Cmb part1

Cosmic Microwave Background Radiation
Lecture 1 : Physics of CMB
Jayanti Prasad
Inter-University Centre for Astronomy & Astrophysics (IUCAA)
Pune, India (411007)
Autumn School on Cosmology (5 - 15th Nov 2013)
BITS PILANI
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Plan of the Talk
Standard Model of Cosmology
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Plan of the Talk
Cosmic Microwave Background
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Plan of the Talk
Theoretical Framework
Statistical Mechanics of photons
Boltzmann Equation
Recombination
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Plan of the Talk
Boltzmann Equation
Recombination
Perturbations
Metric Perturbations
Boltzmann equation for photons
Line of sight integration
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Hot Big Bang Cosmology : Standard Model of Cosmology
Large scale uniformity - Homogeneity and Isotropy - Hubble
expansion
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expansion
Early Universe dense, hot and small - the Big Bang
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expansion
Gravitation only the relevant interaction at large scale -
General Relativity
Gµν =
8πG
c2
Tµ (1)
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expansion
General Relativity
Gµν =
8πG
c2
Tµ (1)
Homogeneous and Isotropic space time - FRW metric:
ds2
= c2
dt2
− a2
(t)
dr2
1 − kr2
+ dθ2
+ sin2
θdφ2
, (2)
only two parameters - scale factor a(t) and spatial curvature
k.
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expansion
General Relativity
Gµν =
8πG
c2
Tµ (1)
ds2
= c2
dt2
− a2
(t)
dr2
1 − kr2
+ dθ2
+ sin2
θdφ2
, (2)
k.
Most of the energy of the Universe is in dark energy (70%)
and darm matter (25%), very less in baryons or atoms (5%).
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expansion
General Relativity
Gµν =
8πG
c2
Tµ (1)
ds2
= c2
dt2
− a2
(t)
dr2
1 − kr2
+ dθ2
+ sin2
θdφ2
, (2)
k.
Most of the energy of the Universe is in dark energy (70%)
and darm matter (25%), very less in baryons or atoms (5%).
Inﬂation
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Friedman Equations
For FRW metric, Einstein equation (1) can be written in
terms of a pair of equations called Friedman equations
˙a2
a2
+
k
a2
=
8πGρ
3
(3)
and
¨a
a
= −
4πG
3
(ρ + 3P) (4)
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Friedman Equations
˙a2
a2
+
k
a2
=
8πGρ
3
(3)
and
¨a
a
= −
4πG
3
(ρ + 3P) (4)
The rate of the expansion of the Universe is given by the
Hubble parameter:
H(t) =
1
a(t)
da(t)
dt
(5)
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Friedman Equations
˙a2
a2
+
k
a2
=
8πGρ
3
(3)
and
¨a
a
= −
4πG
3
(ρ + 3P) (4)
The rate of the expansion of the Universe is given by the
Hubble parameter:
H(t) =
1
a(t)
da(t)
dt
(5)
Energy density of any species is given by the density parameter
Ωρ/ρc where ρc is called the critical density and is deﬁned as:
ρc(t) =
3H2(t)
8πG
(6)
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Distances
Physical distance between objects in an expanding universe
increases in proportion of a(t):
λ(t) =
a(t)
a(t0)
λ(t0) (7)
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Distances
λ(t) =
a(t)
a(t0)
λ(t0) (7)
In comoving coordinate system (which expands with the
universe) distances between objects do not change with time
due to expansion.
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Distances
λ(t) =
a(t)
a(t0)
λ(t0) (7)
due to expansion.
The distance at which two objects in the Universe move away
with each other with the speed of light is called the Hubble
distance dH:
dH =
c
H
(8)
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Distances
λ(t) =
a(t)
a(t0)
λ(t0) (7)
due to expansion.
The distance at which two objects in the Universe move away
with each other with the speed of light is called the Hubble
distance dH:
dH =
c
H
(8)
Comoving size of the Universe is given by η:
η =
cdt
a(t)
=
a
0
da
a
cda
a H(a )
(9)
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Numbers
Hubble parameter h is measured in 100 Km/sec/ Mpc1
H0 =
h
0.98 × 1010year
where 0.5 < h < 1.0 (10)
1
1 Mpc = 3.0856 × 1024
cm
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Numbers
H0 =
h
0.98 × 1010year
where 0.5 < h < 1.0 (10)
Hubble distance :
dH =
c
H0
≈ 9449 Mpc/h (11)
1
1 Mpc = 3.0856 × 1024
cm
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Numbers
H0 =
h
0.98 × 1010year
where 0.5 < h < 1.0 (10)
Hubble distance :
dH =
c
H0
≈ 9449 Mpc/h (11)
Critical density:
ρc =
3H2
0
8πG
= 1.88h2
× 10−29
gm cm−3
= 2.775h−1
× 1011
M /(h−1
Mpc)3
(12)
1
1 Mpc = 3.0856 × 1024
cm
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Numbers
H0 =
h
0.98 × 1010year
where 0.5 < h < 1.0 (10)
Hubble distance :
dH =
c
H0
≈ 9449 Mpc/h (11)
Critical density:
ρc =
3H2
0
8πG
= 1.88h2
× 10−29
gm cm−3
= 2.775h−1
× 1011
M /(h−1
Mpc)3
(12)
Temperature:
TCMB = 2.725K ≈ 2.35 × 10−4
eV (13)
1
1 Mpc = 3.0856 × 1024
cm
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Problem 1
Given that the equation of state parameter for a species is
w = P/ρ show that its energy density will change as
ρ(a) ∝ a−3(1+w), as the universe expands adiabatically.
Problem 2
Show that the Hubble parameter H(a) depends on the energy
densities of various species in the following way:
H2
= H2
0 Ωm
a0
a
3
+ Ωr
a0
a
4
+ ΩΛ + Ωk
a0
a
2
(14)
with Ωk = 1 − ΩTotal.
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Observational Support of the Big Bang Model
Hubble expansion
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Hubble expansion
Big Bang Nucleosynethis
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Hubble expansion
Big Bang Nucleosynethis
Cosmic Microwave Background Radiation
