Exam to General Relativity Winter term 211/212 Prof. V. F. Mukhanov, LMU München 9.2.212 Solution! Problem: 1a 1b 1c 1d 1e 2 3 4 5 Points:
1 Short Questions In the following questions, mark the correct answer with a cross. There is always only one correct answer. a In general relativity, which of the following statements is true in vacuum? The Riemann tensor is zero. The energy-momentum tensor is undetermined. The Ricci tensor is zero. There are no gravitational waves. 1 Point The Ricci tensor is zero according to the vacuum Einstein equations. b Which of the following is the correct transformation law for the energymomentum tensor T ν µ under a general coordinate change x x? T µ ν x γ = Tα β x γ xα x ν x µ x β T µ ν x γ = Tα β x γ xµ x ν x α x β T µ ν x γ = Tα β x γ xα x β x µ x ν T µ ν x γ = Tα β x γ xα x ν x µ x β 1 Point The correct answer is the fourth. c Let W β ɛ α γδ be an arbitrary tensor. Which of the following quantities does transform like a tensor under coordinate transformations? γ W γ δ W β δ W β γ α γδ ɛ α δδ δ α γδ 1 Point d Which of the following is the correct expression for the proper time in a Minkowski metric with signature η αβ = diag+1, 1, 1, 1? dτ 2 dτ 2 = η αβ dx α dx β = η αβ dx α dx β e Which of the following statements is true in General Relativity? 1 Point At an arbitrary point, the gravitational field can be removed by a coordinate transformation. In some finite neighborhood around a point the gravitational field can be removed by a coordinate transformation. 1 Point 2
2 Exotic spacetimes 1 Points Consider the following spacetime, which is described in the local coordinates x α = t, x, y, z by the metric ds 2 = dt + αxdx 2 dx 2 dy 2 dz 2. Here, x is a periodic coordinate, i.e., we identify x x+l, and αx is a periodic function of x only, satisfying αx + L = αx. a Read off and determine the matrices g αβ and g αβ. Compute the nonvanishing Christoffel symbols. b Show that this spacetime is flat by computing the Riemann tensor. 3 Points 2 Points c Make this flatness manifest by showing the existence of a coordinate system in which all metric components are constant. To this end, consider the coordinate transformation t t = t fx. Which conditions need f to satisfy in order to set αx to a constant A? Give an expression for A as an integral over α. d Consider a particle trajectory x α τ = τ1, k,,. 3 Points Assume that αx equals a constant A > 1. Determine a constant k such that this describes a light ray. 2 Points 3
a g αβ = 1 α α α 2 1 1, gαβ = 1 1 α 2 α α 1 1 1 For the Christoffel symbols Γ γ αβ = 1 2 gγδ α g βδ + β g αδ δ g αβ many are manifestly zero. The non-zero candidates are Γ 11 = g 1 g 1 + 1 2 g1 1 g 11 = 1 α 2 α + 1 2 αα2 1 = 1 α 2 α +α 2 α = α and Γ 1 11 = g 1 1 g 1 + 1 2 g11 1 g 11 = αα 1 2 α2 1 =, where = x. The only non-vanishing Christoffel symbols are Γ 11 = α. b From the definition of the Riemann tensor R δ γαβ = α Γ δ βγ β Γ δ αγ + Γ δ αζ Γ ζ βγ Γδ βζ Γ ζ αγ we see that due to the antisymmetrization in α, β there are no nonvanishing components with only Γ 11 non-zero. Thus, this spacetime is flat. c The coordinate transformation implies dt = dt + f xdx dt + αxdx = dt + f x + αxdx Thus, in order to set αx to a constant A by this coordinate transformation, fx needs to satisfy Integration of this equation gives L f x + αx = A f xdx + L αxdx = LA For f to be a valid coordinate transformation it needs to respect the periodicity, i.e., we have f = fl. The first integral therefore vanishes and we conclude A = 1 L αxdx L 4
