Modern Cosmology

Tom Charnock

Contents

1 Introduction 21.1 Observable Features . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 The Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Hubble’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Curved Space Generalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 Light and Redshift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.6 Geodesic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.7 Einstein’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.8 Energy Evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.9 Cosmological Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.10 Observational Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.11 Horizons and Distances in Cosmology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.11.1 Particle Horizon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.11.2 Event Horizon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.12 Luminosity Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.13 Angular Diameter Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Thermal History 142.1 Ideal (Bose and Fermi) Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.1.1 Relativistic Species . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.1.2 Entropy of the Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.1.3 Non-Relativistic (Massive) Species . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.1.4 Several Relativistic Species (Degrees of Freedom) . . . . . . . . . . . . . . . . . . . 16

2.2 Temperature vs. Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Density of the Universe and Dark Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3.1 Dark Matter Candidates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Cosmological Perturbation Theory 193.1 Newtonian Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.1.1 Poisson Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.1.2 Fluid Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Structure Formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2.1 Static Universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2.2 Expanding Universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3 Relativistic Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3.1 Scalar-Vector-Tensor Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3.2 Perturbative Einstein Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.3.3 Observing Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.4 Cosmic Microwave Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.4.1 Anisotropies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.5 Recombination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.5.1 Decoupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

1

Chapter 1

Introduction

1.1 Observable Features

The observable path of the universe, in which the light has been able to travel to the Earth since thebeginning of time, is 3000Mpc ∼ 3.26×106 years = 3×1024cm. On scales greater than 100Mpc then theuniverse is homogeneous and isotropic such that there is no unique place and all features look the samein all directions. The evidence for homogeneity comes from the distribution of large scale structure andfor isotropy it is the distribution of the cosmic microwave background (CMB) photons.

The universe is expanding at speed v = Hd where H is the Hubble constant and d is the distance betweenobjects, such as galaxies. In fact the universe is also accelerating which indicates that there must be someform of dark energy.

There is a uniform distribution of radiation at T ≈2.275K±10−5K which is in the microwave region.Observations of the fluctuations in the CMB indicate that when the universe was roughly a thousandthof the size of today then the fluctuations were as small as 10−5.

The universe contains baryons, about 1 per every 109 photons. There is very little anti-matter. Thebaryonic matter is distributed with 75% hydrogen and roughly 24% helium. The rest is trace amountsof the heavier elements. These elements were not produced in the early universe but instead in stars orsupernovae. Baryons make up about 4% of the total energy density and the rest is in dark “stuff”. Colddark matter makes up roughly 23% of the universe and it is unknown non-electromagnetic interactions,non-relativistic particles. These particles are pressureless. Dark energy makes up the rest of the universe(about 73%). The dark energy has a negative pressure.

1.2 The Metric

In two dimensional flat space then the distance is given by Pythagorus, ∆s2 = ∆x21 + ∆x2

2. If thespace is stretched over time such that it is expanding uniformly then the distance increases as ∆s2 =a2(t)

(∆x2

1 + ∆x22

). a(t) gives the rate of expansion and x1 and x2 are comoving coordinates.

Since time is a coordinate in general relativity then spacetime coordinates are used. There can becurvature and so the infinitesimal separation needs to be used:

ds2 = gµνdxµdxν

gµν are the components of the metric tensor and is a function of the coordinates and so allows spaceto be curved. dxµ and dxν are the infinitesimal coordinate separation. The greek indices take valuesµ = 0, 1, 2, 3. The time coordinate is given by x0 and the spatial coordinates are xi = (x1, x2, x3).

Vectors are components with raised indices, Aµ = (A0, Ai) with i = 1, 2, 3. Aµ and Aµ are differentobjects, Aµ is called a one-form. Changing between vectors and one-forms can be done using the metrictensor:

Aµ = gµνAν and Aµ = gµνAν

2

gµν is the inverse metric defined by gµνgνβ = δνβ . This arises since raising the indices of the metric gives:

gµν = gµαgνβgαβ

So gνβgαβ = 1 when ν = α and gνβgαβ = 0 when v 6= α.

gµν is symmetric and so it has 4 diagonal and 6 off-diagonal components.

In special relativity then Minkowski spacetime is used where xµ = (ct, xi) and the metric in this case is:

ηµν =

−1 · · ·· 1 · ·· · 1 ·· · · 1

For an expanding universe then a scale factor a(t) can be included, so for a flat space time the metric is:

gµν =

−1 · · ·· a2(t) · ·· · a2(t) ·· · · a2(t)

This means the line element is:

ds2 = gµνdxµdxν

= −c2dt2 + a2(t)(dx2 + dy2 + dz2

)(

invariantinterval

)2

= −(

timeinterval

)2

+

(scalefactor

)2

×(

comovinginterval

)2

1.3 Hubble’s Law

Because the expansion is uniform then the physical distance can be written as:

r(t) = a(t)× x(t)

Where x(t) is the comoving distance. Differentiating this with respect to time gives:

r = H(t)r(t) + a(t)t

Here H = a/a is the Hubble parameter. The ax term is the peculiar velocity, or the local dynamics. TheHr term is the dynamics due to the expansion of the universe. This term is key for cosmology since itshows how the expansion rate relates the separation and velocity of recession of distant galaxies. Hubble’slaw is:

v˜(t) = H(t)r˜(t)The convention is to use subscript 0 to denote todays values, so Hubble’s parameter today is:

H0 = 100h km s−1Mpc−1

h is a measure of the uncertainty and, as of 2012, is:

h = 0.744± 0.025

If h = 0.72 then for a galaxy travelling at vexp = 7200km s−1 then the separation between the earth andthe galaxy is 100Mpc. Uncertainty in h feeds into all the cosmological estimations. To estimate the agesof the universe then:

H−10 = 9.77h−1 × 109years

This is not a bad estimate. The evolution of the scale factor, a(t), depends on the density of matter.

