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General Relativity 2012 – Solutions

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General Relativity 2012 – Solutions General Relativity 2012 { Solutions Yichen Shi Easter 2014 1. (a) Explain the following terms: covariant derivative, connection components. A covariant derivative r on a manifold M is a map sending every pair of smooth vector fields X, Y to a smooth vector field rX Y , with properties rfX+gY Z = frX Z + grY Z rX (Y + Z) = rX Y + rX Z rX (fY ) = frX Y + (rX f)Y µ In a basis eµ the connection components Γνρ are defined by µ rρeσ = reρ eσ = Γρσeµ The Christoffel symbols are the coordinate basis components of the Levi-Civita connection, which is defined on any manifold with a metric. (b) What does it mean for a connection to be torsion-free? Define the Levi-Civita connection and derive the formula 1 Γµ = gµσ(@ g + @ g − @ g ): νρ 2 ρ σν ν σρ σ νρ A connection r is torsion-free if [ra; rb]f = 0 for any function f. For a torsion-free connection, if X and Y are vector fields then rX Y − rY X = [X; Y ]: On a manifold with a metric, the metric singles out a preferred connection. Let M be a manifold with a metric g. There exists a unique torsion-free connection r, the Levi-Civita connection, which is metric compatible, i.e., rg = 0. Hence we have l l rigjk = 0 = @igjk − Γijgkl − Γikgjl l l ) @igjk = Γijgkl + Γikgjl l l¨¨ l Hl l¨¨ Hl l ) @igjk + @jgik − @kgij = Γijgkl +¨Γikgjl + Γjigkl + ΓjkHgHil −¨Γkigjl − ΓkjHgHil = 2Γijgkl; k since torsion free implies Γ[ij] = 0. 1 ) Γl g = (@ g + @ g − @ g ) ij kl 2 i jk j ik k ij 1 ) Γl = gkl(@ g + @ g − @ g ): ij 2 i jk j ik k ij 1 (c) A 2d Riemannian manifold has metric ds2 = dr2 + f(r)2dφ2 where φ is periodically identified with period 2π (so curves of constant r are circles). (i) Determine the necessary and sufficient conditions on f(r) for the circle r = r0 to have the property that all vectors are invariant under parallel transport around the circle. We want that for any vector field X, a b µ ν µ ν ρ rV X = 0; or, V raX = 0 ) V @µX + V ΓµρX = 0; @ for some V indicating \around the circle". Since @φ is tangent to the circle, we take @ V = : @φ Hence our equation reduces to ν ν ρ @φX + ΓφρX = 0: To find the Γ’s, we start with the Lagrangian L =r _2 + f(r)2φ_2: The equations of motion are 2f 0 f 0 f 2φ¨ + 2ff 0r_φ_ = 0 ) φ¨ + r_φ_ = 0 ) Γφ = Γφ = ; f rφ φr f 0 _2 r 0 r¨ − ff φ = 0 ) Γφφ = −ff : φ φ r ) ν = φ : @φX + ΓφrX = 0; r r φ ) ν = r : @φX + ΓφφX = 0; which give f 0 @ Xφ + Xr = 0; @ Xr − ff 0Xφ = 0: φ f φ r f 2 φ The first gives @φX = − f 0 @φX . Substituting it into the second, we have 2 φ 02 φ @φX + f X = 0: φ 1 2 r And the second gives @φX = ff 0 @φX . Substituting it into the first, we have 2 r 02 r @φX + f X = 0: Hence Xφ(φ) = Xφ(0) cos(f 0φ) + X0φ(0) sin(f 0φ); Xr(φ) = Xr(0) cos(f 0φ) + X0r(0) sin(f 0φ): So we require Xφ(0) = Xφ(2π) and Xr(0) = Xr(2π). This is true iff 0 f = n; n 2 Z: 2 (ii) Deduce that there is a 2-parameter family of functions f(r) for which all circles of constant r have this property. From (i), we have f(r) = nr + c; c = const. (iii) For this 2-parameter family, show that the metric is locally isometric to the Euclidean metric. Is it globally isometric? Locally isometric means we can find coordinates such that the metric looks like ds2 = dr~2 +r ~2dφ~2; r~ =r ~(r); φ~ = φ~(φ): Indeed we have ds2 = dr2 + (nr + c)2dφ2 = dr~2 +r ~2dφ~2 where we have setr ~ = r + c=n and φ~ = nφ. Hence we have a local isometry. We have a global isometry only when n = 1, since now φ~ is periodically identified with period 2πn. 2. Let Ta be tangent to a 1-parameter family of timelike geodesics of the Levi-Civita connection, parame- terized by proper time. Let Sa be a deviation