Conformal transformations + Schwarzschild

Transcription

1 Intoduction to Geneal Relativity Solutions of homewok assignments 5 Confomal tansfomations + Schwazschild 1. To pove the identity, let s conside the fom of the Chistoffel symbols in tems of the metic tenso deivatives: Γ µ αβ = 1 gµρ [g αρ,β + g ρβ,α g αβ,ρ ]. 1) Then, unde confomal tansfomation g µν = φ g µν, whee φ = φx), we have: g µν,ρ = g µν,ρ φ + g µν φφ,ρ, ) Moeove, g µν = g µν ) 1 and then g µν = φ g µν. 3) Replacing the fome expessions into Eqns. 1) we obtain: Γ µ αβ = 1 gµρ [ g αρ,β + g ρβ,α g αβ,ρ ] = 1 φ g µρ [g αρ,β φ + g αρ φφ,β + g ρβ,α φ + g ρβ φφ,α g αβ,ρ φ g αβ φφ,ρ ] = 1 gµρ [g αρ,β + g ρβ,α g αβ,ρ ] + φ 1 g µρ [g αρ φ,β + g ρβ φ,α g αβ φ,ρ ] = Γ µ αβ + φ 1 [g µ αφ,β + g µ βφ,α g µρ g αβ φ,ρ ] = Γ µ αβ + φ 1 [δ µ αφ,β + δ µ βφ,α g µρ g αβ φ,ρ ] = Γ µ αβ + [δ µ α ln φ,β + δ µ β ln φ,α g µρ g αβ ln φ,ρ ]. 4) Simila expessions can be deived fo the confomal Riemann and Ricci tenso, knowing thei fom in tems of the Chistoffels. Fo the 1

2 cuvatue R µ αβγ we obtain: R µ αβγ = Γ µ αγ,β Γ µ αβ,γ + Γ µ ρβ Γ ρ αγ Γ µ ργ Γ ρ αβ =R µ αβγ + 1 [δµ α ln φ,γ,β + δ µ γ ln φ,α,β + g µσ g αγ ln φ,σ ),β ] 1 [δµ α ln φ,β,γ + δ µ β ln φ,α,γ + g µσ g αβ ln φ,σ ),γ ] [δµ ρ ln φ,β + δ µ β ln φ,ρ + g µσ g ρβ ln φ,σ ] [δ ρ α ln φ,γ + δ ρ γ ln φ,α + g ρσ g αγ ln φ,σ ] 1 4 [δµ ρ ln φ,γ + δ µ γ ln φ,ρ + g µσ g ργ ln φ,σ ] [δ ρ α ln φ,β + δ ρ β ln φ,α + g ρσ g αβ ln φ,σ ] =R µ αβγ + δ µ [γ ln φ,β],α + g µσ g α[γ ln φ,σ ),β] + 1 [ g µσ g α[γ ln φ,β] ln φ,σ + δ µ [βδ ρ γ] ln φ,ρ ln φ,α +δ µ [βg γ]α g ρσ ln φ,ρ ln φ,σ + g µσ ln φ,σ g ρ[β δ ρ ] γ] ln φ,α. 5) Finally, the Ricci tenso R αγ = R µ αµγ eads: R αγ =R αγ + δ µ [γ ln φ,µ],α + g µσ g α[γ ln φ,σ ),µ] + 1 [ g µσ g α[γ ln φ,µ] ln φ,σ + δ µ [µδ ρ γ] ln φ,ρ ln φ,α +δ µ [µg γ]α g ρσ ln φ,ρ ln φ,σ + g µσ ln φ,σ g ρ[µ δ ρ ] γ] ln φ,α =R αγ + δ µ [γ ln φ,µ],α + g µσ g α[γ ln φ,σ ),µ] + [g γα g ρσ ln φ,ρ ln φ,σ + ln φ,γ ln φ,α ] =R αγ 4 ln φ,γ,α + g µσ g αγ ln φ,σ ),µ + [g γα g ρσ ln φ,ρ ln φ,σ + ln φ,γ ln φ,α ]. 6). The Schwazschild metic, defined by the line element dt + d 1 M + dθ + sin θdφ, 7) is chaacteised by two singula points: = 0 and = M. specifically in the fist case: Moe and in the second: 0 g tt and g tt 0, 8) M g tt 0 and g tt. 9)

