Cosmological Perturbation Theory, 26.4.2011 version
Cosmological Perturbation Theory, 26.4.2011 version
Cosmological Perturbation Theory, 26.4.2011 version
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About these lecture notes:I lectured a course on cosmological perturbation theory at the University of Helsinki in thespring of 2003, and then again in the fall of 2010. These are the lecture notes for the lattercourse.There is an unfortunate variety in the notation employed by various authors. My notationis the result of first learning perturbation theory from Mukhanov, Feldman, and Brandenberger(Phys. Rep. 215, 213 (1992)) and then from Liddle and Lyth (<strong>Cosmological</strong> Inflation andLarge-Scale Structure, Cambridge University Press 2000, Chapters 14 and 15), and represents acompromise between their notations.
17 Fluid Components 4017.1 Division into Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4017.2 Gauge Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4117.3 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4118 Adiabatic and Isocurvature <strong>Perturbation</strong>s in a Simplified Universe 4318.1 Background solution for radiation+matter . . . . . . . . . . . . . . . . . . . . . . 4318.2 <strong>Perturbation</strong>s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4518.3 Initial Epoch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4718.3.1 Adiabatic modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4718.3.2 Isocurvature Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4818.3.3 Other <strong>Perturbation</strong>s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4918.4 Full evolution for large scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5019 Effect of a Smooth Component 5120 The Real Universe 5121 Early Radiation-Dominated Era 5321.1 Neutrino Adiabaticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5521.2 Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5621.2.1 The Completely Adiabatic Solution . . . . . . . . . . . . . . . . . . . . . 5721.2.2 Baryon and CDM Entropy <strong>Perturbation</strong>s . . . . . . . . . . . . . . . . . . 5721.3 Neutrino perturbations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5822 Superhorizon Evolution and Relation to Original <strong>Perturbation</strong>s 5823 Gaussian Initial Conditions 5824 Large Scales 5825 Sachs–Wolfe Effect 5926 Matter Power Spectrum 61A General <strong>Perturbation</strong> 62
2 THE BACKGROUND UNIVERSE 2all close to each other, for the perturbed spacetime, for which (1.1) holds. The choice amongthese coordinate systems is called the gauge choice, to be discussed a little later.In first-order (or linear) perturbation theory, we drop all terms from our equations whichcontain products of the small quantities δg µν , δg µν,ρ and δg µν,ρσ . The field equation (1.5)becomes then a linear differential equation for δg µν , making things much easier than in full GR.In second-order perturbation theory, one keeps also those terms with a product of two (but nomore) small quantities. In these lectures we only discuss first-order perturbation theory.The simplest case is the one where the background is the Minkowski space. Then ḡ µν = η µν ,and Ḡµ ν = ¯T µ ν = 0.In cosmological perturbation theory the background spacetime is the Friedmann–Robertson–Walker universe. Now the background spacetime is curved, and is not empty. While it ishomogeneous and isotropic, it is time-dependent. In these lectures we shall only consider thecase where the background is the flat FRW universe. This case is much simpler than the openand closed ones 3 , since now the t = const time slices (“space at time t”) have Euclidean geometry.This will allow us to do 3-dimensional Fourier-transformations in space.The separation into the background and perturbation is always done so that the spatialaverage of the perturbation is zero, i.e., the background value (at time t) is the spatial averageof the full quantity over the time slice t = const.2 The Background UniverseOur background spacetime is the flat Friedmann–Robertson–Walker universe FRW(0). Thebackground metric in comoving coordinates (t,x,y,z) isds 2 = ḡ µν dx µ dx ν = −dt 2 + a 2 (t)δ ij dx i dx j = −dt 2 + a 2 (t)(dx 2 + dy 2 + dz 2 ), (2.1)where a(t) is the scale factor, to be solved from the (flat universe) Friedmann equations,H 2 =where H ≡ ȧ/a is the Hubble parameter,(ȧ ) 2= 8πG ¯ρ (2.2)a 3äa= −4πG (¯ρ + 3¯p) (2.3)3· ≡ d dt ,and the ¯ρ and ¯p are the homogeneous background energy density and pressure. Another <strong>version</strong>of the second Friedmann equation isḢ ≡ ä (ȧ ) 2a − = −4πG(¯ρ + ¯p). (2.4)abyWe shall find it more convenient to use as a time coordinate the conformal time η, defineddη = dta(t)(2.5)3 For cosmological perturbation theory for open or closed FRW universe, see e.g. Mukhanov, Feldman, Brandenberger,Phys. Rep. 215, 203 (1992).
2 THE BACKGROUND UNIVERSE 3so that the background metric isThat is,whereandds 2 = ḡ µν dx µ dx ν = a 2 (η) [ −dη 2 + δ ij dx i dx j] = a 2 (η)(−dη 2 + dx 2 + dy 2 + dz 2 ). (2.6)ḡ µν = a 2 (η)η µν ⇒ ḡ µν = a −2 (η)η µν . (2.7)Using the conformal time, the Friedmann equations are (exercise)( aH 2 ′) 2= = 8πGa 3 ¯ρa2 (2.8)H ′ = − 4πG3 (¯ρ + 3¯p)a2 , (2.9)′≡ddη = a d dt= a( )˙ (2.10)H ≡ a′= aH = ȧ (2.11)ais the conformal, or comoving, Hubble parameter. Note that( aH ′ ′) ′ (= = a′′ a′) 2a a − = (aȧ)˙− ȧ 2 = aä = a 2äaa = a2 (Ḣ + H2 ). (2.12)The energy continuity equationbecomes just˙¯ρ = −3H(¯ρ + ¯p) (2.13)¯ρ ′ = −3H(¯ρ + ¯p) ≡ −3H(1 + w)¯ρ . (2.14)For later convenience we define the equation-of-state parameterw ≡ ¯p¯ρ(2.15)and the “speed of sound squared” 4c 2 s ≡ ˙¯p˙¯ρ ≡ ¯p′¯ρ ′ . (2.16)These two quantities always refer to the background values.From the Friedmann equations (2.8,2.9) and the continuity equation (2.14) one easily derivesadditional useful background relations, likeandH ′ = − 1 2 (1 + 3w)H2 , (2.17)w ′1 + w = 3H(w − c2 s), (2.18)¯p ′ = w¯ρ ′ + w ′¯ρ = −3H(1 + w)c 2 s ¯ρ. (2.19)Eq. (2.17) shows that w = − 1 3 corresponds to constant comoving Hubble length H−1 = const.For w < − 1 3 the comoving Hubble length shrinks with time (“inflation”), whereas for w > −1 3it grows with time (“normal” expansion). When w = const., we have c 2 s = w.4 This turns out to be the speed of sound if our ρ and p describe ordinary fluid. Even if they do not, wenevertheless define this quantity, although the name is then misleading.
3 THE PERTURBED UNIVERSE 43 The Perturbed UniverseWe write the metric of the perturbed (around FRW(0)) universe asg µν = ḡ µν + δg µν = a 2 (η µν + h µν ), (3.1)where h µν , as well as h µν,ρ and h µν,ρσ are assumed small. Since we are doing first-order perturbationtheory, we shall drop from all equations all terms with are of order O(h 2 ) or higher, andjust write “=” to signify equality to first order in h µν . Here the perturbation h µν is not a tensorin the perturbed universe, neither is η µν , but we defineh µ ν ≡ η µρ h ρν , h µν ≡ η µρ η νσ h ρσ . (3.2)One easily finds (exercise), that the inverse metric of the perturbed spacetime isg µν = a −2 (η µν − h µν ) (3.3)(to first order).We shall now give different names for the time and space components of the perturbedmetric, defining[ ]−2A −Bi[h µν ] ≡(3.4)−B i −2Dδ ij + 2E ijwhereD = − 1 6 hi i (3.5)carries the trace of the spatial metric perturbation h ij , and E ij is traceless,δ ij E ij = 0. (3.6)Since indices on h µν are raised and lowered with η µν , we immediately have[ ][h µν −2A +Bi] ≡+B i −2Dδ ij + 2E ij(3.7)On B i and E ij we do not raise/lower indices (or if we do, it is just the same thing, B i = δ ij B j =B i ).The line element is thusds 2 = a 2 (η) { −(1 + 2A)dη 2 − 2B i dηdx i + [(1 − 2D)δ ij + 2E ij ] dx i dx j} . (3.8)The function A(η,x i ) is called the lapse function, and B i (η,x i ) the shift vector.4 Gauge TransformationsThe association between points in the background spacetime and the perturbed spacetime isvia the coordinate system {x α }. As we noted earlier, for a given coordinate system in thebackground, there are many possible coordinate systems in the perturbed spacetime, all closeto each other, that we could use. In GR perturbation theory, a gauge transformation means acoordinate transformation between such coordinate systems in the perturbed spacetime. (It maybe helpful to temporarily forget at this point what you have learned about gauge transformationsin other field theories, e.g. electrodynamics, so that you can learn the properties of this concepthere with a fresh mind, without preconceptions.)
4 GAUGE TRANSFORMATIONS 5In this section, we denote the coordinates of the background by x α , and two different coordinatesystems in the perturbed spacetime (corresponding to two “gauges”) by ˆx α and ˜x α . 5 Thecoordinates ˆx α and ˜x α are related by a coordinate transformation˜x α = ˆx α + ξ α , (4.1)where ξ α and the derivatives ξ α ∂ξα,βare first-order small. The difference between and ∂ξα is∂ˆx β ∂˜x βsecond-order small, and thus ignored, so we can write just ξ α ,β . In fact, we shall think of ξα asliving on the background spacetime.The situation is illustrated in Fig. 2. The coordinate system {ˆx α } associates point ¯P in thebackground with ˆP, whereas {˜x α } associates the same background point ¯P with another point˜P. The association is by˜x α ( ˜P) = ˆx α ( ˆP) = x α ( ¯P). (4.2)Figure 2: Gauge transformation. The background spacetime is on the left and the perturbed spacetime,with two different coordinate systems, is on the right. The “const” for the coordinate lines refer to thesame constant for the two coordinate systems, i.e., ˜η = ˆη = const ≡ η( ¯P) and ˜x i = ˆx i = const ≡ x i ( ¯P).The coordinate transformation relates the coordinates of the same point in the perturbedspacetime, i.e.,˜x α ( ˜P) = ˆx α ( ˜P) + ξ α˜x α ( ˆP) = ˆx α ( ˆP) + ξ α . (4.3)Now the difference ξ α ( ˜P) − ξ α ( ˆP) is second-order small. Thus we write just ξ α and associate itwith the background point:ξ α = ξ α ( ¯P) = ξ α (x β ).Using Eqs. (4.2) and (4.3), we get the relation between the coordinates of the two differentpoints in a given coordinate system,ˆx α ( ˜P) = ˆx α ( ˆP) − ξ α˜x α ( ˜P) = ˜x α ( ˆP) − ξ α . (4.4)5 Ordinarily (when not doing gauge transformations) we write just x α for both the background and perturbedspacetime coordinates.
4 GAUGE TRANSFORMATIONS 6Let us now consider how various tensors transform in the gauge transformation. We have,of course, the usual GR transformation rules for scalars (s), vectors (w α ), and other tensors,s = sw˜α = X ˜αˆβ w ˆβA˜α˜β= X ˜αˆγ Xˆδ˜βAˆγ˜δB˜α˜β= Xˆγ˜α Xˆδ˜βBˆγ ˆβ(4.5)whereX ˜αˆβXˆδ˜β≡≡∂˜xα∂ˆx β = δα β + ξα ,β∂ˆxδ∂˜x β = δδ β − ξδ ,β . (4.6)These rules refer to the values of these quantities at a given point in the perturbed spacetime.However, this is not what we want now! Please, pay attention, since the following is central tounderstanding GR perturbation theory!We shall be interested in perturbations of various quantities. In the background spacetimewe may have various 4-scalar fields ¯s, 4-vector fields ¯w α and tensor fields Āα β , ¯Bαβ . In theperturbed spacetime we have corresponding perturbed quantities,s = ¯s + δsw α = ¯w α + δw αA α β = Āα β + δAα βB αβ = ¯B αβ + δB αβ . (4.7)Consider first the 4-scalar s. The full quantity s = ¯s + δs lives on the perturbed spacetime.However, we cannot assign a unique background quantity ¯s to a point in the perturbed spacetime,because in different gauges this point is associated with different points in the background, withdifferent values of ¯s. Therefore there is also no unique perturbation δs, but the perturbation isgauge-dependent. The perturbations in different gauges are defined aŝδs(x α ) ≡ s( ˆP) − ¯s( ¯P)˜δs(x α ) ≡ s( ˜P) − ¯s( ¯P). (4.8)The perturbation δs is obtained from a subtraction between two spacetimes, and we consider itas living on the background spacetime. It changes in the gauge transformation. Relate now ̂δsto ˜δs:s( ˜P) = s( ˆP) + ∂s [∂ˆx α( ˆP) ˆx α ( ˜P) − ˆx α ( ˆP)]= s( ˆP) − ∂s∂ˆx α( ˆP)ξ α = s( ˆP) − ∂¯s∂x α( ¯P)ξ α ,where we approximated ∂s∂ˆx( ˆP) ≈∂¯sα ∂x( ¯P), since the difference 6 between them is a first orderαperturbation, and multiplication by ξ α makes it second order.Since our background is homogeneous, ¯s = ¯s(η,x i ) = ¯s(η) only, and∂¯s∂x α( ¯P)ξ α = ∂¯s∂η ( ¯P)ξ 0 = ¯s ′ ξ 0 .6 This difference is the perturbation of the covariant vector s ,α. We are assuming perturbations are first ordersmall also in quantities derived by covariant derivation from the “primary” quantities.
4 GAUGE TRANSFORMATIONS 7Thus we getand our final result for the gauge transformation of δs iss( ˜P) = s( ˆP) − ¯s ′ ξ 0 , (4.9)˜δs(x α ) = s( ˆP) − ¯s ′ ξ 0 − ¯s( ¯P) = ̂δs(x α ) − ¯s ′ ξ 0 . (4.10)In analogy with (4.8), the perturbations in vector and tensor fields in the two gauges aredefinedand̂δw α (x β ) ≡ wˆα ( ˆP) − ¯w α ( ¯P)˜δw α (x β ) ≡ w˜α ( ˜P) − ¯w α ( ¯P). (4.11)̂δA α β(x γ ) ≡˜δA α β(x γ ) ≡̂δB αβ (x γ ) ≡Aˆαˆβ( ˆP) − Āα β ( ¯P)A˜α˜β( ˜P) − Āα β ( ¯P)Bˆαˆβ( ˆP) − ¯B αβ ( ¯P)and˜δB αβ (x γ ) ≡ B˜α˜β( ˜P) − ¯B αβ ( ¯P). (4.12)Consider the case of a type (0,2) 4-tensor field. We haveBˆµˆν ( ˜P) = Bˆµˆν ( ˆP) + ∂Bˆµˆν [∂ˆx α ˆx α ( ˜P) − ˆx α ( ˆP)]= Bˆµˆν ( ˆP) − ∂ ¯B µν∂x α ( ¯P)ξ α (4.13)B˜µ˜ν ( ˜P) = X ˆρ˜µ Xˆσ˜ν Bˆρˆσ( ˜P) = (δ ρ µ − ξρ ,µ )(δσ ν − ξσ ,ν ) [= Bˆµˆν ( ˆP) − ξ ρ ,µ Bˆρˆν( ˆP) − ξ σ ,ν Bˆµˆσ( ˆP) − ∂ ¯B µν∂x α ( ¯P)ξ αBˆρˆσ ( ˆP) − ∂ ¯B ρσ∂x α ( ¯P)ξ α ]= Bˆµˆν ( ˆP) − ξ ρ ,µ ¯B ρν ( ¯P) − ξ σ ,ν ¯B µσ ( ¯P) − ∂ ¯B µν∂x α ( ¯P)ξ α , (4.14)where we can replace Bˆµˆσ ( ˆP) with ¯B µσ ( ¯P) in the two middle terms, since it is multiplied by afirst-order quantity ξ σ ,ν and we can thus ignore the perturbation part, which becomes secondorder.Subtracting the background value at ¯P we get the gauge transformation rule for the tensorperturbation δB µν ,˜δB µν ≡ B˜µ˜ν ( ˜P) − ¯B µν ( ¯P)= Bˆµˆν ( ˆP) − ¯B µν ( ¯P) − ξ ρ ,µ ¯B ρν ( ¯P) − ξ σ ,ν ¯B µσ ( ¯P) − ∂ ¯B µν∂x α ( ¯P)ξ α= ̂δB µν − ξ ρ ,µ ¯B ρν − ξ σ ,ν ¯B µσ − ¯B µν,α ξ α . (4.15)In a similar manner we obtain the gauge transformation rules for 4-vector perturbations,and perturbations of type (1,1) 4-tensors (exercise),˜δw α = ̂δw α + ξ α ,β ¯wβ − ¯w α ,β ξβ (4.16)˜δA µ ν = ̂δA µ ν + ξµ ,ρĀρ ν − ξσ ,νĀµ σ − Āµ ν,α ξα . (4.17)
4 GAUGE TRANSFORMATIONS 8Since the background is isotropic and homogeneous, and our background coordinate systemfully respects these properties, the background 4-vectors and tensors must be of the form[ ]¯w α = ( ¯w 0 ,⃗0) Ā α β = Ā0 0 01 , (4.18)03 δi jĀk and they depend only on the (conformal) time coordinate η. Using these properties we can writethe gauge transformation rules for the individual components of 4-scalar, 4-vector and type (1,1)4-tensor perturbations (we now drop the hats from the first gauge),˜δs = δs − ¯s ′ ξ 0˜δw 0 = δw 0 + ξ 0 ,0 ¯w 0 − ¯w 0 ,0ξ 0˜δw i = δw i + ξ i ,0 ¯w 0˜δA 0 0 = δA 0 0 − Ā0 0,0 ξ0 (4.19)˜δA 0 i = δA 0 i + 1 3 ξ0 ,iĀk k − ξ0 ,iĀ0 0˜δA i 0= δA i 0 + ξ i ,0Ā0 0 − 1 3 ξi ,0Āk k˜δA i j = δA i j − 1 3 δi jĀk k,0 ξ0 . (4.20)The following combinations (the trace and the traceless part of ˜δA i j) are also useful:˜δA k k= δA k k − Āk k,0 ξ0˜δA i j − 1 3 δi j ˜δA k k = δA i j − 1 3 δi jδA k k . (4.21)Thus the traceless part of δA i jis gauge-invariant!4.1 Gauge Transformation of the Metric <strong>Perturbation</strong>sApplying the gauge transformation equation (4.15) to the metric perturbation, we have˜δg µν = δg µν − ξ ρ ,µḡ ρν − ξ σ ,νḡ µσ − ḡ µν,0 ξ 0 , (4.22)where we have replaced the sum ḡ µν,α ξ α with ḡ µν,0 ξ 0 , since the background metric dependsonly on the time coordinate x 0 = η, and dropped the hats from the first gauge. Rememberingḡ µν = a 2 (η)η µν from Eq. (2.7), we haveandFrom Eqs. (3.1) and (3.4) we haveḡ µν,0 = 2a ′ aη µν (4.23)˜δg µν = δg µν + a 2 [−ξ ρ ,µ η ρν − ξ σ ,ν η µσ − 2 a′a η µνξ 0 ]. (4.24)[δg µν ] = a 2 [ −2A −Bi−B i −2Dδ ij + 2E ij](4.25)
5 SEPARATION INTO SCALAR, VECTOR, AND TENSOR PERTURBATIONS 9Applying the gauge transformation law (4.24) now separately to the different metric perturbationcomponents, we get first˜δg 00 ≡ −2a 2 Ã= δg 00 + a 2 (−ξ ρ ,0 η ρ0 − ξ σ ,0η 0σ − 2 a′a η 00ξ 0 )from which we obtain the gauge transformation law= −2a 2 A + a 2 (+ξ 0 ,0 + ξ 0 ,0 + 2 a′a ξ0 ), (4.26)Similarly, from δg 0i we obtainà = A − ξ 0 ,0 − a′a ξ0 . (4.27)˜B i = B i + ξ i ,0 − ξ 0 ,i , (4.28)and from δg ij ,− ˜Dδ ij + Ẽij = −Dδ ij + E ij − 1 2 (ξi ,j + ξ j ,i ) − a′a ξ0 δ ij . (4.29)The trace of 1 2 (ξi ,j + ξj ,i ) is ξk ,k, so we can write12 (ξi ,j + ξ j ,i ) = 1 3 δ ijξ k ,k + 1 2 (ξi ,j + ξ j ,i ) − 1 3 δ ijξ k ,k , (4.30)where the last two terms are the traceless part, and we can separate Eq. (4.29) into˜D = D + 1 3 ξk ,k + a′a ξ0Ẽ ij = E ij − 1 2 (ξi ,j + ξ j ,i ) + 1 3 δ ijξ k ,k . (4.31)5 Separation into Scalar, Vector, and Tensor <strong>Perturbation</strong>sIn GR perturbation theory there are two kinds of coordinate transformations of interest. Oneis the gauge transformation just discussed, where the coordinates of the background are keptfixed, but the coordinates in the perturbed spacetime are changed, changing the correspondencebetween the points in the background and the perturbed spacetime.The other kind is one where we keep the gauge, i.e, the correspondence between the backgroundand perturbed spacetime points, fixed, but do a coordinate transformation in the backgroundspacetime. This then induces a corresponding coordinate transformation in the perturbedspacetime. Our background coordinate system was chosen to respect the symmetries ofthe background, and we do not want to lose this property. In cosmological perturbation theorywe have chosen the background coordinates to respect its homogeneity property, which gives usa unique slicing of the spacetime into homogeneous t = const. spacelike slices. Thus we do notwant to change this slicing. This leaves us:1. homogeneous transformations of the time coordinate, i.e., reparameterizations of time, ofwhich we already had an example, when we switched from cosmic time t to conformal timeη,
5 SEPARATION INTO SCALAR, VECTOR, AND TENSOR PERTURBATIONS 102. and transformations in the space coordinates 7where X i′kx i′ = X i′ k xk , (5.1)is independent of time; which is the case we consider in this section.We had chosen the coordinates for our background, FRW(0), so that the 3-metric was Euclidean,g ij = a 2 δ ij , (5.2)and we want to keep this property. This leaves us rotations. The full transformation matricesare then [ ] [ ][ ]1 0 1 0X µ′ ρ = =and X µ 1 00 X i′ k0 R i′ ρ= ′k0 R i , (5.3)k ′where R i′ k is a rotation matrix8 , with the property R T R = I, or R i′ k Ri′ l = (RT R) kl = δ kl . ThusR T = R −1 so that R i′ k = Rk i ′ .This coordinate transformation in the background induces the corresponding transformation,into the perturbed spacetime. Here the metric isx µ′ = X µ′ρx ρ , (5.4)g µν = a 2 [ −1 − 2A −Bi−B i (1 − 2D)δ ij + 2E ij]= a 2 η µν + a 2 [ −2A −Bi−B i −2Dδ ij + 2E ij]. (5.5)Transforming the metric,we get for the different componentsg ρ ′ σ ′ = Xµ ρ ′ X ν σ ′g µν , (5.6)g 0 ′ 0 ′ = Xµ 0X ν ′ 0 ′g µν = X 0 0 ′X0 0 ′g 00 = g 00 = a 2 (−1 − 2A)g 0 ′ l ′ = Xµ 0X ν ′ l ′g µν = X 0 0 ′Xj lg ′ 0j = −a 2 R j( lB ′ j)g k ′ l ′ = Xi k ′Xj lg ′ ij = a 2 −2Dδ ij R i k ′Rj l+ 2E ′ ij R i k ′Rj( l)′= a 2 −2Dδ kl + 2E ij R i k ′Rj l, (5.7)′from which we identify the perturbations in the new coordinates,A ′D ′= A= DB l ′= R j l ′ B jE k ′ l ′ = Ri k ′Rj l ′ E ij . (5.8)Thus A and D transform as scalars under rotations in the background spacetime coordinates,B i transforms as a 3-vector, and E ij as a 3-d tensor. While staying in a fixed gauge, we can thusthink of them as scalar, vector, and tensor fields on the 3-d Euclidean background space. We7 In this section we use ′ to denote the other coordinate system. Do not confuse with ′ ≡ d/dη in the othersections.8 In this notation R i′ j and R i j ′ are two different matrices, corresponding to opposite rotations; the position ofthe ′ indicates which way we are rotating. We have put the first index upstairs to follow the Einstein summationconvention—but we could have written R i ′ j and R ij ′ just as well.