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The cosmic microwave background (CMB) was discovered by
Wilson & Penzias [Penzias & Wilson (1965)] in 1965 and for
this discovery they were awarded 1978 Nobel Prize in Physics.
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CMB was theoretically predicted in the context of synthesis
(nuclear) of elements by Alpher and Herman [Alpher &
Herman (1948)] and Gamow [Gamow (1948)] in late 1940s
and again later rediscovered by Zelodovich, Dicke, Peebles
[Dicke et al. (1965)].
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In early 1990s the COBE mission of NASA discovered [Smoot
et al. (1992)] that the temperature of CMB is not the same
along diﬀerent direction, or there are anisotropies and for this
John C. Mather and George F. Smoot were awarded 2006
Noble prize in physics.
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In early 1990s the COBE mission of NASA discovered [Smoot
et al. (1992)] that the temperature of CMB is not the same
along diﬀerent direction, or there are anisotropies and for this
John C. Mather and George F. Smoot were awarded 2006
Noble prize in physics.
WMAP and Planck have further measured CMB anisotropies
with great precision.
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What we know about CMB ?
CMB is a perfect blackbody radiation with temperature 2.725
degree Kelvin so its speciﬁc intenisty is given by
Iν =
2h3
c2
1
ehν/kB T − 1
(15)
Largest anisotropy 0−3 in the CMB sky is due to the motion
of the solar system with respect to the rest frame of CMB
(dipole) :
∆T
T
=
v
c
cos θ (16)
for v=370 km/sec we get ∆T = 3.358 × 10−3 Kelvin.
Ignoring the dipole anisotropy, CMB anisotropies are of the
order of 10−3.
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CMB Black Body spectrum
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Expansion of the universe, ﬂuctuations in space time metric -
General Relativity
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General Relativity
Change in the densities of particles due to interactions and
expansion - Statistical mechanics (Boltzmann Equation)
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General Relativity
Synthesis of light elements - Nuclear and particle physics.
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General Relativity
Synthesis of light elements - Nuclear and particle physics.
Origin of density ﬂuctuations in Inﬂation - Quantum Field
theory
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The phase space density of photons is given by:
f (p) =
1
epc/kB T − 1
(17)
[Kolb & Turner (1990); Dodelson (2003); Weinberg (2008)]
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f (p) =
1
epc/kB T − 1
(17)
Number density is given by:
nγ = 2
d3p
(2π )3
f (p) = 8π
kBT
hc
3 ∞
0
x2dx
ex − 1
(18)
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f (p) =
1
epc/kB T − 1
(17)
Number density is given by:
nγ = 2
d3p
(2π )3
f (p) = 8π
kBT
hc
3 ∞
0
x2dx
ex − 1
(18)
Energy density:
ργ = 2
d3p
(2π )3
(pc)f (p) =
8π5k4
B
15h3c3
T4
= aT4
=
4σ
c
T4
(19)
with
∞
0
x2dx
ex − 1
= 2ξ(3) = 2.404 and
∞
0
x3dx
ex − 1
= 6ξ(4) =
π2
15
(20)
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Problem 3
Given that the CMB is a black body distribution with temperature
2.725 K show that:
number density of CMB photons is around 440 /cc and
energy density Ωγ ≈ 2.47 × 10−4/h2
photon to baryon ratio is around 109.
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Boltzmann Equation
Boltzmann equation describes the evolution of phase space
density f (t, x, p) in the phase space:
df
dt
= C[f ] (21)
where the RHS is the collision terms which represent the
change in the phase space density due to emission, absorption
and scattering.
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Boltzmann Equation
df
dt
= C[f ] (21)
and scattering.
We can write the LHS explicitly as:
df
dt
=
∂f
∂t
+
∂f
∂xi
dxi
dt
+
∂f
∂pi
dpi
dt
(22)
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Boltzmann Equation
df
dt
= C[f ] (21)
and scattering.
We can write the LHS explicitly as:
df
dt
=
∂f
∂t
+
∂f
∂xi
dxi
dt
+
∂f
∂pi
dpi
dt
(22)
In the absence of scattering, emission or absorption the
Boltzmann equation is simply:
df
dt
= 0 (23)
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Boltzmann Equation
Let us consider the following reaction:
1 + 2 ←→ 3 + 4 (24)
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Boltzmann Equation
1 + 2 ←→ 3 + 4 (24)
The number density n1 of particle type ’1’:
increase due to reaction between particle ’3’ and ’4’ and
decrease due to annihilation with particle ’2’
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Boltzmann Equation
1 + 2 ←→ 3 + 4 (24)
The number density n1 of particle type ’1’:
increase due to reaction between particle ’3’ and ’4’ and
decrease due to annihilation with particle ’2’
If the phase-space densities of particles 1, 2, 3 and 4 and
f1, f2, f3 and f4 respectively then from the Boltzmann
Equation:
a−3 d(a3
n1)
dt
=
d3
p1
(2π)32E1
d3
p2
(2π)32E2
d3
p3
(2π)32E3
d3
p4
(2π)32E4
× δD(E1 + E2 − E3 − E4)δD(p1 + p2 − p3 − p4)M2
× {f3f4[1 ± f1][1 ± f 2] − f1f2[1 ± f3][1 ± f 4]} (25)
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We are interested in a limit in which the exponential term is
far greater than the unity, so the Bose-Fermion diﬀerence can
be ignored:
f = eµ/T
e−E/T
(26)
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We are interested in a limit in which the exponential term is
far greater than the unity, so the Bose-Fermion diﬀerence can
be ignored:
f = eµ/T
e−E/T
(26)
We can replace the ﬁrst two lines of equation (25) by < σv >
so we get:
a−3 d(a3n1)
dt
= n
(0)
1 n
(0)
2 < σv >
n3n4
n
(0)
3 n
(0)
4
−
n1n2
n
(0)
1 n
(0)
2
(27)
where
ni = gi e
µ/T d3
p
(2π)3
e
−E/T
(28)
and
ni (0) =