3

1.4 Curved Space Generalisation

The cosmological principle implies that the universe is isotropic and homogeneous on large scales. Thismeans that the spatial part of the metric must have constant curvature. The most general form of thethree dimensional space with constant curvature (in spherical polars) is:

ds23 = a2

(dr2

1−Kr2+ r2(dϑ2 + sin2 ϑdφ2)

)Here a2 > 0 and from now on dΩ2 = dϑ2 + sin2 ϑdφ2. K measures the curvature of space. If K > 0 thenthe space is spherical (or closed), if K = 0 then space is flat and if K < 0 then space is hyperbolic (oropen). The full spacetime metric is:

ds2 = −c2dt2 + a2(t)

(dr2

1−Kr2+ r2dΩ2

)This is the Robertson-Walker metric. Taking the spatial section and introducing a new variable dχ2 =dr2/1−K2, then normalising K = ±1, 0 and integrating gives:

χ = arcsinhr K = −1χ = r K = 0

0 ≤ χ ≤ ∞

χ = sin r K = 1 0 ≤ χ ≤ π

This means that the metric for the spatial part can be written as:

ds23 = a2(dξ2 + S2

K(χ)dΩ2)

Where SK(χ) is:

SK(χ) = sinhχ K = −1SK(χ) = χ K = 0SK(χ) = sinχ K = 1

For the three dimensional sphere (K = 1) then the distance element on the surface of the two spherewith radius χ is:

ds2 = a2 sin2 ξdΩ

This is equivalent to the line element on a sphere of radius R = a sinχ in flat three dimensional space.The surface area of the two dimensional surface is:

S2D(χ) = 4πR2

= 4πa2 sin2 χ

As the radius χ increases then the surface area grows to a maximum of χ = π/2 and then decreases untilit vanishes at χ = π. The volume is then obtained by integrating dV = S2D(χ)adχ:

V (χ0) = 4πa3

∫ χ0

0

sin2 χdχ

= 2πa3

(χ0 −

1

2sin 2χ0

)

1.5 Light and Redshift

The line element in general for a Friedmann-Robertson-Walker universe is:

ds2 = −c2dt2 + a2(t)

(dr2

1−Kr2+ r2dΩ2

)If there are two galaxies A and B then the light emitted from B has a wavelength which could be changedby the expansion of the universe when observed at A. Light follows null geodesics ds2 = 0. For a radialtrajectory dΩ = 0 so the change in time between emision and observation is:∫ rB

rA

dr√1−Kr2

=

∫ tobs

temit

cdt

a(t)

4

A second peak of the wave is emitted at temit + λemit/c and is observed at tobs + λobs/c and since theintegral over r is just a constant then:∫ tobs

temit

cdt

a(t)=

∫ tobs+λobs/c

temit+λemit/c

cdt

a(t)

This reveals the result:vemit

vobs=

aobs

aemit

Since vemit/vobs = 1 + z then observing shifts in spectral lines allows the relative size of the universe tobe determined when the light was emitted. This is key for observational cosmology.

1.6 Geodesic Equations

The geodesics is the trajectory in curved space. It is the generalisation of the straight line in Minkowskispace in the absence of external forces, i.e. Newton’s Law generalised to an expanding universe.

Since time is a coordinates then a new evolution parameter is needed. For this role, λ is introduced,which increases monotonically along the particles path. The geodesic equation is:

d2xµ

dλ2+ Γµαβ

dxα

dλ

dxβ

dλ= 0

Γµαβ are the Christoffel symbols which are defined as:

Γµαβ =1

2gµν (gαν,β + gβν,α − gαβ,ν)

Given the FRW metric with K = 0 then:

Γ0ij = aa′δij and Γi0j = Γij0 =

a′

aδij

Where a′ = da/cdt = a/c. All the other Christoffel symbols vanish.

1.7 Einstein’s Equations

Einstein’s equation relate the components of the Einstein tensor (geometry) to the energy-momentumtensor (matter). The Einstein tensor is:

Gµν = Rµν −1

2gµνR

Rµν is the Ricci tensor and is symmetric given by:

Rµν = Γαµν,α − Γαµα,ν + ΓαβαΓβµν − ΓαβνΓβµα

R is the Ricci scalar obtained by contracting the inverse metric with the Ricci tensor R = gµνRµν .Einstein’s equations are:

Gµν =8πG

c4Tµν − Λgµν

Λ is the cosmological constant and Tµν is the energy-momentum tensor, which is symmetric. For a perfectisotropic fluid then the energy-momentum tensor is:

T µν =

−%c2 · · ·· p · ·· · p ·· · · p

%(t) is the energy density and p(t) is the pressure of the fluid. By substituting in the Christoffel symbolsfound in section 1.6 into the Ricci Tensor then it can be seen that:

R00 = −3a′′

a

Rij = δij(2a′2 + aa′′

)5

The Ricci scalar can also be found:

R = gµνRµν

= −R00 +Riia2

= 6

(a′′

a+

(a′

a

)2)

The Friedmann equation is the 00 component of Einstein’s equations:

R00 −1

2g00R =

8πG

c4T00 − Λg00

(a

a

)2

=8πG

3%(t) +

Λc2

3

When considering curvature then the Christoffel symbols obtain curvature terms and so the Friedmannequation is: (

a

a

)2

=8πG

3%(t)− Kc2

a2+

Λc2

3

Now considering the ij components of Einstein’s equations gives the acceleration equation:

a

a= −4πG

3

(%(t) +

3p(t)

c2

)+

Λc2

3

The cosmological constant can be absorbed by making:

%→ %+Λc2

8πGand p→ p− Λc2

8πG

Λc2/8πG can be thought of as an energy density %λ or a pressure −pΛ so that p = −%c2. This is importantfor the evolution of the universe and inflation since substituting p = −%c2 into the absorbed accelerationequation shows that if % + 3p/c2 < 0 then a > 0. The acceleration and Friedmann equations with thecosmological constant absorbed are:

a

a= −4πG

3

(%(t) +

3p(t)

c2

)

H2 =8πG

3%(t)− Kc2

a2

The critical density can now be defined as the density for which the universe is flat, i.e. K = 0. Thisdensity is a function of time and is continually changing:

%C(t) =3H2(t)

8πG

The density parameter is then defined as:

Ω =%(t)

%C(t)

=8πG%

3H2

The density today is %0 = 1.878×10−2Ω0h2kgm−3. The density is extremely low but in terms of different

units then %0 = 2.775× 1011Ω0h2MMpc−3. This is on the galactic scale and says that there is roughly

1011 solar masses in the galaxy which is approximately correct. If % = %C then Ω = 1 for a spatially flatuniverse.