5 SEPARATION INTO SCALAR, VECTOR, AND TENSOR PERTURBATIONS 11are, however, not yet satisfied. We can extract two more scalar quantities and one more vectorquantity from B i and E ij .We know from Euclidean 3-d vector calculus, that a vector field can be divided into twoparts, the first one with zero curl, the second one with zero divergence,⃗B = ⃗ B S + ⃗ B V , with ∇ × ⃗ B S = 0 and ∇ · ⃗B V = 0, (5.9)and that the first one can be expressed as (minus) a gradient of some scalar field 9In component notation,⃗B S = −∇B . (5.10)B i = −B ,i + B V i , where δ ij B V i,j= 0. (5.11)In like manner, the symmetric traceless tensor field E ij can be divided into three parts,E ij = E S ij + E V ij + E T ij , (5.12)where E S ij and EV ij can be expressed in terms of a scalar field E and a vector field E i,E S ij = ( ∂ i ∂ j − 1 3 δ ij∇ 2) E = E ,ij − 1 3 δ ijδ kl E ,kl (5.13)E V ij = − 1 2 (E i,j + E j,i ), where δ ij E i,j = ∇ · ⃗E = 0 (5.14)and δ ik E T ij,k = 0, δij E T ij = 0. (5.15)We see that Eij S is symmetric and traceless by construction. EV ij is symmetric by construction,and the condition on E i makes it traceless. The tensor Eij T is assumed symmetric, and the twoconditions on it make it transverse and traceless. The meaning of “transverse” and the natureof the above construction becomes clearer in the next section when we do this in Fourier space.Under rotations in background space,A ′ = A, B ′ = B , D ′ = D , E ′ = E ,B V l ′ = Rj l ′ B V j , E l ′ = R j l ′ E j ,E T k ′ l ′ = Ri k ′Rj l ′ E T ij . (5.16)The metric perturbation can thus be divided into1. a scalar part, consisting of A, B, D, and E,2. a vector part, consisting of B V i and E i ,3. and a tensor part E T ij .The names “scalar”, “vector”, and “tensor” refer to their transformation properties under rotationsin the background space. 10The Einstein tensor perturbation δG µ ν and the energy tensor perturbation δT µ ν can likewisebe divided into scalar+vector+tensor; the scalar part of δG µ ν coming only from the scalar partof δg µν and so on.9 This sign convention corresponds to thinking of the scalar function B as a “potential”, where the vector field⃗B S “flows downhill”. We use the same letter B here for both the original vector field B i and this scalar potentialB. There should be no confusion since the vector field always has an index (or⃗). Same goes for the E.10 Thus “scalar” does not mean, e.g., that the perturbation would be invariant under gauge transformations—scalar perturbations are not gauge-invariant, as we have already seen, e.g. in Eqs. (4.27) and (4.31).
6 PERTURBATIONS IN FOURIER SPACE 12The important thing about this division is that the scalar, vector, and tensor parts do notcouple to each other (in first-order perturbation theory), but they evolve independently. Thisallows us to treat them separately: We can study, e.g., scalar perturbations as if the vectorand tensor perturbations were absent. The total evolution of the full perturbation is just alinear superposition of the independent evolution of the scalar, vector, and tensor part of theperturbation.We imposed one constraint on each of the 3-vectors BiV and E i , and 3 + 1 = 4 constraintson the symmetric 3-d tensor Eij T leaving each of them 2 independent components. Thus the10 degrees of freedom corresponding to the 10 components of the metric perturbation h µν aredivided into1 + 1 + 1 + 1 = 4 scalar2 + 2 = 4 vector2 = 2 tensor (5.17)degrees of freedom.The scalar perturbations are the most important. They couple to density and pressure perturbationsand exhibit gravitational instability: overdense regions grow more overdense. Theyare responsible for the formation of structure in the universe from small initial perturbations.Vector perturbations couple to rotational velocity perturbations in the cosmic fluid. Theytend to decay in an expanding universe, and are therefore probably not important in cosmology.We have done all of the above in a fixed gauge. It turns out that gauge transformationsaffect scalar and vector perturbations, but tensor perturbations are gauge-invariant. Tensorperturbations are nothing but gravitational waves, this time 11 in an expanding universe. Whenthey are extracted from the total perturbation by the above separation procedure, they areautomatically in the “transverse traceless gauge”. (This expression is clarified in the nextsection.) Tensor perturbations have cosmological importance since, if strong enough, they havean observable effect on the anisotropy of the cosmic microwave background.6 <strong>Perturbation</strong>s in Fourier SpaceBecause our background space is flat we can Fourier expand the perturbations. For an arbitraryperturbation f = f(η,x i ) = f(η,⃗x), we writef(η,⃗x) = ∑ ⃗ kf ⃗k (η)e i⃗ k·⃗x . (6.1)(Using a Fourier sum implies using a fiducial box with some volume V . At the end of the day wecan let V → ∞, and replace remaining Fourier sums with integrals.) In first-order perturbationtheory each Fourier component evolves independently. We can thus just study the evolution ofa single Fourier component, with some arbitrary wave vector ⃗ k, and we drop the subscript ⃗kfrom the Fourier amplitudes.Since ⃗x = (x 1 ,x 2 ,x 3 ) is a comoving coordinate, ⃗ k is a comoving wave vector. The comoving(or coordinate) wave number k ≡ | ⃗ k| and wavelength λ = 2π/k are related to the physicalwavelength and wave number of the Fourier mode byk phys = 2πλ phys= 2πaλ = a−1 k . (6.2)11 Contrasted to perturbation theory around Minkowski space, which is the way gravitational waves are usuallyintroduced in GR.
6 PERTURBATIONS IN FOURIER SPACE 13Thus the wavelength λ phys of the Fourier mode ⃗ k grows in time as the universe expands.In the separation into scalar+vector+tensor, we follow Liddle&Lyth and include an additionalfactor k ≡ | ⃗ k| in the Fourier components of B and E i , and a factor k 2 in E, so that wehave, e.g.,B(η,⃗x) = ∑ B ⃗k (η)e i⃗k·⃗x . (6.3)k⃗ kThe purpose of this is to make them have the same dimension and magnitude as B S i , ES ij andE V ij . 12 That is,B i = B S i + B V i , and E ij = E S ij + E V ij + E T ij , (6.4)whereand the conditionsbecomeBi S = −B ,i becomes Bi S = −i k ik BEij S = ( (∂ i ∂ j − 1 3 δ ij∇ 2) E becomes Eij S = − k )ik jk 2 + 1 3 δ ij E ,E V ij = − 1 2 (E i,j + E j,i ) becomes E V ij = − i2k (k iE j + k j E i ), (6.5)δ ij B V i,j = 0, δij E i,j = 0, and δ ik E T ij,k = δij E T ij = 0 (6.6)δ ij k j Bi V = ⃗ k · ⃗B V = 0, δ ij k j E i = ⃗ k · ⃗E = 0, and δ ik k k Eij T = δij Eij T = 0. (6.7)To make the separation into scalar+vector+tensor parts as clear as possible, rotate thebackground coordinates so that the z axis becomes parallel to ⃗ k,(ẑ denoting the unit vector in z direction.) Thenand⎡ ⎤ ⎡10Eij S = 3 E ⎤ ⎡⎣ 0 ⎦ + ⎣ 13 E ⎦ = ⎣1−E3 Eand we can write the scalar part of δg µν as⎡−2Aδgµν S = a2 ⎢ 2(−D + 1 3 E) ⎣2(−D + 1 3 E)+iBFor the vector part we have then⃗ k = kẑ = (0,0,k) (6.8)B S i = (0,0, −iB) (6.9)13 E 13 E − 2 3 E ⎤⎦ (6.10)⎤+iB⎥⎦ (6.11)2(−D − 2 3 E)⃗ k · ⃗ B V = 0 ⇒ ⃗ B V = (B 1 ,B 2 ,0) (6.12)⃗ k · ⃗ E = 0 ⇒ ⃗ E = (E1 ,E 2 ,0) (6.13)12 Powers of k cancel in Eqs. (6.5). The metric perturbations, A, B, D, E, B V i , E i, and E T ij will then all havethe same dimension in Fourier space, which facilitates comparison of their magnitudes.
6 PERTURBATIONS IN FOURIER SPACE 14andso that the vector part of δg µν isFor the tensor part,⎡Eij V = −i2k (k iE j + k j E i ) = − i ⎣2⎡δgµν V = a 2 ⎢⎣−B 1 −B 2⎤−B 1 −iE 1−B 2 −iE 2⎤E 1E 2⎦ , (6.14)E 1 E 2⎥⎦ . (6.15)δ ik k k E T ij ≡ ∑ ik i E T ij = 0 ⇒ ET 3j ≡ ET j3 = 0 (6.16)so that⎡Eij T = ⎣E11 T E12TE12 T −E11T⎤⎦ . (6.17)The tensor part of δg µν becomes⎡δgµν T = a2 ⎢⎣2E11 T 2E12T2E12 T −2E11T⎤ ⎡⎥⎦ = a2 ⎢⎣⎤h + h ×⎥h × −h +⎦ , (6.18)where we have denoted the two gravitational wave polarization amplitudes by E T 11 = 1 2 h + andE T 12 = 1 2 h ×.We see how the scalar part of the perturbation is associated with the time direction, thewave direction ⃗ k and the trace. The vector part is associated with the two remaining spacedirections, those transverse to the wave vector. Thus the vectors have only two independentcomponents. The tensor part is also associated with these two transverse directions; being alsosymmetric and traceless, it thus has only two independent components.Putting all together, the full metric perturbation (for a Fourier mode in the z direction) is⎡δg µν = a 2 ⎢⎣⎤−2A −B 1 −B 2 +iB−B 1 2(−D + 1 3 E) + h + h × −iE 1−B 2 h × 2(−D + 1 3 E) − h + −iE 2+iB −iE 1 −iE 2 2(−D − 2 3 E)6.1 Gauge Transformation in Fourier Space⎥⎦ . (6.19)Consider then how the gauge transformation appears in Fourier space. We need now the Fouriertransform of the gauge transformation vectorξ µ (η,⃗x) = ∑ ⃗ kξ µ ⃗ k(η)e i⃗ k·⃗x . (6.20)
7 SCALAR PERTURBATIONS 15For a single Fourier mode, the gauge transformation Eqs. (4.27,4.28,4.31) becomeà = A − (ξ 0 ) ′ − a′a ξ0 (6.21)˜B i = B i + (ξ i ) ′ − ik i ξ 0 (6.22)˜D = D + 1 3 ik iξ i + a′a ξ0 (6.23)Ẽ ij = E ij − 1 2 i( k i ξ j + k j ξ i) + 1 3 iδ ijk k ξ k . (6.24)For illustration, consider again a mode in the z direction, ⃗ k = (0,0,k). Now the new partadded into the matrix of Eq. (6.19) is⎡⎤2(ξ 0 ) ′ + 2 a′a ξ0 −(ξ 1 ) ′ −(ξ 2 ) ′ −(ξ 3 ) ′ + ikξ 0a 2 −(ξ 1 ) ′ −2 a′⎢a ξ0 −ikξ 1⎣ −(ξ 2 ) ′ −2 a′a ξ0 −ikξ 2 ⎥ (6.25)⎦−(ξ 3 ) ′ + ikξ 0 −ikξ 1 −ikξ 2 −2 a′a ξ0 − 2ikξ 3(note the cancelations on the diagonal from the D and E ij parts). We see that no new tensorpart is introduced. Thus the tensor part of the metric perturbation is gauge-invariant 13 . But wesee that the components ξ 0 and ξ 3 are responsible for a new scalar part and the components ξ 1and ξ 2 are responsible for a new vector part.For Fourier modes in an arbitrary direction, the above means that the time component ξ 0and the component of the space part ⃗ ξ parallel to ⃗ k are responsible for a change in the scalarperturbation and the transverse part of ⃗ ξ is responsible for a change in the vector perturbation.This freedom of doing gauge transformations can be used, e.g., to set 2 of the 4 scalarquantities of scalar perturbations and 2 of the 4 independent components of vector perturbationsto zero. Thus only two of the degrees of freedom are real physical degrees of freedom in eachcase. Thus there are in total 6 physical degrees of freedom, 2 scalar, 2 vector, and 2 tensor.The other 4 (of the total 10) are just gauge degrees of freedom, representing perturbing just thecoordinates, not the spacetime.If the perturbation can be completely eliminated by a gauge transformation, we say theperturbation is “pure gauge”, i.e., it is not a real perturbation of spacetime, just a perturbationin the coordinates.7 Scalar <strong>Perturbation</strong>sFrom here on (except for the beginning of Sec. 9, where we discuss perturbations in the energytensor) we shall consider scalar perturbations only. They are the ones responsible for thestructure of the universe (i.e. the deviation from the homogeneous and isotropic FRW universe).The metric is nowds 2 = a(η) 2 { −(1 + 2A)dη 2 + 2B ,i dηdx i + [(1 − 2ψ)δ ij + 2E ,ij ] dx i dx j} , (7.1)where we have defined 14 the curvature perturbationψ ≡ D + 1 3 ∇2 E . (7.2)13 To make the meaning of this statement clear: Consider a perturbation that is initially purely tensor, anddo an arbitrary gauge transformation. The parts that get added to the perturbation are of scalar and/or vectornature. Thus this perturbation is not gauge-invariant; but its tensor part—because of the way we have definedthe tensor part of a perturbation—is.14 In my spring 2003 lecture notes (CMB Physics / <strong>Cosmological</strong> <strong>Perturbation</strong> <strong>Theory</strong>) I was using the symbolψ for what I am now denoting D. The present notation is better in line with common usage.
7 SCALAR PERTURBATIONS 16In Fourier space this readsThe components of h µν areh µν =ψ ⃗k = D ⃗k − 1 3 E ⃗ k. (7.3)[ −2A B,iB ,i −2ψδ ij + 2E ,ij]. (7.4)Exercise: Curvature of the spatial hypersurface. The hypersurface η = const. is a 3-dimensional curved manifold. Calculate the connection coefficients (3) Γ i jkand the scalar curvature(3) R ≡ g ij(3) R ij of this 3-space for a scalar perturbation in terms of ψ and E.Now if we start from a pure scalar perturbation and do an arbitrary gauge transformation,represented by the field ξ µ = (ξ 0 ,ξ i ), we may introduce also a vector perturbation. This vectorperturbation is however, pure gauge, and thus of no interest. Just like we did for the shift vectorB i earlier, we can divide ξ i into a part with zero divergence (a transverse part) and a part withzero curl, expressible as a gradient of some function ξ,ξ i = ξ i tr − δij ξ ,j = ⃗ ξ tr − ∇ξ where ξ i tr,i = ∇ · ⃗ξ tr = 0. (7.5)The part ξtr i is responsible for the spurious vector perturbation, whereas ξ0 and ξ ,j change thescalar perturbation. For our discussion of scalar perturbations we thus lose nothing, if we decidethat we only consider gauge transformations, where the ξtr i part is absent. These “scalar gaugetransformations” are fully specified by two functions, ξ 0 and ξ,˜η = η + ξ 0 (η,⃗x)˜x i = x i − δ ij ξ ,j (η,⃗x) (7.6)and they preserve the scalar nature of the perturbation.Applied to scalar perturbations and gauge transformations, our transformation equations(4.27,4.28,4.31) becomeà = A − ξ 0 ′ a ′−a ξ0˜B = B + ξ ′ + ξ 0˜D = D − 1 3 ∇2 ξ + a′a ξ0Ẽ = E + ξ , (7.7)where we use the notation ′ ≡ ∂/∂η for quantities which depend on both η and ⃗x. The quantityψ defined in Eq. (7.2) is often used as the fourth scalar variable instead of D. For it, we get7.1 Bardeen Potentials˜ψ = ψ + a′a ξ0 = ψ + Hξ 0 . (7.8)We now define the following two quantities, called the Bardeen potentials,Φ ≡ A + H(B − E ′ ) + (B − E ′ ) ′Ψ ≡ D + 1 3 ∇2 E − H(B − E ′ ) = ψ − H(B − E ′ ). (7.9)These quantities are invariant under gauge transformations (exercise).
