Chapter 4. COSMOLOGICAL PERTURBATION THEORY

Chapter 4. COSMOLOGICAL PERTURBATION THEORY 4.1. NEWTONIAN PERTURBATION THEORY Newtonian gravity is an adequate description on small scales (< H 1 ) and for non-relativistic matter (CDM + baryons after decoupling). The equations for a fluid with mass density ρ, pressure P and velocity u are: t ρ = r (ρu) (C) Continuity equation [conservation of mass] ( t + u r ) u = rp ρ r Φ (E) Euler equation [conservation of momentum] F=ma 2 r Φ = 4πGρ (P ) Poisson equation We wish to consider small fluctuations around a homogeneous background: and linearise the fluid equations. ρ(t, r) = ρ(t) + δρ(t, r),... 1) Static space without gravity The background solution is ρ = const., P = const. and ū = 0. The linearised evolution equations are t δρ = r ( ρu) ρ t u = r δp (C) (E) Combining t (C) and r (E), we get 2 t δρ 2 r δp = 0.

speed of sound For adiabatic fluctuations (see later), we have δp = c 2 s δρ and hence ( ) 2 t c 2 s r 2 δρ = 0 wave equation Solution: δρ = Ae i(ωt k r), where ω = c s k, with k k. fluctuations oscillate 2) Static space with gravity For Φ 0, we get ( ) 2 t c 2 s r 2 δρ = 4πG ρ δρ from ρ 2 r δφ Solution: δρ = Ae i(ωt k r), where ( ) ω 2 = c 2 sk 2 4πG ρ c 2 s k 2 kj 2 pressure gravity Jeans scale: k J 4πG ρ/c s On small scales, k > k J, fluctuations oscillate On large scales, k < k J, fluctuations grow exponentially 2

3) Expanding space In an expanding space, we have r = a(t)x and u ṙ = Hr + v Rather than labelling events by t and r, it is convenient to use t and x: For spatial gradients, this means using r = a 1 x ( ) For time derivatives, this means using ( ) ( ) ( ) ( ) x = + x = r x r ( ) = x x ( (a 1 ) (t)r) + Hx x Notation alert: we will drop the subscripts x from now on! x ( ) x With this in mind, we look at the fluid equations in an expanding universe: Continuity equation: Substituting ( ) and ( ) into (C), we get [ ] [ ρ(1 Hx ] 1 + δ) + a [ ρ(1 + δ)(hax + v) ] = 0, where δ δρ/ ρ is the density contrast. At zeroth-order, we find ρ + 3H ρ = 0, where we used x x = 3. As expected, we have ρ a 3. At first-order, we get [ ρ + 3H ρ ] } {{ } = 0 δ + ρ δ + ρ a v = 0. 3

Hence, we find δ + 1 a v = 0 (C) where δ ( δ ) x. Euler equation: Similar manipulations of the Euler equation lead to v + Hv = 1 a ρ δp 1 a δφ (E) For δp = δφ = 0, this implies v a 1 (cf. Ch. 1). Poisson equation: It takes no work to show that 2 δφ = 4πGa 2 ρ δ (P ) Combining t (C) with (E) and (P ), we get δ + 2H δ c2 s a 2 2 δ = 4πG ρδ. Hubble friction The Jeans scale k J (t) 4πG ρ(t)/c s (t) is time-dependent. k > k J : fluctuations oscillate with decreasing amplitude k < k J : fluctuations grow as a power law Ex: Show that long-wavelength fluctuations in a matter-dominated universe, 4πG ρ = 3 2 H2 Ω m, have the following power-law solutions δ = a, a 3/2. 4

4.2. RELATIVISTIC PERTURBATION THEORY GR is required on large scales and for relativistic fluids. Perturb metric: g µν = ḡ µν + δg µν Perturb matter: ρ = ρ + δρ,... Linearise evolution equations. 4.2.1. PERTURBED METRIC The most general perturbations around FRW are ] ds 2 = a 2 (τ) [(1 + 2A)dτ 2 2B i dx i dτ (δ ij + h ij )dx i dx j, where A(τ, x), etc. Convention: Latin indices are raised and lowered with δ ij : e.g. h i i = δ ij h ij. Write B i i B + ˆB i scalar vector : i ˆBi = 0 h ij 2Cδ ij + 2 i j E + 2 (i Ê j) + 2Êij scalar scalar vector tensor where i j E ( i j 1 3 δ ij 2) E (i Ê j) 1 2 ( iêj + j Ê i ) i Ê i = 0 i Ê ij = Êi i = 0 10 = 4 scalar modes : A, B, C, E + 4 vector modes : ˆBi, Êi + 2 tensor modes : Ê ij Theorem: At first order, scalars, vectors, and tensors don t mix! We can treat them separately. 5

