We wrote down the Boltzmann equation for photons last time; it is:

Transcription

1 1 Objetives In this leture we will take the photon multipole equations derived last time, and onvert them into Fourier-multipole spae. This will be onvenient for linear perturbation theory, sine eah Fourier mode evolves independently, and for the CMB, sine the superposition of CMB flutuations is simpler in multipole than in angle spae. 2 Fourier transform We wrote down the Boltzmann equation for photons last time; it is: Θ = ˆp i Θ A ˆpi xi x i ˆpiˆp j B i x j Ḋ ˆpiˆp j Ė ij an e σ T (Θ v b ˆp) π an eσ T [1 + (ˆp ˆp ) 2 ]Θ(ˆp i )d 2ˆp i. (1) We will now take the Fourier transform of all the perturbation variables. For example, Θ(k, ˆp, η) = Θ(x i, ˆp, η)e ikix i d 3 x i. (2) R 3 The advantage of this approah is that the partial derivative operator / x i simply beomes multipliation: ik i. We an therefore transform Eq. (1) into: Θ = ik iˆp i Θ ik iˆp i A ik j ˆp iˆp j B i Ḋ ˆpiˆp j Ė ij + τ(θ v b ˆp) 3 16π τ [1 + (ˆp ˆp ) 2 ]Θ(ˆp i )d 2ˆp i. (3) (Definition: τ = an e σ T is the optial depth per unit onformal time this is negative so that we will later be able to define τ as the optial depth from the observer to a partiular redshift.) In this equation, Θ, A, B i, D, E ij, and v b have been Fourier-transformed. General rule: perturbation variables are Fourier-transformed; bakgrounds (e.g. n e ) or independent variables (e.g. ˆp i ) are not. Note that there are no terms that ouple different Fourier modes, as appropriate for a homogeneous bakground. 3 Multipole deomposition A further simplifiation is possible if we deompose the perturbations in multipoles. We begin by swithing to a oordinate system in whik k points along the 3-axis. (Superposing different Fourier modes will be neessary to get e.g. the CMB power spetrum, and will involve some bookkeeping and rotation matries, but no new physis.) So we an simply write kê 3 in plae of k. In this oordinate system, one an write ˆp in spherial oordinates: ˆp = (sin θ osφ, sin θ sinφ, os θ), (4) 1

2 with µ = osθ = ˆp ˆk. We will then do a multipole deomposition: Θ(k, θ, φ, η) = lm ( i) l 4π(2l + 1)Θ lm (k, η)y lm (θ, φ), (5) whih has inverse: Θ lm (k, η) = i l 4π(2l + 1) Y lm(θ, φ)θ(k, θ, φ, η)sin θ dθ dφ. (6) In the speial ase of m = 0 we may write this in terms of the Legendre polynomials: Θ l0 (k, η) = il P l (µ)θ(k, θ, φ, η)dµ dφ. (7) 4π The advantage of the multipole deomposition is that it dramatially simplifies Eq. (3). Taking the time derivative of Eq. (6), we get: Θ lm (k, η) = i l 4π(2l + 1) Y lm(θ, φ) Θ(k, θ, φ, η)sin θ dθ dφ. (8) Therefore if we evaluate eah term in the Boltzmann equation, and substitute into Eq. (8), we an get the multipole-spae Boltzmann equation. This is our next goal. Free-streaming term. First let s onsider the free-streaming term, ik iˆp i Θ. We an simplify k iˆp i = kµ. So the ontribution to Θ lm from this term is, from Eq. (8): Θ lm fs = i l 4π(2l + 1) i l = ikµ 4π(2l + 1) = ikµ 2l + 1 i l l 2l + 1 Θ l m l m Ylm (θ, φ)( ikµ)θ(θ, φ)sin θ dθ dφ Ylm (θ, φ) ( i) l 4π(2l + 1)Θ l m Y l m (θ, φ)sin θ dθ dφ l m Ylm (θ, φ)µy l m (θ, φ)sin θ dθ dφ. (9) We ve simplified the problem down to an integral over spherial harmonis. This integral omes with a simple seletion rule, namely that by azimuthal symmetry in φ it is only nonzero if m = m. It is also straightforward to see that under the symmetry θ π θ (mu µ) we must have l l odd. Finally, µ is a spherial harmoni of order 1, so the triangle rule for addition of angular momenta implies l l = 1, 0, +1 (and we just learned that 0 is ruled out). So all of this boils down to the two ases of l = l 1, m = m and l = l + 1, m = m. The relevant integral is (reall dipole radiation from atomi physis!): Y lm(θ, φ)µy l 1,m (θ, φ)sin θ dθ dφ = (l + m)(l m) (2l + 1)(2l 1). (10) 2