8 CONFORMAL–NEWTONIAN GAUGE 178 Conformal–Newtonian GaugeWe can use the gauge freedom to set the scalar perturbations B and E equal to zero. FromEq. (7.7) we see that this is accomplished by choosingξ = −Eξ 0 = −B + E ′ . (8.1)Doing this gauge transformation we arrive at a commonly used gauge, which has many names:the conformal-Newtonian gauge (or sometimes, for short, just the Newtonian gauge), the longitudinalgauge, and the zero-shear gauge. We shall denote quantities in this gauge with the suborsuperscript N. Thus B N = E N = 0, whereas you immediately see thatA N= ΦD N = ψ N = Ψ . (8.2)Thus the Bardeen potentials are equal to the two nonzero metric perturbations in the conformal-Newtonian gauge.From here on (until otherwise noted) we shall calculate in the conformal-Newtonian gauge.The metric is thus justds 2 = a(η) 2 [ −(1 + 2Φ)dη 2 + (1 − 2Ψ)δ ij dx i dx j] , (8.3)org µν = a 2 [ −1 − 2Φ(1 − 2Ψ)δ ij][ ]and g µν = a −2 −1 + 2Φ, (8.4)(1 + 2Ψ)δ ijor[ ]−2Φh µν =−2Ψδ ij[ ]and h µν −2Φ=. (8.5)−2Ψδ ij8.1 <strong>Perturbation</strong> in the Curvature TensorsFrom the conformal-Newtonian metric (8.3) we get the connection coefficients[Γ 0 00 = a′a + Φ′ Γ 0 0k = Φ ,k Γ 0 ij = a′a δ ij − 2 a′a]δ (Φ + Ψ) + Ψ′ ijand the sumsΓ i 00 = Φ ,i Γ i 0j = a′a δi j − Ψ′ δ i jΓ i kl = − ( Ψ ,l δk i + Ψ ,kδli ) (8.6)+ Ψ,i δ klΓ α 0α = 4a′ a + Φ′ − 3Ψ ′Γ α iα = Φ ,i − 3Ψ ,i (8.7)where we have dropped all terms higher than first order in the small quantities Φ and Ψ. Thusthese expressions contain only 0 th and 1 st order terms, and separate into the background andperturbation, accordingly:Γ α βγ = ¯Γ α βγ + δΓα βγ , (8.8)where¯Γ 0 00 = H ¯Γ0 0k= 0 ¯Γ0 ij = Hδ ij¯Γ i 00 = 0¯Γi 0j = Hδ i j¯Γ i kl = 0 (8.9)
8 CONFORMAL–NEWTONIAN GAUGE 18andδΓ 0 00 = Φ′ δΓ 0 0k = Φ ,k δΓ 0 ij = − [2H(Φ + Ψ) + Ψ′ ] δ ijδΓ i 00 = Φ ,i δΓ i 0j = −Ψ′ δj i δΓ i kl = − ( Ψ ,l δk i + Ψ ,kδl) i + Ψ,i δ kl .The Ricci tensor is(8.10)R µν = Γ α νµ,α − Γ α αµ,ν + Γ α αβ Γβ νµ − Γ α νβ Γβ αµ= ¯R µν + δΓ α νµ,α − δΓ α αµ,ν + ¯Γ α αβ δΓβ νµ + ¯Γ β νµδΓ α αβ − ¯Γ α νβ δΓβ αµ − ¯Γ β αµδΓ α νβ . (8.11)Calculation givesR 00 = −3H ′ + 3Ψ ′′ + ∇ 2 Φ + 3H(Φ ′ + Ψ ′ )R 0i = 2(Ψ ′ + HΦ) ,iR ij= (H ′ + 2H 2 )δ ij+ [ −Ψ ′′ + ∇ 2 Ψ − H(Φ ′ + 5Ψ ′ ) − (2H ′ + 4H 2 )(Φ + Ψ) ] δ ij+ (Ψ − Φ) ,ij (8.12)Next we raise an index to get R µ ν. Note that we can not just raise the index of the backgroundand perturbation parts separately, sinceR µ ν = g µα R αν = (ḡ µα + δg µα )( ¯R αν + δR αν ) = ¯R µ ν + δg µα ¯Rαν + ḡ µα δR αν . (8.13)We getR 0 0 = 3a −2 H ′ + a −2 [ −3Ψ ′′ − ∇ 2 Φ − 3H(Φ ′ + Ψ ′ ) − 6H ′ Φ ]R 0 i = −2a −2 ( Ψ ′ + HΦ ) ,iR i 0 = −R 0 i = 2a −2 ( Ψ ′ + HΦ ) ,iR i j= a −2 (H ′ + 2H 2 )δ i jand summing for the curvature scalar+ a −2 [ −Ψ ′′ + ∇ 2 Ψ − H(Φ ′ + 5Ψ ′ ) − (2H ′ + 4H 2 )Φ ] δ ij+ a −2 (Ψ − Φ) ,ij . (8.14)R = R 0 0 + R i i= 6a −2 (H ′ + H 2 )+ a −2 [ −6Ψ ′′ + 2∇ 2 (2Ψ − Φ) − 6H(Φ ′ + 3Ψ ′ ) − 12(H ′ + H 2 )Φ ] . (8.15)And, finally, the Einstein tensorG 0 0G 0 i= R 0 0 − 1 2 R= −3a −2 H 2 + a −2 [ −2∇ 2 Ψ + 6HΨ ′ + 6H 2 Φ ]= R 0 iG i 0 = R0 i = −Ri 0 = −G 0 i= Rj i − 1 2 δi j RG i j= a −2 (−2H ′ − H 2 )δ i j+ a −2 [ 2Ψ ′′ + ∇ 2 (Φ − Ψ) + H(2Φ ′ + 4Ψ ′ ) + (4H ′ + 2H 2 )Φ ] δ i j+ a −2 (Ψ − Φ) ,ij . (8.16)
9 PERTURBATION IN THE ENERGY TENSOR 19Note the background (written first) and perturbation parts in all these quantities. Since thebackground ¯R µ ν and Ḡµ ν are diagonal, the off-diagonals contain just the perturbation, and wehaveR 0 i = G0 i = δR0 i = δG0 i . (8.17)9 <strong>Perturbation</strong> in the Energy TensorConsider then the energy tensor 15 .The background energy tensor is necessarily of the perfect fluid form 16¯T µν = (¯ρ + ¯p)ū µ ū ν + ¯pḡ µν¯T µ ν = (¯ρ + ¯p)ū µ ū ν + ¯pδ µ ν . (9.1)Because of homogeneity, ¯ρ = ¯ρ(η) and ¯p = ¯p(η). Because of isotropy, the fluid is at rest,ū i = 0 ⇒ ū µ = (ū 0 ,0,0,0) in the background universe. Sinceū µ ū µ = ḡ µν ū µ ū ν = a 2 η µν ū µ ū ν = −a 2 (ū 0 ) 2 = −1, (9.2)we haveū µ = 1 a (1, ⃗0) and ū µ = a(−1,⃗0). (9.3)The energy tensor of the perturbed universe isT µ ν = ¯T µ ν + δT µ ν . (9.4)Just like the metric perturbation, the energy tensor perturbation has 10 degrees of freedom, ofwhich 6 are physical and 4 are gauge. It can likewise be divided into scalar+vector+tensor, with4+4+2 degrees of freedom, of which 2+2+2 are physical. The perturbation can also be dividedinto perfect fluid + non-perfect, with 5+5 degrees of freedom.The perfect fluid degrees of freedom in δT µ ν are those which keep T µ ν in the perfect fluid formT µ ν = (ρ + p)u µ u ν + pδ µ ν . (9.5)Thus they can be taken as the density perturbation, pressure perturbation, and velocity perturbationρ = ¯ρ + δρ, p = ¯p + δp , and u i = ū i + δu i = δu i ≡ 1 a v i . (9.6)The δu 0 is not an independent degree of freedom, because of the constraint u µ u µ = −1. Weshall callv i ≡ au i (9.7)the velocity perturbation. It is equal to the coordinate velocity, since (to first order)dx idη = uiu 0 = uiū 0 = aui = v i . (9.8)15 This section could actually have been earlier. We do not specify a gauge here, and the restriction to scalarperturbations is done only in the end.16 The “imperfections” can only show up in the energy tensor if there is inhomogeneity or anisotropy. Whetheran observer would “feel” the ¯p as pressure is another matter, which depends on the interactions of the fluidparticles. But gravity only cares about the energy tensor.
9 PERTURBATION IN THE ENERGY TENSOR 20It is also equal to the fluid velocity observed by a comoving (i.e., one whose x i = const.) observer,since the ratio of change in comoving coordinate dx i to change in conformal time dη equals theratio of the corresponding physical distance adx i to the change in cosmic time dt = adη.To express u µ and u ν in terms of v i , write them asu µ = ū µ + δu µ ≡ ( a −1 + δu 0 ,a −1 v 1 ,a −1 v 2 ,a −1 v 3)u ν = ū ν + δu ν ≡ (−a + δu 0 ,δu 1 ,δu 2 ,δu 3 ) . (9.9)These are related by u ν = g µν u ν and u µ u µ = −1. Usingwe getg µν = a 2 [ −1 − 2A −Bi−B i (1 − 2D)δ ij + 2E ij], (9.10)u 0 = g 0µ u µ= a 2 (−1 − 2A)(a −1 + δu 0 ) − δ ij a 2 B i a −1 v j= −a − a 2 δu 0 − 2aA (9.11)(where we dropped higher than 1 st order quantities, like B i v j ), from which followsδu 0 = −a 2 δu 0 − 2aA. (9.12)Likewiseδu i = u i = g iµ u µ = −aB i + av i . (9.13)We solve the remaining unknown, δu 0 fromu µ u µ = ... = −1 − 2aδu 0 − 2A = −1 ⇒ δu 0 = − 1 a A (9.14)Thus we have for the 4-velocityu µ = 1 a (1 − A,v i) and u µ = a(−1 − A,v i − B i ). (9.15)Inserting this into Eq. (9.5) we getT ν µ = ¯T ν µ + δT νµ [ −¯ρ 0=0 ¯pδ ji] [ −δρ (¯ρ + ¯p)(vi − B+i )−(¯ρ + ¯p)v i δpδji]. (9.16)There are 5 remaining degrees of freedom in the space part, δTj i , corresponding to perturbationsaway from the perfect fluid from. We write them as( )δTj i = δpδi j + Σ ij ≡ ¯p δp¯p + Π ij . (9.17)Here Σ ij and Π ij ≡ Σ ij /¯p are symmetric and traceless, which makes the separation intoandδp ≡ δT k k (9.18)Σ ij ≡ δT i j − 1 3 δi j δT k k (9.19)unique (the trace and the traceless part of δT i j ). Σ ij is called anisotropic stress (or anisotropicpressure). For a perfect fluid Σ ij = 0.
9 PERTURBATION IN THE ENERGY TENSOR 219.1 Separation into Scalar, Vector, and Tensor PartsThe energy tensor perturbation δT µ ν is built out of the scalar perturbations δρ, δp, the 3-vector⃗v = v i and the traceless 3-tensor Π ij . Just like for the metric perturbations, we can extract ascalar perturbation out of ⃗v:and a scalar + a vector perturbation out of Π ij :wherev i = v S i + v V i , where v S i = −v ,iand ∇ · ⃗v V = 0. (9.20)Π S ijΠ ij = Π S ij + Π V ij + Π T ij , (9.21)= ( ∂ i ∂ j − 1 3 δ ij∇ 2) ΠΠ V ij = − 1 2 (Π i,j + Π j,i ) and (9.22)δ ik Π T ij,k = 0. (9.23)We see that perfect fluid perturbations (Π ij = 0) do not have a tensor perturbation component.For Fourier components of v i and Π ij we use the same (Liddle&Lyth) convention as for B iand E ij (see Sec. (6)).In the early universe, we have anisotropic pressure from the cosmic neutrino backgroundduring and after neutrino decoupling, and from the cosmic microwave background during andafter photon decoupling. <strong>Perturbation</strong>s in the metric will make the momentum distribution ofnoninteracting particles anisotropic (this is anisotropic pressure). If there are sufficient interactionsamong the particles, these will isotropize the momentum distribution. Decoupling meansthat the interactions become too weak for this. If we aim for high precision in our calculations,we need to take this anisotropic pressure into account. For a more approximate treatment, theperfect fluid approximation can be made, which simplifies the calculations significantly.9.2 Gauge Transformation of the Energy Tensor <strong>Perturbation</strong>s9.2.1 General RuleUsing the gauge transformation rules from Sect. 4 we have˜δT 0 0 = − ˜δρ = δT 0 0 − ¯T 0 0,0 ξ0 = −δρ + ¯ρ ′ ξ 0˜δT i 0 = −(¯ρ + ¯p)ṽ i = δT i 0 + ξi ,0 ( ¯T 0 0 − 1 3 ¯T k k )= −(¯ρ + ¯p)v i − ξ i ,0(¯ρ + ¯p)1 ˜δT k 3 k = ˜δp = 1 3 (δT k k − ¯T k,0 k ξ0 ) = δp − ¯p ′ ξ 0˜δT i j − 1 3 δi j ˜δT k k = ¯p˜Π ij = δT i j − 1 3 δi j δT k k = ¯pΠ ij (9.24)and we get the gauge transformation laws for the different parts of the energy tensor perturbation:˜δρ = δρ − ¯ρ ′ ξ 0 (9.25)˜δp = δp − ¯p ′ ξ 0 (9.26)ṽ i = v i + ξ,0 i (9.27)˜Π ij = Π ij . (9.28)
10 FIELD EQUATIONS FOR SCALAR PERTURBATIONS IN THE NEWTONIAN GAUGE22Thus the anisotropic stress is gauge-invariant (being the traceless part of δTj i ). Note that theδρ and δp equations are those of a perturbation of a 4-scalar, as they should be, as ρ and p are,indeed, 4-scalars.9.2.2 Scalar <strong>Perturbation</strong>sFor scalar perturbations, v i = −v ,i and ξ i = −ξ ,i , so that we haveṽ = v + ξ ′˜Π = Π. (9.29)These hold both in coordinate space and Fourier space (we use the same Fourier convention forξ as for v and B).9.2.3 Conformal-Newtonian GaugeWe get to the conformal-Newtonian gauge by ξ 0 = −B + E ′ and ξ = −E. Thusδρ N = δρ + ¯ρ ′ (B − E ′ ) = δρ − 3H(1 + w)¯ρ(B − E ′ )δp N = δp + ¯p ′ (B − E ′ ) = δp − 3H(1 + w)c 2 s ¯ρ(B − E′ )v N = v − E ′Π = Π. (9.30)9.3 Scalar <strong>Perturbation</strong>s in the Conformal-Newtonian GaugeFrom here on we shall (unless otherwise noted)1. consider scalar perturbations only, so that v i = −v ,i and B i = −B ,i2. use the conformal-Newtonian gauge, so that B = 0.Thus the energy tensor perturbation has the formδT µ ν = [ −δρN−(¯ρ + ¯p)v N ,i(¯ρ + ¯p)v N ,i δp N δ i j + ¯p(Π ,ij − 1 3 δ ij∇ 2 Π)]. (9.31)10 Field Equations for Scalar <strong>Perturbation</strong>s in the NewtonianGaugeWe can now write the Einstein equationsδG µ ν = 8πGδT µ ν (10.1)for scalar perturbations in the conformal-Newtonian gauge. We have the left-hand side δG µ νfrom Sect. 8.1 and the right-hand side δT µ ν from Sect. 9.3:δG 0 0= a −2 [ −2∇ 2 Ψ + 6H(Ψ ′ + HΦ) ] = −8πGδρ NδG 0 i = −2a −2 ( Ψ ′ + HΦ ) ,i= −8πG(¯ρ + ¯p)v N ,iδG i 0 = 2a −2 ( Ψ ′ + HΦ ) ,i= 8πG(¯ρ + ¯p)v N ,iδG i j= a −2 [ 2Ψ ′′ + ∇ 2 (Φ − Ψ) + H(2Φ ′ + 4Ψ ′ ) + (4H ′ + 2H 2 )Φ ] δ i j+ a −2 (Ψ − Φ) ,ij = 8πG [ δp N δ i j + ¯p(Π ,ij − 1 3 δ ij∇ 2 Π) ] . (10.2)
10 FIELD EQUATIONS FOR SCALAR PERTURBATIONS IN THE NEWTONIAN GAUGE23Separating the δG i j equation into its trace and traceless part (the trace of δi jof (Ψ − Φ) ,ij is ∇ 2 (Ψ − Φ)) the full set of Einstein equations isis 3, and the trace3H(Ψ ′ + HΦ) − ∇ 2 Ψ = −4πGa 2 δρ N (10.3)(Ψ ′ + HΦ) ,i = 4πGa 2 (¯ρ + ¯p)v N ,i (10.4)Ψ ′′ + H(Φ ′ + 2Ψ ′ ) + (2H ′ + H 2 )Φ + 1 3 ∇2 (Φ − Ψ) = 4πGa 2 δp N (10.5)The off-diagonal part of the last equation givesIn Fourier space this reads(∂ i ∂ j − 1 3 δi j ∇2 )(Ψ − Φ) = 8πGa 2¯p(∂ i ∂ j − 1 3 δi j ∇2 )Π. (10.6)(Ψ − Φ) ,ij = 8πGa 2¯pΠ ,ij for i ≠ j . (10.7)−k i k j (Ψ ⃗k − Φ ⃗k ) = − k ik jk 2 8πGa2¯pΠ ⃗k for i ≠ j . (10.8)(with the Liddle&Lyth Fourier convention for Π). Since we can always rotate the backgroundcoordinate system so that more than one of the components of ⃗ k are non-zero, this means thatk 2 (Ψ ⃗k − Φ ⃗k ) = 8πGa 2¯pΠ ⃗k for ⃗ k ≠ ⃗0 . (10.9)The 0 th Fourier component represents a constant offset. But the split into a background and aperturbation is always chosen so that the spatial average of the perturbation vanishes (and thebackground value thus represents the spatial average of the full perturbed quantity).Thus we have (going back to ⃗x-space)Ψ − Φ = 8πGa 2¯pΠ. (10.10)Likewise, since the spatial average of a perturbation is always zero, the equality of gradientsof two perturbations means the equality of those perturbations themselves. Thus Eq. (10.4) saysthatΨ ′ + HΦ = 4πGa 2 (¯ρ + ¯p)v N = 3 2 H2 (1 + w)v N . (10.11)Inserting this into Eq. (10.3) gives∇ 2 Ψ = 4πGa 2¯ρ [ δ N + 3H(1 + w)v N] , (10.12)where we have defined the relative energy density perturbationδ ≡δρ¯ρ. (10.13)(and w ≡ ¯p/¯ρ).The final form of the Einstein equations can be divided into two constraint equations∇ 2 Ψ = 4πGa 2¯ρ [ δ N + 3H(1 + w)v N] (10.14)Ψ − Φ = 8πGa 2¯pΠ (10.15)that apply to any given time slice, and to two evolution equationsΨ ′ + HΦ = 3 2 H2 (1 + w)v N (10.16)Ψ ′′ + H(Φ ′ + 2Ψ ′ ) + (2H ′ + H 2 )Φ + 1 3 ∇2 (Φ − Ψ) = 4πGa 2 δp N . (10.17)
11 ENERGY-MOMENTUM CONTINUITY EQUATIONS 24that determine how the metric perturbation evolves in time.In Fourier space the Einstein equations can be written asH −1 Ψ ′ + Φ + 1 3( kH) 2Ψ = − 1 2 δNH −1 Ψ ′ + Φ =3 2 (1 + w)H k vNH −2 Ψ ′′ + H −1 ( (Φ ′ + 2Ψ ′) +(1 + 2H′ k 2 )ΦH 2 − 1 3(Φ − Ψ) =H) 3 δp N2¯ρ( ) k 2(Ψ − Φ) = 3wΠ, (10.18)Hwhere the powers of H are arranged so that the distance scales k −1 and time scales dη are alwaysrelated to the conformal Hubble scale H −1 , and we used the background relationwhich follows directly from the Friedmann equation (2.2).4πGa 2¯ρ = 3 2 H2 , (10.19)11 Energy-Momentum Continuity EquationsWe know that from the Einstein equation, G µ ν = 8πGT ν µ , the energy-momentum continuityequations,= 0, (11.1)T µ ν;µfollow. Just like for the background universe, we may use the energy-momentum continuityequations instead of some of the Einstein equations.CalculatingT µ ν;µ = T µ ν,µ + Γ µ αµT α ν − Γ α νµT µ α = 0 (11.2)to first order in perturbations, one obtains the 0 th order (background) equation¯ρ ′ = −3H(¯ρ + ¯p) (11.3)and the 1 st order (perturbation) equations, which for scalar perturbations in the conformal-Newtonian gauge are (exercise)(δρ N ) ′ = −3H(δρ N + δp N ) + (¯ρ + ¯p)(∇ 2 v N + 3Ψ ′ ) (11.4)(¯ρ + ¯p)(v N ) ′ = −(¯ρ + ¯p) ′ v N − 4H(¯ρ + ¯p)v N + δp N + 2 3 ¯p∇2 Π + (¯ρ + ¯p)Φ (11.5)Note that v N is the velocity potential, ⃗v N = −∇v N . It is easy to interpret the various terms inthese equations.In the energy perturbation equation (11.4), we have first the effect of the background expansion,then the effect of velocity divergence (local fluid expansion) and then the effect of theexpansion/contraction in the metric perturbation.In the momentum perturbation equation (11.5), the lhs and the first term on the rightrepresent the change in inertia×velocity. The second on the right is the effect of background expansion.The third and last terms represent forces due to gradients in pressure and gravitationalpotential.
