (b Find a mapping x = φ y = g(λ from the sphere to a plane that preserves area, da sphere = da flat. (c Starting with the distance between points on the sphere, show that the distance between points on the projected map is given by ds 2 = (1 y 2 dx y 2 2 (d Find the geodesics in terms of an integral. The integral is elliptic so do not try to do it. 2. Solution: (a In terms of the latitude λ the surface area element is trivially cos λdλdφ. We do not need to worry about the minus sign. (b If x = φ y = g(λ then dx = g (λdλdφ. Thus if area is preserved g (λ = cos λ. Thus g(λ = sin λ. (c Thus dx = dφ = cos λdλ = 1 y 2 dλ. Thus since ds 2 = dλ 2 + cos 2 λdφ 2 this gives ds 2 = (1 y 2 dx y

4 (d The equivalent Lagrangian is L = 1 2 [(1 y2 ẋ y 2 ẏ2 ] There are two first integrals associated with t x P = (1 y 2 ẋ we also use which combine to give E = 1 2 [(1 y2 ẋ y 2 ẏ2 ] dx = ẏ ẋ. E = This can be rearranged to give or separating variables to give x = ± We know that a great circle is 1 2(1 y 2 P 2 1 [1 + (1 y 2 2 ( dx 2 ] dx = ±(1 y2 2(1 y P 2 E P 2 P (1 y 2 2(1 y 2 E P. 2 tan λ = cot θ = α sin(φ + β So in terms of x y we have. y ± 1 y 2 = α sin(x + β one can show that this satisfies the DE given above. 43

5 3. Geodesics on a Torus (a Indicate by a geometrical discussion how the parametric equations x = (a b cos θ cos φ y = (a b cos θ sin φ z = b sin θ where 0 < b < a, with 0 θ 2π 0 φ 2π describe a torus. Give the interpretation of a b. Cylindrical coordinates might help you interpret what is going on. (b Show that the line element is given by (c Find the geodesic equation. ds 2 = b 2 dθ 2 + (a b cos θ 2 dφ 2 3. Solution: (a Let s find the equation connecting x, y z. Start with ρ 2 = x 2 + y 2 = (a b cos θ 2 where ρ is the cylindrical radial coordinate. Thus (a ρ 2 + z 2 = b 2 This is the equation of a circle of radius b whose centre lies at a distance a from the origin. The rotational symmetry about the z axis, turns this into a torus, so b is the minor radius a is the major radius. Now, dx = (a b cos θ sin φdφ + b sin θ cos φdθ = (a b cos θ cos φdφ + b sin θ sin φdθ dz = b cos θdθ Taking the sum of squares gives as required. (b The equivalent Lagrangian is ds 2 = (a b cos θ 2 dφ 2 + b 2 dθ 2 L = 1 2 [(a b cos θ2 φ2 + b 2 θ2 ] Now the stard geodesic equations would just be the E-L equations for the above Lagrangian, but more progress can be made from the two conserved quantities J = (a b cos θ 2 φ E = 1 2 [(a b cos θ 2 J 2 + b 2 θ2 ] 44

6 This last result is a first-order DE for θ. Separation of variables will give you an integral that is too nasty to do yet. Alternatively if we want the shape we can use derive the equation dθ dφ = θ φ J 2 E = 1 2 [(a b cos θ2 + b 2 ( dθ dφ 2 ] (a b cos θ 4 Separation of variables here also will give you an integral that is too nasty to do yet. 4. Geodesics on a hyperbolic paraboloid Show that if a, b c are constants the parameters u v satisfy u > b 2 v > a 2 then the equations a x = 2 (u + a 2 (v a 2 c 2 (a 2 + b 2, b y = 2 (u b 2 (v + b 2 c 2 (a 2 + b 2, z = 1 2c (u v + c2 + a 2 b 2 give a parametrisation of part of the hyperbolic paraboloid x 2 a 2 y2 b 2 + 2z c = 1. Find, in terms of u v, expressions for the coefficients E, F G of the line element ds 2 = E du 2 + 2F du dv + G dv Solution: There is nothing to fear, but fear itself! Let s begin x 2 a 2 = (u + a2 (v a 2 c 2 (a 2 + b 2 Subtracting gives y 2 b 2 = (u b2 (v + b 2 c 2 (a 2 + b 2 x 2 a 2 y2 b 2 = (u + a2 (v a 2 (u b 2 (v + b 2 c 2 (a 2 + b 2 = (v u + b2 a 2 (a 2 + b 2 c 2 (a 2 + b 2 = (v u + b2 a 2 c 2 2z c 2 = u v + c2 b 2 + a 2 c 2 Thus adding gives the required result. 45

7 Now, while dx = = a 2 (v a 2 c 2 (a 2 + b 2 (u + a 2 b 2 (v + b 2 c 2 (a 2 + b 2 (u b 2 du 2 + du 2 + dz = 1 (du dv 2c a 2 (u + a 2 c 2 (a 2 + b 2 (v a 2 b 2 (u b 2 c 2 (a 2 + b 2 (v + b 2 dv 2 dv 2 Thus, E = F = G = a 2 (v a 2 4c 2 (a 2 + b 2 (u + a 2 + b 2 (v + b 2 4c 2 (a 2 + b 2 (u b c 2 a 2 2c 2 (a 2 + b 2 + b 2 2c 2 (a 2 + b 2 1 2c 2 a 2 (u + a 2 4c 2 (a 2 + b 2 (v a 2 + b 2 (u b 2 4c 2 (a 2 + b 2 (v + b c 2 These simplify to E = u(u + v 4c 2 (u b 2 (u + a 2 F = 0 G = v(u + v 4c 2 (v a 2 (v + b 2 46