3 By swapping l and l 1 we may obtain the other integral, Ylm(θ, (l m)(l + 1 m) φ)µy l+1,m (θ, φ)sin θ dθ dφ =. (11) (2l + 3)(2l + 1) Putting it all together: [ ] (l + m)(l m) (l m)(l + 1 m) Θ lm fs = k Θ l 1,m Θ l+1,m. 2l + 1 2l + 1 (12) Gravitational soures. We next onsider the terms in the Boltzmann equation from the metri perturbations A, B, D, and E. Take A as an example: Θ lm A = i l 4π(2l + 1) Ylm (θ, φ)( ikµa)sin θ dθ dφ. (13) But we an fator out ika and deompose µ as a spherial harmoni, 4π µ = osθ = 3 Y 10(θ, φ), (14) so that orthonormality implies: Θ lm A = 1 3 kaδ l1δ m0. (15) Thus the A metri perturbation only soures a photon dipole, and even then only the m = 0 omponent. The same exerise an be repeated for the other gravitational soure terms B, D, and E. The overall result is: Θ 00 grav = 1 3 ikb 3 Ḋ; Θ 10 grav = 1 3 ka; Θ 1,±1 grav = 0; Θ 20 grav = 2 15 ikb Ė33; Θ 2,±1 grav = ik(±b 1 ib 2 ) (±Ė13 iė23); 1 Θ 2,±2 grav = 5 6 (Ė11 Ė22 2iĖ12); Θ lm,l 3 grav = 0. (16) The fat that there are only loal soures for l 2 is a diret onsequne of gravity being a spin-2 field. 3

4 Sattering terms. The sattering term has three parts. The first (S1) is the τθ term, whih trivially tansforms into multipole spae as: Θ lm S1 = τθ lm. (17) The seond (S2) is the baryon veloity term τv b ˆp, whih an be transformed the same way we handled the gravitational soures: Θ 10 S2 = 1 3 i τv b3; Θ 1,±1 S2 = i τ( v b1 + iv b2 ). (18) (All others vanish sine veloity is a dipole.) The third term (S3) is: i Θ l lm S3 = Ylm(ˆp) 3 4π(2l + 1) 16π τ[1 + (ˆp ˆp ) 2 ]Θ(ˆp )d 2ˆp d 2ˆp. (19) Its evaluation involves expanding the dot produt, and separating out the ˆp integral. This will be a homework exerise, but the answer is: Θ 00 S3 = τθ 00 ; Θ 2m S3 = 1 10 τθ 2m, (20) with all others vanishing. Salars, vetors, and tensors. Note that the different values of m don t talk to eah other; this is a onsequene of rotational invariane around the k axis. So we an onsider eah ase independently. We ll fous on the m = 0 modes, i.e. perturbations that are invariant under rotations around the k axis. These are alled salar modes in osmology, whih distinguishes them from vetor modes (m = ±1) and tensor modes (m = ±2). The salars are the only primordial modes that have been deteted; we ll disuss tensors later but not vetors (in standard osmology they have no soure). The salar modes. If we put all of this together, we get for the m = 0: Θ 00 = kθ ikb 3 Ḋ; Θ 10 = k Θ 00 2Θ 20 3 Θ 20 = k 2Θ 10 3Θ 30 [ 5 Θ l0 = k + τθ i τv b ka; τθ l 2l + 1 Θ l 1,0 l + 1 2l + 1 Θ l+1,0 15 ikb ] 5Ė33; + τθ l0 (l 3). (21) The usual omputation ours in the Newtonian gauge (A = Ψ, D = Φ, B = E = 0) in whih ase, dropping the 0 subsripts: Θ 0 = kθ 1 Ψ; 4