6

1.8 Energy Evolution

The energy-momentum conservation is given by:

Tµν;µ = Tµν,µ + ΓµαµTαν − ΓανµT

µα

= 0

When ν = 0 then Tµ0,µ + ΓµαµTα0 − Γα0µT

µα = 0, and assuming isotropy then T 0

i vanishes since there is nostrain and thus placing in the Christoffel symbols for the FRW universe reveals:

%+3a

a

(%+

p

c2

)= 0

This is the fluid equation. The pressure and energy density need to be related and is done so using theequation of state p = p(%). For most fluids (with no torsion) then p = w% where w = (γ−1) is a constant.The fluid equation has a solution:

% ∝ a−3(1+w)

The Friedmann equation can be solved to give:

a ∝ t2/(3(1+w))

These can be used to find the time evolution versions of the energy density and scale factor.

Matter Domination When w = 0 then p = 0 and since matter is a pressureless dust. When Ωm = 1then the universe is Einstein-de Sitter and the values of the energy density and the scale factor are:

%m(t) = %m0

(a0

a

)3

and a(t) = a0

(t

t0

)2/3

Radiation Domination Radiation has a w = 1/3 and a universe with Ωr = 1 is called a Tolmanuniverse. The energy density and scale factor evolve as:

%r(t) = %r0

(a0

a

)4

and a(t) = a0

(t

t0

)1/2

Cosmological Constant For the cosmological constant then w = −1 and so the scale factor has tobe expanded as a Taylor series to find its evolution. If the universe has a cosmological constant then itis de Sitter and has energy density and a scale factor:

%Λ(t) = %0 and a(t) = a0eH(t−t0)

Here H = 8πG%0/3.

The total energy density is given by % = %m + %r + %v where %v is the vacuum energy density where w isnot confined to −1 as for a cosmological constant. If w varies with time then it provides the force knownas quintessence. The Friedmann equation can now be solved using the time evolution energy densities:

H2 =8πG

3(%m + %r + %v)−

Kc2

a2

=8πG

3

(%m0(1 + z)3 + %r0(1 + z)4 + %v0(1 + z)3(1+w)

)− Kc2

a2

=8πG

3H20

H20

(%m0(1 + z)3 + %r0(1 + z)4 + %v0(1 + z)3(1+w)

)− Kc2

a2

= H20

(Ωm0(1 + z)3 + Ωr0(1 + z)4 + Ωv0(1 + z)3(1+w)

)− Kc2

a2

7

Since K is not observable then it should be eliminated by solving the Friedmann equations for theconstants known today:

Kc2 = a20H

20 (Ω0 − 1)

= a20H

20 (Ωm0 + Ωr0 + Ωv0)

Substituting this back into the Friedmann equation gives:

H2 = H20 (1 + z)2

(Ωm0((1 + z)− 1) + Ωr0((1 + z)2 − 1) + Ωv0((1 + z)(1+3w) − 1)

)This is very powerful since t(z) can be found by integrating and the integrals are numerically straightforward.

1.9 Cosmological Solutions

The universe has lots of fluids and may not be flat. Conformal time can be introduced so that thepropertime becomes a(t)τ . This conformal time, η is given by cddt = a(η)dη so that:

η =

∫cdt

a(t)

For c = 1 then the solutions to different cosmologies can be calculated in terms of η. It is useful to notethat a′ = da/dη. The Friedmann equation is now written as:

a′2 =8πG

3%a4 −Ka2

And the acceleration equation is:

a′′ =4πG

3(%− 3p)a3 −Ka

For a single fluid, say radiation with p = %/3 then the acceleration equation is simply:

a′′ +Ka = 0

This is simple to solve since it is of the form of a harmonic oscillator (depending on K). The solutionsare:

a(η) = cr sinh η K = −1a(η) = crη K = 0

0 ≤ η ≤ ∞

a(η) = cr sin η K = 1 0 ≤ η ≤ π

The second integration constant is fixed by a(0) = 0. The physical time is then:

t =

∫a(η)dη

For the three different curvature situations then:

t(η) = cr(cosh η − 1) K = −1

t(η) =crη

2

2K = 0

t(η) = cr(1− cos η) K = 1

These are parametric solutions, i.e. a(η) and t(η). This give a(t) but not in such an accessible form,except for the trivial K = 0 case.

For dust then the same steps can be taken by now p = 0.

Some unusual solutions can be found. Take the radiation-Λ solution (in K = 0). the Friedmann equationfor radiation-Λ is:

H2(z) = H20

(Ωr0

(a0

a

)4

+ ΩΛ0

)

8

Square rooting both sides and rearranging gives:

d

dta2 = 2H0

√Ωr0a

20

(1 + α2

(a

a0

)4)1/2

Here α = ΩΛ0/Ωr0. Introducing y = a2/a20 then:

y = 2H0

√Ωr0

(1 + α2y2

)1/2Integrating this gives a solution to a(t):

a(t) = a0

(Ωr0ΩΛ0

)1/4 (sinh(2H0

√ΩΛ0t)

)1/2

Where a(0) = 0 has been used as the boundary condition. At early times (t 1) then sinh t ≈ t so

a(t) ∝ t1/2 which is radiation dimension. At late times, t 1 then sinht ≈ et/2 so a(t) ∝ eH0

√ΩΛ0t

which is an inflationary solution. This model could be a precursor before the onset of inflation in theearly universe.

