Covariant Perturbation Theory
• Covariant= takes sameform in all coordinate systems
• Invariant= takes the samevaluein all coordinate systems
• Fundamental equations:Einstein equations, (covariant)conservationof stress-energy tensor:
Gµν = 8πGTµν
∇µTµν = 0
• how manydegrees of freedom: 10 for a symmetric 4× 4 tensor
• perturbations aresmall: look at CMB where fluctuations are∼ 10−5
Metric Tensor
• Expand metric tensor around thegeneral 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 − 2HijT)
• (1) A ≡ a scalarpotential; (3) Bi a vectorshift, (1) HL aperturbation in the spatialcurvature; (5) Hij
T a tracelessdistortionto spatial metric =(10)
Matter Tensor
• Similarly expand the matterstress energytensor aroundhomogeneous densityρ and pressurep:
T00 = −ρ − δρ,
T0i = (ρ + p)(vi − Bi),
T i0 = −(ρ + p)vi,
T ij = (p + δp)δi
j + pΠij,
• (1) δρ adensity perturbation; (3) vi a vector velocity;(1) δp apressure perturbation; (5) Πij ananisotropic stressperturbation
• So far this isfully generaland applies to any type of matter orcoordinate choice
Gauge
• Metric and matter fluctuations take ondifferent valuesindifferent coordinate systems
• Generalcoordinate transformation:
η = η + T
xi = xi + Li
free to choose(T, Li) to simplify equations or physics.
Newtonian Gauge
• A coordinate system isfully specifiedif 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 CMBandlensingworkBad: numericallyunstable
Newtonian Gauge
• Metric perturbed at first order
g00 = −1−2Ψ(x, t), g0i = 0, gij = a2δij(1+2Φ(x, t))
• Consider onlyCMB photons: find how photon distribution ismodified by the presence ofmetric perturbation
• Boltzmann formalismis 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
• ConsidercollisionlessBoltzmann equation withC[f ] = 0, i.e.Liouville equation
• Photon distribution functionf may change due to the presence ofmetric perturbation: even without interactions
• Evaluate[
∂
∂t+
dqdt
∂
∂q+
dxdt
∂
∂x
]
f
up to first order in perturbationsΨ andΦ
• f = f (x, t, Pµ): at zero-th order does not depends onx and ondirection of momentumq
LHS of Boltzmann Equation
• Remember that
Pµ =dxµ
dλ
and 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 perturbationΨ
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
dt, P0 ≡ dt
dλ⇒ dxi
dt=
dxi
dλ
dλ
dt=
Pi
P0
• find Pi in terms ofp andp. AnsatzPi = Cpi
p2 = gijPiPj = gijp
ipjC2 = 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) ⇒ dp
dt(1− Ψ) = p
∂Ψ
∂t
• 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 forα = 0 which gives−2∂βΨ
• Next
−gαβ,tPαPβ
p= −g00,t
P0P0
p− gij,t
PiPj
p
= 2p∂tΨ − gij,tPiPj
p
• At first ordergij = a2(t)(1 + 2Φ)δij
gij,t = δij
(
2aa + 4aaΦ + 2a2Φ)
= a2δij [2H(1 + 2Φ) + 2∂tΦ]
• But
δijPiPj
p= δijp
21− 2Φ
a2ppipj = p
1− 2Φ
a2
LHS of Boltzmann Equation
• We need to expand at 1st order photon distribution functionf
• Photons arebosonsi.e. Bose-Einstein distribution
f (x, t, p, p) =
[
exp
pT(t)[1 + Θ(x, p, t)]
− 1
]
−1
whereΘ = δT/T
• In homogeneous and isotropic UniverveT dependsonly on t
• At first order dependences onposition (non homogeneity)anddirection (anisotropy)
• SinceΘ is small (CMB anisotropies≃ 10−5) expandf 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 neglectΦ andΨ
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 wrtT
∂f (0)
∂t=
∂f (0)
∂TdTdt
= −dT/dtT
p∂f (0)
∂p
• Finally
−p∂f (0)
∂p
[
−dT/dtT
− da/dta
]
= 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-streamingi.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 elementM contains physics of interaction (Comptonscattering)
• (Lorentz invariant) phase space element
∫
d(phase space) = Πigi
(2π)3
∫
d3qi
2Ei
• Conservation:(2π)4δ(4)(q1 + q2 + . . . )
RHS of Boltzmann Equation
• FIRAS givesT0 = 2.725± 0.002 K for CMB with BB spectrumdown to0.02%
• At recombinationzrec ≈ 1100 we haveT ∼ 3000 K∼ few eV
Compton Scattering
• Photons and electrons interact via Compton scatteringγ′ + e−′ ↔ γ + e− and Coulomb interaction link photons tobaryons
C[f ] =1
2E(q)
∫
DqeDq′eDq′(2π)4δ(4)(q + qe − q′ − q′e)
[fe(q′
e)f (q′) − fe(qe)f (q)]|M|2
• Small energy transfer implies:non-relativisticComptonscattering;q ≃ q′ andqe ≫ q, q′
• Breakδ(4) into energy and momentum conservation
• Expand at first order around electron energy change
• Scattering Matrix|M|2 = 8πσT m2e: no angular dependence
|M|2
• |M|2 = 8πσT m2e is wrong for two reasons
• There is an angular dependence|M|2 ∝ (1 + cos[q · q′]): smalldifference (down to 1% level)
• |M|2 has also a polarisation dependence∝ |ǫ · ǫ′| whereǫ andǫ′
are polarization of incoming and outgoing photons
• When Compton is efficient this dependence is removed (youaverage on all possible directions uniformly)
• Nearrecombinationthis is no more true→ quadrupoleanisotropyis produced leaving a netlinear polarizationin theCMB
Collision Term: Take One
C[f ] =2π2σT
p
∫
d3qe
(2π)3 fe(qe)
∫
d3q′
(2π)3q′δ(q − q′)[f (q′) − f (q)]
+
∫
d3qe
(2π)3 fe(qe)
∫
d3q′
(2π)3q′(q − q′)qe
me
∂δ(q − q′)∂q′
[f (q′) − f (q)]
• integral offe(qe) givesne while integral offe(qe)qe givesnevb
• Take photons distribution up to first order
• Move to polar coordinatesd3q′ → dq′dΩ′
• Integration indq′ involves two important terms:Θ(q′) andq′ · 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 observerwithin his own horizon
• Finally
C[f ] =neσT
p
∫
dq′q′
δ(q − q′)
(
−q∂f (0)
∂q′Θ0 + q
∂f (0)
∂qΘ(q)
)
+ q · vb∂δ(q − q′)
∂q′
(
f (0)(q′) − f (0)(q))
Collision Term: Final Take
• Integrating by parts (using properly theδ in the second integral)we get
C[f ] = −q∂f (0)
∂qneσT [Θ0 − Θ(q) + q · vb]
which has no zero-th order terms as expected
• Collisions driveΘ to Θ0 and all other moments are suppressed
• The full BE reads
∂Θ
∂t+
qi
a∂Θ
∂xi +∂Φ
∂t+
qi
a∂Ψ
∂xi = neσT(Θ0 − Θ + q · vb)
• Solving will give anisotropies on the LSS, then we need toproject them today