5 Θ 1 = k Θ 0 2Θ 2 + τθ i τv b kψ; Θ 2 = k 2Θ 1 3Θ τθ 2; [ l Θ l = k 2l + 1 Θ l 1 l + 1 ] 2l + 1 Θ l+1 + τθ l (l 3). (22) The sequene of multipole moments is alled the Boltzmann hierarhy. The vetor modes. One an write a similar hierarhy for the vetor modes, m = ±1. In this ase there is no monopole (l = 0). One finds: Θ 1 = 1 3 kθ 2 + τθ 1 i τ 3 2 ( v b1 + iv b2 ); 3Θ1 2 2Θ 3 Θ 2 = k ±i(b 1 iė13) + (B 2 iė23) τθ 2; [ ] (l 1)(l + 1) l(l + 2) Θ l = k Θ l 1 2l + 1 2l + 1 Θ l+1 + τθ l (l 3). (23) Inflation does not generate vetor perturbations, and they are deaying anyway (as we ll see later it requires us to understand baryon veloity). So we won t do muh with them. In exoti senarios (osmi strings) one an have vetors. The tensor modes. Finally we onsider the tensor modes, m = ±2. These have no monopole or dipole. They annot be generated in linear perturbation theory by density flutuations, but one an make them during inflation. (Primordial gravitational waves!) The equations are: Θ 2 = 1 kθ τθ (Ė11 Ė22 2iĖ12); [ ] (l 2)(l + 2) (l 1)(l + 3) Θ l = k Θ l 1 Θ l+1 + τθ l (l 3).(24) 2l + 1 2l Neutrino equations Neutrinos are just like photons with two exeptions: (i) they are fermions, and (ii) they don t have Thomson sattering. (Atually they have mass but that doesn t onern us yet.) For a fermion that was initially in a Fermi-Dira distribution with zero hemial potential, the unperturbed phase spae density is: f (0) (x i, p, ˆp i ; η) = 1 e p/tν + 1. (25) In analogy to the photons, we an define a neutrino temperature perturbation N: { } 1 f(x i, p, ˆp i p ; η) = exp T ν (η)[1 + N(x i, p, ˆp i ; η)] + 1. (26) 5

6 If we had detetors that ould see milli-ev neutrinos then we ould make maps of N just as the CMB observers make maps of Θ. One an go through the mahinery of Fourier-transforming N and then doing a multipole deomposition. It works the same way as for photons, exept that it is simpler (no sattering term). We get, for the salars in Newtonian gauge: et. N 0 5 Dark matter = kn 1 Ψ; N 1 = k N 0 2N kψ; [ N l l = k 2l + 1 N l 1 l + 1 ] 2l + 1 N l+1 (l 2), (27) Next we ome to the dark matter. Usually in osmology we will assume that it is old dark matter (CDM), whih means that initially the dark matter partiles all move at the same veloity. Late in the history of the Universe when galaxies form, one may have CDM partiles whose orbits ross and hene at a given point there may be a veloity dispersion, but this is not part of linear perturbation theory. There was at one time a theory of hot dark matter (HDM) where initially the dark matter partiles were moving rapidly ( rapidly = an move a perturbation wavelength in less than the age of the universe). We ould do this but it s very ompliated. So here we go with CDM: the dark matter at eah point is desribed by a density ρ and a veloity v (subsript is for CDM). We an determine their equations of motion from the ontinuity equation, µ T µν = 0. (28) So all we need to do is write T µν in terms of ρ and v, and we re done. The stress-energy tensor in the normal observer s frame, is: T ˆ0ˆ0 = ρ ; T ˆ0î = ρ v i ; T îĵ = 0. (29) We an onvert this into a ontravariant tensor by: T µν = T ˆ0ˆ0 u µ u ν + T ˆ0î [u µ (eî) ν + u ν (eî) µ ] + T îĵ (eî) µ (eĵ) ν. (30) Reall that u µ = a 1 (1 A, B i ); then: T 00 = ρ (1 2A); a2 T 0i = ρ a 2 (B i + v i ); T ij = 0. (31) 6