d The 4-velocity is given by u α = dxα dτ = 1, k,, and the trajectory describes a light ray if the 4-velocity is null, = g αβ u α u β = g + 2kg 1 + k 2 g 11 This gives a quadratic equation for k, which is solved by 1 + 2Ak + A 2 1k 2 = k = 1 1 + A or k = 1 1 A 5
3 Differential Bianchi identities 1 Points a Prove that for any torsion-free connection the commutator of covariant derivatives gives the Riemann tensor, [ Dα, D β Vγ := D α D β D β D α V γ = R δ γαβ V δ. 1 b Prove that the differential Bianchi identity 4 Points D α R βγδζ + D β R γαδζ + D γ R αβδζ = 2 follows as a consequence of the Jacobi identity [[ Dα, D β, Dγ + [[ Dβ, D γ, Dα + [[ Dγ, D α, Dβ =. 3 4 Points Hint: Act with the left-hand side of 3 on a vector V δ and use 1. You may use the algebraic Bianchi identity R δ γαβ + R δ αβγ + R δ βγα =. c Derive the contracted differential Bianchi identities D α R αβγδ D γ R δβ + D δ R γβ =, D α R αβ 1 2 Rg αβ = where R αβ is the Ricci tensor and R the Ricci scalar. 2 Points 6
a D α D β V γ α β = α D β V γ Γ δ αβd δ V γ Γ δ αγd β V δ α β = α β V γ Γ δ βγv δ Γ δ αγ β V δ Γ ζ βδ V ζ α β = α Γ δ βγ β Γ δ αγ + Γ δ αζ Γ ζ βγ Γδ βζ Γ ζ αγ Vδ = R δ γαβv δ b = [Dα, D β, Dγ + cycl. V δ = [ D α, D β Dγ V δ D γ [ Dα, D β Vδ + cycl. = R ζ γαβd ζ V δ R ζ δαβd γ V ζ + D γ R ζ δαβ V ζ + cycl. = R ζ γαβd ζ V δ + D γ R ζ δαβ Vζ + cycl. The cyclic sum of the first term vanishes due to the algebraic Bianchi identity. Thus the cyclic sum of the second term has to vanish, implying the differential Bianchi identity c Contracting in 4 α, ζ we obtain D α R ζ δβγ + D β R ζ δγα + D γ R ζ δαβ = 4 Contracting further with g δγ we get D α R αδβγ D β R δγ + D γ R δβ = = D α R αβ D β R + D γ R γβ = 2D α R αβ 1 2 g αβr 7
4 Metric Perturbations 1 Points a Prove that the transformation law for metric perturbations δg αβ in an arbitrary curved background spacetime g αβ under an infinitesimal coordinate transformation x µ x µ = x µ + ξ µ is given by δ g αβ = δg αβ g αβ,γ ξ γ g γβ ξ γ,α g αγ ξ γ,β 4 Points Hint: Split the metric into background and perturbation parts as g αβ = g αβ + δg αβ and use the usual transformation law for tensors, i.e. Keep only the terms linear in δg and ξ. g αβ x = xµ x α x ν x β g µνx. 5 b Show that the transformation law derived in a can be written in a compact form as δ g αβ = δg αβ g µβ ξ µ ;α g αµ ξ µ ;β 6 where the semicolon denotes the covariant derivative with respect to the background metric g αβ. 3 Points Hint: Write out the covariant derivatives in 6 explicitly and compare with your result from a. Use the Christoffel symbols Γ µ αβ = 1 2 g µν g αν,β + g βν,α g αβ,ν. c Consider the Friedmann metric g αβ dx α dx β = a 2 η dη 2 δ ij dx i dx j 7 with small perturbations, g αβ = g αβ + δg αβ. Using the transformation law derived in a find the explicit form of the transformed metric perturbations δ g, δ g i, δ g ij in the new coordinate system { x µ }. For this split the infinitesimal vector as ξ µ ξ, ξ i + ζ,i with ξ,i i =. Also use the explicit form of the background metric given in 7. 3 Points Hint: For comparison we give the correct answer for δ g = δg 2a aξ where x = η. 8