For a matter-Λ solution (a good way of describing the universe today since ΩΛ0 ≈ 0.7 and Ωm0 ≈ 0.23and so are comparable) then when K = 0 the Friedmann equation is:

H2(z) = H20

(Ωm0

(a0

a

)3

+ ΩΛ0

)Now y = (a/a0)3/2 and solving for a(t) gives:

a(t) = a0

(Ωm0

ΩΛ0

)1/2(sinh

(3

2H0

√ΩΛ0t

))2/3

Again, at early times then a(t) ∝ t2/3 which is usual for matter domination and once again, for large t

then sinh t ≈ et and so a(t) ∝ eH0

√ΩΛ0t which is inflationary or dark energy.

1.10 Observational Parameters

Hubbles parameter (constant - which is not constant) - H0 = 100hkms−1Mpc−1 with h = 0.744± 0.025.

Deceleration parameter - q0 = −a(t0)/(a(t0)H20 ). This is measurable directly on large scales. Type 1a

supernovae show that q0 ≈ −0.6 and so the observation show that the universe is accelerating.

1.11 Horizons and Distances in Cosmology

The FRW universe metric is:

ds2 = −cdt2 + a2(t)

(dr2

1−Kr2+ r62

(dϑ2 + sin2 ϑdφ2

))A light ray travels from r = 0 and t = tem to r = r0 and t = t along a radial null direction so thatds = dϑ = dφ = 0. It is useful to use conformal time and dχ = dr/

√1−Kr2 so that the metric becomes:

0 = −dη2 + dχ2

Integrating through these gives ξ(η) = ±η + constant. The solutions are straight lines at angle at 45o inthe η − χ plane and is true of all geometries, K = ±1, 0.

9

1.11.1 Particle Horizon

For a universe of finite age then light can only travel a finite distance in that time. Moreover, since lighthas only travelled a finite distance then there is a finite volume of space within which signals could bereceived. The boundary of this volume is the particle horizon which is expected to be of order c×age ofthe universe. The maximum distance that light can propagate is given by the difference in |chi at thebeginning and end of the universe. Given that χ(η) = ±η+constant then:

χ(η) = η − ηi

=

∫ t

ti

cdt

a

At η events at χ > χp(η) are inaccessible to an observer located at χ = 0. It is usually alright to chooseηi = ti = 0 especially when there is an initial singularity. The physical size of the particle horizon is:

R(t) = a(t)χp

= a(t)∫titcdt

a(t)

This is the radius of the observable universe at some time t. It could be infinite or finite depending on thecosmology. If the cosmology is decelerating then R(t) is finite, i.e. a(t) ∝ tα (0 < α < 1) has a radius:

R(t) =ct

1− αThis is finite at any given t. The Einstein-de Sitter universe has α = 2/3 and so the particle horizon isat R(t) = 3ct0 > ct0.

1.11.2 Event Horizon

The event horizon can be thought of as the compliment of the particle horizon. It encloses a set of pointsfrom which signals sent at a given moment t(η) will never be received by an observer in the future. Thesecorrespond to the points:

χev(t) = ηmax − η

=

∫ tmax

t

cdt′

a(t′)

ηmax is the final moment of conformal time. The physical size of the event horizon is:

Rev(t) = ca(t)

∫ tmax

t

dt′

a(t′)

If the universe expands forever then tmax =∞ and when K = 0,−1 then these are decelerating universesso that χev and Rev → ∞ and so there is no event horizon. If the universe is accelerating then Rev isfinite even for K = 0,−1 and so there is an event horizon. For a flat de Sitter expansion, a(t) ∝ eHt

where H is a constant then:

Rev(t) = ceHt∫ ∞t

dt′

eHt′

= cH−1

This is a size of order the scale of the curvature of the universe.

1.12 Luminosity Distance

In Minkowski space then the definition scale is no problem so if a source was emitting with absoluteluminosity Ls then the flux of light, F , received at a distance d is given by:

F =Ls

4πd2

10

In the expanding case then the true distance is not known and so the Minkowski result is used as thenew definition of distance, called the Luminosity Distance:

d2L =

Ls4πF

An object has an absolute luminosity Ls which is located at comoving distance χS from an observer atχ = 0. The energy of the light emitted within time interval ∆t1 is ∆E1 and it reaches the observer on asphere of radius χS as ∆E0 over a period ∆t0. ∆E ∝ ν and since c is invariant then c = ν1λ1 = ν0λ0.Relating this all to the red shift gives:

1 + z =λ0

λ1=ν1

ν0=

∆t0∆t1

=∆E1

∆E0

The relation between the frequency and the time interval, ν0∆t0 = ν1∆t1, has been used. The luminositiesare:

Ls =∆E1

∆t1and L0 =

∆E0

∆E1

These can be rearranged using the previous relations to give:

Ls = L0(1 + z)2

The two factors of (1 + z) arise from the fact that each photon loses energy and that the number ofphotons arriving each second decreases over time as the universe expands.

Light travels along the χ direction (dϑ = dφ = ds = 0) then:

χs =

∫ χs

0

dχs

=

∫ t0

0

cdt

a(t)

=c

a0

∫ z

0

dz′

H(z′)

This relation has been obtained from z = −H(1 + z) and 1 + z = a0/a. The area of a sphere at t = t0 isgiven by S = 4π(a0Sk(χs))

2 as seen on page 4 and from this result the observed flux is:

F =L0

4π(a0Sk(χs))2

The luminosity distance d2L = Ls/F can now be found by substituting in relations to get:

dL = a0χs(1 + z)

The consequence of this relation is that distant objects appear further away than they really are becauseof redshift. Comparing the luminosity distance to the physical distance is:

dphys = a(t)

∫ χs

0

dχ

= a0χS

So dL = dphys(1 + z) which means that for z 1 then dL ≈ dphys.