7 Going bak to the ontinuity equation, we an write the ν = 0 omponent: T 00 + i T 0i + Γ µ µ0 T 00 + Γ µ µi T 0i + Γ 0 00 T Γ 0 0i T 0i + 2Γ 0 ij T ij = 0. (32) In first order perturbation theory, T ij = 0 so we drop the last term. Also T 0i, Γ 0 0i, and Γµ µi are all first order, so their produts are seond-order and an be dropped: T 00 + i T 0i + (Γ µ µ0 + Γ0 00)T 00 = 0. (33) We an now use the Christoffel symbols: Then we have: ρ a 2 Γ µ µ0 = 0 ln g = 4aH + Ȧ + 3Ḋ; Γ0 00 = ah + Ȧ. (34) { 2Ȧ + [ ρ 2aH](1 2A) + i (B i + v ρ ) i + (5aH + 2Ȧ + 3Ḋ)(1 2A) Simplifying: } = 0. (35) (1 2A) ρ + i (B i + v i ) + 3aH(1 2A) + 3Ḋ = 0. (36) ρ Divide by 1 2A and solve for ρ : ρ ρ = 3(aH + Ḋ) i(b i + v i ). (37) Note that in the unperturbed ase, this is 3aH, in aordane with the usual saling ρ a 3. It is onventional to define the frational density perturbation, Its derivative is, to first order, δ = η (ln δ ) δ (0) The seond term is 3aH, so we get: δ = ρ 1. (38) ρ (0) = η lnδ η lnδ (0). (39) δ = 3Ḋ i(b i + v i ). (40) Now we need to get an equation for the dark matter veloity. This omes from momentum onservation, i.e. the ν = i omponents of µ T µν = 0: T 0i + j T ij + Γ µ µ0 T 0i + Γ µ µj T ji + Γ i 00 T Γ i 0j T 0j + Γ i jk T jk = 0. (41) Again, the purely spaelike omponents of T vanish at first order, so we have: T 0i + Γ µ µ0 T 0i + Γ i 00 T Γ i 0j T 0j = 0. (42) 7

8 Now T 0j is a first-order quantity, so we only need Γ µ µ0 and Γi 0j to zeroeth order: The other Christoffel symbol we need to first order: Γ µ µ0 = 4aH; Γi 0j = ahδ i j. (43) Γ i 00 = A,i Ḃi ahb i. (44) With this we find: {( ρ a 2 2aH + ρ ) } (B i + v i ρ ) + Ḃi + v i + 4aH(B i + v i ) + A,i Ḃi ahb i + 2aH(B i + v i ) = 0. (45) Now ρ /ρ multiplies a perturbation so we an substitute the unperturbed value 3aH. This simplifies our equation to: or: ahv i + v i + A,i = 0, (46) v i = ahv i A,i. (47) Fourier deomposition. If we take the Fourier transform of the density and veloity equations, we get: δ = 3Ḋ ik(b 3 + v 3 ); v 3 = ahv 3 ika; v i = ahv i (i = 1, 2). (48) In the ase of the photons and neutrinos, we deomposed the perturbations into salars (invariant under rotations around the 3 axis) and vetors (whih pik up a fator of e ±i φ under suh rotations). In this ase δ and v 3 are salars. The vetors are: v (±1) = v1 + iv2. (49) 2 The numbers v (+1), v 3, and v( 1) desribe the 3 omponents of veloity. The equations of motion for the vetors are: v (±1) = ahv (±1), (50) whih have solution v (±1) a 1. This means that the vetor perturbations in the CDM are purely deaying and (aside from nonlinear effets) are not expeted today. In the Newtonian gauge, the salar equations are: δ = 3 Φ ikv ; v = ahv ikψ. (51) 8

9 6 Baryons Cold baryons. Finally we ome to the baryoni matter. We will treat the baryons an be treated as pressureless ( old ) just like the dark matter. The ondition under whih this is justified is that a pressure (sound) wave not be able to propagate a perturbation wavelength in a Hubble time, i.e. s ah k 1. (52) We must evaluate the left hand side in order to establish the range of sales over whih the baryons may be treated as old. At reombination (z = 1100), the temperature of the baryons is 3000 K, implying a sound speed of , and s ah 0.003h Mp 1. (53) The perturbations that we see in the mirowave bakground have wavenumbers k < 0.05h Mp 1. In the ase of galaxy lustering, we may use wavenumbers as high as 0.3 Mp 1. Thus we see that for the purposes of the CMB, or for setting up the initial onditions for galaxy lustering alulations, we may treat the baryons as old. (Of ourse baryon pressure has a large influene on the proess of galaxy formation itself.) The equations. For old baryons, the perturbation variables are δ b and v b, exatly analogous to the CDM variables. The differene is that baryons are ated on by an external fore: radiation pressure. The law of onservation of energy-momentum for baryons is then: µ T µν = F ν, (54) where F ν is the 4-momentum per unit 4-volume delivered to the baryons. This is a 4-vetor, and it an be deomposed into F ˆ0 (the power delivered by the radiation per unit volume) and F î (the 3-fore per unit volume). We investigated the power delivered by the radiation earlier when we onsidered the Compton effet; it was: F ˆ0 = 4π 15 n etγσ 4 T γ T m T = 3 m e 2 n T m Compton. (55) This power orresponds to the heating of baryons by the CMB and the onsequent hange in T ˆ0ˆ0. However as long as we are treating the baryons as old we an neglet this effet and take F ˆ0 0. (56) Of greater interest is the momentum exhange between the baryons and photons. We found in the letures on Compton sattering that the net fore experiened by an eletron was F e = σ T j γ, (57) 9