12 PERFECT FLUID SCALAR PERTURBATIONS IN THE NEWTONIAN GAUGE 25With manipulations involving background relations, these can be worked (exercise) into theform(δ N ) ′ = (1 + w) ( )∇ 2 v N + 3Ψ ′) + 3H(wδ N − δpN(11.6)¯ρ(v N ) ′ = −H(1 − 3w)v N − w′1 + w vN + δpN¯ρ + ¯p + 2 3w1 + w ∇2 Π + Φ . (11.7)12 Perfect Fluid Scalar <strong>Perturbation</strong>s in the Newtonian Gauge12.1 Field EquationsFor a perfect fluid, things simplify a lot, since now Π = 0 and thus for a perfect fluidΨ = Φ , (12.1)and we have only one degree of freedom in the scalar metric perturbation. We can now replaceΨ with Φ in the field equations. The original set becomesand the reworked set becomes∇ 2 Φ − 3H(Φ ′ + HΦ) = 4πGa 2 δρ N (12.2)(Φ ′ + HΦ ) ,i= 4πGa 2 (¯ρ + ¯p)v N ,i (12.3)Φ ′′ + 3HΦ ′ + (2H ′ + H 2 )Φ = 4πGa 2 δp N , (12.4)∇ 2 Φ = 4πGa 2¯ρ [ δ N + 3H(1 + w)v N]= 3 2 H2 [ δ N + 3H(1 + w)v N] (12.5)Φ ′ + HΦ = 4πGa 2 (¯ρ + ¯p)v N = 3 2 H2 (1 + w)v N (12.6)Φ ′′ + 3HΦ ′ + (2H ′ + H 2 )Φ = 4πGa 2 δp N = 3 2 H2 δp N /¯ρ , (12.7)where we have used Eq. (2.8).If we change the time variable from conformal time η to cosmic time t, they read∇ 2 Φ = 4πGa 2¯ρ [ δ N + 3aH(1 + w)v N] (12.8)˙Φ + HΦ = 4πGa(¯ρ + ¯p)v N (12.9)¨Φ + 4H ˙Φ(+ 2Ḣ + 3H2) Φ = 4πGδp N . (12.10)We define the total entropy perturbation as( δpS ≡ H¯p ′ − δρ )¯ρ ′(≡ H δṗ¯p − δρ˙¯ρ ). (12.11)From the gauge transformation equations (9.25,9.26) we see that it is gauge invariant.Using the background relations ¯ρ ′ = −3H(1 + w)¯ρ and ¯p ′ = c 2 s ¯ρ ′ we can also write(1S =3(1 + w)δρ¯ρ − 1 )δpc 2 , (12.12)s ¯ρfrom which we getδp = c 2 s [δρ − 3(¯ρ + ¯p)S] , (12.13)
13 SCALAR PERTURBATIONS IN THE MATTER-DOMINATED UNIVERSE 26which holds in any gauge.Using the entropy perturbation, we can now write in the rhs of Eq. (12.7)δp N /¯ρ = c 2 [s δ N − 3(1 + w)S ] , (12.14)where, from (12.5) and (12.6),δ N = −3H(1 + w)v N + 23H 2 ∇2 Φ = − 2 H (Φ′ + HΦ) + 23H 2 ∇2 Φ . (12.15)With some use of background relations, the evolution equation (12.7) becomesH −2 Φ ′′ + 3(1 + c 2 s )H−1 Φ ′ + 3(c 2 s − w)Φ = c2 s H−2 ∇ 2 Φ − 9 2 c2 s (1 + w)S . (12.16)12.2 Adiabatic perturbations<strong>Perturbation</strong>s where the total entropy perturbation vanishes, 17S = 0 ⇔ δp = c 2 sδρ (12.17)are called adiabatic perturbations. 18 For adiabatic perturbations, Eq. (12.16) becomes (going toFourier space)( )H −2 Φ ⃗ ′′k+ 3(1 + c 2 s)H −1 Φ ⃗ ′ k+ 3(c 2 cs k 2s − w)Φ ⃗k = − Φ ⃗k , (12.18)Hfrom which Φ ⃗k (η) can be solved, given the initial conditions. From Eq. (12.6) we then get v N ⃗ k(η),and after that, from Eq. (12.5), δ N ⃗ k(η).12.3 Continuity EquationsFor a perfect fluid, Eqs. (11.6) and (11.7) become(δ N ) ′ = (1 + w) ( )∇ 2 v N + 3Φ ′) + 3H(wδ N − δpN¯ρ(12.19)(v N ) ′ = −H(1 − 3w)v N − w′1 + w vN + δpN¯ρ + ¯p + Φ . (12.20)13 Scalar <strong>Perturbation</strong>s in the Matter-Dominated UniverseLet us now consider density perturbations in the simplest case, the matter-dominated universe.By “matter” we mean here non-relativistic matter, whose pressure is so small compared toenergy density, that we can ignore it here. 1920 In general relativity this is often called “dust”.According to our present understanding, the universe was radiation-dominated for the firstfew ten thousand years, after which it became matter-dominated. For our present discussion, we17 For pressureless matter, where δp = ¯p = w = c 2 s = 0, Eq. (12.11) is not defined, but δp = c 2 sδρ holds always(0 = 0). We use then this latter definition for adiabaticity; and the perturbations are necessarily adiabatic.18 Warning: In cosmology, it is customary to speak about adiabatic perturbations, when one means perturbationswhich were initially adiabatic. Such perturbations do not usually stay adiabatic as the universe evolves.19 Likewise, we can ignore its anisotropic stress. Thus nonrelativistic matter is a perfect fluid for our purposes.20 Note that there are situations where we can make the matter-dominated approximation at the backgroundlevel, but for the perturbations pressure gradients are still important at small distance scales (large k). Here we,however, we make the approximation that also pressure perturbations can be ignored.
13 SCALAR PERTURBATIONS IN THE MATTER-DOMINATED UNIVERSE 27take “matter-dominated” to mean that matter dominates energy density to the extent that wecan ignore the other components. This approximation becomes valid after the first few millionyears.Until late 1990’s it was believed that this matter-dominated state persists until (and beyond)the present time. But new observational data points towards another component in theenergy density of the universe, with a large negative pressure, resembling vacuum energy, ora cosmological constant. This component is called “dark energy”. The dark energy seems tohave become dominant a few billion (10 9 ) years ago. Thus the validity of the matter-dominatedapproximation is not as extensive as was thought before; but anyway there was a significantperiod in the history of the universe, when it holds good.We now make the matter-dominated approximation, i.e., we ignore pressure,¯p = w = 0 and δp = Π = 0. (13.1)This is our first example of solving a perturbation theory problem. The order of work isalways:1. Solve the background problem.2. Using the background quantities as known functions of time, solve the perturbation problem.In the present case, the background equations are( aH 2 ′) 2= = 8πGa 3 ¯ρa2 (13.2)H ′ = − 4πG3 ¯ρa2 , (13.3)from which we have2H ′ + H 2 = 0. (13.4)The background solution is the familiar k = 0 matter-dominated Friedmann model, a ∝ t 2/3 .But let us review the solution in terms of conformal time. Since ¯ρ ∝ a −3(1+w) ∝ a −3 , the solutionof Eq. (13.2) givesa(η) ∝ η 2 . (13.5)Since dt = adη, or dt/dη = a, we get t(η) ∝ η 3 ∝ a 3/2 or a ∝ t 2/3 .From a ∝ η 2 we getH ≡ a′a = 2 and H ′ = − 2 ηη 2 (13.6)Thus, from Eq. (13.2),4πGa 2¯ρ = 3 2 H2 = 6 η 2 . (13.7)The perturbation equations (12.5), (12.6), and (12.7) with ¯p = w = δp = 0, areUsing Eq. (13.4), Eq. (13.10) becomes∇ 2 Φ = 4πGa 2¯ρ [ δ N + 3Hv N] (13.8)Φ ′ + HΦ = 4πGa 2¯ρv N (13.9)Φ ′′ + 3HΦ ′ + (2H ′ + H 2 )Φ = 0. (13.10)Φ ′′ + 3HΦ ′ = Φ ′′ + 6 η Φ′ = 0 , (13.11)
13 SCALAR PERTURBATIONS IN THE MATTER-DOMINATED UNIVERSE 28whose solution isΦ(η,⃗x) = C 1 (⃗x) + C 2 (⃗x)η −5 . (13.12)The second term, ∝ η −5 is the decaying part. We get C 1 (⃗x) and C 2 (⃗x) from the initial valuesΦ in (⃗x), Φ ′ in (⃗x) at some initial time η = η in,asΦ in (⃗x) = C 1 (⃗x) + C 2 (⃗x)ηin −5(13.13)Φ ′ in (⃗x) = −5C 2(⃗x)ηin −6(13.14)C 1 (⃗x) = Φ in (⃗x) + 1 5 η inΦ ′ in(⃗x) (13.15)C 2 (⃗x) = − 1 5 η6 inΦ ′ in(⃗x) (13.16)Unless we have very special initial conditions, conspiring to make C 1 (⃗x) vanishingly small, thedecaying part soon becomes ≪ C 1 (⃗x) and can be ignored. Thus we have the important resultΦ(η,⃗x) = Φ(⃗x), (13.17)i.e., the Bardeen potential Φ is constant in time for perturbations in the flat (k = 0) matterdominateduniverse.Ignoring the decaying part, we have Φ ′ = 0 and we get for the velocity perturbation fromEq. (13.9)v N HΦ 2Φ= =4πGa2¯ρ 3H = 1 3 Φη ∝ η ∝ a1/2 ∝ t 1/3 . (13.18)and Eq. (13.8) becomes∇ 2 Φ = 4πGa 2¯ρ [ δ N + 2Φ ] = 3 2 H2 [ δ N + 2Φ ] (13.19)orIn Fourier space this readsδ N = −2Φ + 23H 2 ∇2 Φ . (13.20)δ N ⃗ k(η) = −[2 + 2 3( kH) 2]Φ ⃗k . (13.21)Thus we see that for superhorizon scales, k ≪ H, or k phys ≪ H, the density perturbation staysconstant,δ N ⃗ k= −2Φ ⃗k = const. (13.22)whereas for subhorizon scales, k ≫ H, or k phys ≫ H, they grow proportional to the scale factor,δ ⃗k = − 2 3( kH) 2Φ ⃗k ∝ η 2 ∝ a ∝ t 2/3 . (13.23)Since the comoving Hubble scale H −1 grows with time, various scales k are superhorizonto begin with, but later become subhorizon as H −1 grows past k −1 . (In physical terms, in theexpanding universe λ phys /2π = k −1phys ∝ a grows slower than the Hubble scale H−1 ∝ a 3/2 .) Wesay that the scale in question “enters the horizon”. (The word “horizon” in this context refersjust to the Hubble scale 1/H, and not to other definitions of “horizon”.) We see that densityperturbations begin to grow when they enter the horizon, and after that they grow proportional
14 PERFECT FLUID SCALAR PERTURBATIONS WHEN P = P(ρ) 29to the scale factor. Thus the present magnitude of the density perturbation at comoving scalek should be a 0 /a k times its primordial valueδ ⃗k (t 0 ) ∼ a 0a kδ N ⃗ k,pr, (13.24)where a 0 is the present value of the scale factor, and a k is its value at the time the scale k“entered horizon”. The “primordial” density perturbation δ N ⃗ k,prrefers to the constant value ithad at superhorizon scales, after the decaying part of Φ had died out. Of course Eq. (13.24)is valid only for those (large) scales where the perturbation is still small today. Once theperturbation becomes large, δ ∼ 1, perturbation theory is no more valid. We say the scale inquestion “goes nonlinear” 21 .If we take into account the presence of “dark energy”, the above result is modified toδ ⃗k (t 0 ) ∼ a DEa kδ N ⃗ k,pr, (13.25)where a DE a 0 /2 is the value of a when dark energy became dominant, since that stops thegrowth of density perturbations.One has to remember that these results refer to the density and velocity perturbationsin the conformal-Newtonian gauge only. In some other gauge these perturbations, and theirgrowth laws would be different. However, for subhorizon scales general relativistic effects becomeunimportant and a Newtonian description becomes valid. In this limit, the issue of gaugechoice becomes irrelevant as all “sensible gauges” approach each other, and the conformal-Newtonian density and velocity perturbations become those of a Newtonian description. TheBardeen potential can then be understood as a Newtonian gravitational potential due to densityperturbations. (Eq. (13.8) acquires the form of the Newtonian law of gravity as the second termon the right becomes small compared to the first term, δ N . The factor a 2 appears on the rightsince ∇ 2 on the left refers to comoving coordinates.)14 Perfect Fluid Scalar <strong>Perturbation</strong>s when p = p(ρ)We shall next discuss a somewhat more complicated case, where pressure can not be ignored,but the equation of state has the formp = p(ρ), (14.1)i.e., there are no other thermodynamic variables than ρ that the pressure would depend on. Inthis case the perturbations are guaranteed to be adiabatic, since now(Also, we keep assuming the fluid is perfect.)δpδρ = ¯p′¯ρ ′ = dpdρ = c2 s . (14.2)21 Since in our universe this happens only at subhorizon scales, the nonlinear growth of perturbations can betreated with Newtonian physics. First the growth of the density perturbation (for overdensities) becomes muchfaster than in linear perturbation theory, but then the system “virializes”, settling into a relatively stable structure,a galaxy or a galaxy cluster, where further collapse is prevented by the conservation of angular momentum, asthe different parts of the system begin to orbit the center of mass of the system. For underdensities, we have ofcourse ρ ≥ 0 ⇒ δ ≥ −1 always, so the underdensity cannot “grow” beyond that.
15 OTHER GAUGES 30This discussion will also apply to adiabatic perturbations of general perfect fluids. 22 Thatis, when the fluid in principle may have more state variables, but these other degrees of freedomare not “used”. Let us show that the adiabaticity of perturbations,δp = c 2 sδρ ≡ ¯p′¯ρ ′δρ, (14.3)implies that a unique relation (14.1) holds everywhere and -when:Note first, that in the background solution, where ¯p = ¯p(t) and ¯ρ = ¯ρ(t), we can (assuming¯ρ decreases monotonously with time) invert for t(¯ρ), and thus ¯p = ¯p(t(¯ρ)), defining a function¯p(¯ρ), whose derivative is c 2 s. Now Eq. (14.3) guarantees that also p = ¯p + δp and ρ = ¯ρ + δρsatisfy this same relation.We note a property, which illuminates the nature of adiabatic perturbations: A small regionof the perturbed universe is just like the background universe at a slightly earlier or later time.We can thus think of adiabatic perturbations as a perturbation in the “timing” of the differentparts of the universe. (In adiabatic oscillations, this “corresponding background solution time”may oscillate back and forth.)(Perhaps I will finish this section some day...)15 Other GaugesDifferent gauges are good for different purposes, and therefore it is useful to be able to work indifferent gauges, and to switch from one gauge to another in the course of a calculation. 23 Whenone uses more than one gauge it is important to be clear about to which gauge each quantityrefers. One useful gauge is the comoving gauge. Particularly useful quantities which refer tothe comoving gauge, are the comoving density perturbation δ C and the comoving curvatureperturbationR ≡ −ψ C . (15.1)This is often just called the curvature perturbation, however that term is also used to refer to ψin any other gauge, so beware! (There are different sign conventions for R and ψ. In my signconventions, positive R, but negative ψ, correspond to positive curvature of the 3-dimensionalη = const. slice.Gauges are usually specified by giving gauge conditions. These may involve the metricperturbation variables A, D, B, E (e.g., the Newtonian gauge condition E = B = 0), theenergy-momentum perturbation variables δρ, δp, v, or both kinds.15.1 Slicing and ThreadingThe gauge corresponds to the coordinate system {x µ } = {η,x i } in the perturbed spacetime.The conformal time η gives the slicing of the perturbed spacetime into η = const. time slices(3-dim spacelike hypersurfaces). The spatial coordinates x i give the threading of the perturbedspacetime into x i = const. threads (1-dim timelike curves). See Fig. 3. Slicing and threadingare orthogonal to each other if and only if the shift vector vanishes, B i = 0.22 Thus we could have called this section “Adiabatic perfect fluid scalar perturbations”. The reason we did not,is that mentioned in the earlier footnote. In this section we require the perturbations stay adiabatic the wholetime. <strong>Perturbation</strong>s of general fluids, which are initially (when they are at superhorizon scales) adiabatic, acquireentropy perturbations when they approach and enter the horizon.23 One may even combine quantities of different gauges in the same equation to simplify the equations. Thismay sound dangerous but is not, when one knows what one is doing.