gi
mi T
2π
3/2
e−mi /T
, ifT << mi
, gi
T3
π2 ifT >> mi
(29)
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Number density of a species can be computed by solving the
ordinary diﬀerential equation ( 27).
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The left hand side of Equation (27) is of order n1/t or of the
order of n1H and the right hand side is of the order of
n1n2 < σv >.
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n1n2 < σv >.
If the reaction rate n2 < σv >>> H then the RHS will be
much larger and the particles can be in equilibrium.
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n1n2 < σv >.
If the reaction rate n2 < σv >>> H then the RHS will be
much larger and the particles can be in equilibrium.
We can maintain the equality if the individual terms in RHS
cancel each other.
n3n4
n
(0)
3 n
(0)
4
=
n1n2
n
(0)
1 n
(0)
2
(30)
This equation is called Nuclear Statistical Equilibrium (NSE)
or Saha equation.
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Applications of Boltzmann Equations
Till 1010.5K = 2.7 Mev neutrinos are kept in thermal
equilibrium by weak interaction:
ν + ¯ν ←→ e+
+ e−
(31)
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ν + ¯ν ←→ e+
+ e−
(31)
Creation and annihilation of electron-positron pair stops at
1010K ≈ 1 Mev.
e+
+ e−
←→ 2γ (32)
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ν + ¯ν ←→ e+
+ e−
(31)
Creation and annihilation of electron-positron pair stops at
1010K ≈ 1 Mev.
e+
+ e−
←→ 2γ (32)
Above temperature 3500 K or 0.3 eV photons are hot enough
to ionize any hydrogen atom which forms:
e−
+ p+
←→ H + γ (33)
however once temperature of photons falls below 0.3 eV they
decouple and this event is called decoupling, recombination,
or last scattering.
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Recombination
Up to temperature 1 eV, photons remain tightly coupled to
electrons via Compton scattering and electrons to protons via
Coulomb scattering.
For e− + p = H + γ to be in equilibrium : we need
nenp
nH
=
n
(0)
e n
(0)
p
n
(0)
H
(34)
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Recombination
Up to temperature 1 eV, photons remain tightly coupled to
electrons via Compton scattering and electrons to protons via
Coulomb scattering.
For e− + p = H + γ to be in equilibrium : we need
nenp
nH
=
n
(0)
e n
(0)
p
n
(0)
H
(34)
Deﬁning :
Xe =
ne
ne + nH
=
np
ne + nH
(35)
equation (34) can be written as:
X2
e
1 − Xe
=
1
ne + nH
meT
2π
3/2
e− 0/T
(36)
where 0 = me + mp − mH is the Binding energy of hydrogen
atom.
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We can express ne + nH ≈ nb in terms of baryon-photon
ration η i.e., nb = ηnγ.
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Using the fact that nγ ∝ T3 equation (36) can be written as:
X2
e
1 − Xe
≈ 109 me
2πT
3
e− 0/T
≈ 1015
when T = 0 (37)
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X2
e
1 − Xe
≈ 109 me
2πT
3
e− 0/T
≈ 1015
when T = 0 (37)
Since the RHS becomes very large so the equation is satisﬁed
only when Xe is close to unity or all the atoms are ionized.
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X2
e
1 − Xe
≈ 109 me
2πT
3
e− 0/T
≈ 1015
when T = 0 (37)
Since the RHS becomes very large so the equation is satisﬁed
only when Xe is close to unity or all the atoms are ionized.
Saha equation (34) correctly predicts the epoch of
recombination but for fails when electron fraction drops and
the equation goes out of equilibrium and we need solve the
full Boltzmann equation numerically :
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a−3 d(a3
ne)
dt
= n(0)
e n(0)
p < σv >
nH
n
(0)
H
−
n2
e
n
(0)
e n
(0)
p
= nb < σv > (1 − Xe)
meT
2π
3/2
e− 0/T
− X2
e nb (38)
or
dXE
dt
= (1 − Xe)β − X2
e nbα(2)
(39)
with ionization rate β and the recombination rate α(2) are given by:
β =
meT
2π
3/2
e− 0/T
(40)
and
α(2)
=< σv > (41)
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Recombination
There is a superscript 2 on the recombination rate because
recombination to the ground (n=1) is not useful since it leads
to production of reionizing photon.
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Recombination
The only way for recombination to proceed is via capture to
one of the excited states of hydrogen.
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Recombination
The change in the number density of free electrons is
important from the point of view of observational cosmology
since recombination at z∗ ≈ 1000 is directly related to the
decoupling of CMB photons.
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Recombination
The change in the number density of free electrons is
important from the point of view of observational cosmology
since recombination at z∗ ≈ 1000 is directly related to the
decoupling of CMB photons.
Decoupling of CMB photons occurs roughly when the rate for
photons to Compton scatter oﬀ electrons becomes smaller
than the expansion rate.
neσT = XenbσT = 7.477 × 10−30
cm−1
XeΩbh2
a−3
(42)
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Recombination
Dividing the recombination rate by expansion rate (radiation
dominated):
H
H0
= Ω
1/2
m a−3/2
[1 + a/eeq]1/2
(43)
gives:
neσT
H
= 113Xe
Ωbh2
0.02
.15
Ωmh2
1/2
1 + z
1000
3/2
1 +
1 + z
3600
0.15
Ωmh2
−1/2
(44)
Problem 4
Derive equation (44)
Show that decoupling will eventually happen whether
recombination takes place or not.
Find the redshift of decoupling for Xe = 10−2 and Xe = 1.0.
How zeq, zdec and zrec are related and ﬁnd their values.
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CMB Theory : Metric perturbations
Geometric structure of a homogeneous and isotropic Universe
is given by the Friedman-Robertson-Walker (FRW) metric and
for spatially ﬂat case this can be written as:
ds2
= −c2
dt2
+ a2
(t)δij dxi
dxj
(45)
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ds2
= −c2
dt2
+ a2
(t)δij dxi
dxj
(45)
Since gµν is a second rank symmetric tensor and so have 10
components.
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ds2
= −c2
dt2
+ a2
(t)δij dxi
dxj
(45)
components.
There is a theorem called the decomposition theorem which
says that perturbations to the metric can be divided up into
three types: scalar, vector, and tensor and of these type
evolves independently.
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ds2
= −c2
dt2
+ a2
(t)δij dxi
dxj
(45)
components.
There is a theorem called the decomposition theorem which
says that perturbations to the metric can be divided up into
three types: scalar, vector, and tensor and of these type
evolves independently.
If some physical process in the early universe sets up tensor
perturbations, these do not induce scalar perturbations and
vice versa.
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Scalar Perturbations
Scalar perturbation to metric are represented by (in conformal
Newtonian Gauge) by two functions Ψ(x, t) and which Φ(x, t)
which represents perturbations in Newtonian potential and spatial
curvature respectively.
ds2
= −[1 + 2Ψ(x, t)]c2
dt2
+ a2
(t)δij [1 + 2Φ(x, t)]dxi
dxj
(46)
We can compute the Einstein tensor for the metric given above:
gµν −→ Γ −→ (R, Rµν) −→ Gµν
Problem 5
Show that for the metric given by (46) Ricci Tensor:
R00 = −3
¨a
a
−
k2
a2
Ψ − 3Ψ,00 + 3H(Ψ,0 − 2Φ,0)
Rij = δij (2a2
H2
+ a¨a)(1 + 2Φ − 2Ψ) + a2
H2
(6Φ,0 − Ψ,0)
+a2
Φ,00 + k2
Φ + ki kj (Φ + Ψ) (47)
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In order to solve for the potential Ψ and Φ we use the
Einstein equation. The temporal part:
δG0
0 = 8πGT0
0 (48)
and for spatial component we use only the traceless part:
ˆki
ˆkj
−
1
3
δj
i Gi
j = 8πG ˆki
ˆkj
−
1
3
δj
i Ti
j (49)
Note that non-relativistic particles, such as baryons and dark
matter, do not contribute anisotropic stress.
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In order to solve for the potential Ψ and Φ we use the
Einstein equation. The temporal part:
δG0
0 = 8πGT0
0 (48)
and for spatial component we use only the traceless part:
ˆki
ˆkj
−
1
3
δj
i Gi
j = 8πG ˆki
ˆkj
−
1
3
δj
i Ti
j (49)
Note that non-relativistic particles, such as baryons and dark
matter, do not contribute anisotropic stress.
The energy momentum tensor is given by:
T0
0 (x, t) = − gi
d3p
(2π)3
Ei (p)fi (p, x, t) = −ργ(1 + 4Θ0)
(50)
where Θ0 is the monopole part:
Θ0(x, t) =
1
4π
dΩ Θ(ˆp , ˆx, t) (51)
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Temporal part of Einstein equation gives:
k2
Φ+3
˙a
a
˙Φ − Ψ
˙a
a
= 4πGa2
[ρdmδdm+ρbδb+4ργΘ0+4ρνN0]
(52)
note that here dot represent derivative with conformal time
and N0 is the monopole term for neutrinos.
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Temporal part of Einstein equation gives:
k2
Φ+3
˙a
a
˙Φ − Ψ
˙a
a
= 4πGa2
[ρdmδdm+ρbδb+4ργΘ0+4ρνN0]
(52)
note that here dot represent derivative with conformal time
and N0 is the monopole term for neutrinos.
Second Einstein equation gives:
k2
(Φ + Ψ) = −32πGa2
(ργΘ2 + ρνN2) (53)
the two gravitational potentials are equal and opposite unless
the photons or neutrinos have appreciable quadruple
moments.
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Boltzmann Equation Photons
Boltzmann equation is given by:
df
dt
= C[f ] (54)
where C[f ] is the collision term which corresponds to
Compton scattering of photons with electrons:
e−
(q) + γ(p) ←→ e−
(q ) + γ(p ) (55)
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df
dt
= C[f ] (54)
e−
(q) + γ(p) ←→ e−
(q ) + γ(p ) (55)
We can explicitly write:
df
dt
=
∂f
∂t
+
∂f
∂x
dx
dt
+
∂f
∂p
dp
dt
(56)
34 / 50

