Lecture 29: Cosmology Cosmology Reading: Weinberg, Ch A metric tensor appropriate to infalling matter In general (see, eg, Weinberg, Ch ) we may write a spherically symmetric, time-dependent metric in the form (dτ) 2 = B(r, t) (dt) 2 A(r, t) (dr) 2 r 2 (dθ) 2 + r 2 sin 2 θ (dϕ) 2 (29) and from this we deduce R tt = B 2A + B A 4 A A + B B Ȧ r + 2A Ȧ Ȧ 4 A A + B B R rr = B 2B B A 4 B A + B B A ra + A 2B Ȧ Ȧ 4 B A + B B r B R θθ = + 2A B A A + A (292t) (292r) (292θ) R ϕϕ = R θθ sin 2 θ (292ϕ) R tr R rt = Ȧ ra (293) with all other terms = 0 We now want to solve the Einstein equations in the following cases: Empty space Equation 293 implies that Ȧ = Ȧ = 0, hence A is time-independent By the methods used earlier, we find AB =, therefore B is time-independent also Thus we recover the Schwarzschild metric, A(r) = 2MG r (294) This reflects a specific choice of time coordinate, as before 37
A metric tensor appropriate to infalling matter This proves Birkhoff s Theorem----a spherically symmetric metric in empty space is time-independent Thus there can be no gravitational radiation from a spherical source 2 Dust to dust Let us redefine r and t to get co-moving coordinates, appropriate to an oberver falling freely with some particular piece of matter: (dτ) 2 = (dt) 2 U(r, t) (dr) 2 (r, t) (dθ) 2 + sin 2 θ (dϕ) 2 (295) then with this metric, U R tt = 2U + U 2 4 U 2 2 2 2 R rr = 2 R θθ = + 2U U U + 2 U + U 2 U 4U 2 2 U 4U 4U U 2 (296t) (296r) (296θ) R ϕϕ = R θθ sin 2 θ R tr = 2 2 U 2U and all other components vanish (296ϕ) (296tr) Dust can be defined as a group of particles with energy density but no pressure An example is a large cluster of well-spaced galaxies The energy-momentum tensor of dust is T µν = ρ U µ U ν (297) The invariant volume element is dt dr dθ dϕ g = dt dr dθ dϕ sinθ U (298) In co-moving coordinates, there is no local motion of a particle, hence U t U µ U r 0 = U θ = 0 U ϕ 0 Of the 4 equations T µν ; ν = 0, (299) the space components are automatically satisfied: T kν ; ν = 0 (290) leaving 38
Lecture 29: Cosmology or T 0ν ; ν = 0 (29) g x ν (T tν g) + t σ ν T σν = U t (ρ U) + ρ t t t = 0 (292) and since t t t = 2 gtt t g tt = 0, we have (ρ U) = 0 (293) t The gravitational field equations become U R tt = 2U + U 2 4 U 2 2 2 2 = 4πGρ R rr = 2 R θθ = + 2U (294t) U U + 2 U + U 2 4U U = 4πGρ (294r) 2 U 4U 2 2 U = 4πGρ (294θ) 4U R ϕϕ = R θθ sin 2 θ = 4πGρ sin 2 θ (294ϕ) R tr = 2 2 U 2U = 0 To solve these, multiply Eq 294tr by and divide by, so d dt ln( ) d 2 dt ln(u) = 0 or (294tr) = F(r) U (295) where F(r) is some arbitrary----for the moment----function of r Next add Eq 294r and Eq 294t and subtract twice Eq 294θ, to get 2 2U 2 + 2 + 2 2 2 2 = 0 (296) Combining Eq 296 with Eq 295 we get 2 F2 (r) + 2 + 2 2 2 = 0 39