15 OTHER GAUGES 31Figure 3: Slicing and threading the perturbed spacetime.In the gauge transformation,˜x α = x α + ξ α , (15.2)or˜η = η + ξ 0x i = x i + ξ i , (15.3)ξ 0 changes the slicing, and ξ i changes the threading. From the 4-scalar transformation law,δ˜s = δs − ¯s ′ ξ 0 , we see that perturbations in 4-scalars, e.g., δρ and δp depend only on the slicing.15.2 Comoving GaugeWe say that the slicing is comoving, if the time slices are orthogonal to the fluid 4-velocity.For scalar perturbations such a slicing always exists. This condition turns out to be equivalentto the condition that the fluid velocity perturbation v i equals the shift vector B i . For scalarperturbations,Comoving slicing ⇔ v = B . (15.4)From the gauge transformation equations (7.7) and (9.29),ṽ = v + ξ ′˜B = B + ξ ′ + ξ 0 (15.5)we see that we get to comoving slicing by ξ 0 = v − B.We say that the threading is comoving, if the threads are world lines of comoving (with thefluid) observers, i.e., the velocity perturbation vanishes, v i = 0. For scalar perturbations,Comoving threading ⇔ v = 0. (15.6)
15 OTHER GAUGES 32We get to comoving threading by the gauge transformation ξ ′ = −v. Note that comoving threadsare usually not geodesics, since pressure gradients cause the fluid flow to deviate from free fall.The comoving gauge is defined by requiring both comoving slicing and comoving threading.ThusComoving gauge ⇔ v = B = 0. (15.7)(We assume that we are working with scalar perturbations.) The threading is now orthogonalto the slicing. We denote the comoving gauge by the sub- or superscript C. Thus the statementv C = B C = 0 is generally true, whereas the statement v = B = 0 holds only in the comovinggauge.We get to the comoving gauge from an arbitrary gauge by the gauge transformationξ ′= −vξ 0 = v − B . (15.8)This does not fully specify the coordinate system in the perturbed spacetime, since only ξ ′ isspecified, not ξ. Thus we remain free to do time-independent transformations˜x i = x i − ξ(⃗x) ,i , (15.9)while staying in the comoving gauge. This, however, does not change the way the spacetimeis sliced and threaded by the coordinate system, it just relabels the threads with differentcoordinate values x i .Applying Eq. (15.8) to the general scalar gauge transformation Eqs. (7.7,7.8,9.25,9.26,9.29),we get the rules to relate the comoving gauge perturbations to perturbations in an arbitrarygauge. For the metric:For the energy tensor:A C = A − (v − B) ′ − H(v − B) (15.10)B C = B − v + (v − B) = 0D C = − 1 3 ∇2 ξ + H(v − B)E C= E + ξψ C ≡ −R = ψ + H(v − B).δρ C = δρ − ¯ρ ′ (v − B) = δρ + 3H(1 + w)¯ρ(v − B) (15.11)δp C = δp − ¯p ′ (v − B) = δp + 3H(1 + w)c 2 s ¯ρ(v − B)δ C = δ + 3H(1 + w)(v − B)v C = v − v = 0Π C = Π.Because of the remaining gauge freedom (relabeling the threading) left by the comoving gaugecondition (only ξ ′ is fixed, not ξ), D C and E C are not fully fixed. However, ψ C is, and likewiseE C ′ = E ′ − v . (15.12)In particular, we get the transformation rule from the Newtonian gauge (where A = Φ,B = 0, ψ = D = Ψ, E = 0) to the comoving gauge. For the metric:A C = Φ − v N ′ − Hv N (15.13)R = −Ψ − Hv NE C ′ = −v N .
15 OTHER GAUGES 33For the energy tensor:δρ C = δρ N + 3H(1 + w)¯ρv N (15.14)δp C = δp N + 3H(1 + w)c 2 s ¯ρv Nδ C = δ N + 3H(1 + w)v N .Thus, for example, we see that in the matter-dominated universe discussed in Sect. 13,Eqs. (13.18,13.20) lead toδ C = δ N + 3Hv N = −2Φ + 23H 2 ∇2 Φ + 2Φ = 23H 2 ∇2 Φ ∝ H −2 ∝ a (15.15)both inside and outside (and through) the horizon.Note that in Fourier space we have included an extra factor of k in v, so that, e.g., Eq. (15.14)reads in Fourier space asδ C = δ N + 3H(1 + w)k −1 v N . (15.16)15.3 Mixing GaugesWe see that Eq. (15.14c) is the rhs of Eq. (10.14), which we can thus write in the shorter form:∇ 2 Ψ = 4πGa 2¯ρδ C . (15.17)This is an equation which has a Newtonian gauge metric perturbation on the lhs, but a comovinggauge density perturbation on the rhs. Is it dangerous to mix gauges like this? No, if we knowwhat we are doing, i.e., in which gauge each quantity is, and the equations were derived correctlyfor this combination of quantities.We could also say, that instead of working in any particular gauge, we work with gaugeinvariantquantities, that the quantity that we denote by δ C is the gauge-invariant quantitydefined byδ C = δ + 3H(1 + w)(v − B)which just happens to coincide with the density perturbation in the comoving gauge. Just likeΨ happens equal the metric perturbation ψ in the Newtonian gauge.Switching to the comoving gauge for δ and δp, but keeping the velocity perturbation in theNewtonian gauge, the Einstein and continuity equations can be rewritten as (Exercise):∇ 2 Ψ = 3 2 H2 δ C (15.18)Ψ − Φ = 3H 2 wΠ (15.19)Ψ ′ + HΦ = 3 2 H2 (1 + w)v N (15.20)Ψ ′′ + (2 + 3c 2 s)HΨ ′ + HΦ ′ + 3(c 2 s − w)H 2 Φ + 1 3 ∇2 (Φ − Ψ) = 3 2 H2δp C¯ρ(15.21)δ C ′ − 3Hwδ C = (1 + w)∇ 2 v N + 2Hw∇ 2 Π (15.22)v N ′ + Hv N = δp C¯ρ + ¯p + 2 w31 + w ∇2 Π + Φ . (15.23)
15 OTHER GAUGES 3415.4 Comoving Curvature <strong>Perturbation</strong>The comoving curvature perturbationR ≡ −ψ C = −ψ − H(v − B) (15.24)turns out to be a useful quantity fro discussing superhorizon perturbations.From the second Einstein equation in the Newtonian gauge, Eq. (10.16), we have thatv N =23H 2 (1 + w) (Ψ′ + HΦ), (15.25)so that the relation of the comoving curvature perturbation and the Bardeen potentials isR = −Ψ −2 (H −1 Ψ ′ + Φ ) 2= −Ψ −3(1 + w)3(1 + w) Φ − 23(1 + w) H−1 Ψ ′ . (15.26)Derivating Eq. (15.26),R ′ = −Ψ ′ 2w ′ (+H −13(1 + w) 2 Ψ ′ + Φ ) )2−(− H′3(1 + w) H 2Ψ′ + H −1 Ψ ′′ + Φ ′= − 2H−13(1 + w) Ψ′′ − 4 + 6c2 s 23(1 + w) Ψ′ −3(1 + w) Φ′ + 2H w − c2 s1 + w Φ ,(where we used some background relations), and using the Einstein equations (10.18) and(10.18), we get an evolution equation for R,− 3 2 (1 + w)H−1 R ′= H −2 Ψ ′′ + H −1 ( Φ ′ + 2Ψ ′) + 3c 2 (s H −1 Ψ ′ + Φ ) − 3wΦ= H −2 Ψ ′′ + (2 + 3c 2 s)H −1 Ψ ′ + H −1 Φ ′ + 3(c 2 s − w)Φ) 2(Φ − Ψ) + 3 δp N2= 3c 2 (s H −1 Ψ ′ + Φ ) ( k+ 1 3H( ) k 2 ( k= −c 2 s Ψ + 1H3H¯ρ) 2 ( δp(Φ − Ψ) + 3 N)2¯ρ − c2 sδ N . (15.27)From Sec. 12 we haveS =13(1 + w)(δ − 1 c 2 s)δp¯ρ⇒ δp = c 2 s[δρ − 3(¯ρ + ¯p)S] (15.28)(valid in any gauge), so that Eq. (15.27) becomes the important result( ) k 2− 3 2 (1 + w)H−1 R ′ [= c2H s Ψ + 1 3 (Ψ − Φ)] + 9 2 c2 s (1 + w)S . (15.29)As an evolution equation, this does not appear very useful, since it mixes metric perturbationsfrom two different gauges. However, the importance of this equation comes from the two observationswe can now make immediately: 1) For adiabatic perturbations, S = 0, the second termon the rhs vanishes. 2) For superhorizon perturbations, i.e., for Fourier modes whose wavelengthis much larger than the Hubble distance, k ≪ H, we can ignore the first term on the rhs. Thus:For adiabatic perturbations, the comoving curvature perturbation stays constant outside thehorizon.Adiabatic (scalar) perturbations have only one degree of freedom, and thus their full evolutionis captured in the evolution of R.
15 OTHER GAUGES 35A general perturbation at a given time can be decomposed into an adiabatic component,which has S = 0, and an isocurvature component, which has R = 0. Because of the linearityof first order perturbation theory, these components evolve independently, and the evolution ofthe general perturbation is just the superposition of the evolution of these two components, andthus they can be studied separately. Beware, however, that the ”adiabatic” component doesnot necessarily remain adiabatic in its evolution, and the ”isocurvature” component does notnecessarily maintain zero comoving curvature. We later show that adiabatic perturbations stayadiabatic while they are well outside the horizon.Adiabatic perturbations are important, since the simplest theory for the origin of structureof the universe, single-field inflation, produces adiabatic perturbations. Using the constancyof R, it is easy to follow the evolution of these perturbations while they are well outside thehorizon.Using Eq. (15.21) we can write the second line of Eq. (15.27) as− 3 2(1 + w)HR′ =Ψ ′′ + (2 + 3c 2 s )HΨ′ + HΦ ′ + 3(c 2 s − w)H2 Φ= 3 2 H2δp C¯ρ − 1 3 ∇2 (Φ − Ψ) (15.30)or (using Eq. 15.19)H −1 R ′( )= − δpC¯ρ + ¯p − 2 wδ31 + w ∇2 Π = −c 2 Cs1 + w − 3S − 2 w31 + w ∇2 Π (15.31)which is completely in the comoving gauge.15.5 Perfect Fluid Scalar <strong>Perturbation</strong>s, AgainWith Π = 0 ⇒ Ψ = Φ, Eqs. (15.18,15.19,15.20,15.21,15.22,15.23) become∇ 2 Φ = 3 2 H2 δ C (15.32)Φ ′ + HΦ = 3 2 H2 (1 + w)v N (15.33)Φ ′′ + 3(1 + c 2 s )HΦ′ + 3(c 2 s − w)H2 Φ = 3 2 H2δp C¯ρ(15.34)δ ′ C − 3Hwδ C = (1 + w)∇ 2 v N (15.35)v ′ N + Hv N = δp C¯ρ + ¯p + Φ . (15.36)and Eqs. (15.26,15.31) becomeRH −1 R ′2 (= −Φ − Φ ′ + HΦ ) (15.37)3(1 + w)H( )= − δpC δC¯ρ + ¯p = −c2 s1 + w − 3S (15.38)and Eq. (15.34) asH −2 Φ ′′ + 3(1 + c 2 s )H−1 Φ ′ + 3(c 2 s − w)Φ = 3 2 c2 s [δC − 3(1 + w)S]. (15.39)which is Eq. (12.16), with just ∇ 2 Φ replaced by δ C using Eq. (15.32). (I seem to be repeatinghere some material from Sec. 12.) Our aim is to get a differential equation with preferably just
15 OTHER GAUGES 36one perturbation quantity to solve from, so this might seem a step backwards, replaceing one ofthe Φ with δ C , but we can now go ahead and replace also all the other Φ with δ C :Taking the Laplacian of this (Exercise), we get the Bardeen equationH −2 δ ′′ C + ( 1 − 6w + 3c 2 s)H −1 δ ′ C − 3 (2 1 + 8w − 6c2s − 3w 2) δ C = c 2 sH −2 ∇ 2 [δ C − 3(1 + w)S] ,(15.40)a differential equation from which we can solve the evolution of the comoving density perturbationfor superhorizon scales (when one can ignore the rhs) and for adiabatic perturbations at allscales (when S = 0).For the general case we need also an equation for S. To be able to do this, we need to takea closer look at the fluid, see Sec. 17 and beyond.15.5.1 Adiabatic Perfect Fluid <strong>Perturbation</strong>s at Superhorizon ScalesRewrite Eq. (15.37) as23 H−1 Φ ′ + 5 + 3w Φ = −(1 + w)R. (15.41)3If we have a period in the history of the universe, where we can approximate w = const.,then, for adiabatic perturbations at superhorizon scales, Eq. (15.41) is a differential equationfor Φ with w = R = const. for that period. This equation has a special solutionThe corresponding homogeneous equation isΦ = − 3 + 3w5 + 3w R.H −1 Φ ′ + 5 + 3w Φ = 02so that the general solution to Eq. (15.41) is⇒ a dΦda= − 5 + 3w Φ2⇒ Φ = Ca −5+3w 2Φ = − 3 + 3w5 + 3w R + Ca−5+3w 2 . (15.42)If w ≈ const. for a long enough time, the second part becomes negligible, and we haveIn particular, we have the relationsΦ = Ψ = − 3 + 3w R = const. (15.43)5 + 3wΦ k = − 2 3 R k (adiab., rad.dom, w = 1 3, k ≪ H) (15.44)Φ k = − 3 5 R k (adiab., mat.dom, w = 0, k ≪ H). (15.45)While the universe goes from radiation domination to matter domination, w is not a constant,so Eq. (15.43) does not apply, but we know from Eq. (15.45) that, Φ k changes from − 2 3 R k to− 3 5 R k, i.e, changes by a factor 9/10 (assuming k ≪ H the whole time, so that R k stays constant.
16 SYNCHRONOUS GAUGE 3716 Synchronous GaugeThe synchronous gauge is defined by the requirement A = B i = 0. Synchronous gauge can beused both for scalar and vector perturbations. In this section we consider only scalar perturbations,where it means thatA Z = B Z = 0. (16.1)(We use Z instead of S to denote synchronous gauge, so as not to confuse it with the scalar partof a perturbation.)One gets to synchronous gauge from an arbitrary gauge with a gauge transformation ξ µ thatsatisfiesξ 0′ + Hξ 0 = A (16.2)ξ ′ = −ξ 0 − B . (16.3)This is a differential equation from which to solve ξ 0 . Thus it is not easy to switch from anothergauge to synchronous gauge. We see that, like for comoving gauge, only the time derivative ofξ is determined.In the synchronous gauge the threads are orthogonal to the slices and the metric perturbationis only in the space part of the metric,h ij = −2D Z δ ij + 2 ( E Z ,ij − 1 3 ∇2 E Z δ ij)= −2ψ Z δ ij + 2E Z ,ij . (16.4)From Eq. (A.2), setting A = B = 0, the Christoffel symbols areΓ 0 00 = H (16.5)Γ 0 0i = Γ i 00 = 0Γ 0 ij= H [(1 − 2ψ)δ ij + 2E ,ij ] − δ ij ψ ′ + E ′ ,ijΓ i 0j= Hδ ij − δ ij ψ ′ + E ′ ,ijΓ i jk = −δ ij ψ ,k − δ ik ψ ,j + δ jk ψ ,i + E ,ijk .From Γ 0 0i = Γi 00 = 0 follows that the space coordinate lines are geodesics. Namely, the geodesicequationsẍ 0 + Γ 0 αβẋα ẋ β = ẍ 0 + Hẋ 0 ẋ 0 + Γ 0 ijẋ i ẋ j = 0 (16.6)ẍ i + Γ i αβẋα ẋ β = ẍ i + 2Γ i 0jẋ 0 ẋ j + Γ i jkẋj ẋk = 0 (16.7)are satisfied by x i = const. ⇒ ẋ i = 0 and ẋ 0 ≡ dη/dτ ≡ dt/adτ = a ⇒ dt = dτ. Thusone can construct the synchronous gauge coordinate system by choosing an initial spacelikehypersurface, distributing observers carrying space coordinate values on that hypersurface, withtheir initial 4-velocities orthogonal to the hypersurface (i.e., they are at rest with respect to thatsurface), synchronizing the clocks of the observers, and letting the observers then fall freely. Theworld lines of these observers are then the space coordinate lines and their clocks show the timecoordinate (t, not η).We use Ma & Bertschinger[2] (hereafter MB) as our main reference for synchronous gaugeand adopt from their notationh ≡η ≡−6D Z ≡ h iZiψ Z = D Z + 1 3 ∇2 E Zµ ≡ 2E Z , (16.8)
16 SYNCHRONOUS GAUGE 38so thath ij = 1 3 hδ ij + (∂ i ∂ j − 1 3 δ ij∇ 2 )µ= −2ηδ ij + µ ,ij . (16.9)Unfortunately, the MB notation fro the synchronous curvature perturbation is the same as ournotation for conformal time (MB use τ for the latter). Let’s hope this does not lead to confusion.MB discuss the synchronous metric perturbation in terms of the variables h and η, instead ofthe pair h,µ or η,µ. In coordinate space this appears impractical, since to solve µ from h andη requires integration (η = 1 6 (−h + ∇2 µ)). However, MB work entirely in Fourier space, whereη = − 1 6(h + µ) or µ = −h − 6η. Thus, in Fourier space, the metric ish ij= −2D Z δ ij − 2ˆk iˆkj E Z + 2 3 EZ δ ij= 1 3 hδ ij − ˆk iˆkj µ + 1 3 µδ ij= ˆk iˆkj h + (ˆk iˆkj − 1 3 δ ij)6η , (16.10)which is MB Eq. (4).From Eq. (8.1), we get from the synchronous gauge to the Newtonian gauge byFrom Eq. (7.9), the Bardeen potentials areξ Z→N = −E Z = − 1 2 µ (16.11)ξ 0 Z→N = EZ ′ = −ξ ′ Z→N .Φ = −HE Z ′ − E Z ′′ = − 1 2 Hµ′ − 1 2 µ′′Ψ = ψ Z + HE Z′ = η + 1 2 Hµ′ . (16.12)One has to be careful when Fourier transforming equations which employ different Fourierconventions for different terms. Thus, in Fourier space, Eq. (16.11) readsand Eq. (16.12) readsξ Z→N = − 12k µ = 1 (h + 6η) (16.13)2kξ 0 Z→N = − 1 k ξ′ Z→N (16.14)Φ = 12k 2 (−Hµ ′ − µ ′′) = 12k 2 [h ′′ + 6η ′′ + H(h ′ + 6η ′ ) ]Ψ = η − 12k 2 H(h′ + 6η ′ ), (16.15)which is MB Eq. (18).The opposite transformation, from Newtonian to synchronous gauge is, of course, justξ 0 N→Z = −ξ0 Z→N ξ N→Z = −ξ Z→N , (16.16)so we can easily express it in synchronous gauge quantities, which is actually what we typicallywant.