vector for this family. (a) Explain the term deviation vector. Explain why [S; T ] = 0. A 1-parameter family of geodesics is a map γ : I × I0 !M where I and I0 both are open intervals in R, such that for fixed s, γ(s; t) is a geodesic with affine parameter t, and s is the parameter that labels the geodesic. Further, the map (s; t) ! γ(s; t) is smooth and one-to-one with a smooth inverse. A deviation vector Sa is the displacement of two objects travelling along two such neighbouring geodesics such that xa(s + δs; t) ' xa(s; t) + δsSa(s; t) a a a @x (s;t) a @x (s;t) where S (s; t) = @s . Note that T (s; t) = @t . The family of geodesics forms a two-dimensional surface Σ ⊂ M. Treating s and t as coordinates, we can extend them to (s; t; u; :::) defined in a neighbourhood of Σ. This gives a coordinate chart on Σ where S = @=@s and T = @=@t are vector fields satisfying [S; T ] = 0: 3 (b) State and prove the geodesic deviation equation. You may assume the definition of the Riemann curvature tensor R(X; Y )Z = rX rY Z − rY rX Z − r[X;Y ]Z: The geodesic deviation equation is rT rT S = R(T;S)T if r has vanishing torsion. Proof: Vanishing torsion implies rX Y − rY X = [X; Y ]; we have from above [S; T ] = 0; and we have rT T = 0 since T is tangent to affinely parametrised geodesics. Hence R(T;S)T = rT rST − rSrT T − r[T;S]T = rT rT S: (c) Prove that if Sa and T a are orthogonal at one point along a geodesic γ (belonging to the 1-parameter family) then they are orthogonal everywhere along γ. 1 1 r (T · S) = (r T ) · S + T · (r S) = T r S = T r T = r (T · T ) = r (-1) = 0 T T T T S 2 S 2 S since T is timelike. Hence there is no change in the value of T · S along the geodesic. (d) Suppose that spacetime is four-dimensional with Riemann tensor 1 R = R(g g − g g ) abcd 12 ac bd ad bc Show that R must be constant. We first contract the given Riemann tensor to find 1 1 R = R(4g − g ) = Rg : bd 12 bd bd 4 bd Using this, we have the Einstein tensor 1 1 G = R − Rg = − Rg : ab ab 2 ab 4 ab a But r Gab = 0: Hence raR = 0; i.e., R is constant. a a a (e) Assume that S is orthogonal to T . Let f = SaS . Show that, in the above spacetime, 1 K = (r S) (r S)a − Rf T a T 12 is constant along a geodesic γ in the family. 4 1 1 r K = r (r S · r S) − (r R)S · S − Rr (S · S) T T T T 12 T 12 T 1 = 2r r S · r S − RS · r S (since R here is constant) T T T 6 T 1 = 2R(T;S)T · r S − RS · r S T 6 T 1 = 2R T bT cSdT er Sa − RSaT er S abcd e 6 e a 1 1 = R(g g − g g )T bT cSdT er Sa − RSaT er S 6 ac bd ad bc e 6 e a 1 1 1 = RT T SdT er Sa − RT T cS T er Sa − RS T er Sa 6 d a e 6 c a e 6 a e 1 1 1 = R(T Sd)T T er Sa − R(T T c)S T er Sa − RS T er Sa 6 d a e 6 c a e 6 a e 1 1 = RS T er Sa − RS T er Sa (since T Sd = 0 by orthogonality and T is timelike) 6 a e 6 a e d = 0: (f) Obtain a second order differential equation for the evolution of f along γ. Hence deduce that if R > 0 then geodesics which are close initially will diverge exponentially. What happens if R < 0? rT rT f = rT rT (S · S) = 2rT (S · rT S) = 2(rT S) · (rT S) + 2S · rT rT S 1 = 2(K + Rf) + 2S · R(T;S)T 12 1 1 = 2K + Rf + 2Sa R(g g − g g )T bT cSd 6 12 ac bd ad bc 1 1 1 = 2K + Rf + RSTT cSd − RS T T cSd 6 6 c d 6 d c 1 = 2K + Rf (since SdT = 0): 3 d So the differential equation for the evolution of f is d2f 1 − Rf − 2K = 0; dt2 3 from which we can see that if R > 0, the late-time solution is dominated by an exponential growth ∼ exp(pR=3 t), but if R < 0 there are oscillations. 3. In the study of linearised perturbations of Minkowski spacetime, it is assumed that there exist global µ coordinates x with respect to which the metric has components gµν = ηµν + hµν where the components of hµν have absolute values much smaller than 1.
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