The Gauge Problem Perturbations depend on the choice of coordinates ( gauge choice ). Consider an unperturbed FRW universe, ds 2 = a 2 (τ) [ dτ 2 δ ij dx i dx j], ρ = ρ(t),. A change of spatial coordinates, x i x i x i + ξ i (τ, x), implies dx i = d x i τ ξ i dτ k ξ i dx k, which results in fictitious metric perturbations ds 2 = a 2 (τ) [ dτ 2 2ξ id x i dτ ( ) δ ij + 2 (i ξ j) d x i d x j] + O(ξ 2 ) fake perturbations (gauge modes) Similarly, a change of time slicing, τ τ + ξ 0 (τ, x), induces fictitious density perturbations ρ(τ) ρ(τ + ξ 0 ) = ρ(τ) + ρ ξ 0. We need a way to identify true perturbations. Gauge Transformations Consider X µ X µ X µ + ξ µ (τ, x), where ξ 0 T, ξ i i L. To see how the metric transforms, note that ds 2 = g µν (X)dX µ dx ν = g αβ ( X)d X α d X β. Writing d X α = ( X α / X µ )dx µ (and similarly for d X β ), we find g µν (X) = X α X µ X β X ν g αβ( X). 6

As an example, let us take µ = ν = 0: g 00 (X) = X α τ X β τ g αβ( X) ( ) 2 to first order τ g 00( τ X). We can write this as a 2 (τ) ( 1 + 2A ) = ( 1 + T ) 2 a 2 (τ + T ) ( 1 + 2Ã) = ( 1 + 2T + )( a(τ) + a T + )2( 1 + 2 Ã ) = a 2 (τ) ( 1 + 2HT + 2T + 2Ã + ), where H a /a. This implies that A Ã = A T HT. Ex: Show that B B = B + T L C C = C HT 1 3 2 L, E Ẽ = E L. Gauge-Invariant Variables Ex: Show that are gauge-invariant. Ψ A + H(B E ) + (B E ), Φ C H(B E ) + 1 3 2 E, The Bardeen potentials Ψ and Φ represent true perturbations that cannot be removed by a coordinate transformation. 7

Gauge Fixing An alternative solution of the gauge problem is to fix the gauge, follow all perturbations (metric and matter), compute observables. Two popular gauges are: Newtonian gauge: B = E 0. Spatially flat gauge: C = E 0. 4.2.2. PERTURBED MATTER In a homogeneous universe, T µ ν = ( ρ + P )Ū µ Ū ν P δ µ ν, where Ū µ = a 1 δ µ 0 for a comoving observer. In a perturbed universe, T µ ν = T µ ν + δt µ ν = T µ ν + (δρ + δp )Ū µ Ū ν + ( ρ + P )(δu µ Ū ν + Ū µ δu ν ) δp δ µ ν Π µ ν anisotropic stress (mostly negligible) To derive δu µ, we consider g µν U µ U ν = 1: At first order, this implies δg µν Ū µ Ū ν + 2ŪµδU µ = 0. Using Ū µ = a 1 δ µ 0 and δg 00 = 2a 2 A, we find δu 0 = Aa 1. Writing δu i = v i /a, where v i dx i /dτ, we get U µ = a 1 [1 A, v i ]. Ex: Show that U µ = a[1 + A, (v i + B i )]. 8

Hence, we find T 0 0 = ρ + δρ T i 0 = ( ρ + P )v i q i 3-momentum density T 0 j = ( ρ + P )(v j + B j ) T i j = ( P + δp )δj i In a multi-component universe, we have T µν = I T µν I and hence δρ = I δρ I δp = I δp I q i = I q i I (note: velocities don t add!) Gauge Transformations X µ X µ = X µ + (T, i L) implies T µ ν(x) = Xµ X α X β X ν T α β( X). and δρ δρ T ρ, δp δp T P, v v + L, where v i i v. A gauge-invariant density fluctuation is ρ δρ + ρ (v + B). Popular matter gauges are: comoving-gauge density contrast Uniform density gauge: δρ 0 Comoving gauge: q 0 + B 0. 9

Adiabatic Fluctuations are predicted by inflation. What are they? Consider a mixture of fluids, with ρ I, P I, etc. Adiabatic perturbations are perturbations induced by a common, local shift in time of all background quantities: e.g. δρ I (τ, x) = ρ I (τ + δτ(x)) ρ I (τ) = ρ I δτ(x) same for all I. Hence, δτ = δρ I ρ I = δρ J ρ J, for all I and J. Using ρ I = 3H(1 + w I ) ρ I and δ I δρ I / ρ I, we find δ I 1 + w I = δ J 1 + w J. Adiabatic perturbations satisfy δ r = 4 3 δ m. Whatever dominates the background also dominates fluctuations: δρ tot = ρ tot δ tot = I ρ I δ I similar for all I. 10