16 SYNCHRONOUS GAUGE 39We get the Einstein tensor perturbations from Eq. (A.7), setting A = B = 0,δG 0 0 = a −2 [ 2k 2 ψ + 6HD ′] (16.17)δG 0 i = a −2 [ −2ik i ψ ′]δG i 0 = a −2 [ 2ik i ψ ′]δG i j= a −2 [ 2D ′′ + k 2 D + 4HD ′ − 1 3 k2 E + 1 3 E′′ + 2 3 HE′] δji−k i k j a[D −2 − 1 3 E + 1 ]k 2(E′′ + 2HE ′ )In the MB η, h notation,δG i i = a −2 [ 6D ′′ + 2k 2 ψ + 12HD ′] .δG 0 0 = a −2 [ 2k 2 η − Hh ′] (16.18)δG 0 i = a −2 [ −2ik i η ′]δG i 0 = a −2 [ 2ik i η ′]δG i j= a −2 [ − 1 2 h′′ − η ′′ − H(h ′ + 2η ′ ) + k 2 η ] δji−k i k j a[η −2 − 1 ]k 2(1 2 h′′ + 3η ′′ + Hh ′ + 6Hη ′ )The Einstein equations are thusδG i i = a −2 [ −h ′′ + 2k 2 η − 2Hh ′] .k 2 η − 1 2 Hh′ = −4πGa 2 δρ Z = − 3 2 H2 δ Z (16.19)k 2 η ′= 4πGa 2 (¯ρ + ¯p)kv Z = 3 2 H2 (1 + w)kv Zh ′′ + 2Hh ′ − 2k 2 η = −24πGa 2 δp Z = −9H 2δpZ¯ρh ′′ + 6η ′′ + 2Hh ′ + 12Hη ′ − 2k 2 η = −16πGa 2¯pΠ = −6H 2 wΠ,which is MB Eq. (21). Note that MB uses the notationθ ≡ ∇ · ⃗v = −∇ 2 v = kv and (¯ρ + ¯p)σ ≡ 2 3 ¯pΠ. (16.20)We get the continuity equations from Eq. (A.15), setting A = B = 0 and D = − 1 6 h,δρ Z ′ = −3H(δρ Z + δp Z ) − (¯ρ + ¯p)( 1 2 h′ + kv Z ) (16.21)(¯ρ + ¯p)v Z ′= −(¯ρ + ¯p) ′ v Z − 4H(¯ρ + ¯p)v Z + kδp Z − 2 3 k¯pΠ)δ Z ′= −(1 + w)(kv Z + 1 2 h′ ) + 3H(wδ Z − δpZ¯ρv Z ′ = −H(1 − 3w)v Z − w′1 + w vZ + kδpZ¯ρ + ¯p − 2 3w1 + w kΠ.The two last equations are MB Eq. (30).Exercise: Derive the synchronous gauge Einstein equations and continuity equations from the correspondingNewtonian gauge equations by gauge transformation .
17 FLUID COMPONENTS 4017 Fluid Components17.1 Division into ComponentsIn practice, the cosmological fluid consists of many components (e.g., particle species: photons,baryons, CDM, neutrinos, . ..). It is useful to divide the energy tensor into such components:T µ ν = ∑ iT µ ν(i) , (17.1)which have their corresponding background and perturbation parts¯T µ ν = ∑ i¯T µ ν(i)and δT µ ν = ∑ iδT µ ν(i) . (17.2)Here i labels the different components (so I’ll use l and m for space indices).Often the component pressure is a unique function of the component energy density, p i =p i (ρ i ), or can be so approximated (common cases are p i = 0 and p i = ρ i /3), although this is nottrue for the total pressure and energy density.The total fluid and component quantities for the background are related aswhere¯ρ = ∑ ¯ρ i¯p = ∑ ¯p i = ∑ w i¯ρ iw ≡c 2 s ≡ ¯p′¯ρ ′ =¯p¯ρ = ∑ ¯ρ i¯ρ w i∑ ¯p′i¯ρ ′= ∑ ¯ρ ′ i¯ρ ′ c2 i , (17.3)w i ≡ ¯p iand c 2 i¯ρ ≡ ¯p′ ii ¯ρ ′ . (17.4)iThe total fluid and component quantities for the perturbations are related asδρ = ∑ δρ iδp = ∑ δp iδ ≡δρ¯ρ = ∑ ¯ρ i¯ρ δ i , (17.5)where δ i ≡ δρ i /¯ρ i . Fromwe get thatand from(¯ρ + ¯p)v l = ∑ (¯ρ i + ¯p i )v l(i) (17.6)v l = ∑ ¯ρ i + ¯p i¯ρ + ¯p v l(i) (17.7)Σ lm = ∑ Σ lm(i) = ∑ ¯p i Π lm(i)we get thatΠ lm ≡ Σ lm¯p= ∑ ¯p i¯p Π lm(i) = ∑ w i¯ρ iw¯ρ Π lm(i) . (17.8)
17 FLUID COMPONENTS 4117.2 Gauge TransformationsPerhaps I’ll write more here someday ...Note that gauge conditions that refer to fluid perturbations, refer usually to the total fluid.Thus, e.g., in the comoving gauge the total velocity perturbation v vanishes, but the componentvelocities v i do not vanish (unless they happen to be all equal). Thus, the velocity v thatappears in the gauge transformation equations to comoving gauge refers to the total densityperturbation. For example the component gauge tranformation equations that correspond toEq. (15.14) readδρ C i = δρ N i − ¯ρ ′ i vN (17.9)= δp N i − ¯p ′ iv Nδp C iδ C i = δ N i − ¯ρ′ i¯ρ vN .Since the gauge transformations are the same for all components, we find some gauge invariances.The relative velocity perturbation between two components i and j,v i − v j is gauge invariant. (17.10)Like the total anisotropic stress, also the component anisotropic stressesΠ i are gauge invariant. (17.11)We can also define a kind of entropy perturbation( )δρ iS ij ≡ −3H¯ρ ′ − δρ ji¯ρ ′ j(17.12)between two fluid components i and j which turns out to be gauge invariant due to the waythe density perturbations δρ i transform. The above is a special case of a generalized entropyperturbation( δxS xy ≡ H¯x ′ − δy )ȳ ′ (17.13)between any two 4-scalar quantities x and y, which is gauge invariant due to the way 4-scalarperturbations transform. Beware of the many different quantities called “entropy perturbation”!Some of them can be interpreted as perturbations in some entropy/particle ratio. What iscommon to all of them, is that they all vanish in the case of adiabatic perturbations.17.3 EquationsThe Einstein equations (both background and perturbation) involve the total fluid quantities.The metric perturbations are not divided into components due to different fluid components!There is a single gravity, due to the total fluid, which each fluid component obeys.If there is no energy transfer between the fluid components in the background universe, thebackground energy continuity equation is satisfied separately by such independent components,and in that case we can write Eq. (17.3) as¯ρ ′ i = −3H(¯ρ i + ¯p i ), (17.14)c 2 s = ∑ ¯ρ i + ¯p i¯ρ + ¯p c2 i , (17.15)
17 FLUID COMPONENTS 42and Eq. (17.12) asS ij =δρ i δρ j− = δ i− δ j(17.16)(1 + w i )¯ρ i (1 + w j )¯ρ j 1 + w i 1 + w jLikewise, the gauge transformation equations (17.9) becomeδρ C i = δρ N i + 3H(1 + w i )¯ρ i v N (17.17)= δp N i + 3H(1 + w i )c 2 i ¯ρ iv Nδp C iδ C i = δ N i + 3H(1 + w i )v N .But if there is energy transfer between two fluid components, then their component energycontinuity equations acquire an interaction term.Even if there is no energy transfer between fluid components at the background level, theperturbations often introduce energy and momentum transfer between the components. It mayalso be the case that the energy transfer can be neglected in practice, but the momentum transferremains important.In the case of noninteracting fluid components (no energy or momentum transfer), we havethe perturbation energy and momentum continuity equations separately for each such fluidcomponent,= 0. (17.18)T µ ν(i);µFor the case of scalar perturbations in the conformal-Newtonian gauge, they read(δi N ) ′ = (1 + w i ) ( ∇ 2 vi N + 3Ψ ′) ( )+ 3H w i δiN − δpN i¯ρ i(vi N )′ = −H(1 − 3w i )vi N − w′ ivi N + δpN i+ 2 1 + w i ¯ρ i + ¯p i 3In Fourier space they become(δ N i ) ′ = (1 + w i ) ( −kv N i + 3Ψ ′) + 3H(w i δ N iw i− δpN i¯ρ i)(vi N ) ′ = −H(1 − 3w i )vi N − w′ ivi N + kδpN i− 2 1 + w i ¯ρ i + ¯p i 3(17.19)1 + w i∇ 2 Π i + Φ . (17.20)w i(17.21)1 + w ikΠ i + Φ . (17.22)If there are interactions between the fluid components, there will be interaction terms (“collisionterms”) in these equations.Note that one cannot write the mixed gauge equations for fluid components by just replacingthe fluid quantities in equations (15.22,15.23) with component quantities, since these equationswere derived by making a gauge transformation between the comoving and Newtonian gauges,which involved the total fluid velocity.For the case where energy transfer between components can be neglected both at the backgroundand perturbation level, we can use Eqs. (17.14) and (17.19) to find the time derivativeof the entropy perturbation S ij ,S ′ ij = ∇ 2 ( v N i − v N j)+ 3H(c 2 i δρN i − δp N i¯ρ i + ¯p i= ∇ 2 (v i − v j ) − 9H ( c 2 i S i − c 2 j S j),)− c2 j δρN j − δp N j¯ρ j + ¯p j(17.23)where( δpiS i ≡ H¯p ′ − δρ )ii¯ρ ′ i(17.24)
18 ADIABATIC AND ISOCURVATURE PERTURBATIONS IN A SIMPLIFIED UNIVERSE43is the (gauge invariant) internal entropy perturbation of component i. If we have in addition,that for both components the equation of state has the form p i = f i (ρ i ), then the internalentropy perturbations vanish 24 , and we have simplyIn Fourier space Eq. (17.25) readsS ′ ij = ∇ 2 (v i − v j ). (17.25)S ′ ij = −k(v i − v j ) or H −1 S ′ ij = − k H (v i − v j ), (17.26)showing that entropy perturbations remain constant at superhorizon scales k ≪ H. In otherwords, adiabatic perturbations (S ij = 0) stay adiabatic while outside the horizon.We can also obtain an equation for S ij ′′ from Eq. (17.20), if also momentum transfer betweencomponents can be neglected.In the synchronous gauge, Eqs. (17.21,17.22) become(δi Z ) ′ = −(1 + w i ) ( ( )kvi Z + 1 2 h′) + 3H w i δi Z − δpZ i(17.27)¯ρ i(vi Z )′ = −H(1 − 3w i )vi Z − w′ ivi Z + kδpZ i− 2 1 + w i ¯ρ i + ¯p i 3w i1 + w ikΠ i . (17.28)Since E does not appear in the general gauge scalar continuity equations (A.15), the onlydifference in them when going from Newtonian to synchronous gauge (as both have B = 0) isthat Ψ = D N is replaced by − 1 6 h = DZ and Φ = A N is dropped as A Z = 0.18 Adiabatic and Isocurvature <strong>Perturbation</strong>s in a SimplifiedUniverseConsider the case where the energy tensor consists of two perfect fluid components, matter withp = 0 and radiation with p = 1 3ρ, that do not interact with each other, i.e., there is no energy ormomentum transfer between them. (Compared to the real universe, this is a simplification since,while cold dark matter does not much interact with the other fluid components, the baryonicmatter does interact with photons. Also, the radiation components of the real universe, neutrinosand photons, behave like a perfect fluid only until their decoupling.)18.1 Background solution for radiation+matterWe have now two fluid components,whereandρ = ρ r + ρ m ,ρ r ∝ a −4 and ρ m ∝ a −3 ,p m = 0 and p r = 1 3 ρ r .The equation of state and sound speed parameters arew m = c 2 m = 0 and w r = c 2 r = 1 3 . (18.1)24 For baryons this requires that we ignore baryon pressure, since p b = p b (n b , T) = n b T, and ρ b = ρ b (n b , T) =n b (m b + 3 2 T).
18 ADIABATIC AND ISOCURVATURE PERTURBATIONS IN A SIMPLIFIED UNIVERSE44so thatandWe definew =ρ rρ = 11 + y13(1 + y)ρ mρ =y1 + y1 + w = 4 + 3y3(1 + y)y ≡aa eq= ρ mρ r(18.2)ρ r + p rρ + p = 44 + 3yc 2 s = 43(4 + 3y)ρ m + p mρ + p= 3y4 + 3y(18.3)1 − 3c 2 s = 3y4 + 3y . (18.4)whereThe Friedmann equation isH 2 ≡ 1 ( ) da 2a 2 dη= 8πG3 ρa2 = 8πG3 (1 + y)ρ r0a 4 0 a−2⇒dy√ 1 + y= 2Cdη ,The solution isAt the time of matter-radiation equality,√2πGC ≡3 ρ a 2 0r0 .a eqy = C 2 η 2 + 2Cη . (18.5)y = y eq = C 2 η 2 eq + 2Cη eq = 1 ⇒ Cη eq = √ 2 − 1, (18.6)so we can write the solution aswhereη 3 ≡( ) η 2 ( ) ηy = + 2 , (18.7)η 3 η 3η eq(√ )√ = 2 + 1 η eq (18.8)2 − 1is the time when y = 3 (ρ m = 3ρ r ).The Hubble parameter isAt matter-radiation equality it has the valueH ≡ a′a = y′y = η + η 3η 3 η + 1 . (18.9)2η2 H eq = 2√ 2√2 + 11η eq= 4 − 2√ 2η eq= 2√ 2η 3. (18.10)At early times, η ≪ η 3 , the universe is radiation dominated, so thaty ≈ 2ηη 3≪ 1 ⇒ a ∝ η ⇒ H = 1 η ∝ a−1 . (18.11)At late times, η ≫ η 3 , the universe is matter dominated, so thaty ≈( ηη 3) 2≫ 1 ⇒ a ∝ η 2 ⇒ H = 2 η ∝ a−1/2 . (18.12)
18 ADIABATIC AND ISOCURVATURE PERTURBATIONS IN A SIMPLIFIED UNIVERSE45When solving for perturbations it turns out to be more convenient to use y (or log y) as timecoordinate instead of η. Inverting Eq. (18.7), we have that(√ ) √ 1 + y − 1η = 1 + y − 1 η 3 = √ η eq , (18.13)2 − 1andH =√ 1 + yy2η 3=√ 1 + yyH eq√2. (18.14)18.2 <strong>Perturbation</strong>sIn terms of the component perturbations the total perturbations are nowδ =v =and the relative entropy perturbation is11 + y δ r + y1 + y δ m (18.15)44 + 3y v r + 3y4 + 3y v m (18.16)S ≡ S mr = δ m − 3 4 δ r . (18.17)From the pair (18.15,18.17) we can solve δ m and δ r in terms of δ and S:δ m = 3 + 3y4 + 3y δ + 44 + 3y S (18.18)δ r = 4 + 4y4 + 3y δ − 4y4 + 3y S . (18.19)Likewise we can express v m and v r in terms of the total and relative velocity perturbations, vand v m − v r :v m = v + 44 + 3y (v m − v r ) (18.20)v r = v − 3y4 + 3y (v m − v r ). (18.21)We can now also relate the total entropy perturbation S to S:(1S =3(1 + w)δρ¯ρ − 1 )δpyc 2 = ... =s ¯ρ 4 + 3y S = 1 3 (1 − 3c2 s)S . (18.22)The Bardeen equation (15.40) becomes nowH −2 δ ′′ C + ( 1 − 6w + 3c 2 s)H −1 δ ′ C − 3 (2 1 + 8w − 6c2s − 3w 2) δ C( ) k 2[δC − (1 + w)(1 − 3c 2 sH)S]= −c 2 s= −c 2 s( kH) 2 (δ C −y )1 + y S . (18.23)We get the entropy evolution equation by derivation Eq. (17.25),S ′ = −k(v m − v r ) ⇒ S ′′ = −k(v m ′ − v′ r ). (18.24)
18 ADIABATIC AND ISOCURVATURE PERTURBATIONS IN A SIMPLIFIED UNIVERSE46Using the v N ievolution equations (17.22), this becomes (Exercise)S ′′ = Hk(v m − v r ) + Hkv r + 1 4 k2 δ N r . (18.25)Here the δ N r is converted to the comoving gauge using the total fluid velocity (see Eq. 17.17),so that Eq. (18.24) becomesS ′′δ N r = δ C r − 3H(1 + w r)k −1 v N (18.26)( ) 4= Hk (v m − v r ) + 14 + 3y4 k2 δr C . (18.27)Replacing v m − v r by −k −1 S ′ we get (Exercise) the Kodama-Sasaki equationH −2 S ′′ + 4 ( ) k 2 ( 1 + y4 + 3y H−1 S ′ =H 4 + 3y δC − y )4 + 3y S . (18.28)orH −2 S ′′ + 3c 2 s H−1 S ′ = 1 ( ) k 2 ( )13 H 1 + w δ − (1 − 3c2 s )S ,where δ ≡ δ C .The two equations (18.23) and (18.28) form a pair of ordinary differential equations, fromwhich we can solve the evolution of the perturbations δ C ⃗ k(η) and S ⃗k (η). Since the coefficientfunctions of these equations can more easily be expressed in terms of the scale factor y, it may bemore convenient to use y as the time coordinate instead of η. The time derivatives are convertedwith (Exercise)H −1 f ′ = y dfdyand the equations becomey 2 d2 δdy 2 + 3 2 (1 − 5w + 2c2 s)y dδandH −2 f ′′ = y 2d2 fdy 2 + 1 df2(1 − 3w)ydy( kdy − 3 2 (1 + 8w − 6c2 s − 3w 2 )δ = −Hy 2d2 Sdy 2 + 1 2 (1 − 3w + 6c2 s)y dS ( k=dy H(18.29)) 2 (c 2 s δ −y )1 + y S (18.30)) 2 ( 1 + y4 + 3y δC − y )4 + 3y S .For solving the other perturbation quantities, we collect here the relevant equations:Φ = − 3 ( ) H 2δ (18.31)2 k( )v N 2 k (H=−1 Φ ′ + Φ )3(1 + w) H( ) Hδ N = δ − 3 (1 + w)v NkR= −Φ −2 (Φ ′ + HΦ )3(1 + w)HWhen judging which quantities are negligible at superhorizon scales (k ≪ H), one has to exrcisesome care, and not just look blindly at powers of k/H in equations which contain different perturbationquantities. From Eq. (18.31a) one sees that at superhorizon scales, a small comovingδ can still be important and cause a large gravitational potential perturbation Φ. Eq. (15.38),( ) δH −1 R ′ = −c 2 s1 + w − 3S ,
18 ADIABATIC AND ISOCURVATURE PERTURBATIONS IN A SIMPLIFIED UNIVERSE47may seem to contradict the statement that for adiabatic perturbations R stays constant atsuperhorizon scales, but the explanation is, that R is then of the same order of magnitude asΦ, whereas δ is suppressed by two powers of k/H compared to Φ. Using Eqs. (18.31a,18.22) werewrite (15.38) as18.3 Initial EpochH −1 R ′= c 2 s[11 + wFor y ≪ 1, the equations (18.23,18.28,18.30) can be approximated by( ) ]2 k 2Φ + (1 − 3c 2 s3 H)S . (18.32)H −2 δ ′′ − 2δ = − 1 3( kH) 2(δ − yS) (18.33)orH −2 S ′′ + H −1 S ′ = 1 4( kH(y 2 d2 δk 2dy 2 − 2δ = −1 3(δ − yS) = −H) 2 3( )y 2d2 Sk 2dy 2 + ydS = 1dy4(δ − yS) = 1 H2) 2(δ − yS) (18.34)( kH eq) 2(y 2 δ − y 3 S ) (18.35)( kH eq) 2(y 2 δ − y 3 S ) . (18.36)At early times, all cosmologically intersecting scales are outside the horizon, and we start bymaking the approximation, that the rhs of these equations can be ignored. Then the evolutionof δ and S decouple and the general solutions areδ ⃗k = A ⃗k y 2 + B ⃗k y −1 (18.37)S ⃗k = C ⃗k + D ⃗k ln y . (18.38)Thus, for each Fourier component ⃗ k, there are four independent modes. We identify the adiabaticgrowing (A ⃗k ) and decaying (B ⃗k ) modes, and the isocurvature 25 “growing” (C ⃗k ) and decaying(D ⃗k ) modes. The D-mode is indeed decaying, since 0 < y ≪ 1.The decaying modes diverge as y → 0, but we can suppose that our description of theuniverse is not valid all the way to y = 0, but at very early times there is some process that isresponsible for fixing the initial values of δ ⃗k , δ ′ ⃗ k, S ⃗k , and S ′ ⃗ kat some early time y init > 0, andfixing thus A ⃗k , B ⃗k , C ⃗k , and D ⃗k .Let us now consider the effect of the rhs of the eqs. These couple the δ and S. Thus it isnot consistent to assume that the other is exactly zero.18.3.1 Adiabatic modesConsider first the adiabatic modes (C ⃗k = D ⃗k = 0). The coupling forces the existence of a smallnonzero S, which at first is ≪ δ. We can thus keep ignoring S on the rhs, and for the δ equationwe can keep ignoring the whole rhs. But for the S equation we now have very small S on the lhs,and therefore we can not ignore the large δ on the rhs, even though it is suppressed by (k/H) 2 ,or y 2 . Thus for adiabatic modes the pair of equations can be approximated asH −2 δ ′′ − 2δ = 0 (18.39)H −2 S ′′ + H −1 S ′ = 1 4( kH) 2δ (18.40)25 The term “isocurvature” will be explained later.
