Cosmology & CMB. Set2: Linear Perturbation Theory. Davide Maino

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1 Cosmology & CMB Set2: Linear Perturbation Theory Davide Maino

2 Covariant Perturbation Theory Covariant = takes same form in all coordinate systems Invariant = takes the same value in all coordinate systems Fundamental equations: Einstein equations, (covariant) conservation of stress-energy tensor: G µν = 8πGT µν µ T µν = 0 how many degrees of freedom: 10 for a symmetric 4 4 tensor perturbations are small: look at CMB where fluctuations are 10 5

3 Metric Tensor Expand metric tensor around the general FRW metric g 00 = a 2, g ij = a 2 γ ij. Add in general perturbation (Bardeen 1980) g 00 = a 2 (1 2A) g 0i = a 2 B i g ij = a 2 (γ ij 2H L γ ij 2H ij T ) (1) A a scalar potential; (3) B i a vector shift, (1) H L a perturbation in the spatial curvature; (5) H ij T a traceless distortion to spatial metric = (10)

4 Matter Tensor Similarly expand the matter stress energy tensor around homogeneous density ρ and pressure p: T 0 0 = ρ δρ, T 0 i = (ρ + p)(v i B i ), T i 0 = (ρ + p)v i, T i j = (p + δp)δ i j + pπi j, (1) δρ a density perturbation; (3) v i a vector velocity; (1) δp a pressure perturbation; (5) Π ij an anisotropic stress perturbation So far this is fully general and applies to any type of matter or coordinate choice

5 Gauge Metric and matter fluctuations take on different values in different coordinate systems General coordinate transformation: η = η + T x i = x i + L i free to choose (T, L i ) to simplify equations or physics.

6 Newtonian Gauge A coordinate system is fully specified if there is a prescription for (T, L i ) Newtonian: B = H T = 0 Ψ Φ Ã (Newtonian potential) H L (Newtonian curvature) L = H T /k T = B/k + Ḣ T /k 2 Good: intuitive Newtonian like gravity; commonly chosen for analytic CMB and lensing work Bad: numerically unstable

7 Newtonian Gauge Metric perturbed at first order g 00 = 1 2Ψ(x, t), g 0i = 0, g ij = a 2 δ ij (1+2Φ(x, t)) Consider only CMB photons: find how photon distribution is modified by the presence of metric perturbation Boltzmann formalism is the correct way to handle the problem

8 Boltzmann Equation Liouville theorem states that phase space distribution is conserved along a trajectory in the absence of interactions [ Df Dt = t + dq dt q + dx ] f = 0 dt x Boltzmann properly accounts for interactions Heuristically Df Dt = C[f] C[f] = particle sources sinks Collision term: integrate over phase space of incoming particles, connect to outgoing state with some interaction strength

9 LHS of Boltzmann Equation Consider collisionless Boltzmann equation with C[f] = 0, i.e. Liouville equation Photon distribution function f may change due to the presence of metric perturbation: even without interactions Evaluate [ t + dq dt q + dx ] f dt x up to first order in perturbations Ψ and Φ f = f(x, t, P µ ): at zero-th order does not depends on x and on direction of momentum ˆq

10 LHS of Boltzmann Equation Remember that P µ = dxµ dλ and photons are massless so that we can write one components in terms of the others 0 = P 2 g µν P µ P ν = g 00 P 0 P 0 + g ij P i P j = (1+2Ψ)(P 0 ) 2 + p 2 At first order in perturbation Ψ Covariant derivative P 0 = p 1 + 2Ψ = p(1 Ψ) Df Dt = f t + f dx i x i + f dp dt p dt + f dˆp i ˆp i dt

11 dx i dt Remember that P i dxi dt, P0 dt dλ dxi dt find P i in terms of p and ˆp. Ansatz P i = Cˆp i = dxi dλ = Pi dλ dt P 0 p 2 = g ij P i P j = g ijˆp iˆp j C 2 = a 2 (1+2Φ)δ ijˆp iˆp j C 2 = a 2 (1+2Φ)C 2 Finally dx i dt P i = = pˆpi (1 Φ) a pˆp i a 1 + 2Φ = 1 Φ pˆpi a 1 p(1 Ψ) = ˆpi (1 Φ + Ψ) a

12 dp dt Use the geodesic equation dp 0 dλ = Γ0 αβ Pα P β dp0 dt dt dλ dp0 dt d dt [p(1 Ψ)] = Γ0 αβ Now the last term in the RHS P α P β dp (1 + Ψ) p dt = Γ 0 P α P β αβ P 0 (1 Ψ) = p Ψ t Γ 0 P α P β αβ p = g0ν 2 (g να,β + g βν,α g αβ,t ) Pα P β p = g00 2 (2g 0α,β g αβ,t ) Pα P β p 1 + 2Ψ = (2g 0α,β g αβ,t ) Pα P β 2 p