A metric tensor appropriate to infalling matter We see that 2 2 is independent of t This suggests 2 (r, t) = R 2 (t) Γ(r), and thence 2 Γ(r) 2R 2 + 2ṘṘ R 2 F(r) + 2 = 0 (297) 2 ie, R 2 + 2ṘṘ df = constant = k (298) From Eq 295 we have Γ (r) R(t) = F(r) Γ(r) U (299) which suggests U(r, t) = R(t) f(r) We are at liberty to redefine r so that Γ(r) = r 2 ; thus from Eq 297 F 2 (r) = 4 ( kr 2 ) (2920) and from Eq 299 4 f(r) = F 2 (r) = ( kr2 ) (292) The metric now has the form (dτ) 2 = (dt) 2 R 2 (t) (dr) 2 This is called the Robertson-Walker metric kr 2 + r2 (dθ) 2 + sin 2 θ (dϕ) 2 (2922) We suppose that the density varies with time only (since the radial and angular velocity components vanish) Then energy conservation becomes t (ρ U) = r2 f(r) t ρ(t) R3 (t) = 0 (2923) from which we deduce ρ(t) = ρ(0) R3 (0) R 3 (t) (2924) 40
Lecture 29: Cosmology Time evolution of dust By convention, R(0) = From Eq 294t, U R tt = 2U + U 2 4 U 2 2 2 2 = 4πGρ, we find ṘṘ 4πGρ(0) 3 = 4πGρ(t) = R 3 (t) We also had R 2 + 2ṘṘ = k which when combined with Eq 2925 gives k + R 2 = 8πGρ(0) 3R(t) If Ṙ(0) = 0, then we may re-write Eq 2926 as 2 2 (2925) (2926) Ṙ(t) = 8πGρ(0) 3 R(t) (2927) where we choose the negative root in order to describe collapse That is, the dust, initially in some distribution with ρ(0) 0, falls freely inward To simplify Eq 2927, let R(t) = ( + cosψ) 2 Then 2 ψ sinψ = λ cosψ + cosψ 2 λ sinψ + cosψ (2928) which can be integrated simply to give (note ψ(0) = 0 ) (ψ + sinψ) = λt (2929) 2 Eq 2929 describes a cycloid: The time to collapse is evidently T collapse = π 2λ (2930) Since, if k>0, the Robertson-Walker metric demands r 2 < k = 3 8πGρ(0) we see that the collapse time is 4
Outside the ball of dust T = π 2 r max c To recapitulate, a ball of dust (ie p=0), initially at rest, will collapse to a point, under its mutual gravitational attraction, in time T Outside the ball of dust It is possible to put the metric outside in Schwarzschild form (dτ) 2 = B(r ) (dt ) 2 A(r ) (dr ) 2 r 2 (dθ) 2 + sin 2 θ (dϕ) 2 (293) To do this, we need to match up at the surface Let θ = θ, ϕ = ϕ and r = r R(t) Then dr = R dr + r Ṙ dt and we define the outside time by t = ka 2 k 2 S(r, t) S(r, t) = kr 2 ka 2 2 dr ka 2 R R R If we fit at r = a (the radius of the dustball) then we have ka 3 B(a, t ) = ar(t) A(a, t ) = ka 3 ar(t) 2 (2932a) ( R(t)) (2932b) (2933) (2934) This matches the outside solution if 2MG = ka 3 But from Eq 2926 we have k = 8πGρ(0) 3 so the condition is M = 4πa3 3 ρ(0), ----not a very surprising result! see Weinberg, p 345 42
Lecture 29: Cosmology Collapse seen by outside observer We now ask what the collapse looks like to a distant observer We see that if light is emitted radially from the surface of the star at (outside) time t 0, it propagates according to dτ = 0, or dt = A(r ) dr hence r t = t 0 + a R(t) dr 2MG r (2935) We see that as R(t) approaches 2MG a (that is, as the surface approaches the Schwarzschild radius, the time for the light to reach the observer becomes (logarithmically) infinite The gravitational red shift of light reaching the observer becomes df dt z = = dt 0 a Ṙ(t) dt dt 2MG a R(t) (2936) hence as the radius reaches r S, z exp t 2MG (2937) For most of the star s life, r >> r S and t t, ie the redshift is essentially zero But as the end of the collapse approaches, an outside observer sees an exponentially increasing redshift, ie the star disappears into redness, with a time scale