Summary of equations for CMB power spectrum calculations Hans Kristian Eriksen 9. april 2010 1 Definitions and background cosmology Four time variables: t = physical time, η = t 0 ca 1 (t)dt = conformal time, a = scale factor, x = lna Metric: ds 2 = c 2 dt 2 + a 2 (t)δ ij dx i dx j = a 2 (η)( dη 2 + δ ij dx i dx j ) Friedmann s equations: H 1 da a dt = H 0 (Ωm + Ω b )a 3 + Ω r a 4 + Ω Λ (1) H c da a dη = H 0 (Ωm + Ω b )a 1 + Ω r a 2 + Ω Λ a 2 (2) Conformal time as a function of scale factor (to be computed numerically), η(a) = a 2 The perturbation equations 2.1 Overview of involved quantities 0 c a H(a ) da, All quantities are functions of wavenumber k and time x. δ dark matter density perturbation δ b baryonic matter density perturbation v dark matter velocity v b baryonic matter velocity Θ l l th photon moment dη da = c ah (3) Θ P l l th polarized photon moment N l l th neutrino moment Φ gravitational curvature potential Ψ Newtonian gravitational potential 1
2.2 Boltzmann-Einstein equations without neutrinos or polarization All dynamic quantities are functions of x and k. Derivatives are with respect to x: Θ 0 = ck H Θ 1 Φ, Θ 1 = ck 3H Θ 0 2ck 3H Θ 2 + ck 3H Ψ + τ Θ l = lck (2l + 1)H Θ l 1 Θ l = ck H Θ l 1 (l + 1)ck (2l + 1)H Θ l+1 + τ (4) Θ 1 + 1 3 v b, (5) Θ l 1 10 Θ lδ l,2, 2 l < l max (6) (l + 1)c Hη(x) Θ l + τ Θ l, l = l max (7) δ = ck H v 3Φ (8) v = v ck H Ψ (9) δ b = ck H v b 3Φ (10) v b = v b ck H Ψ + τ R(3Θ 1 + v b ) (11) Φ = Ψ c2 k 2 3H 2 Φ + H2 0 2H 2 Ωm a 1 δ + Ω b a 1 δ b + 4Ω r a 2 Θ 0 Ψ = Φ 12H2 0 c 2 k 2 a 2 Ω rθ 2 (13) R = 4Ω r 3Ω b a In the tight coupling regime, the equation for v b must be approximated, and reads q = (1 2R)τ + (1 + R)τ (3Θ 1 + v b ) ck H Ψ + (1 H H ) ck H ( Θ 0 + 2Θ 2 ) ck H Θ 0 (1 + R)τ + H H 1 (15) v b = 1 v b ckh 1 + R Ψ + R(q + ckh ( Θ 0 + 2Θ 2 ) ckh Ψ) (16) The photon moment equations in the same regime read (12) (14) Θ 1 = 1 3 (q v b ) (17) Θ 2 = 4ck 9Hτ Θ 1 (18) Θ l = 2l + 1 Hτ Θ l 1 (19) 2
2.3 Boltzmann-Einstein equations with neutrinos and polarization Θ 0 = ck H Θ 1 Φ, Θ 1 = ck 3H Θ 0 2ck 3H Θ 2 + ck 3H Ψ + τ Θ l = lck (2l + 1)H Θ l 1 (l + 1)ck (2l + 1)H Θ l+1 + τ (20) Θ 1 + 1 3 v b, (21) Θ l 1 10 Πδ l,2, 2 l < l max (22) Θ l = ck H Θ l 1 l + 1 Hη(x) Θ l + τ Θ l, l = l max (23) Θ P0 = ck H Θ P1 + τ Θ P0 1 2 Π (24) Θ Pl = lck (l + 1)ck (2l + 1)H ΘP l 1 (2l + 1)H ΘP l+1 + τ Θ P l 1 10 Πδ l,2, 1 l < l max (25) Θ P,l = ck H ΘP l 1 l + 1 Hη(x) ΘP l + τ Θ P l, l = l max (26) N 0 = ck H N 1 Φ, N 1 = ck 3H N 0 2ck 3H N 2 + ck N l = lck (2l + 1)H N l 1 (27) 3H Ψ (28) (l + 1)ck (2l + 1)H N l+1, 2 l < l max,ν (29) N l = ck H N l 1 l + 1 Hη(x) N l, l = l max,ν (30) δ = ck H v 3Φ (31) v = v ck H Ψ (32) δ b = ck H v b 3Φ (33) v b = v b ck H Ψ + τ R(3Θ 1 + v b ) (34) Φ = Ψ c2 k 2 3H 2 Φ + H2 0 2H 2 Ωm a 1 δ + Ω b a 1 δ b + 4Ω r a 2 Θ 0 + 4Ω ν a 2 N 0 Ψ = Φ 12H2 0 c 2 k 2 a 2 Ω rθ 2 + Ω ν N 2 (36) R = 4Ω r 3Ω b a Π = Θ 2 + Θ P 0 + ΘP 2 (35) (37) (38) 3