df
dt
= C[f ] (54)
e−
(q) + γ(p) ←→ e−
(q ) + γ(p ) (55)
We can explicitly write:
df
dt
=
∂f
∂t
+
∂f
∂x
dx
dt
+
∂f
∂p
dp
dt
(56)
Now we need to compute dx/dt and dp/dt in the perturbed
metric.
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Let us consider that the four momentum of photon is Pµ then
:
Pµ
Pµ
= g00(P0
)2
+ p2
= −(1 + 2Ψ)(P0
)2
+ p2
(57)
or
P0
=
p
√
1 + 2Ψ
≈ p(1 − ψ) (58)
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Let us consider that the four momentum of photon is Pµ then
:
Pµ
Pµ
= g00(P0
)2
+ p2
= −(1 + 2Ψ)(P0
)2
+ p2
(57)
or
P0
=
p
√
1 + 2Ψ
≈ p(1 − ψ) (58)
For spatial part we can write:
Pi
= Cˆpi
(59)
where C is a constant which we can compute in the following
way:
p2
= Pi
Pi = C2
gij ˆpi
ˆpj
= C2
a2
(1 + 2Φ) (60)
and so
C =
p
a
√
1 + 2Φ
(61)
and
Pi
=
p
a
(1 − Φ)ˆpi
(62)
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The velocity can be computed as:
dxi
dt
=
dxi
dλ
dλ
dt
=
Pi
P0
=
1
a
(1 + Ψ − Φ)ˆpi
(63)
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The velocity can be computed as:
dxi
dt
=
dxi
dλ
dλ
dt
=
Pi
P0
=
1
a
(1 + Ψ − Φ)ˆpi
(63)
For momentum we use Geodesic equation:
dPµ
dλ
+ Γµ
αβ
dxα
dλ
dxβ
dλ
= 0 (64)
which for 0 component:
dP0
dλ
+ Γ0
αβ
dxα
dλ
dxβ
dλ
= 0 (65)
we can compute:
Γµ
αβ =
1
2
gµν ∂gνα
∂xβ
+
∂gνβ
∂xα
−
∂gαβ
∂xν
(66)
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Problem 6
Using equation (65) and metric given by equation (46) show that:
1
p
dp
dt
= −H −
∂Ψ
∂t
−
ˆpi
a
∂Ψ
∂xi
(67)
Using the expressions for dxi /dt and dpi /dt we can write:
df
dt
=
∂f
∂t
+
ˆpi
a
∂f
∂xi
− p
df
dp
H +
∂Φ
∂t
+
ˆpi
a
∂Ψ
∂xi
(68)
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Linear Approximation
We can get the evolution equation for Θ(x, ˆp, t) from the
evolution equation for f (t, x, p) by expanding f around its
zeroth order and keeping only the linear terms in Θ(x, ˆp, t):
f = f (0)
− p
∂f (0)
∂p
Θ (69)
where
f (0)
=
1
ep/T − 1
(70)
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Linear Approximation
We can get the evolution equation for Θ(x, ˆp, t) from the
evolution equation for f (t, x, p) by expanding f around its
zeroth order and keeping only the linear terms in Θ(x, ˆp, t):
f = f (0)
− p
∂f (0)
∂p
Θ (69)
where
f (0)
=
1
ep/T − 1
(70)
Keeping only up to linear terms in Θ, the Boltzmann equation
for photons become:
df
dt
= −p
∂f (0)
∂p
∂Θ
∂t
+
ˆpi
a
∂Θ
∂xi
+
∂Φ
∂t
+
ˆpi
a
∂Ψ
∂xi
(71)
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Compton scattering
Scattering (Compton) between free electrons and photons also
change the phase space density of photons:
e−
(q) + γ(p) ↔ e−
(q ) + γ(p), (72)
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Compton scattering
e−
(q) + γ(p) ↔ e−
(q ) + γ(p), (72)
The change in the phase density of photons due to Compton
scattering is given by:
c[f (p)] = −p
∂f (0)
∂p
neσT [Θ0 − Θ(p) + ˆp.vb] (73)
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Compton scattering
e−
(q) + γ(p) ↔ e−
(q ) + γ(p), (72)
The change in the phase density of photons due to Compton
scattering is given by:
c[f (p)] = −p
∂f (0)
∂p
neσT [Θ0 − Θ(p) + ˆp.vb] (73)
The full Boltzmann equation can be written as:
∂Θ
∂t
+
ˆpi
a
∂Θ
∂xi
+
∂Φ
∂t
+
ˆpi
a
∂Ψ
∂xi
= neσT [Θ0 − Θ + ˆp.vb] (74)
This equation is called the Brightness equation [Kurki-Suonio
(2010)]
39 / 50