of minutes The further collapse to R(T) = 0 is invisible to an outside observer A co-moving observer has no difficulty seeing the collapse to R = 0 His time becomes disconnected from that of the outside world after he passes within the Schwarzschild radius The surface r = r S represents a trapped discontinuity that separates inside from outside Stuff can fall in, but it can never get out again, in classical General Relativity Model universes The most appropriate metric for cosmology is the spatially homogeneous Robertson-Walker metric (dτ) 2 = (dt) 2 R 2 (t) (dr) 2 kr 2 + r2 (dθ) 2 + sin 2 θ (dϕ) 2 (2922) that arises automatically from co-moving coordinates The Robertson-Walker metric embodies the idea that at fixed t (spacelike hypersurface) any point is equivalent to any other point The curvature of the 3-dimensional hypersurfaces t = const is K 3 (t) = k R 2 (t) By rescaling r and R(t) k can be assming tidal forces do not exceed his personal Roché limit 43
Positive curvature normalized to ±, if k 0 Thus, a space of positive curvature K 3 kas k = +, and a space of negative curvature has k = Positive curvature When k = +, the proper circumference of the space is L 3 = 2π R(t) and the proper volume is 3 = 2π 2 R 3 (t) (2938a) (2938b) At fixed t the universe is the surface of a 3-sphere of radius R(t) embedded in a Euclidean 4-dimensional manifold, so R(t) is the radius of the universe Space is finite, but unbounded (since (dr) 2 ( kr 2 ) ) Zero curvature When k = 0, we say space is flat (in an average or global sense) Flat space is infinite, since the 3-dimensional hypersurfaces t = const are open Negative curvature When k = space is also infinite because a negatively curved hypersurface K 3 = const < 0 is open Influence of matter In isotropic 3-space T 00 must be scalar with respect to transformations of r, θ, ϕ ; hence T 00 = ρ(t) (2939a) T k0 = 0 T jk = p(t) g jk (2939b) (2939c) We can define a flux of galaxies J G µ : J G 0 = n G (t) (2940a) J G µ = n G U µ (2940b) and then 44
Lecture 29: Cosmology T µν = p g µν + (p + ρ) U µ U ν (294) and as before, U 0 = U k = 0 Conservation of galaxies may be written g 2 t g 2 n G = 0 (2942) and with g = R 6 (t) r4 sin 2 θ kr 2 we find, unsurprisingly, n G (t) R 3 (t) = const (2943) Conservation of energy-momentum, T µν ; ν = 0 implies R 3 (t) p t = t R3 (t) (p + ρ) (2944) If pressure is negligible, then as for the dust model of a collapsing star, ρ(t) R 3 (t) = const (2945) Note that 2945 and 2943 are inequivalent unless we neglect pressure Proper distances Imagine observers in galaxies along a line of sight to some distant galaxy at r n, at some cosmic time t Each measures the distance to the next galaxy by----say----the travel time for a light signal Then the sum of the distances along the line of sight would be n ds n = n 2 (dr n ) 2 R 2 (t) kr 2 (2946) or if we assume the observers closely spaced relative to the overall distance, df r D proper (t) = dr g rr r 2 = R(t) 0 dr kr 2 2 (2947) 0 45