1.13 Angular Diameter Distance

Objects of a given physical size, ` are assumed to be perpendicular to an observers line of sight so thatif they subtend an angle ∆ϑ (which is small in astronomy) then the angular diameter distance is:

ddiam =`

∆ϑ

11

The line element in comoving coordinates is:

ds2 = a2(η)[−dη2 + dχ2 + Sk(χ)(dϑ+ sin2 ϑdφ2)

]At the particle horizon then Sk(χp) = 2/(a0H0Ω0), which is true for any value of Ω0. A dust dominateduniverse is considered so that Ω0 ≈ Ωm. When it comes to the angular diameter distance then only theredshift z of the galaxy when the light was emitted which is being detected.

SK(χemit(z)) =2

Ω20a0H0(1 + z)

(Ω0z + (Ω0 − 2)

(√1 + Ω0z − 1

))Taking the large z limit then SK(χp) is recovered. The large z limit is further back in time. If there isan extended object of given transverse size ` at a comoving distance χemit then it can be aligned so thatφ =constant. Photons are emitted from both sides of the object at temit and propagate along the radialgeodesic and arrive at t0. These photons arrive at an apparent angular separation ∆ϑ. When the lightis emitted then dt = dφ = dχ so the interval between emission events at the end points:

` =√ds2

= a(temit)Sk(χemit)∆ϑ

The angular diameter distance is therefore:

ddiam =`

∆ϑ

= a(temit)Sk(χemit)

= a0(1 + z)−1Sk(χemit)

= dL(1 + z)−2

This means that dL = (1 + z)2ddiam. This is true for all curved spaces. ∆ϑ has some unusual features:

∆ϑ =`

a(temit)Sk(χemit)

=`

a(η0 − χemit)Sk(χemit)

If an object is nearby such that χemit η0 then a(η0 − χemit) ≈ a(η0) and also Sk(χemit) ≈ χemit for allcurvature. It follows from these two results thats:

∆ϑ ≈ `

a(η0)χemit

=`

dphys

This is as expected. Far away objects which are near the particle horizon then η0 − χemit η0 soa(η0 − χemit) a(η0) and Sk(χemit)→ Sk(χp) = 2/(a0Ω0H0) =constant. This means that:

∆ϑ ∝ `

a(η0 − χemit)

As the object gets further away it looks as if it is getting bigger, i.e. ∆ϑ increases with distance. As theobject approaches the particle horizon then then image will cover the whole sky.

The angular size of objects peak nearby but also far away. This can be understood because the universewas smaller in the past so a given physical sized object appears larger.

For a matter dominated flat universe then the angular diameter distance peaks at a redshift of z = 5/4.Using 1 + z = a0/a(temit) then the angular diameter distance is:

∆ϑ =(1 + z)`

a0Sk(χemit(z))

12

χemit can be written in terms of z using dz = (1 + z)H(t)dt:

χemit(z) = η0 − ηemit

=

∫ t0

temit

cdt

a(t)

=1

a0

∫ z

0

cdz

H(z)

Also from this the age of the universe, depending on choice of cosmology, can be calculated:

t(z) =

∫ ∞z

dz

(1 + z)H(z)

A flat dust filled universe, where Sk(χemit) = χemit and Ω0 = Ωm0, has H(z) = H0(z)Ωm0(1 + z)3/2 so:

t(z) =2

3H0

1

(1 + z)3/2and χ(z) =

2c

a0H0

(1− 1√

1 + z

)and ∆ϑ =

`H0

2c

(1 + z)3/2

(1 + z)1/2 − 1

Here ∆ϑ has a minimum at z = 5/4. This corresponds to the maxima in the angular diameter distance.

13

Chapter 2

Thermal History

Assume there is a statistic equilibrium then there will be a number density n, an energy density %c2 anda pressure p, all arising from a distribution function. One type of particles, species A, has a distributionfunction fA(p˜, t) where p˜ is the 3-momentum.

Different species interact with rate Γ(t) > H(t). The interactions are thought to be short range, soprovide the mechanism to maintain thermal equilibrium.

2.1 Ideal (Bose and Fermi) Particles

The equilibrium distribution function is:

fA(p˜, t)d3p˜ =gA

(2π)3

d3p˜eβ(EA−µA) ± 1

β = (kBT )−1 where kB is Boltzmann’s factor, gA is the spin degeneracy which gives the number of

relativistic species at any given temperature TA. µA is the chemical potential and EA(p˜0 =√p˜2c2 +m2c4.

the + sign refers to fermions and the - sign to bosons. The number of particles is conserved so µA = 0.From the distribution function it follows that the number density is:

n =1

~3

∫f(p˜)d3p˜

=g

2π2c3~3

∫ ∞mc2

(E2 −m2c4

eβE ± 1EdE

The energy density is:

%c2 = En

=1

~3

∫E(p˜)f(p˜)d3p˜

=g

2π2c3~3

∫ ∞mc2

(E2 −m2c4

eβE ± 1E2dE

And the pressure is given by:

p =1

~3

∫ ∞mc2

|p˜c2|3E(p˜)f(p˜)d3p˜

=g

6π2c3~3

∫ ∞mc2

(E2 −m2c4)3/2

eβE ± 1dE

14

2.1.1 Relativistic Species

In the relativistic regime then mc2 kBT . Introducing x = βE then the number density is:

n =

(kBT

c

)3g

2π2~3

∫ ∞0

x2dx

ex ± 1

(Bosons) =

(kBT

c~

)3gζ(3)

π2

(Fermions) =

(kBT

c~

)3gζ(3)

π2

(1− 3

4

)

=3

4nB

∝ T 3

The ζ function is given by:

ζ(n) =

∞∑n=1

m−n

=1

Γ(n)

∫ ∞0

un−1du

eu − 1

The fermions can be written as two species of bosons at different temperatures:

1

ex + 1=

1

ex − 1− 2

e2x − 1

The CMB photons have a number density which can be found using the bosonic equation. Since T ≈2.725K and g = 2 for photons then nγ = 4.1× 108m−3. The energy density for bosons is:

%Bc2 =

π2

30c3~3g(kBT )4

And for fermions is:

%F c2 =

7

8%Bc

2

These are both proportional to T 4. The pressure is given by p = %c2/3. Since %B ∝ T 4 and %B ∝ a−4 forrelativistic particles then it can be seen that the temperature is inversely proportional to the scale factorT ∝ 1/a.