18 ADIABATIC AND ISOCURVATURE PERTURBATIONS IN A SIMPLIFIED UNIVERSE48ory 2 d2 δdyy 2d2 Sdy 2 + ydS dy− 2δ = 0 (18.41)2= 1 ( ) k 2y 2 δ . (18.42)2 H eqThe solution for δ remains Eq. (18.37), but we also get thatS ⃗k = 1 ( ) k 2A ⃗k y 4 + 1 ( ) k 2B ⃗k y . (18.43)32 H eq 2 H eqThus for the growing adiabatic modeand for the decaying adiabatic modeS ⃗k = 1 ( ) k 2δ ⃗k (18.44)64 HS ⃗k = 1 4( kH) 2δ ⃗k . (18.45)So you see that, although these are called adiabatic modes, the perturbations are not exactlyadiabatic! The name just means that S → 0 as y → 0. The entropy perturbations remain smallcompared to the density perturbation while the Fourier mode is outside the horizon, but canbecome large near and after horizon entry.18.3.2 Isocurvature ModesConsider then the isocurvature modes (A ⃗k = B ⃗k = 0). Now the coupling causes a small δ ≪ S.We can ignore the rhs of the S equation, but the large S cannot be ignored on the rhs of the δequation. Thus for isocurvature modes the pair of equations can be approximated byorH −2 δ ′′ − 2δ = + 1 3( kH) 2yS (18.46)H −2 S ′′ + H −1 S ′ = 0 (18.47)( )y 2 d2 δk 2dy 2 − 2δ = +2 y 3 S (18.48)3 H eqy 2d2 Sdy 2 + ydS dy= 0. (18.49)The solution for S remains Eq. (18.38), but for the density perturbation we getδ ⃗k = 1 ( ) k 2C ⃗k y 3 + 1 ( ) k 2 (D ⃗k y 3 ln y − 5 )6 H eq 6 H eq 4 y3 . (18.50)(For the decaying isocurvature mode the differential equation for δ is a little more difficult, butyou can check (Exercise) that the above is the correct solution.) For the growing isocurvaturemode we have thus thatδ ⃗k = 1 ( ) k 2yS ⃗k . (18.51)12 H
18 ADIABATIC AND ISOCURVATURE PERTURBATIONS IN A SIMPLIFIED UNIVERSE49So although the relative entropy perturbation stays constant at early times, the density perturbationis growing in this mode. For the decaying isocurvature mode18.3.3 Other <strong>Perturbation</strong>sδ ⃗k = 1 ( k 2 (y −12 H) 5 )yS ⃗k ≈ 1 ( ) k 2yS ⃗k . (18.52)4 ln y 12 HFor the initial eboch, w = 1 3, and Eqs. (18.31) becomeΦ = − 3 2( Hk) 2δ (18.53)For the growing adiabatic mode, we havev N = 1 ( ) k (H −1 Φ ′ + Φ )2 H( ) Hδ N = δ − 4 v NkR = − 3 2 Φ − 1 2 H−1 Φ ′Φ = − 3 2(Heqk) 2A = const. (18.54)R = − 3 2Φ = const. (18.55)v N = − 1 ( √ ( )k 2 kR = − yR (18.56)3 H)3 H eqδ N = −2Φ = 4 3R. (18.57)For the growing isocurvature mode, where S = const. we haveΦ = − 1 8 yS = − H eq8 √ ηS (18.58)21R =4 √ 2 H eqSη = 1 4Sy = −2Φ (18.59)v N = − 1 ( ) k4 √ Sy 2 (18.60)2 H eqδ N = 1 ( ) k 2Sy 3 + 16 H2 Sy ≈ 1 2Sy = 2R. (18.61)eqThus the growing adiabatic mode is characterized by a constant R and the growing isocurvaturemode by a constant S. If both modes are present, and these constants are of a similarmagnitude, then the growing S associated with the adiabatic mode is negligible as long ask ≪ H, and the growing R associated with the isocurvature mode is negligible as long as y ≪ 1.Thus, after the decaying modes have died out, the general (adiabatic+isocurvature) mode ischaracterized by these two constants (for each Fourier mode), which we denote by R ⃗k (rad) and
18 ADIABATIC AND ISOCURVATURE PERTURBATIONS IN A SIMPLIFIED UNIVERSE50S ⃗k (rad). Including the two small growing contributions we have, during the initial epoch,R ⃗k = 9 ( ) 2 HeqA ⃗k + 18 k4 yC ⃗ k= R ⃗k (rad) + 1 4 yS ⃗ k(rad) (18.62)S ⃗k = 1 ( ) k 2A ⃗k y 4 + C ⃗k = S ⃗k (rad) + 1 ( ) k 4R ⃗k (rad)y 432 H eq 36 H eq= S ⃗k (rad) + 1 9( kH) 4R ⃗k (rad). (18.63)18.4 Full evolution for large scalesThe full evolution in the general case is not amenable to analytic solution, and has to be solvednumerically. This is not too difficult since we just have a pair of ordinary differential equationsto solve for each k. The results from Sect. 18.3 can be used to set initial values at some small y.For large scales (k ≪ k eq ≡ H eq ) we can, however, solve the evolution analytically. We nowdrop the decaying modes and consider what happens to the growing modes after the radiationdominatedepoch.Our basic equations are (18.32) and (18.28):[ ( ) ]H −1 R ⃗ ′ k= c 2 1 2 k 2sΦ ⃗k + (1 − 3c 2 s1 + w 3 H)S ⃗ k(18.64)H −2 S ⃗ ′′k+ 3c 2 s H−1 S ⃗ ′ k= 1 ( k 2 []13 H)1 + w δ ⃗ k− (1 − 3c 2 s )S ⃗ k. (18.65)We see that for superhorizon scales, S ⃗k = const. remains a solution even when the universe isno longer radiation dominated. Since the other solution has decayed away, we conclude thatS ⃗k = const. = S ⃗k (rad) for k ≪ H . (18.66)We can also see that, for the adiabatic mode, R ⃗k stays constant for superhorizon scales,since on the rhs, S ⃗k remains negligible for k ≪ H. For scales that enter the horizon during thematter-dominated epoch, the R ⃗k of adiabatic perturbations stays constant even through andafter horizon entry, since c 2 s becomes negligibly small, before k/H and S ⃗k become large. We canthus conclude thatR ⃗k = const. = R ⃗k (rad) for adiabatic modes with k ≪ k eq . (18.67)If the isocurvature mode is present, however, we cannot assume that S ⃗k is negligible forsuperhorizon scales; it’s just constant, and we have, for k ≪ H,orIntegrating this, we getH −1 R ′ ⃗ k= c 2 s (1 − 3c2 s )S ⃗ k, (18.68)dR ⃗kdyR ⃗k = R ⃗k (rad) +=4(4 + 3y) 2S ⃗ k. (18.69)y4 + 3y S ⃗ k(rad) for k ≪ H . (18.70)
19 EFFECT OF A SMOOTH COMPONENT 51For k ≪ k eq , this has reached the final value R ⃗k (rad) + 1 3 S ⃗ k(rad) when the universe has becomematter dominated, before horizon entry. After that, R ⃗k stays constant even through and afterhorizon entry by the same argument as for the adiabatic mode. Thus we conclude thatR ⃗k = const. = R ⃗k (rad) + 1 3 S ⃗ k(rad) for k ≪ k eq and η ≫ η eq . (18.71)For smaller scales, the exact solution has to be obtained numerically. It is customary torepresent the solution in terms of transfer functions T RR (k) and T RS (k), which we can define byR ⃗k = const. ≡ T RR (k)R ⃗k (rad) + 1 3 T RS(k)S ⃗k (rad) for η ≫ η eq , (18.72)so that T RR (k) = T RS (k) = 1 for k ≪ k eq .Finally, we can ask what happens to S ⃗k after horizon entry (Exercise). ...19 Effect of a Smooth ComponentShould add here a section about the effect of a fluid component, whose perturbations we ignore,so that it affects only the background solution. The main application is dark energy. In casedark energy is just a cosmological constant, it has no perturbations. If the dark energy is closeto a cosmological constant (w ∼ −1), its perturbations should be small.20 The Real UniverseAccording to present understanding, the universe contains 5 major “fluid” components: “baryons”(including electrons), cold dark matter, photons, neutrinos, and the mysterious dark energy,ρ = ρ b + ρ c + ρ γ + ρ ν + ρ DE . (20.1)We shall here assume there are no perturbations in the dark energy.We make the approximation that the pressure of baryons and cold dark matter can beignored. 26 Thus p b = p c = 0 (both for background and perturbations). For photons, p γ = ρ γ /3.We assume massless neutrinos, so the same relation holds for them. Thus we haveMoreover,w b = w c = c 2 b = c 2 c = 0w γ = w ν = c 2 γ = c 2 ν = 1 3 . (20.2)δp b = δp c = 0δp γ = 1 3 δρ γδp ν = 1 3 δρ ν . (20.3)Thus we have the happy situation, that for each component we have a unique relationp i = p i (ρ i ), which moreover is very simple, either p i = 0 or p i = ρ i /3. (Also, the simplest kindof dark energy has p DE = −ρ DE .The components are, however, not all independent. Cold dark matter does not interactwith the other components. We can ignore the interactions of neutrinos, since we are now only26 For small distance scales the baryon pressure is important after photon decoupling. If we were interestedin small-scale structure formation, we should include it. Now our main interest is the CMB, where we are notinterested in as small distance scales.
20 THE REAL UNIVERSE 52interested in times much after neutrino decoupling. But the baryons and photons interact viaThomson scattering.Our continuity equations for perturbations are thus (for scalar perturbations in the conformal-Newtonian gauge)δ ′ c = ∇ 2 v c + 3Ψ ′ (20.4)v ′ cδ ′ bv ′ b= −Hv c + Φ= ∇ 2 v b + 3Ψ ′ + (collision term)= −Hv b + Φ + collision termδ ′ γ = 4 3 ∇2 v γ + 4Ψ ′ + (collision term)v ′ γ = 1 4 δ γ + 1 6 ∇2 Π γ + Φ + collision termδ ′ ν = 4 3 ∇2 v ν + 4Ψ ′v ′ ν = 1 4 δ ν + 1 6 ∇2 Π ν + Φ .We have put the collision terms for δ ′ b and δ′ γ in parenthesis, since it will turn out that they canbe neglected, and only momentum transfer between photons and baryons is important.In Fourier space these equations readδ ′ c = −kv c + 3Ψ ′ (20.5)v ′ cδ ′ bv ′ bδ ′ γ= −Hv c + kΦ= −kv b + 3Ψ ′ + (collision term)= −Hv b + kΦ + collision term= − 4 3 kv γ + 4Ψ ′ + (collision term)v ′ γ = 1 4 kδ γ − 1 6 kΠ γ + kΦ + collision termδ ′ ν = − 4 3 kv ν + 4Ψ ′v ′ ν = 1 4 kδ ν − 1 6 kΠ ν + kΦ .(Remember our Fourier convention for v and Π.)These equations are supplemented by 2 Einstein equations (there are 4 Einstein equationsfor perturbations, but since we are also using continuity equations, only two of them remainindependent). Thus we have 10 perturbation equations, but there are 12 perturbation quantitiesto solve. (If we think of the Einstein equations as the equations for Φ and Ψ, the “extra”quantities lacking an equation of their own are the anisotropic stresses Π γ and Π ν . In theperfect fluid approximation these vanish, and the number of quantities equals the number ofequations.) Also, we do not yet have the collision terms.Thus more work is needed. This will lead us to the Boltzmann equations which employ amore detailed description of the fluid components, in terms of distribution functions.In synchronous gauge one can simplify the cold dark matter equations, since cold dark matterfalls freely (in our approximation). Thus we can use cold dark matter particles as the freelyfalling observers that define the synchronous space coordinate, so thatv Z c = 0. (20.6)
21 EARLY RADIATION-DOMINATED ERA 53In synchronous gauge Eqs. (20.5) become thusδ ′ c = − 1 2 h′ (20.7)v c = 0δ ′ bv ′ bδ ′ γ= −kv b − 1 2 h′ + (collision term)= −Hv b + collision term= − 4 3 kv γ − 2 3 h′ + (collision term)v ′ γ = 1 4 kδ γ − 1 6 kΠ γ + collision termδ ′ ν = − 4 3 kv ν − 2 3 h′v ′ ν = 1 4 kδ ν − 1 6 kΠ ν .21 Early Radiation-Dominated EraThe initial conditions for the evolution of the large scale structure and the cosmic microwavebackground can be specified during the radiation-dominated epoch, sufficiently early that allscales k of interest are outside the horizon. We do not want to deal with the electron-positronannihilation or big-bang nucleosynthesis (BBN), so we limit this time period to start after BBN.We assume it is sufficiently far after whatever event created the perturbations in the first place,so that we can assume that all decaying modes have already died out.We are not just specifying “initial values” at some particular instant of time. Rather, we solvethe perturbation equations for this particular epoch, and find that the solutions are characterizedby quantities that remain constant for the whole epoch. Other perturbations quantities arerelated to these constant quantities by some powers of kη. These perturbations during thisepoch we call the “primordial perturbations”.There are different modes of primordial perturbations. In adiabatic perturbations all fluidperturbations are determined by the metric perturbations. However, the metric perturbationsdepend only on the total fluid perturbations δ, δ p , v, and Π. Thus there are additional degreesof freedom in the component fluids: entropy perturbations. For adiabatic perturbations, allcomponent velocity perturbations are equal, v i = v, and the density perturbations are relatedδ i= δ1 + w i 1 + w . (21.1)The (relative) entropy perturbations are defined( )δρ iS ij ≡ −3H¯ρ ′ − δρ ji¯ρ ′ j=δ i1 + w i− δ j1 + w j(21.2)(we assume we can ignore energy transfer between fluid components).For N fluid components, there are N −1 independent entropy perturbations. Often photonsare taken as the reference fluid component for entropy perturbations, so that the independententropy perturbations are taken to beS i ≡δ i1 + w i− 3 4 δ γ , i ≠ γ . (21.3)For this section, we shall work mainly in the Newtonian gauge. Unless otherwise specified,
21 EARLY RADIATION-DOMINATED ERA 54δ ≡ δ N and v = v N (!!!). The relevant equations are the Einstein equations (10.18):( ) k 2H −1 Ψ ′ + Φ + 1 3Ψ = − 1H2 δ (21.4)H −1 Ψ ′ + Φ = 3 2 (1 + w)H k v (21.5)H −2 Ψ ′′ + H −1 ( Φ ′ + 2Ψ ′) (+(1 + 2H′ k 2 )ΦH 2 − 1 3(Φ − Ψ) =H) 3 δp2¯ρ( ) k 2(Ψ − Φ) = 3wΠ, (21.7)H(21.6)the fluid equations (20.5), and we also want to refer to the comoving curvature perturbation⇒2R = −Ψ −3(1 + w) Φ − 23(1 + w) H−1 Ψ ′23 H−1 Ψ ′ + 5 + 3w Ψ = −(1 + w)R + 2 (Ψ − Φ) (21.8)33(from Eq. 15.26).The early radiation-dominated era has 4 properties which simplify the solution of the perturbationequations:1. All scales of interest are outside the horizon, k ≪ H. This allows us to drop some (butnot necessarily all) of the gradient terms (those with k) from the perturbation equations.2. Radiation domination, ρ γ , ρ ν ≫ ρ b , ρ c , ρ DE⇒ w = c 2 s = 1 3⇒ the background solution is H = 1 η , (21.9)and we can ignore the baryon, CDM, and DE contributions to the total fluid perturbation.We can also ignore the collision term in the photon velocity equation (but not in the baryonvelocity equation) since the momentum the baryonic fluid can transfer to the photon fluidis negligible compared to the inertia density of the photon fluid.3. This era is before recombination and photon decoupling, so baryons and photons are tightlycoupled⇒ v b = v γ (21.10)and the continuous interaction with baryons (really the electrons) keeps the photon distributionisotropic⇒ Π γ = 0. (21.11)4. m ν ≪ T ⇒ We can approximate neutrinos to be massless. This helps in solvingthe evolution of the neutrino momentum distribution. We do not discuss this here; thisbelongs to the course of CMB Physics, and we need to take one result from there.Thus we haveδ = (1 − f ν )δ γ + f ν δ νΠ = f ν Π ν . (21.12)wheref ν ≡ρ νρ γ + ρ ν= const. ∼ 0.405. (21.13)
21 EARLY RADIATION-DOMINATED ERA 55andThe relevant equations thus becomeH −1 δ γ ′ = − 4 ( kv γ + 4H3 H)−1 Ψ ′H −1 v γ ′ = 1 ( ( k kδ γ + Φ4 H)H)H −1 δ ν ′ = − 4 ( kv ν + 4H3 H)−1 Ψ ′H −1 v ν ′ = 1 ( kδ ν −4 H)1 ( ( k kΠ ν + Φ .6 H)H)H −1 Ψ ′ + Φ = − 1 2 δ (21.14)H −1 Ψ ′ + Φ = 2 H k v (21.15)H −2 Ψ ′′ + H −1 ( Φ ′ + 2Ψ ′) − Φ = 1 2( ) δ (21.16)k 2(Ψ − Φ) = f ν Π ν (21.17)H21.1 Neutrino Adiabaticity23 H−1 Ψ ′ + 2Ψ = − 4 3 R + 2 (Ψ − Φ), (21.18)3We consider now the simpler case, when there are no neutrino entropy perturbations,S ν = 0 ⇒ δ ν = δ γ = δ and v ν = v γ = v . (21.19)We allow for the presence of baryon and CDM entropy perturbations. However, during theradiation-dominated epoch their effect on metric perturbations is negligible. Thus the evolutionof metric perturbations are as if the perturbations were adiabatic.In the simpler case discussed in Sec. (18), where we had Φ = Ψ, we found that the growingadiabatic mode had Φ = Ψ = const. Guided by that, we now try the ansatzΦ = const. and Ψ = const. (21.20)and check that it is a solution. The Einstein and R equations are satisfied withδ ν = δ γ = δ = −2Φ = const. (21.21)v ν = v γ = v = 1 ( ( k kΦ = −2 H)1 4δ =H)1 2kηΦ (21.22)Π ν = 1 f ν( kH) 2(Ψ − Φ) (21.23)R = −(Ψ + 1 2Φ) = const. (21.24)In the δ ′ γ and δ ′ ν equations we can now ignore the (k/H) terms, and we see that they are alsosatisfied. Using H = 1/η the velocity equations becomeηv ′ γ = 1 4 kηδ γ + kηΦηv ′ ν = 1 4 kηδ ν − 1 6 kηΠ ν + kηΦ .