In terms of conformal time the full Boltzmann equation can
be written as:
˙Θ + ˆpi ∂Θ
∂xi
+ ˙Φ + ˆpi ∂Ψ
∂xi
= neσT a[Θ0 − Θ + ˆp.vb] (75)
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In terms of conformal time the full Boltzmann equation can
be written as:
˙Θ + ˆpi ∂Θ
∂xi
+ ˙Φ + ˆpi ∂Ψ
∂xi
= neσT a[Θ0 − Θ + ˆp.vb] (75)
In Fourier space equation (75) becomes a algebraic equation:
˙˜Θ + ikµ˜Θ + ˙˜Φ + ikµ˜Ψ = − ˙τ[˜Θ0 − ˜Θ + µ˜vb] (76)
where :
Θ(ˆx) =
d3k
(2π)3
exp[ik.x]˜Θ(ˆk) (77)
and the optical depth τ is deﬁned as:
τ(η) =
η0
η
dη neσT a (78)
where −neσT a = ˙τ and the direction of propagation of
photon µ = ˆk.ˆp.
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Note that if we take into account that the Compton scattering
between photons and electrons depend on the direction also and
temperature and polarization ﬁelds are coupled to each other, we
get the following Boltzmann equation for photons:
˙˜Θ + ikµ˜Θ + ˙˜Φ + ikµ˜Ψ = − ˙τ ˜Θ0 − ˜Θ + µ˜vb −
1
2
P2(µ)Π (79)
where P2(µ) = (3µ2 − 1)/2 is the second Legendre polynomial and
Π is deﬁned as:
Π = Θ2 + ΘP2 + ΘP0 (80)
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Boltzmann Equations
Considering that the Universe at the time of decoupling
consists photons, neutrinos, baryons and dark matter, we have
the following set of seven equations for the evolution of
Θ, ΘP, δ, v, δb, vb and neutrino temperature N
1
2
P2(µ)Π (81)
˙˜ΘP + ikµ˜ΘP = − ˙τ −˜ΘP +
1
2
(1 − P2(µ))Π (82)
˙δ + ikv = −3 ˙Φ (83)
˙v +
˙a
a
= −ikΨ (84)
˙δb + ikvb = −3 ˙Φ (85)
˙vb +
˙a
a
vb = ikΨ +
˙τ
R
[vb + 3iΘ1] (86)
˙N + ikµN = − ˙Φ − ikµΨ (87)
42 / 50

