Cosmology & CMB

Set2: Linear Perturbation Theory

Davide Maino

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:

Gµν = 8πGTµν∇µTµν = 0

• how many degrees of freedom: 10 for a symmetric 4× 4 tensor• perturbations are small: look at CMB where fluctuations are∼ 10−5

Metric Tensor

• Expand metric tensor around the general FRW metric

g00 = −a2, gij = a2γij .

• Add in general perturbation (Bardeen 1980)

g00 = −a−2(1− 2A)

g0i = −a−2Bi

gij = a−2(γij − 2HLγij − 2Hij

T)

• (1) A ≡ a scalar potential; (3) Bi a vector shift, (1) HL aperturbation in the spatial curvature; (5) Hij

T a traceless distortionto spatial metric = (10)

Matter Tensor

• Similarly expand the matter stress energy tensor aroundhomogeneous density ρ and pressure p.

• Starting from a fluid expression Tµν = (ρ+ p)uµuν + pδµν we get

T00 = −ρ− δρ,

T0i = (ρ+ p)(vi − Bi),

T i0 = −(ρ+ p)vi,

T ij = (p + δp)δi

j + pΠij,

• (1) δρ a density perturbation; (3) vi a vector velocity; (1) δp apressure perturbation; (5) Πij an anisotropic stress perturbation

• So far this is fully general and applies to any type of matter orcoordinate choice

Gauge

• Metric and matter fluctuations take on different values indifferent coordinate systems

• General coordinate transformation:

η = η + T

xi = xi + Li

free to choose (T,Li) to simplify equations or physics.

Newtonian Gauge

• A coordinate system is fully specified if there is a prescriptionfor (T,Li)

• Newtonian:

B = HT = 0

Ψ ≡ A (Newtonian potential)

Φ ≡ HL (Newtonian curvature)

L = −HT/k

T = −B/k + HT/k2

Good: intuitive Newtonian like gravity; commonly chosen foranalytic CMB and lensing workBad: numerically unstable

Newtonian Gauge

• Metric perturbed at first order

g00 = −1−2Ψ(x, t), g0i = 0, gij = a2δij(1+2Φ(x, t))

• Consider only CMB photons: find how photon distribution ismodified by the presence of metric perturbation

• Boltzmann formalism is the correct way to handle the problem

Boltzmann Equation

• Liouville theorem states that phase space distribution isconserved along a trajectory in the absence of interactions

DfDt

=

[∂

∂t+

dqdt

∂

∂q+

dxdt

∂

∂x

]f = 0

• Boltzmann properly accounts for interactions

DfDt

= C[f ]

• Heuristically

C[f ] = particle sources − sinks

• Collision term: integrate over phase space of incoming particles,connect to outgoing state with some interaction strength

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 ofmetric perturbation: even without interactions

• Evaluate [∂

∂t+

dqdt

∂

∂q+

dxdt

∂

∂x

]f

up to first order in perturbations Ψ and Φ

• f = f (x, t,Pµ): at zero-th order does not depends on x and ondirection of momentum q

LHS of Boltzmann Equation

• Remember thatPµ =

dxµ

dλand photons are massless so that we can write one components interms of the others

0 = P2 ≡ gµνPµPν = g00P0P0 +gijPiPj = −(1+2Ψ)(P0)2 +p2

• At first order in perturbation Ψ

P0 =p√

1 + 2Ψ= p(1−Ψ)

• Covariant derivative

DfDt

=∂f∂t

+∂f∂xi

dxi

dt+∂f∂p

dpdt

+∂f∂pi

dpi

dt

dxi

dt

• Remember that

Pi ≡ dxi

dλ, P0 ≡ dt

dλ⇒ dxi

dt=

dxi

dλdλdt

=Pi

P0

• find Pi in terms of p and p. Ansatz Pi = Cpi

p2 = gijPiPj = gijpipjC2 = a2(1 + 2Φ)δijpipjC2 = a2(1 + 2Φ)C2

• Finally

Pi =ppi

a√

1 + 2Φ= ppi 1− Φ

a

dxi

dt=

ppi(1− Φ)

a1

p(1−Ψ)=

pi

a(1−Φ + Ψ)

dpdt

• Use the geodesic equation

dP0

dλ= −Γ0

αβPαPβ ⇒ dP0

dtdtdλ⇒ dP0

dt= −Γ0

αβ

PαPβ

P0

ddt

[p(1−Ψ)] = −Γ0αβ

PαPβ

p(1 + Ψ)