2.1.2 Entropy of the Background

Entropy and energy are both extensive quantities, so are proportional to the number of particles in thesystem, such that:

∂S

∂V=S

Vand

∂E

∂V=E

V

Starting with the first law of thermodynamics and substituting in dE and dS gives:

dE = TdS − pdV

∂E

∂TdT +

∂E

∂VdV = T

(∂S∂T dT + ∂S

∂V dV)− pdV

This is true for arbitrary dT and dV and so collecting the terms gives:

dT :∂E

∂T= T

∂S

∂T

dV : S =E + pV

T

15

In the relativistic limit where kBT mc2 then the entropy density is obtain by substituting in the resultsfrom section 2.1.1:

s ≡ S

V

=4

3

%Bc2

T

=2π2kB45c3~3

g(kBT )3

There would be a factor of 7/8 for fermions. The number density, nB ∝ T 3 for relativistic particles andsince s scales in the same way then it also counts the number of particles. This is why cosmologists saythat the ration nγ/nbaryon is the entropy per baryon. nγ/nbaryon ≈ 109 today.

2.1.3 Non-Relativistic (Massive) Species

The non-relativistic (massive) species have mc2 kBT . In this case the number density is:

n ≈ g

~3

(mkBT

2π

)3/2

e−βmc2

It is Boltzmann suppressed. The energy density is % = mn and the pressure is p = n(kT ). This ismuch less than the energy density so p ≈ 0. The baryon to photon ratio is η = nγ/nbaryon. The photonnumber is nγ ≈ 411cm−3 and the baryon number is nB ≈ %B/mB ≈ ΩB%c/mB . The critical density is%C ≈ 1.9× 10−26kgm−3 and the mass of a proton is mp ≈ 1.67× 10−24g so that:

η ≈ 5.5× 10−10

(ΩBh

2

0.02

)This means there is of the order of one billion photons for every baryon. The number of baryons istherefore nB ≈ 0.22m−3.

2.1.4 Several Relativistic Species (Degrees of Freedom)

The g factor used throughout section 2.1 describes the number of species. If there is a collection ofmultiple species of relativistic particle each at equilibrium at a different temperature, Ti, then the totalenergy is:

%Relc2 =

k4BT

4γ

2π2~3c3

∑i

gi

(TiTγ

)4 ∫ ∞xi

u3du

eu ± 1

Tγ is the temperature of the CMB photons. The total energy density can be written is a compact fromby doing the integral in the limit xi → 0 giving ζ(4):

%Relc2 =

(kTγ)4

30~3c3π3g∗

g∗ is the effective number of degrees of freedom:

g∗ =∑

bosons

gi

(TiTγ

)4

+7

8

∑fermions

gi

(TiTγ

)4

As the temperature of the background photons decreases the effective number of degrees of freedom inradiation also decreases since massive particles become non-relativity, mic

2 > kTi.

T 1MeV

The only relativistic particles are the three neutrinos and the photon. The photon is a boson and has twopolarisation states (two degrees of freedom) and the neutrinos are fermions with two degrees of freedom

16

(spin up and spin down). Neutrinos at this scale are decoupled from the thermal bath and are slightlycolder Tν ≈ (4/11)1/3Tγ . The effective number of degrees of freedom is:

g∗ = 2 +7

3× 3× 2×

(4

11

)4/3

≈ 3.36

1Mev < T < 100MeV

The electron and the positron have masses ∼0.5MeV and so both of these are relativistic. The differencebetween Tγ and Tν is due to the electron-positron annihilation, but when the electrons and positrons arerelativistic then this process shuts down and so Tγ = Tν . The electrons now need to be included in thecalculation of the effective number of degrees of freedom.

g∗ = 2 +7

8(3× 2 + 2× 2)

= 10.75

T ≤ 300GeV

At high energies such as at the electroweak unification at the scale of the standard model particles theng∗ ≈ 106.75.

2.2 Temperature vs. Time

For photons then the radiation density parameter is:

Ωr =%r%c

=π2

30c5~3%cg(kT )4

≈ 2.47× 10−5h−2

The neutrino density parameter is:

Ων ≈ 3× 7

8×(

4

11

)4/3

Ωr

≈ 0.68Ωr

≈ 1.68× 10−5h−2

If the only relativistic particles today are photons and neutrinos then the relativistic contributions are:

Ωrel = Ωr + Ων

≈ 4.15× 10−5h−2

Since Ωmo ≈ 0.25 then all the relativistic particles are negligible compared to non-relativistic matter.The evolution of these quantities can be determined using %rel ∝ a−4 and %m ∝ a−3:

Ωrel

Ωm=

4.15× 10−5

Ωm0h2

(a0

a

)=

4.15× 10−5

Ωm0h2(1 + z)

17

There is a value of z when Ωrel = Ωm which is a period called equality. This redshift is at:

(1 + zeq) = 24074(Ωmoh2)

(T0

2.725K

)4

For T > Teq then the universe is radiation dominated and when T < Teq then the universe is matterdominated.

Decoupling is a period when the formulation of the CMB occurred because of the photons decouplingfrom the electrons as atoms formed.