10 where j γ is the radiation momentum density in the baryon rest frame. This is related to the momentum density in the normal frame by a Lorentz transformation, (j γ) ī = T (e 0)ˆµ (γ)ˆµˆν (e )ˆν ī, (58) where ī denotes baryon frame omponents. The baryon frame basis vetors are to first order in the veloity, so (e 0)ˆµ = (1, v i b ) and (eī )ˆν = (v i b, δ ij) (59) (j γ) ī = T ˆ0î (γ) T ˆ0ˆ0 (γ) vi b T îĵ (γ) vj b. (60) Now sine v b is first order, we may replae T ˆ0ˆ0 îĵ (γ) and T(γ) with their unperturbed values ρ γ and ρ γ /3: (j γ) ī = T ˆ0î (γ) 4 3 ρ γvb. i (61) The omponent T ˆ0î (γ) exists at first order in the photon perturbations and annot be ignored. It is the momentum density in the normal frame so we an write it as an integral over phase spae: T ˆ0î (γ) = 2 p2 dp d 2ˆp (2π) 3 f(p, bfp)pˆp ˆ i. (62) Substitute in the equation for f in terms of Θ: T ˆ0î 1 (γ) = 4π 3 d 2ˆp ˆp i p 3 dp e p/[tγ0(1+θ)] 1 = = ρ γ π 1 4π 3 d 2ˆp ˆp i π4 15 T γ0[ Θ(ˆp i )] d 2ˆp ˆp i Θ(ˆp i ). (63) Now eah ˆp i is a linear ombination of spherial harmonis of order 1: 2π ˆp 1 = 3 [ Y 11(ˆp) + Y 1, 1 (ˆp)] 2π ˆp 2 = 3 i[y 11(ˆp) + Y 1, 1 (ˆp)] 4π ˆp 3 = 3 Y 10(ˆp). (64) This implies that: T ˆ0î (γ) = 4iρ γ ( Θ11 + Θ 1, 1, i Θ ) 11 + Θ 1, 1, Θ 10. (65)

11 The fore per unit volume F î is then n e times the fore per eletron Eq. (57), F î = 4 ( ) 3 n eσ T ρ γ vb 1 3iΘ 11 Θ 1, 1, vb 2 + 3Θ 11 + Θ 1, 1, vb 3 + 3iΘ (66) The inlusion of this fore in the energy-momentum onservation equation gives no hange in the baryon density equation: δ b = 3Ḋ i(b i + v i b). (67) However for the veloity equation, we now have: v b i = ahvi b A,i 4 ( 3 an eσ T ρ γ vb i 4an eσ T ρ γ i Θ 11 + Θ 1, 1, Θ ) 11 + Θ 1, 1, iθ If we do the Fourier-multipole deomposition, we get: (68) δ b = 3Ḋ ik(b 3 + v 3 b) (69) and v 3 b = ahv 3 b ika 4 3 an eσ T ρ γ (v i b + 3iΘ 10 ) (70) for the salars, and v (±1) b = ahv (±1) b 4 3 an eσ T ρ γ [v (±1) b + 3iΘ 1,±1 ] (71) for the vetors. We often define R 3ρ b /4ρ γ, whih simplifies these equations. Note that R = 0 at early times, but by reombination R 0.4. This gives us, for the salars, For the vetors, δ b = 3 Φ ikv b ; v b = ahv b ikψ + τ R (v b + 3iΘ 1 ). (72) v (±1) b = ahv (±1) b + τ R [v(±1) b + 3iΘ 1,±1 ]. (73) 11