21 EARLY RADIATION-DOMINATED ERA 56Here the Π ν term can be ignored since Π ν is suppressed by (k/H) 2 compared to Ψ and Φ, andwe then see that these equations are also satisfied.We are still missing a piece of information that would tell us what Φ −Ψ, or, in other words,what Π ν is. The neutrino anisotropy Π ν depends on the neutrino momentum distribution.Before neutrino decoupling, interactions kept Π ν = 0. After neutrinos decoupled, neutrinoshave been “freely streaming”, i.e., moving without collisions through the perturbed universe. InCMB Physics we derive a so-called Boltzmann hierarchy of equations that relates the evolutionof the different moments of the neutrino momentum distribution to each other. Using theseone can show that, starting from Π ν = 0, one obtains for superhorizon scales a “decreasinghierarchy”, where the effect of the higher moments on lower moments can be ignored. Theevolution of the “second moment” Π ν depends then only on the “first moment” v ν , and therelevant equation isH −1 Π ′ ν = 8 ( kv ν . (21.25)5 H)This finally allows us to solve (Exercise):Ψ = (1 + 2 5 f ν)Φ ≈ 1.162Φ (21.26)Π ν = 2 5 (kη)2 Φ (21.27)R = − 3 (1 + 4 )2 15 f ν Φ = const. (21.28)Φ = − 2 13 1 + 415 f R ≈ −0.6017Rν(21.29)Ψ = − 2 1 + 2 53 1 + 415 f R ≈ −0.6992R.ν(21.30)(We see that the perfect fluid approximation, which gave Ψ = Φ = − 2 3R, led to a 10% errorin Φ and to a 5% error in Ψ.)21.2 MatterDuring the early radiation-dominated era, the metric and the radiation perturbations do notcare about matter perturbations, but matter perturbations will become important later, andtherefore we are interested in their “initial” behavior in the radiation-dominated era. Thecontinuity equations for baryons and CDM are( ) kH −1 δ c ′ + v c − 3H −1 Ψ ′ = 0 (21.31)H( ) kH −1 δ b ′ + v b − 3H −1 Ψ ′ = 0H( kH −1 v c ′ + v c − Φ = 0H)( kH −1 v b ′ + v 4ρ γb − Φ = an e σ T (v γ − v b ),H)3ρ bwhere the collision term in the last equation is derived in CMB Physics, σ T is the Thomsoncross section for photon-electron scattering, and n e is the free electron number density. Wellbefore photon decoupling, an e σ T (4ρ γ )/(3ρ b ) is very large, and the collision term forces v b = v γ(baryons are tightly coupled to photons).
21 EARLY RADIATION-DOMINATED ERA 57For the above photon+neutrino adiabatic growing mode solution, these becomeηδ ′ c + kηv c = 0 (21.32)ηδ ′ b + kηv b = 0ηv ′ c + v c − kηΦ = 0ηv ′ b + v b − kηΦ = an e σ T4ρ γ3ρ b(v − v b ).21.2.1 The Completely Adiabatic SolutionOne solution for Eq. (21.32) is the completely adiabatic solution:δ c = δ b = 3 4 δ = −3 2Φ = const. (21.33)v c = v b = v = 1 2 kηΦ .To check this, substitute Eq. (21.33) into Eq. (21.32). This givesfor the δ equations, and0 + 1 2 (kη)2 Φ = 012 kηΦ + 1 2kηΦ − kηΦ = 0,so the equations are indeed satisfied. The δ equations are satisfied to accuracy (kη) 2 ≪ 1, i.e.,in a Hubble time, δ i will deviate from its initial value − 3 2 Φ by about −1 2 (kη)2 Φ, a negligiblechange.21.2.2 Baryon and CDM Entropy <strong>Perturbation</strong>sThere are three independent entropy perturbations: the neutrino, baryon, and CDM entropyperturbations,S ν ≡ 3 4 (δ ν − δ γ ) S b ≡ δ b − 3 4 δ γ S c ≡ δ c − 3 4 δ γ . (21.34)Their evolution equations areS ′ ν = −k(v ν − v γ ) S ′ b = −k(v b − v γ ) S ′ c = −k(v c − v γ ). (21.35)The relative entropy perturbation stays constant unless there is a corresponding relative velocityperturbation.Entropy perturbations also tend to stay constant at superhorizon scales 27 , as( ) kH −1 S i ′ = − (v i − v γ ). (21.36)HAssume now the neutrino-adiabatic growing mode solution of Sec. 21.1. This assumes S ν = 0,but we may still have baryon and CDM entropy perturbations.The baryon and neutrino density perturbations areδ b = 3 4 δ + S b and δ c = 3 4 δ + S c . (21.37)27 This property is not as general as the constancy of R at superhorizon scales: The result (21.36) relies on twoassumptions: 1) no interaction terms in the component energy continuity equations, and 2) the component fluidshave a unique relation p i = p i(ρ i). Note also that this does not hold for the “total entropy perturbation” S.
22 SUPERHORIZON EVOLUTION AND RELATION TO ORIGINAL PERTURBATIONS58Since baryons are tightly coupled to photons, v b = v γ = v, we haveS b ′ = 0 ⇒ S b = const. ⇒ δ b = const. (21.38)For CDM we do not have this constraint. Writev rel ≡ v c − v ⇒ v c = 1 2 kηΦ + v rel . (21.39)Eq. (21.32c) becomesη 1 2 kΦ + ηv′ rel + 1 2 kηΦ + v rel − kηΦ = 0⇒ ηv rel ′ = −v rel ⇒ v rel ∝ η −1 .Thus we havev c = 1 2 (kη)Φ + Cη−1 . (21.40)from which we identify a growing mode and a decaying mode. As time goes on, the decayingmode decays away, and v c → v. Ignoring the decaying mode, we havev c = v ⇒ S c = const. ⇒ δ c = const. (21.41)Thus (assuming neutrino adiabaticity), the “initial conditions” at the early radiation-dominatedepoch can be specified by giving three constants for each Fourier mode ⃗ k: Φ k (rad), S c ⃗ k(rad),and S b ⃗ k(rad). The general perturbation is a superposition of three modes, where two of theseconstants are zero:(Φ,S c ,S b ) = (Φ,0,0) adiabatic mode (AD) (21.42)(Φ,S c ,S b ) = (0,S c ,0)CDM isocurvature mode (CDI)(Φ,S c ,S b ) = (0,0,S b ) baryon isocurvature mode (BI) .In the following we shall use R instead of Φ (see Eq. 21.28) as the first initial value constant(since it is better in staying constant also later).21.3 Neutrino perturbationsPerhaps I will do this someday ...22 Superhorizon Evolution and Relation to Original <strong>Perturbation</strong>sSee Section I3.2 of CMB Physics 2004.23 Gaussian Initial ConditionsSee Section I4 of CMB Physics 2004.24 Large ScalesSee Section I5 of CMB Physics 2004.
25 SACHS–WOLFE EFFECT 5925 Sachs–Wolfe EffectConsider photon travel in the perturbed universe. The geodesic equation isd 2 x µdu 2 + dx α dx βΓµ αβdu du= 0, (25.1)where u is an affine parameter of the geodesic. For photons, we choose u so that the photon4-momentum isp µ = dxµdu , (25.2)which allows us to write the geodesic equation asDividing by p 0 = dη/du, this becomesdp µdu + Γµ αβ pα p β = 0. (25.3)dp µdη + p α p βΓµ αβp 0 = 0. (25.4)In the following, we need only the time component of this equation,dp 0dη + Γ0 00 p0 + 2Γ 0 0k pk + Γ 0 p i p jijp 0 = 0. (25.5)Assuming scalar perturbations and using the Newtonian gauge (the Γ µ αβfrom Eq. (8.6)), thisbecomesdp 0dη + ( H + Φ ′) p 0 + 2Φ ,k p k + [ H − 2H(Φ + Ψ) − Ψ ′] δ ij p i p jp 0 = 0. (25.6)These 4-momentum components p µ are in the coordinate frame. What the observer interpretsas the photon energy and momentum are the components pˆµ in his local orthonormal frame.Since the metric is diagonal, the con<strong>version</strong> is easy, pˆµ = √ |g µµ |p µ (for a comoving observer):E ≡ pˆ0pî= a √ 1 + 2Φ p 0 = a(1 + Φ)p 0= a √ 1 − 2Ψ p i = a(1 − Ψ)p i . (25.7)Since photons are massless, E 2 = δ ij pîpĵ.In the background universe, the photon energy redshifts as Ē ∝ a−1 ⇔ aĒ presence of perturbations, q ≡ aE ≠ const. Thus we define q and ⃗q,= const. In theq ≡ aE = a 2 (1 + Φ)p 0 ⇒ p 0 = a −2 (1 − Φ)qq i ≡ apî = a 2 (1 − Ψ)p i ⇒ p i = a −2 (1 + Ψ)q i (25.8)where q 2 = δ ij q i q j , as suitable quantities to track the perturbation in the redshift.Rewriting Eq. (25.6) in terms of q and ⃗q (and dropping 2nd order terms) gives (exercise)(1 − Φ) dqdη = qdΦ dη − qΦ′ − 2q k Φ ,k + qΨ ′ . (25.9)Here the rhs is 1st order small, therefore dq/dη is also 1st order small, and we can drop thefactor (1 − Φ). Dividing by q we get1 dqq dη = dΦdη − Φ′ + Ψ ′ ⃗q · ∇Φ− 2 . (25.10)q
25 SACHS–WOLFE EFFECT 60andso thatHere the total derivative along the photon geodesic is⃗qqso that Eq. (25.10) becomesddη = ∂ ∂η + dxk ∂dη ∂x k (25.11)(1 − Ψ)pk=(1 + Φ)p 0 ≈ pkp 0 = dxkdηto 0th order (25.12)(⃗q · ∇Φ−2 = −2 dxk ∂Φ dΦq dη ∂x k = −2 dη − ∂Φ ), (25.13)∂η1 dqq dη = −dΦ dη + Φ′ + Ψ ′ (25.14)The relative perturbation in the photon energy, δE/Ē, that the photon accumulates whentraveling from x ∗ to x obs in the perturbed universe is thusδEĒ = δq¯q ∫ ∫ ∫ dq(Φ = q = − dΦ + ′ + Ψ ′) dη∫ ηobs( ∂Φ= Φ(x ∗ ) − Φ(x obs ) +∂η + ∂Ψ )dη , (25.15)∂ηwhere the integrals are along the photon path.For a thermal distribution of photons, a uniform relative photon energy perturbation correspondsto a temperature perturbation of the same amount:( ) δT= δĒ Tjour E . (25.16)Here “jour” refers to the temperature perturbation the photon distribution accumulates on thejourney between x ∗ and x obs .The other contributions to the observed CMB temperature anisotropy are due to the localphoton energy density perturbation and photon velocity perturbation at the origin of the photon,on the last scattering surface:( ) δT= 1T4 δ γ(x ∗ ) − ⃗v(x ∗ ) · ˆn , (25.17)intrwhere ˆn is the direction the observer is looking at.For a given observer, the Φ(x obs ) part is common to photons from all directions, and theobserver interprets it as part of the mean (background) photon temperature. Thus the observedCMB temperature anisotropy isδTT = 1 4 δN γ (x ∗) − ⃗v N (x ∗ ) · ˆn + Φ(x ∗ ) +η ∗∫ ηobsη ∗( ∂Φ∂η + ∂Ψ )dη . (25.18)∂η(There is also a contribution from the motion of the observer that causes a dipole pattern inthe observed anisotropy. To get rid of that, the dipole of the observed anisotropy is subtractedaway from the observations before any cosmological analysis.)For large scales, much larger than the horizon size at photon decoupling, the Doppler effect−⃗v N (x ∗ ) · ˆn is small compared to the other terms. The contribution( ) δT= 1T4 δN γ (x ∗ ) + Φ(x ∗ ) (25.19)SW
26 MATTER POWER SPECTRUM 61is called the ordinary Sachs–Wolfe effect, and the contribution( ) ∫ δT ηobs( ∂Φ=T∂η + ∂Ψ )dη (25.20)∂ηISWis called the integrated Sachs–Wolfe effect.For more, see Section I5.3 of CMB Physics 2004.26 Matter Power Spectrumη ∗See Chapter M of CMB Physics 2004, which is based on Chapter 7 of Dodelson[3]. Note that themultipole moments of the photon brightness function Θ l are related to the Θ m lof CMB Physics2007 and to perturbations of the photon energy tensor byΘ 0 ≡ Θ 0 0 ≡ 1 4 δ γ Θ 1 ≡ 1 3 Θ0 1 ≡ 1 3 v γ . (26.1)
A GENERAL PERTURBATION 62A General <strong>Perturbation</strong>From Eq. (3.8) we have that the general perturbed metric (around the flat Friedmann model) isds 2 = a 2 (η) { −(1 + 2A)dη 2 − 2B i dηdx i + [(1 − 2D)δ ij + 2E ij ] dx i dx j} .(A.1)The Christoffel symbols areΓ 0 00 = H + A ′ (A.2)Γ 0 0i = −HB i + A ,iΓ 0 ij= H [(1 − 2A − 2D)δ ij + 2E ij ] + 1 2 (B i,j + B j,i ) − δ ij D ′ + E ′ ijΓ i 00 = −HB i − B ′ i + A ,iΓ i 0j= Hδ ij + 1 2 (B j,i − B i,j ) − D ′ δ ij + E ′ ijΓ i jk = Hδ jk B i − δ i jD ,k − δ i k D ,j + δ jk D ,i + E ij,k + E ik,j − E jk,i .and we have the Christoffel sumsΓ µ 0µ = 4H + A ′ − 3D ′ (A.3)Γ µ iµ= A ,i − 3D ,i .Note that the Christoffel sums contain only scalar perturbations. Thus for vector and tensorperturbations, these sums contain only the background value Γ µ 0µ = 4H.The Einstein tensor isG 0 0 = −3a −2 H 2 + a −2 [ −2∇ 2 D + 6HD ′ + 6H 2 ]A − 2HB i,i − E ik,ik (A.4)G 0 i = a −2 [ −2D ,i ′ − 2HA ,i − 1 2 (B i,kk − B k,ik ) − E ik,k′ ]G i 0 = a −2 [ 2D ,i ′ + 2HA ,i + 1 2 (B i,kk − B k,ik ) + 2H ′ B i − 2H 2 B i + E ik,k′ ]G i j= a −2 ( −2H ′ − H 2) δ ij+a −2 [ 2D ′′ − ∇ 2 (D − A) + H(2A ′ + 4D ′ ) + (4H ′ + 2H 2 )A − B k,k ′ − 2HB ]k,k − E kl,kl δij+a −2 [ (D − A) ,ij + 1 2 (B′ i,j + B′ j,i ) + H(B i,j + B j,i ) + E ij ′′ − ∇2 E ij + E ik,jk + E jk,ik + 2HE ij′ ]
A GENERAL PERTURBATION 63For scalar perturbations, the perturbation in the Einstein tensor becomesδG 0 0 = a −2 [ −2∇ 2 ψ + 6HD ′ + 6H 2 A + 2H∇ 2 B ] (A.5)δG 0 i = a −2 [ −2ψ ′ ,i − 2HA ,i]δG i 0 = a −2 [ 2ψ ′ ,i + 2HA ,i − 2H ′ B ,i + 2H 2 B ,i]δG i j= a −2 [ 2D ′′ − ∇ 2 (D − A) + H(2A ′ + 4D ′ ) + (4H ′ + 2H 2 )A + ∇ 2 B ′ + 2H∇ 2 B ] δ i j+a −2 [ − 1 3 ∇2 (∇ 2 E) − 1 3 ∇2 E ′′ − 2 3 H∇2 E ′] δ i j+a −2 ( D − A − B ′ − 2HB + E ′′ + 1 3 ∇2 E + 2HE ′) ,ij .The trace of the space part isδG i i = a−2 [ 6D ′′ − 2∇ 2 ψ + 2∇ 2 A + 3H(2A ′ + 4D ′ ) + (12H ′ + 6H 2 )A + 2∇ 2 B ′ + 4H∇ 2 B ] .(A.6)In Fourier space these read asδG 0 0 = a −2 [ 2k 2 ψ + 6HD ′ + 6H 2 A − 2HkB ] (A.7)δG 0 i = a −2 [ −2ik i ψ ′ − 2iHk i A ]δG i 0 = a −2 [ 2ik i ψ ′ + 2iHk i A − 2iH ′ k i B + 2iH 2 k i B ]δG i j= a −2 [ 2D ′′ + k 2 (D − A) + H(2A ′ + 4D ′ ) + (4H ′ + 2H 2 )A − kB ′ − 2HkB ] δ i j+a −2 [ − 1 3 k2 E + 1 3 E′′ + 2 3 HE′] δji−k i k j a[D −2 − A − 1 k B′ − 2 H k B − 1 3 E + 1 ]k 2(E′′ + 2HE ′ )δG i i = a −2 [ 6D ′′ + 2k 2 ψ − 2k 2 A + H(6A ′ + 12D ′ ) + (12H ′ + 6H 2 )A − 2kB ′ − 4HkB ] .
REFERENCES 64The general perturbed energy tensor isThe energy continuity equationsT0 0 = −¯ρ − δρ (A.8)Ti 0 = (¯ρ + ¯p)(v i − B i )= −(¯ρ + ¯p)v iT i 0T i j = ¯pδ i j + δpδi j + ¯pΠ ijT µ ν;µ ≡ T µ ν,µ + Γµ αν T α ν − Γα νµ T µ α = 0(A.9)become the background equationand the fluid perturbation equations¯ρ ′ + 3H(¯ρ + ¯p) = 0(A.10)δρ ′ = −3H(δρ + δp) + (¯ρ + ¯p)(3D ′ − ∇ · ⃗v) (A.11)(¯ρ + ¯p)(v i − B i ) ′ = −(¯ρ + ¯p) ′ (v i − B i ) − 4H(¯ρ + ¯p)(v i − B i )−δp ,i − ¯pΠ ij,j − (¯ρ + ¯p)A ,iFor scalar perturbations, the fluid perturbation equations becomeδρ ′ = −3H(δρ + δp) + (¯ρ + ¯p)(3D ′ + ∇ 2 v) (A.12)(¯ρ + ¯p)(v − B) ′ = −(¯ρ + ¯p) ′ (v − B) − 4H(¯ρ + ¯p)(v − B)+δp + 2 3 ¯p∇2 Π + (¯ρ + ¯p)A.Using δ ≡ δρ/¯ρ and background relations, these can be writtenδ ′ = (1 + w) ( ∇ 2 v + 3D ′) (+ 3H wρ − δp¯ρ)(A.13)(v − B) ′ = −H(1 − 3c 2 s)(v − B) + δp¯ρ + ¯p + 2 w31 + w ∇2 Π + A.The factor H(1 − 3c 2 s ) can also be written as (1 − 3w)H + w′ /(1 + w).In Fourier space these areδρ ′ = −3H (δρ + δp) + (¯ρ + ¯p)(3D ′ − kv) (A.14)(¯ρ + ¯p)(v − B) ′ = −(¯ρ + ¯p) ′ (v − B) − 4H(¯ρ + ¯p)(v − B)Referencesδ ′+kδp − 2 3k¯pΠ + k(¯ρ + ¯p)A(= (1 + w)(−kv + 3D ′ ) + 3H wδ − δp¯ρ)(v − B) ′ = −H(1 − 3w)(v − B) − w′ kδp(v − B) +1 + w ¯ρ + ¯p − 2 3wkΠ + kA.1 + w[1] A.R. Liddle and D.H. Lyth: <strong>Cosmological</strong> Inflation and Large-Scale Structure (CambridgeUniversity Press 2000).[2] C.-P. Ma and E. Bertschinger, ApJ 455, 7 (1995).[3] S. Dodelson: Modern Cosmology (Academic Press 2003).