Boltzmann Equations
Considering that the Universe at the time of decoupling
consists photons, neutrinos, baryons and dark matter, we have
the following set of seven equations for the evolution of
Θ, ΘP, δ, v, δb, vb and neutrino temperature N
1
2
P2(µ)Π (81)
˙˜ΘP + ikµ˜ΘP = − ˙τ −˜ΘP +
1
2
(1 − P2(µ))Π (82)
˙δ + ikv = −3 ˙Φ (83)
˙v +
˙a
a
= −ikΨ (84)
˙δb + ikvb = −3 ˙Φ (85)
˙vb +
˙a
a
vb = ikΨ +
˙τ
R
[vb + 3iΘ1] (86)
˙N + ikµN = − ˙Φ − ikµΨ (87)
We have two equations from Einstein’s equation for potential
Ψ and Φ:
k
2
Φ + 3
˙a
a
˙Φ − Ψ
˙a
a
= 4πGa
2
[ρdmδdm + ρbδb + 4ργ Θ0 + 4ρν N0] (88)
k
2
(Φ + Ψ) = −32πGa
2
(ργ Θ2 + ρν N2) (89)
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Boltzmann Equations
In order to solve the set of 9 ﬁrst order diﬀerential
(Boltzmann-Einstein) equations we need initial conditions.
43 / 50