4.2.3. LINEARISED EVOLUTION EQUATIONS See: www.damtp.cam.ac.uk/user/db275/cosmology/cpt.nb (thanks to Yi Wang) We will work in Newtonian gauge: g µν = a 2 ( 1 + 2Ψ 0 0 (1 2Φ)δ ij ), and ignore anisotropic stress: Π ij = 0. Inverse metric Christoffel symbols At first order, we find g µν = 1 a 2 ( 1 2Ψ 0 0 (1 + 2Φ)δ ij Ex: Show that Γ 0 0i = i Ψ, ) + O(Ψ 2, Φ 2 ). Γ 0 00 = 1 2 g00 0 g 00 = 1 2a 2(1 2Ψ) 0[a 2 (1 + 2Ψ)] Γ i 00 = δ ij j Ψ, = H + Ψ. Γ 0 ij = Hδ ij [Φ + 2H(Φ + Ψ)] δ ij, Γ i j0 = Hδ i j Φ δ i j, Γ i jk = 2δ i (j k)φ + δ jk δ il l Φ. 11

PERTURBED CONSERVATION EQUATIONS Consider µ T µ ν = 0 = µ T µ ν + Γ µ µαt α ν Γ α µνt µ α. The ν = 0 component leads to the continuity equation: At zeroth-order, we find ρ = 3H( ρ + P ). At first-order, we get δρ = 3H(δρ + δp ) + 3Φ ( ρ + P ) q dilution by expansion perturbed expansion: fluid flow a eff a(1 Φ) For adiabatic perturbations, we can write this as δ + (1 + w) ( v 3Φ ) + 3H(c 2 s w)δ = 0 (C) For non-relativistic matter (w = c 2 s 1) on sub-hubble scales (Φ v Ch. 5), this reproduces the Newtonian continuity equation δ + v 0. The ν = i component leads to the Euler equation: v = Hv redshifting + 3H P ρ v δp Ψ ρ + P relativistic correction For adiabatic perturbations, we get pressure gravity v + H(1 3c 2 s)v = δ Ψ (E) 1 + w For non-relativistic matter (w = c 2 s 1), this reproduces the Newtonian Euler equation. c2 s 12

PERTURBED EINSTEIN EQUATIONS Γ λ µν(φ, Ψ) R µν G µν R µν 1 2 g µνr Ex: Show that (see lecture notes) G 00 = 3H 2 + 2 2 Φ 6HΦ G 0i = 2 i (Φ + HΨ) G ij = (2H + H 2 )δ ij + i j (Φ Ψ) + [ 2 (Ψ Φ) + 2Φ + 2(2H + H 2 )(Φ + Ψ) + 2HΨ + 4HΦ ] δ ij Consider the trace-free part of the ij-einstein equation: Next, look at the 00-equation: i j (Φ Ψ) = 0 Φ = Ψ. G 00 = 8πGT 00 3H 2 + 2 2 Φ 6HΦ = 8πGg 0µ T µ 0 = 8πG ( ) g 00 T 0 0 + g 0i T i 0 = 8πGa 2 (1 + 2Φ)( ρ + δρ) = 8πGa 2 ρ(1 + 2Φ + δ). At zeroth-order, we recover the Friedmann equation: 3H 2 = 8πGa 2 ρ. At first-order, we get 2 Φ = 4πGa 2 ρδ + 3H(Φ + HΦ) ( ) Moving on to the 0i-equation: G 0i = 8πGT 0i Φ + HΦ = 4πGa 2 q ( ) 13

Substituting ( ) into ( ), we find 2 Φ = 4πGa 2 ρ Poisson equation where is the comoving-gauge density contrast. Finally, we look at the trace of the ij-equation: G i i = 8πGT i i. At zeroth-order, we recover the second Friedmann equation: 2H + H 2 = 8πGa 2 P. At first-order, we get an evolution equation for Φ: Φ + 3HΦ + (2H + H 2 )Φ = 4πGa 2 δp. 14

4.3. CONSERVED CURVATURE PERTURBATION A very important quantity is the comoving curvature perturbation R. It is conserved on superhorizon scales (k H = ah) and therefore provides the link between fluctuations created by inflation (Ch. 6) and those observed in the late universe (Ch. 5). Consider the perturbed 3-metric γ ij = a 2 [(1 + 2C)δ ij + 2E ij ], and the corresponding 3-curvature a 2 R (3) = 4 (C 2 1 ) 3 2 E. The comoving curvature perturbation is [ R C 1 ] 3 2 E q i = B i = 0 or, in gauge-invariant form R = C 1 3 2 E + H(v + B). Evaluated in Newtonian gauge, this is R = Φ + Hv = Φ H(Φ + HΦ) 4πGa 2 ( ρ + P ). Using the Einstein equations, we can show that (see lecture notes) 4πGa 2 ( ρ + P )R = 4πGa 2 H (δp P ) ρ δρ + H P ρ 2 Φ The first term on the r.h.s. vanishes for adiabatic perturbations and we get ( ) 2 d ln R k d ln a k H 0 H i.e. R is conserved on superhorizon scales! 15