Cosmic red shift Cosmic red shift The equation of motion of light is dτ = 0, or dr dt = R(t) kr 2 (2948) hence if the light leaves r at t and arrives at r=0 (Earth) at time t 0 we have t 0 r dt R(t) = 0 t dr kr 2 = f(r ) (2949) The right side of Eq 2949 is independent of time For nearby galaxies, kr 2 << so f(r ) r Assume the next wave crest leaves at t + δt and arrives at t 0 + δt 0 ; then so t 0 + δt 0 t + δt δt 0 R(t 0 ) t 0 dt R(t) = t dt R(t) = f(r ) (2950) δt R(t ) = 0 (295) But since the time between successive wave crests, at a fixed location, is df δt = ν = λ c, we may write λ 0 = R(t 0 ) λ R(t ) and thus the red-shift z defined in 2936 above becomes z = λ 0 λ = R(t 0 ) R(t ) (2952) λ R(t ) In an expanding universe, t < t 0 so that R(t ) < R(t 0 ) and we get a cosmological red-shift, z > 0 For a nearby galaxy, we should say the proper distance is D(t) R(t) r and that its radial velocity is therefore v rad = Ḋ(t) = Ṙ(t) r (2953) But since R(t 0 ) R(t ) Ṙ(t 0 ) t 0 t, and t 0 t R(t 0 ) we find r 46
Lecture 29: Cosmology ie z Ṙ(t 0 ) R(t 0 ) t 0 t R (t0 ) R(t 0 ) r R(t 0 ) = v r (2954) v r = Ṙ(t 0 ) R(t 0 ) D(t 0 ) (2955) Thus we may identify Ṙ(t 0 ) R(t 0 ) as the Hubble constant, H 0 Deceleration parameter Assuming the cosmic scale parameter R(t) to be well-enough behaved, we may expand it in Taylor s series, measuring time from the present: R(t) R(t 0 ) Ṙ(t 0 ) + R(t 0 ) t + Ṙ (t 0 ) 2 R(t 0 ) t2 + (2957) Now, the deceleration parameter is defined as df Ṙ (t 0 ) q 0 = H 2 0 R(t 0 ) (note that deceleration corresponds to q 0 > 0 ) Equation 2957 can then be written in standard format R(t) R(t 0 ) + H 0 t 2 2 q 0 H 0 t + (2958) Our isotropic model universe satisfies the equations Ṙ R = 4πG 3 (ρ + 3p) R2 (2959t) Ṙ R + 2R 2 + 2k = 4πG (ρ p) R 2 (2959r) and thence R 2 + k = 8πG 3 ρ R2, (2960) and (from energy-momentum conservation) R 3 (t) p = t t R3 (t) (p + ρ) There are two obvious cases: p << ρ (dust) then R 3 (t) ρ const and from Eq 2959t, Ṙ R 2 (296) 47
Deceleration parameter which leads to a power-law behavior for R(t): R t 2 3 (2962) 2 p = ρ (ultrarelativistic gas) 3 Now, from Eq 296, ρ R 4 = const R(t) t 2 (2963) The (dimensionless) deceleration parameter q 0 can be related to the average density of mass-energy in the universe; hence the question of whether the universe is open (k 0) or closed (k > 0) can in principle be answered by measuring q 0 Unfortunately, while the observational evidence that q 0 > 0 is good, we cannot say more than that at present And recently new evidence has been obtained that may indicate q 0 < 0, which would mean the expansion of the universe is accelerating From Eq 2960, we may obtain k R 2 (t 0 ) + H 0 2 = 8πG 3 ρ(t 0 ) (2964) and from Eq 2959t and the definition of q 0 we find q 0 H 0 2 = 4πG 3 ρ 0 + 3p 0 (2965) From Eq 2965 and Eq 2964 we can then derive an expression for the pressure now: k p 0 = 8πG R 2 + H 0 2 ( 2q 0 ) (2966) 0 Moreover, from Eq 2964 we determine that k = R 2 8πG 0 3 ρ 0 ρ crit (2967) so that the criterion for closure of the metric is ρ 0 > ρ crit Moreover, if the present pressure is 0, then setting Eq 2966 to 0 gives ρ 0 2q 0 ρ crit (2968) hence q 0 > 2 ρ 0 > ρ crit and consequently, k > 0 48