• Let us make explicit the time derivative

dpdt

(1−Ψ) = pdΨ

dt− Γ0

αβ

PαPβ

p(1 + Ψ)

• Multiply by (1 + Ψ), drop quadratic terms

dpdt

= p∂Ψ

∂t+

pi

a∂Ψ

∂xi

− Γ0

αβ

PαPβ

p(1 + 2Ψ)

dpdt

• Now the last term in the RHS

Γ0αβ

PαPβ

p=

g0ν

2(gνα,β + gβν,α − gαβ,t)

PαPβ

p

=g00

2(2g0α,β − gαβ,t)

PαPβ

p

=−1 + 2Ψ

2(2g0α,β − gαβ,t)

PαPβ

p

dpdt

• Given the metric: g0α 6= 0 only for α = 0 which gives −2∂βΨ

• Next

−gαβ,tPαPβ

p= −g00,t

P0P0

p− gij,t

PiPj

p

= 2p∂tΨ− gij,tPiPj

p

• At first order gij = a2(t)(1 + 2Φ)δij

gij,t = δij

(2aa + 4aaΦ + 2a2Φ

)= a2δij [2H(1 + 2Φ) + 2∂tΦ]

• But

δijPiPj

p= δijp2 1− 2Φ

a2ppipj = p

1− 2Φ

a2

LHS of Boltzmann Equation

• Putting together all those pieces we get

Γ0αβ

PαPβ

p= (1− 2Ψ)

[−∂Ψ

∂tp− 2

∂Ψ

∂xippi

a− p

∂Φ

∂t+ H

]• and collecting all terms we arrive

1p

dpdt

= −H − ∂Φ

∂t− pi

a∂Ψ

∂xi

• and the BE is

dfdt

=∂f∂t

+pi

a∂f∂xi − p

∂f∂p

[H +

∂Φ

∂t+

pi

a∂Ψ

∂xi

]

LHS of Boltzmann Equation

• We need to expand at 1st order photon distribution function f• Photons are bosons i.e. Bose-Einstein distribution

f (x, t, p, p) =

[exp

p

T(t)[1 + Θ(x, p, t)]

− 1]−1

where Θ = δT/T• 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

f (x, t, p, p) =1

exp(p/T)− 1+

(∂

∂T[exp(p/T)− 1]−1

)TΘ

= f (0) − p∂f (0)

∂pΘ T

∂f (0)

∂T= −p

∂f (0)

∂p

Zero-th order Equation

• At zero-th order neglect Φ and Ψ

Df (0)

Dt=∂f (0)

∂t− pH

∂f (0)

∂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)

∂TdTdt

= −dT/dtT

p∂f (0)

∂p

• Finally

−p∂f (0)

∂p

[−dT/dt

T− da/dt

a

]= 0⇒ T ∝ a−1

First order Equation

• Eliminating the zero-th order equation from the LHS we getretaining only 1st order terms

Df (1)

Dt= −p

∂f (0)

∂p

[∂Θ

∂t+

pi

a∂Θ

∂xi +∂Φ

∂t+

pi

a∂Ψ

∂xi

]• free-streaming i.e. anisotropies down to small scales• gravity effect on photon distribution

RHS of Boltzmann Equation

• Form:

C[f ] =

∫d(phase space)[energy − momentum conservation]

× |M|2[emission − absorption]

• Matrix element M contains physics of interaction (Comptonscattering)

• (Lorentz invariant) phase space element∫d(phase space) = Πi

gi

(2π)3

∫d3qi

2Ei

• Conservation:(2π)4δ(4)(q1 + q2 + . . . )

RHS of Boltzmann Equation

• FIRAS gives T0 = 2.725± 0.002 K for CMB with BB spectrumdown to 0.02%

• At recombination zrec ≈ 1100 we have T ∼ 3000 K ∼ 0.3 eV

Compton Scattering

• Photons and electrons interact via Compton scatteringγ′ + e−′ ↔ γ + e− and Coulomb interaction link photons tobaryons

C[f ] =1

2E(p)

∫DqDq′Dp′(2π)4δ(4)(p + q− p′ − q′)

[fe(q′)f (p′)− fe(q)f (p)]|M|2

• Small energy transfer implies: non-relativistic Comptonscattering; p ' p′ and q p, p′