Boltzmann Equations
Since variables depend on each other so we do not need initial
conditions for all.
43 / 50

Boltzmann Equations
conditions for all.
In fact when considering Ψ = −Φ we need just one initial
condition i.e., for Φ.
43 / 50

Boltzmann Equations
conditions for all.
Inﬂation which explain large scale uniformity of the CMB sky
also provides a mechanism to create perturbations in Φ.
43 / 50

Boltzmann Equations
conditions for all.
Inﬂation which explain large scale uniformity of the CMB sky
also provides a mechanism to create perturbations in Φ.
In the very early universe kη << 1 i.e., modes are outside
horizon, these equations become quite simple since we can
ignore terms which have k and higher power of k.
43 / 50

Line of sight integral
Problem 7
Show that the Boltzmann equation for photons can be solved as:
Θ(k, µ, η0) =
η0
0
dη˜S(k, µ, η)eikµ(η−η0)−τ(η)
(90)
where
˜S = − ˙Φ − ikµΨ − ˙τ Θ0 + µvb −
1
2
P2(µ)Π (91)
Hint: write
˙Θ + (ikµ − ˙τ)Θ = e
−ikµη d
dη
[Θe
ikµη−τ
] (92)
Rather than solving equation (90) for Θ(k, µ, η0) we solve for each
multipole Θl (k, η0) which becomes challenging since modes are
coupled. [Seljak & Zaldarriaga (1996)]
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Multipole expansion
Non-relativistic particles like dark matter and baryons can be
characterized by their densities δ(x, t) and velocities v(x, t)
(which are equivalent to monopole and dipole).
In Fourier space the evolution of densities and velocities for
non-relativistic species depend on the magnitude of k.
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Multipole expansion
The scalar velocities here are the components parallel to k;
these are the only ones that are cosmologically relevant.
45 / 50

Multipole expansion
The scalar velocities here are the components parallel to k;
these are the only ones that are cosmologically relevant.
We need much more information to specify relativistic particle
like photons since they have not only a monopole perturbation
and a dipole but also a quadrupole, octopole, and higher
moments as well.
45 / 50

Multipole expansion
When there is azimuthal symmetry then we can write:
Θ(k, η, µ) =
l
(−i)l
(2l + 1)Θl (k, η) (93)
where
Θl (k, η) =
1
(−i)l
1
−1
dµ
2
Pl (µ)Θ(k, η, µ) (94)
and where Pl is the Legendre polynomial of order l.
46 / 50