The tight coupling equations read q = (1 2R)τ + (1 + R)τ (3Θ 1 + v b ) ck H Ψ + (1 H H ) ck H ( Θ 0 + 2Θ 2 ) ck H Θ 0 (1 + R)τ + H H 1 (39) v b = 1 v b ckh 1 + R 0 + 2Θ 2 ) ckh Ψ) (40) Θ 2 = 8ck 15Hτ Θ 1 (41) Θ l = 2l + 1 Hτ Θ l 1 (42) Θ P 2 = 1 4 Θ 2 Θ P l = 2l + 1 2.4 Initial conditions Hτ ΘP l 1 The integration of the above equations starts at x i = lna i, with a i = 10 8. Initial conditions are the following, (43) (44) Φ = 1 (45) δ = δ b = 3 2 Φ (46) v = v b = ck 2H Φ (47) Θ 0 = 1 2 Φ (48) Θ 1 = ck 6H Φ (49) { 8ck Θ 2 = 15Hτ Θ 1, (with polarization) 4ck (50) 9Hτ Θ 1, (without polarization) Θ l = 2l + 1 Hτ Θ l 1 (51) For polarization and neutrinos, the corresponding initial conditions are Θ P 0 = 5 4 Θ 2 (52) Θ P 1 = ck 4Hτ Θ 2 (53) Θ P 2 = 1 4 Θ 2 (54) Θ P l = 2l + 1 Hτ ΘP l 1 (55) N 0 = 1 2 Φ (56) N 1 = ck 6H Φ (57) N 2 = c2 k 2 a 2 Φ 1 12H0 2Ω 5 ν 2f ν + 1 (58) ck N l = (2l + 1)H N l 1, l 3 (59) 4
3 Recombination and the visibility function (Note: Units of c, k B and are missing in this section it is your job to get these right!) Define the optical depth, The visibility function is η0 0 τ(η) = η0 η τ = n eσ T a H n e σ T adη (60) (61) g(η) = τe τ(η) = Hτ e τ(x) = g(x) (62) g(x) = τ e τ = g(x) H, (63) g(η)dη = (Need the free electron fraction, X e ne n H = ne n b, as a function of time.) 0 At early times, when X e > 0.99, use the Saha equation, X 2 e 1 X e = 1 H g(x)dx = 1. (64) ( ) 3/2 me T b e ǫ0/t b, (65) 2π where n b = Ω bρ c m h a 3, ρ c = 3H2 0 8πG, T b = T r = T 0 /a = 2.725K/a, and ǫ 0 = 13.605698eV. During and after recombination, use the Peebles equation, where dx e dx = C r(t b ) β(t b )(1 X e ) n H α (2) (T b )Xe 2, (66) H C r (T b ) = Λ 2s 1s + Λ α Λ 2s 1s + Λ α + β (2) (T b ), (67) Λ 2s 1s = 8.227s 1 (68) Λ α = H (3ǫ 0) 3 (8π) 2 n 1s (69) n 1s = (1 X e )n H (70) β (2) (T b ) = β(t b )e 3ǫ0/4T b (71) ( ) 3/2 β(t b ) = α (2) me T b (T b ) e ǫ0/t b (72) 2π α (2) (T b ) = 64π α 2 ǫ0 27π m 2 φ 2 (T b ) (73) e T b φ 2 (T b ) = 0.448 ln(ǫ 0 /T b ) (74) 5
4 The CMB power spectrum 1. Define and set Θ P l = 0 if you are not interested in polarization. 2. Compute the source function, S(k, x) = g d H d dx dx (H gπ) Π = Θ 2 + Θ P 2 + Θ P 0 (75) Θ 0 + Ψ + 14 Π + e τ Ψ + Φ 1 d ck dx (H gv b) + 3 d 4c 2 k 2 H d dx dx (H gπ) = d(hh ) gπ + 3HH ( gπ + gπ ) + H 2 ( g Π + 2 g Π + gπ ), (77) dx Π = 2ck H 5H H Θ 1 + Θ 1 + 3 10 τ Π + τ Π (78) 3ck H 5H H (Θ 3 + Θ p 1 + ΘP 3 ) + (Θ 3 + Θ P1 + Θ P3) (79) 3. Compute the multipole expansion of the transfer function observed today Θ l (k, x = 0) = 0 (76) S(k, x)j l k(η 0 η(x))dx (80) 4. Compute the CMB power spectrum (for a power law inflationary spectrum with spectral index n), C l = 0 ( k H 0 ) n 1 Θ 2 l (k) dk k (81) 5. Normalize spectrum such that maximum value (in units of l(l + 1)/2π) equals the WMAP best-fit spectrum of 5570µK 2. 6