a g αβ x = xµ ξ µ x α x ν ξ ν x β g µν x + δg µν = δ µ α α ξ µ δ ν β β ξ ν g µν x + δg µν = g αβ x + δg αβ g µβ α ξ µ g αµ β ξ µ + Oξ 2, δg ξ Then = g αβ x + δ g αβ = g αβ x + µ g αβ ξ µ + δ g αβ δ g αβ = δg αβ µ g αβ ξ µ g µβ α ξ µ g αµ β ξ µ. b δ g αβ = δg αβ g µβ ξ µ,α + Γ µ αλ ξλ g αµ ξ µ,β + Γ µ βλ ξλ = δg αβ g µβ ξ µ,α g µβ ξ λ 1 g µν g αν,λ + g λν,α g αλ,ν 2 g µα ξ µ,β 1 g µα g µν ξ λ g βν,λ + g λν,β g βλ,ν 2 = δg αβ g µβ ξ µ,α g µα ξ µ,β 1 2 ξλ g αβ,λ + g λβ,α g αλ,β + + g βα,λ + g αλ,β g βλ,α = δg αβ µ g αβ ξ µ g µβ α ξ µ g αµ β ξ µ. c δ g = δg ξ µ µ g g µ ξ µ g µ ξ µ = δg ξ g 2 g ξ = δg ξ 2aa 2a 2 ξ = δg 2a aξ δ g i = δg i ξ µ,i g µ ξ µ, g µi ξ λ g i,λ =δg i ξ,i g ξ j, g ij =δg i ξ,ia 2 ξ j + ζ,j δ ij a 2 [ =δg i + a 2 ξ i + ζ ξ,i where I have used the convention ξ i ξ i and ζ,i ζ,i. d δ g ij = δg ij g ik ξ k,j g kj ξ k,i ξ g ij, =δg ij + a 2 ξ i,j + a 2 ξ j,i + ξ 2aa δ ij [ =δg ij + a 2 2 a a δ ijξ + 2ζ,ij + ξ i,j + ξ j,i 9
5 Perturbations of the energy-momentum tensor 5 Points The background metric of a flat FLRW universe can be written as: g αβ dx α dx β = a 2 η dη 2 δ ij dx i dx j, 8 where η is the conformal time. For this metric, the Einstein tensor is: G = 3H a 2 G i =, where H = a /a and x = η. G i j = 1 a 2 2H + Hδ i j, 9 a Consider fluctuations of the energy-momentum tensor around a background, Tβ α = Tβ α + δt β α. As in exercise 4, the fluctuations transform under infinitesimal coordinate transformations x µ x µ = x µ + ξ µ as δ T α β = δt α β T α β,γ ξ γ T α γ ξ γ,β + T γ β ξα,γ. 1 Determine the transformation rules of δt, δti and δtj i the parameter ξ µ ξ, ξ i + ζ,i with ξ,i i =. with respect to Hint: Use the background Einstein equations G α β = 8πG Tβ α. 2 Points b The metric fluctuations transform as follows B B ζ + ξ E E ζ 11 Verify that the quantities δt = δt + T B E δt i = δti + T Tk k /3 B E,i 12 are gauge invariant. δ = δ + B E Solution: 3 Points a By using the background Einstein equations, we can immediately infer from eqs.9 that: T i =, δ i j 13 and that the spatial derivatives of T β α vanish. We can now calculate the components of the energy-momentum tensor fluctuation from eqs. 1: δ T = δt T,γ ξ γ T γ ξ γ, + T γ ξ,γ 1
= δt T ξ T ξ + T ξ = δt T ξ 14 δ T i = δt i T i,γ ξ γ T γ ξ γ,i + T γ i ξ,γ = δti T ξ,i + 1 Tk k δ j i 3 ξ,j = δti T Tk k /3 ξ,i 15 δ = δ,γ ξ γ T i γ ξ γ,j + T γ j ξi,γ = δ ξ T i k ξ k,j + T k j ξ i,k = δ ξ 16 b Applying the transformation rules derived above and eqs.11, we find: δt δt T ξ + B T ζ + ξ E + ζ = δt + T B E 17 δt i δti T Tk k /3 ξ,i + B T Tk /3 k ζ + ξ E + ζ,i = δti + T Tk k /3 B E,i 18 δ δ ξ + B ζ + ξ E + ζ = δ + B E 19 11