Multipole expansion
Θ(k, η, µ) =
l
(−i)l
(2l + 1)Θl (k, η) (93)
where
Θl (k, η) =
1
(−i)l
1
−1
dµ
2
Pl (µ)Θ(k, η, µ) (94)
Rather than using Θ(k, η, µ) to specify the CMB anisotropy in
Fourier space, we can use the multipole moments Θl (k, η)
and study their evolution.
46 / 50

Multipole expansion
Θ(k, η, µ) =
l
(−i)l
(2l + 1)Θl (k, η) (93)
where
Θl (k, η) =
1
(−i)l
1
−1
dµ
2
Pl (µ)Θ(k, η, µ) (94)
Rather than using Θ(k, η, µ) to specify the CMB anisotropy in
Fourier space, we can use the multipole moments Θl (k, η)
and study their evolution.
Note that before recombination since photons and baryons
were tightly coupled so only monopole Θ0(k, η) terms were
signiﬁcant.
46 / 50

Inhomogeneities to anisotropies
If recombination happens instantaneously then the CMB
anisotropy Θ(ˆn) is related to the inhomogeneity at the last
scattering surface:
Θ(ˆn) = dDΘ(x)δD(D − D∗) (95)
where D∗ is the comoving distance of the recombination.
47 / 50

scattering surface:
Θ(ˆn) = dDΘ(x)δD(D − D∗) (95)
We can expand the inhomogeneity Θ(x) in Fourier space:
Θ(x) =
d3k
(2π)3
˜Θ(k)eik.x
(96)
47 / 50

scattering surface:
Θ(ˆn) = dDΘ(x)δD(D − D∗) (95)
We can expand the inhomogeneity Θ(x) in Fourier space:
Θ(x) =
d3k
(2π)3
˜Θ(k)eik.x
(96)
Translational and rotational invariance of Θ(x) leads:
< ˜Θ(k)˜Θ(k ) >= (2π)3
δD(k − k )PT (K) (97)
where PT (K) is the power spectrum.
47 / 50

We generally expand CMB anisotropy in spherical harmonics:
Θ(ˆn) =
lmax
l=0
m=l
m=−l
almYlm(ˆn) (98)
where Ylm(ˆn) are spherical harmonics basis and follow the
orthogonality relations:
dˆnYlm(ˆn)Yl m (ˆn) = 2πδll δmm (99)
48 / 50

We generally expand CMB anisotropy in spherical harmonics:
Θ(ˆn) =
lmax
l=0
m=l
m=−l
almYlm(ˆn) (98)
where Ylm(ˆn) are spherical harmonics basis and follow the
orthogonality relations:
dˆnYlm(ˆn)Yl m (ˆn) = 2πδll δmm (99)
In Fourier space:
Θ(ˆn) =
d3k
(2π)3
˜Θ(k)eik.D∗ˆn
(100)
we can expand plane wave in spherical harmonics:
eik.D∗ˆn
= 4π
lm
il
jl (kD∗)Y ∗
lm(k)Ylm(ˆn) (101)
48 / 50

Problem 8
Show that in the multipole expansion CMB multipole alm and
the Fourier amplitude of the inhomogeneity ˜Θ(k) are related
in the following way:
alm =
d3k
(2π)3
˜Θ(k)4πil
jl (kD∗)Y ∗
lm(k) (102)
Given that ∆2
T (k) = k3P(k)/2π2 is slowly varying and
∞
0 j2
l (x)dlnx = 1/(2l(2l + 1)) show that the angular power
spectrum Cl can be written in the following form:
< almal m >= (2π)3
δll δmm Cl (103)
with
Cl =
2π
l(l + 1)
∆2
T (l/D∗) (104)
49 / 50

References
Alpher, R. A., & Herman, R. C. 1948, Physical Review, 74, 1737
Dicke, R. H., Peebles, P. J. E., Roll, P. G., & Wilkinson, D. T. 1965,
Astrophys. J. , 142, 414
Dodelson, S. 2003, Modern cosmology (San Diego, U.S.A.: Academic
Press)
Gamow, G. 1948, Physical Review, 74, 505
Kolb, E. W., & Turner, M. S. 1990, The early universe.
Kurki-Suonio, H. 2010, ArXiv e-prints
Penzias, A. A., & Wilson, R. W. 1965, Astrophys. J. , 142, 419
Seljak, U., & Zaldarriaga, M. 1996, Astrophys. J. , 469, 437
Smoot, G. F., et al. 1992, Astrophys. J. Lett. , 396, L1
Weinberg, S. 2008, Cosmology (Oxford University Press)
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