• Break δ(4) into energy and momentum conservation• Expand at first order around electron energy change• Scattering Matrix |M|2 = 8πσTm2

e : no angular dependence

|M|2

• |M|2 = 8πσTm2e is wrong for two reasons

• There is an angular dependence |M|2 ∝ (1 + cos[p · p′]): smalldifference (down to 1% level)

• |M|2 has also a polarisation dependence ∝ |ε · ε′| where ε and ε′

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 theCMB

Collision Term

C[f ] =1p

∫d3q

(2π)32Ee(q)

∫d3q′

(2π)32Ee(q′)

∫d3p′

(2π)32E(p′)|M|2(2π)4

× δ3[p + q− p′ − q′] δ[E(p) + Ee(q)− E(p′)− Ee(q′)]

×

fe(q′)f (p′)− fe(q)f (p)

• E(p) = p and Ee(q) = me + q2/(2me) ' me

• We use δ3 to solve integral on q′ (meaning that q′ = q + p− p′)

C[f ] =π

4m2ep

∫d3q

(2π)3

∫d3p′

(2π)3p′δ

[p +

q2

2me− p′ − (q + p− p′)2

2me

]× |M|2

fe(q + p− p′)f (p′)− fe(q)f (p)

Collision Term: Take One

C[f ] =2π2σT

p

∫d3qe

(2π)3 fe(qe)

∫d3p′

(2π)3p′δ(p− p′)[f (p′)− f (p)]

+

∫d3qe

(2π)3 fe(qe)

∫d3p′

(2π)3p′(p− p′)qe

me

∂δ(p− p′)∂p′

[f (p′)− f (p)]

• integral of fe(qe) gives ne while integral of fe(qe)qe gives nevb

• Take photons distribution up to first order• Move to polar coordinates d3p′ → dp′dΩ′

• Integration in dp′ involves two important terms: Θ(p′) and p′ · vb

Collision Term: Take Two

• Introduce Monopole perturbation

Θ0(x, t) ≡ 14π

∫dΩ′Θ(x, t, q)

which is deviation of the mean temperature for each observer withinhis own horizon

Collision Term: Take Two

• Monopole Θ0 simplifies the matter (also in view of the furtherexpansion in multipoles of Θ)

C[f ] =neσT

p

∫dp′p′

δ(p− p′)

(−p′

∂f (0)

∂p′Θ0 + p

∂f (0)

∂pΘ(p)

)

+ p · vb∂δ(p− p′)

∂p′

(f (0)(p′)− f (0)(p)

)

Collision Term: Final Take

• Integrating by parts (using properly the δ in the second integral)we get

C[f ] = −p∂f (0)

∂pneσT [Θ0 −Θ(p) + p · vb]

which has no zero-th order terms as expected• Collisions drive Θ to Θ0 and all other moments are suppressed• The full BE reads

∂Θ

∂t+

pi

a∂Θ

∂xi +∂Φ

∂t+

pi

a∂Ψ

∂xi = neσT(Θ0 −Θ + p · vb)

• Solving will give anisotropies on the LSS, then we need toproject them today

Boltzmann Equation

• We rewrite BE using conformal time

Θ + pi∂iΘ + Φ + pi∂iΨ = aneσT(Θ0 −Θ + p · vb)

• Define µ = k · p/k where the wave-vector moves along thechange of temperature: µ = 1 along k while µ = 0 alongconstant temperature

• Optical depth for Compton Scattering

τ =

∫ η0

ηdη′neaσT τ = −neaσT

• Moving to Fourier space the BE reads

Θ + ikµΘ + Φ + ikµΨ = −τ(Θ0 −Θ + µvb)

Harmonic Expansion of BE

• Usually work out with Θ(k) = Θ(k, η, µ) with a furtherharmonic expansion

Θ`(k, η) ≡ (−i)−`∫ 1

−1

dµ2

P`(µ)Θ(k, η, ν)

with P`(µ) the Legendre polynomials• Dipole moment

Θ1 ≡ i∫ 1

−1

dµ2µΘ(µ)

• It is possible to demonstrate that Θ` ∝ Θ`−1/τ and since τ 1we can neglect high-order expansion: only monopole and dipolemodes are activated

Full set of BE for different species

γΘ + ikµΘ + Φ + ikµΨ = −τ(Θ0 −Θ + µvb)

DMδ + ikv = −3Φ

v +aa

v = −ikΨ

Baryonsδb + ikvb = −3Φ

vb +aa

vb = −ikΨ +τ

R[vb + 3iΘ1]

νN + ikµN = −Φ− ikµΨ

where 1R ≡

4ργ3ρb