13 dp dt Given the metric: g 0α 0 only for α = 0 which gives 2 β Ψ Next P α P β g αβ,t p P 0 P 0 P i P j = g 00,t g ij,t p p P i P j = 2p t Ψ g ij,t p At first order g ij = a 2 (t)(1 + 2Φ)δ ij ) g ij,t = δ ij (2aȧ + 4aȧΦ + 2a 2 Φ = a 2 δ ij [2H(1 + 2Φ) + 2 t Φ] But δ ij P i P j p = δ ijp 2 1 2Φ a 2 p ˆpiˆp j = p 1 2Φ a 2

14 LHS of Boltzmann Equation We need to expand at 1 st order photon distribution function f Photons are bosons i.e. Bose-Einstein distribution [ { f(x, t, p,ˆp) = exp where Θ = δt/t p T(t)[1 + Θ(x,ˆp, t)] } ] 1 1 In homogeneous and isotropic Univerve T depends only on t At first order dependences on position (non homogeneity) and direction (anisotropy) Since Θ is small (CMB anisotropies 10 5 ) expand f in Θ ( ) 1 f(x, t, p,ˆp) = exp(p/t) 1 + [exp(p/t) 1] 1 TΘ T = f (0) p f (0) f (0) Θ T p T = p f (0) p

15 Zero-th order Equation At zero-th order neglect Φ and Ψ Df (0) Dt = f (0) t f (0) ph p = 0 The RHS is zero: collision-less BE, but collisions are first order (Compton scattering is very efficient to establish equilibrium) Rewrite time derivative at derivative wrt T f (0) t = f (0) T dt dt = dt/dt (0) p f T p Finally f (0) [ p dt/dt da/dt ] = 0 T a 1 p T a

16 First order Equation Eliminating the zero-th order equation from the LHS we get retaining only 1 st order terms Df (1) Dt [ f (0) Θ = p + ˆpi Θ p t a x i + Φ ] + ˆpi Ψ t a x i free-streaming i.e. anisotropies down to small scales gravity effect on photon distribution

17 RHS of Boltzmann Equation Form: C[f] = d(phase space)[energy momentum conservation] M 2 [emission absorption] Matrix element M contains physics of interaction (Compton scattering) (Lorentz invariant) phase space element g i d(phase space) = Π i (2π) 3 Conservation:(2π) 4 δ (4) (q 1 + q ) d 3 q i 2E i

18 RHS of Boltzmann Equation FIRAS gives T 0 = ± K for CMB with BB spectrum down to 0.02% At recombination z rec 1100 we have T 3000 K few ev

19 Compton Scattering Photons and electrons interact via Compton scattering γ + e γ + e and Coulomb interaction link photons to baryons 1 C[f] = Dq e Dq e 2E(q) Dq (2π) 4 δ (4) (q + q e q q e ) [f e (q e )f(q ) f e (q e )f(q)] M 2 Small energy transfer implies: non-relativistic Compton scattering; q q and q e q, q Break δ (4) into energy and momentum conservation Expand at first order around electron energy change Scattering Matrix M 2 = 8πσ T m 2 e: no angular dependence

20 M 2 M 2 = 8πσ T m 2 e is wrong for two reasons There is an angular dependence M 2 (1 + cos[q q ]): small difference (down to 1% level) M 2 has also a polarisation dependence ˆǫ ˆǫ where ˆǫ and ˆǫ are polarization of incoming and outgoing photons When Compton is efficient this dependence is removed (you average on all possible directions uniformly) Near recombination this is no more true quadrupole anisotropy is produced leaving a net linear polarization in the CMB

21 Collision Term: Take One C[f] = + { 2π2 σ T d 3 q e p (2π) 3 f e(q e ) d 3 q e (2π) 3 f e(q e ) d 3 q (2π) 3 q δ(q q )[f(q ) f(q)] d 3 q (2π) 3 q (q q )q e δ(q q } ) m e q [f(q ) f(q)] integral of f e (q e ) gives n e while integral of f e (q e )q e gives n e v b Take photons distribution up to first order Move to polar coordinates d 3 q dq dω Integration in dq involves two important terms: Θ(ˆq ) and q v b

22 Collision Term: Take Two Introduce Monopole perturbation Θ 0 (x, t) 1 4π dω Θ(x, t,ˆq) which is deviation of the mean temperature for each observer within his own horizon Finally C[f] = n eσ T p ( f (0) dq q {δ(q q ) q q Θ 0 + q δ(q q ) ( ) } + q v b q f (0) (q ) f (0) (q) ) f (0) q Θ(ˆq)

23 Collision Term: Final Take Integrating by parts (using properly the δ in the second integral) we get f (0) C[f] = q q n eσ T [Θ 0 Θ(ˆq) + ˆq v b ] which has no zero-th order terms as expected Collisions drive Θ to Θ 0 and all other moments are suppressed The full BE reads Θ t + ˆqi Θ a x i + Φ + ˆqi Ψ t a x i = n e σ T (Θ 0 Θ + ˆq v b ) Solving will give anisotropies on the LSS, then we need to project them today

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