3 Howeve, it can be shown that = 0 and = M have diffeent natue, being a physical o cuvatue) and a coodinate singulaity, espectively. In the second case, the divegence of one of the metic tenso components is due to a wong choice of coodinate fame, and can be spoiled by finding a moe appopiate one. To check whethe a point is a genuine cuvatue singulaity, we should compute the scalas obtained fom the Riemann tenso, and study thei behaviou. Fo the Schwazschild metic the non vanishing components of R µ αβγ ae given by: R t t = M 3 1 M ) 1, 10) R t θtθ = 1 sin θ Rt φtφ = M 5, 11) R θ φθφ = M 5 sin θ, 1) R θθ = 1 sin θ R φφ = M 5. 13) The fome expessions divege both fo = 0 and = M. Howeve, if we compute the Ketschmann invaiant K = R αβγδ R αβγδ : K = 48M 6, 14) we find that it diveges only at = To pove that the spacetime descibed by the line element ds = dt + 4 ) 9M /3 ) 9M /3 d + t) dω, 15) 9 t) is not dynamical, we will make a coodinate change in ode to tun it into the Schwazschild geomety. We fist define the new vaiable ) 9M 1/3 = t) t) = 3 9M, 16) fom which we get 3 t = 9M dt = d 3 d = d d. 17) 9M M Replacing the fome expession into the metic 15) we obtain: [ ] ds = d M d + 4 [ 9 M ] 1/3 9M 9 3 d + dω = 1 M ) d M d + M dd + dω. 18) 3

4 We need now to make a futhe change of coodinate to cancel the off-diagonal component dd. To this aim, we assume the following ansatz: = τ + f), 19) whee f is a geneic function of the new coodinate, and then d = dτ + f d whee f = df d. 0) Using this elation, Eq. 18) becomes: dτ + f d) M d + dω + M dτ + f d)d = 1 M ) [ dτ + M f 1 M ) ] f ) d M + dω + [ f 1 M ) ] + ddτ. 1) M We immediately note fom the last expession that to cancel the offdiagonal component dτd, the function f) is equied to satisfy the identity f = 1 M ) 1. ) M With this substitution the expession within the squae backets on the thid line of Eq. 1) also eads: M f 1 M ) f ) M = 1 1 M, 3) and the metic takes the final fom: dτ + + d 1 M + dω. 4) The latte descibes the Schwazschild spacetime in the coodinate chat x µ = τ,, θ, φ), ad it is static, as it does not depend explicitly on the time coodinate. 4. Given the Schwazschild metic in spheical coodinates x µ = t,, θ, φ), we conside the equations of motion fo a massless paticle in the equatoial plane θ = π/: E ṫ = 1 M, φ = L, θλ) = π, 5) 4

5 ṙ = E L 1 M ), 6) whee the dot epesents diffeentiation with espect the affine paamete λ which paametises the geodesics. E and L ae the enegy at infinity and the angula momentum of the test paticle. Fist, we ewite the equation fo the adial coodinate as: ṙ = E V ) whee V ) = L 1 M ). 7) We immediately note that V ) 0 fo and V ) fo 0. Moeove, it has a maximum at MAX = 3M, fo which V MAX = L 7M. 8) In ode to study the motion, we also compute the adial acceleation ṙ = dv ) ṙ = 1 dv ) d d, 9) and we assume that the paticle moves fom with ṙ < 0. Let conside thee diffeent scenaios. a) E > V MAX. In this case ṙ > 0 always, and the paticle will move with inceasing velocity, falling into the cental object. b) E = V MAX. The paticle eaches the edge of the potential with deceasing velocity, and such that ṙ MAX = 0. Fom Eq. 9) we see that the acceleation also vanishes, and then the paticle is tapped into a cicula obit with adius = MAX. Howeve this is an unstable obit. In this case unde a small petubation the object will fall into the cental body o escape, depending on wethe it is displaced to < MAX o > MAX. c) E < V MAX. The paticle moves towads the cental object until it eaches a distance 0 > MAX, whee ṙ 0 = 0. This is a tuning point, whee the paticle cannot penetate the potential baie. Since the adial acceleation is positive at = 0, the body is foced to invet its motion and escapes to infinity on a open tajectoy. Summaising, we have found that fo massless paticle thee in only one unstable cicula obit fo which E = L /7M ). 5