Gravitation & Cosmology. Exercises # µ x = 0 (1)

Transcription

1 Gravitation & Cosmology. Exercises # Geoesics a) Show that the Euler-Lagrange equations for the Lagrangian L τ ẋ L µ x = 0 (1) µ L = 1 2 g µνẋ µ ẋ ν (2) are the geoesic equations where, as usual, 2 x µ τ 2 + Γµ νλẋν ẋ λ = 0 (3) Γ µ νλ = gµρ Γ ρνλ Γ µνλ = 1 2 (g µν,λ + g µλ,ν g νλ,µ ) (4) b) Show that L is constant along any geoesic, i.e. that for x µ (τ) a solution of the geoesic equation. τ (g µνẋ µ ẋ ν ) = 0 (5) c) Show that the Euler-Lagrange equations of the Lagrangian L = 1 2 ( θ 2 + sin 2 (θ) φ 2 ) (6) coincie with the geoesic equations for S 2 erive in Lecture 4. a) L x = 1 2ẋρ ẋ ν µ µ g ρν (7) L ẋ = g ρνẋ ρ µ ẋ µ ẋν = g ρν ẋ ρ δµ ν = g ρµ ẋ ρ (8) 1

2 ( ) L = g τ ẋ µ ρµ ẍ ρ + ẋ ρ ẋ ν ν g ρµ = g ρµ ẍ ρ (ẋρ ẋ ν ν g ρµ + ẋ ν ẋ ρ ρ g νµ ) (9) The Euler-Lagrange equations becomes : [E. L.] = g µρ ẍ ρ ( νg ρµ + ρ g νµ µ g ρν )ẋ ν ẋ ρ = g µρ ẍ ρ + Γ µνρ ẋ ρ ẋ ν = 0 (10) an they can be written in the usual (geoesic equations) form by multiplying by g λµ to move the inex µ up : g λµ (g µρ ẍ ρ + Γ µνρ ẋ ρ ẋ ν ) = ẍ λ + Γ λ νρẋ ρ ẋ ν = 0 (11) b) First we compute : ( ) τ L = 1 2 τ g µνẋ µ ẋ ν = 1 2g µν ẍ µ ẋ ν + (ẋ ρ ρ g µν ẋ µ ẋ ν ) 2 (12) Now using the ientity : ρ g µν = Γ µνρ + Γ νµρ (13) together with the fact that x µ (τ) is a solution to the geoesic equation, which means that we also have : ẍ µ = Γ µ νρẋ ν ẋ ρ (14) leaves us with : τ L = 1 2 ( 2g µνγ µ λρẋλ ẋ ρ ẋ ν + ẋ ρ (Γ µνρ + Γ νµρ )ẋ µ ẋ ν ) = 1 2 ( 2Γ νλρẋ λ ẋ ρ ẋ ν + (Γ µνρ + Γ νµρ )ẋ µ ẋ ν ẋ ρ ) = 0 which is obviously zero if we relabel the inices. (15) c) Using the Euler-Lagrange equations with L = 1 2 ( θ 2 + sin 2 (θ) φ 2 ) we get: τ τ ( L L θ θ = 1 2 τ 2 θ 2 sin(θ) cos(θ) φ 2 ) = 0 L L φ φ = τ (sin2 (θ) φ) = sin 2 (θ) φ + 2 cos(θ) sin(θ) θ φ = 0 (16) which are precisely the equations erive in Lecture 4. 2

3 4.2 - Killing vector fiels a) In flat Minkowski spacetime fin ten Killing vectors that are linearly inepenent. b) Give a physical interpretation of the what you foun in a). c) Show that Lorentz transformations of the Killing vector fiels liste in b) give rise to linear recombinations of the same fiels with constant coefficients. a) In the Minkowski spacetime all the Levi-Civita connection coefficients vanish. The Killing equation becomes µ X ν + ν X µ = 0 (17) From this equation it follows that X µ is, at most, of the first orer in x 1. Then, we can have the constant solutions X µ (i) = δµ i (0 i 3) (18) Next, let X µ = a µν x ν, a µν being constant. Equation (17) implies that a µν is anti-symmetric with respect to µ ν. There are six inepenent solutions of this forms, three given by while the others X (j)0 = 0 X (j)m = ε jmn x n (1 j, m, n 3) (19) X (k)0 = x k X (k)m = δ km x 0 (1 k, m 3) (20) Let s check that these are really solutions of the Killing equation. For (19) we only have to check the case in which µ, ν 0 since all the other cases are trivial p X q + q X p = p ε jqn x n + q ε jpn x n = ε jqn δ n p + ε jpn δ n q = ε j(pq) = 0 (21) For (20) we have to check the case in which µ = 0, ν 0 an viceversa 0 X p + p X 0 = δ kp + δ kp = 0 (22) 1 In 2D with coorinates (t, x) the Killing equation can be expresse by t ξ t = 0, x ξ x = 0 an t ξ x + x ξ t = 0. This means that the Killing vector fiels satisfy the following relations ξ t = f(x), ξ x = g(t) an f (x) + ġ(t) = 0 which prove the claim. 3