(1 + zdec) ≈ 103

2.3 Density of the Universe and Dark Matter

The critical density today is %c(t0) ≈ 1.88h2 × 1026kgm−3. For stars then Ωstars ≈ 0.005 → 0.01, forbaryons from nucleosynthesis (tightest constraints) 0.016 ≤ ΩBh

2 ≤ 0.024. For h ∼ 0.7 then ΩB ≥0.03 Ωstars which implies there is baryonic material in the universe which is not visible in stars.

The dynamics of galaxies such as rotation curves give a density parameter Ωhalo. Going further to clustersof galaxies and looking at the virial dynamics and peculiar velocities then Ωm0 ∼ 0.3h1/2. The currentbest bound on the contribution of matter today is Ωm0 ∼ 0.25. There is also gravitational lensing ofdistant quasars which gives Ωm0 ∼ 0.25. Numerical simulations can be done and tend to suggest thatΩΛ0 ∼ 0.2 to get the models to work.

All of these suggest that ωm0 < 1. Ω0 is of the order unity and so there must be some dark energy outthere. CMB anisotropies are used to probe the geometry of the universe and hence obtain a value ofΩ0. The WMAP7 data suggests that ΩK = −0.015+0.02

−0.016 and so is consistent with a flat universe. When

combined with the Hubble space telescope data then ΩK = −0.01+0.016−0.009 and Ωm0 = 0.72 ± 0.04. Today

it is believed that ΩK ≈ 0, Ωm0 ∼ 0.3 0.04 from nucleosynthesis and so most of the matter is in darkmatter. ΩΛ0 ∼ 0.7 is the dark energy. More evidence of dark matter is present in the Bullet Cluster.

2.3.1 Dark Matter Candidates

There are three basic types of dark matter candidates:

Hot Dark Matter The particles decouple from the background radiation when they are relativistic.Their number density is nHDM ∼ nγ where the typical mass scale is m ∼ O(ev) such as neutrinos.Ωrel ∝ mν . These are not currently favoured since free-streaming leads to erased structures on largescales.

Weak Dark Matter These relic particles decouple early enough such that there is a similar abundanceto photons but can be boosted by annihilations other than just through electron-positron annihilation.There are ∼ 100 distinct particles species which have mWDM ∼ 1 − 10keV . These fell out of favourbecause of free streaming but they have recently come back in vogue as a way of addressing dwarf galaxyproblems in large scale structure formation.

Cold Dark Matter The particles decouple when they are non-relativistic. The number density istherefore suppressed nCDM ∼ e−βmc

2

. The freeze out temperature is ∼ O(MeV ) where the mass scalesof the particles are mCDM ∼ 10GeV → O(TeV ). The main candidates are called WIMPs (weaklyinteracting massive particles) and come from supersymmetry, i.e. neutralinos or gravitinos. There arealso axions which are a class of particles invented to solve the ϑ problem in QCD, but can also be usedas dark matter candidates. They have masses 10−6eV < mAxion < 10−3eV . Primordial black holes havea mass 1015g ≤ m ≤ 1026g where the lower bound comes from the black hole evaporation rate and theupper bound from microlensing. These have not been ruled out.

Maybe dark matter and dark energy do not exist and what is really happening is a modification of gravityon large scales. These ideas still need to be explored.

18

Chapter 3

Cosmological Perturbation Theory

The idea behind perturbation theory is to use Taylor expansion and use the leading orders in the theory.If there is a function at a point X0, f(X0) = f0 then a point close to this is given by f(X) = f) + f1Xwhere f1X is small. The full set equations would have FWR background plus a small fluctuation anddropping second order (fluctutation)2 terms.

Units Natural unit are used so that c = ~ = kB = 1. Using this all units can be measured in units ofenergy.

[Energy] = [Mass] = [Temperature] =1

[Length]=

1

[Time]

The unit of energy which will be used is the GeV∼ 1.6× 10−10J.

1GeV ≈ 1.56× 1038Mpc−1

1pc ≈ 3 light years1kg ≈ 5.6× 1026GeV

≈ 8.77× 1064Mpc−1

1K ≈ 8.6× 10−14GeV≈ 1.3× 1025Mpc−1

1m ≈ 5.07× 1015GeV−1

≈ 3.24× 10−23Mpc1s ≈ 1.5× 1024GeV−1

≈ 9.7× 10−15Mpc≈ 2.997925× 108m

Hubble’s constant is:

H0 = 100hkm s−1Mpc−1

=100h× 103m

3× 108m Mpc

∼ 3.33h× 10−4Mpc−1

∼ 2.1h× 10−42GeV

The current size of the universe is:

1

H0≈ 1

3 × 104Mpc

≈ 3000Mpc

3.1 Newtonian Gravity

F˜ = −Gm1m2

r2r

19

To work out the dynamics of the universe then the force caused by each particle needs to be taken intoconsideration. This is obviously impossible. Tho rectify this then the particles are modelled as a fluid. Ifthe total volume is V and there is a small volume δV which still contains a large number, N , of particlesthen the mass density is:

%δV = limδV→0

Nm

δV

By doing this for many small volumes to make up the total universe then the energy density is obtained%(t, x˜). Each particle has its own velocity and so taking a small volume of fluid then average out thevelocities of all the particles to get the fluid velocity v˜(t, x˜). The pressure can be defined in a similar way,%(t, x˜).