4 b) The ten inepenent Killing fiels correspon to the ten generators of the Poincaré algebra: four translations, three rotations, an three boosts. Given in terms of the basis vectors e a, we the Killing vector fiels are therefore Translations : e t, e x, e y, e z ; Rotations : y e x x e y, z e y y e z, x e z z e x ; Boosts : x e t + t e x, y e t + t e y, z e t + t e z ; Each of these ten vector fiels manifestly satisfies Killing s equation. Remark In m-imensional Minkowski spacetime (m 2), there are m(m + 1)/2 Killing vector fiels, m of which generate translations, (m 1), boosts an (m 1)(m 2)/2, space rotations. Those spaces (or spacetimes) which amit m(m + 1)/2 Killing vector fiels are calle maximally symmetric spaces. c) Since every Lorentz transformation can be built from infinitesimal ones, it is sufficient to emonstrate the claim for infinitesimal Lorentz transformations. An this makes our work exceptionally easy. Infinitesimal Lorentz transformations are simply the ientity plus a constant multiple of the generators of the Lorentz algebra; but (the last six of) the Killing fiels liste in b) are nothing but these Lorentz generators. Therefore, any infinitesimal Lorentz transformation of the Killing fiels liste in b) is a linear combination of those same Killing fiels with constant coefficients. An by extension, the same is true for any finite Lorentz transformation. 4

5 4.3 - Killing vector fiels II a) Show that if ζ a (x) is a Killing vector fiel an p a (λ) is the tangent vector to a geoesic curve γ(λ), then p a ζ a (x) is constant along γ. b) Show that if ζ a an η a are Killing fiels an α, β constants, then αζ a + βη a is Killing. a) The erivative of p a ζ a along γ is p b b (p a ζ a ) = p a p b b ζ a + ζ a p b b p a (23) The first term vanishes because p a p b is symmetric while b ζ a is antisymmetric (because it is Killing). The secon term vanishes because p a is the tangent of a geoesic, which practically by efinition implies that it obeys the geoesic equation, p b b p a = 0. b) By irect check: b (αζ a + βη a ) = α b ζ a + β b η a = α a ζ b β a η b = a (αζ b + βη b ) (24) because, being constants, α, β commute with the graient an ζ a, η a are Killing. Therefore equation (24) implies that (αζ a + βη a ) is Killing. 5

6 Extra Consier a pair of twins are born somewhere in spacetime. One of the twins ecies to explore the universe; she leaves her twin brother behin an begins to travel in the x-irection with constant acceleration a = 10m/s 2 as measure in her rocket frame. After ten years accoring to her watch, she reverses the thrusters an begins to accelerate with a constant a for a while. a) At what time on her watch shoul she again reverse her thrusters so she ens up home at rest? b) Accoring to her twin brother left behin, what was the most istant point on her trip? c) When the sister returns, who is oler, an by how much? a) There is an obvious symmetry in this problem: if it took her 10 years by her watch to go from rest to her present state, then 10 years of reverse acceleration will bring her to rest, at her farthest point from home. Because of the constant negative acceleration, after reaching her estination at 20 years, she will begin to accelerate towars home again. In 10 more years, when her watch reas 30 years, she will be in the same state as when her watch rea 10 years, only going in the opposite irection. Therefore, at 30 years, she shoul reverse her thrusters again so she arrives home in her homes rest frame. b) To o this, we nee only to solve the equations for the travelling twins position an time as seen in the stationary twins frame. This was one in a previous exercise class but, in brief, we know that her four-acceleration is normal to her velocity: a α u α = 0 everywhere along her trip, an a α a α = a 2 is constant. This leas us to conclue that a t = ut τ = aux a x = ux τ = aut (25) where τ is the proper time as observe by the travelling twin. This system is quickly solve for an appropriate choice of origin 2 : t = 1 a sinh(aτ) x = 1 cosh(aτ) (26) a This is vali for the first quarter of the twins trip. All four legs can be given explicitly by gluing together segments built out of the above. For the purposes of calculating, it is necessary to make aτ imensionless. This is one simply by 2 We consier the twin to begin at (t = 0, x = 1). a = 10 m s 2 = year 1 (27) 6

7 An approximate result 3 coul have been obtaine by thinking of c = m/s an s = 1 year. So the istance at 10 years is simply x(10 yr) = 1 cosh(10.53) = light years. (28) The maximum istance travelle by the twin as observe by her (long-ecease) brother is therefore twice this istance, max(x) = light years. (29) c) Well, in the brothers rest frame, his sisters trip took four legs, each requiring which means that t(10 yr) = 1 sinh(10.53) = years. (30) t total = years (31) In contrast, his sisters time was simply her proper time, or 40 years. Therefore the brother who staye behin is now years oler than his twin sister. 3 Since know c an the number of secons per year to rather high-precision, there is no reason not to use the correct a. Inee, the approximate value of a 1 year 1 gives an answer almost 40% below our answer. 7