3.1.1 Poisson Equation

The force between the two regions is calculated using Newtons gravitation equation:

δ2F˜ = −Gδmδmi

|r˜− r˜ i|2 (r − ri)

This is now done for all of the particles effecting just the particle at r. The force density can also bedefined f˜ = δF˜/δV :

δF˜ = −G∑i

δmδmi

|r˜− r˜ i|2 (r − ri)

= f˜δVSince %(t, r˜)δV = δm then:

f˜ = −G∑i

%(t, r˜)%(t, r˜ i)|r˜− r˜ i|2 δVi(r − ri)

Take the small volume limit so that δV → d3r and the sum becomes an integral then:

f˜ = −G%(t, r˜)∫d3r′

%(t, r˜′)|r˜− r˜′|2 (r − r′)

This unit vector can be made into a scalar using the the gradient operator:

∇|r˜− r˜′| =

∂x∂y∂z

√(x− x′)2 + (y − y′)2 + (z − z′)2

=1

2√

(x− x′)2 + (y − y′)2 + (z − z′)2

2(x− x′)2(y − y′)2(z − z′)

=

r˜− r˜′|r˜− r˜′|= r − r′

This means that:r − r′

r˜− r˜′ = −∇(

1

|r˜− r˜′|)

The force density can therefore be written as:

f˜(t, r˜) = G%(t, r˜)∫d3r′%(t, r˜)∇

(1

|r˜− r˜′|)

20

This is equivalent to Newtons second law, F˜ = m · a˜→ f˜ = %(t, r˜) · g˜. The acceleration due to gravity is:

g˜(t, r˜) = G

∫d3r′%(t, r˜)∇

(1

|r˜− r˜′|)

The Newtonian potential is defined as φ(t, r˜) and since˜= dφ/dr then:

φ(t, r˜) = −G∫d3r′

%(t, r˜)|r˜− r˜′|This means that g˜ = −∇φ. Taking the scalar product of this gives:

∇ · g˜ = −∇2φ

= G

∫d3r′%(t, r˜)∇2

(1

|r˜− r˜′|)

= G

∫d3r′%(t, r˜)(−4πδ3(r˜− r˜′)

= −4πG%(t, r˜)This is the Poisson equation:

∇2φ = 4πG%

This is the analogue of Newtons law for continuous media.

3.1.2 Fluid Equations

The mass in a volume V is time dependent:

m(t) =

∫d3r%(t, r˜)

The rate of decrease of this mass out of the volume is therefore:

−dmdt

= −∫d3r

∂%

∂t

The change in mass is related to the flux of particles into or our of the surface of V . This depends onthe velocity of the particles. The flux is maximal when the velocity vector is parallel to the normal tothe surface of the volume, ds˜, and is zero when the flux is perpendicular. This suggests that the flux isproportional to v˜ · d− s˜. Using dimensional analysis then it can be seen that:

[v˜] = LT−1

= 0

[ds˜] = L2

[v˜ · ds˜] = L2

[%] = ML−3

= L−4

[∂m

∂t

]= ML−1

= L−2

So the required quantity for the rate of change in mass is %v˜ · ds˜ since its dimension is L−2. Integratingthis over the surface gives: ∫

S

%v˜ · ds˜ = −∫V

d3r∂%

∂t

21

Using Stokes theorem then: ∫S

%v˜ · ds˜ =

∫V

d3r∇ · (%v˜)Substituting this into the previous equation gives:

−∫V

d3r∂%

∂t=

∫d3r∇ · (%v˜)

Since this holds for any volume then:

∇ · (%v˜) +∂%

∂t= 0

Using the product rule it can be seen that this is just the conservation of mass equation:

0 =D

Dt%+ %∇ · v˜

=∂%

∂t+ v˜ · ∇%+ %∇ · v˜

Here D/Dt is the convective or substantial derivative. Newtons second law of motion can be used to getthe rate of change of velocity. The second law written in terms of the force per unit volume is:

f˜ = % · a˜The velocity now depends not only on time but also on position and so the new form of a˜ needs to befound. Consider A(t, r˜) then a small change in A is given by:

δA =∂A

∂tδt+

∂A

∂xδx+

∂A

∂yδy +

∂A

∂zδz

Now divide by δt and take the limit where t→ 0 so δt→ dt:

dA

dt=

∂A

∂t+∂A

∂x

dx

dt+∂A

∂y

dy

dt+∂A

∂z

dz

dt

=∂A

∂t+ v˜ · ∇A

This is just the convective derivative. Using this it can be seen that a˜ = Dv˜/Dt and so the force densityis:

f˜ = % · a˜= %

[∂v˜∂t

+ v˜ · ∇v˜]

The force density due to gravity is:

f˜ grav = % · g˜= −%∇φ

The pressure is the hydrodynamical force and is given by the force per area. The force on a slab whosetop has a point A and whose bottom has a point B is ∆Fz = FA−FB . The force at A is the pressure atA per area FA = pA∆A = pA∆X∆Y and the force of B is FB = pB∆X∆Y . So ∆F = (pA−pB)∆X∆Y .For a small ∆Z then the pressures at A and B can be Taylor expanded:

pB = pA +∂p

∂z∆z

The difference in the forces in the z direction is:

∆Fz = −∂p∂z

∆X∆Y∆Z

= −∂p∂z

∆V

22

Repeating for the forces along the x and y directions reveals that the total ∆F is given by:

∆F = ∆Fx + ∆Fy + ∆Fz

=

(−∂p∂x− ∂p

∂y− ∂p

∂z

)∆V

= −∇p∆V

So the force density is just:f˜ hydro = −∇p

The total force density is finally:

f˜ = −%∇φ−∇p

= %[∂v

∂t + v˜ · ∇v˜]The Euler equation is then obtained:

∂v˜∂t

+ v˜ · ∇v˜+∇pp

+∇φ = 0

The three equations which describe fluids are:

0 =∂v˜∂t

+ v˜ · ∇v˜+∇pp

+∇φ

0 =∂%

∂t+ v˜ · ∇%+ %∇ · v˜

4πG% = ∇2φ

There are three equations for four unknowns, %, p, v˜ and φ. What is needed is an equation of state to

relate p and %. They are related by the speed of sound, c2s:

∇p = c2s∇%

In this case Eulers equation becomes:

∂v˜∂t

+ v˜ · ∇v˜+c2s∇%%

+∇φ = 0

These are impossible to solve analytically. Certain approximations need to be made.

3.2 Structure Formation

3.2.1 Static Universe

First assume that the fluid is homogeneous such that any ∇ terms are zero. This leaves ∂%/∂t = 0 so that% = % is constant in time and space. The same is also true for the velocity. In fact a frame of referencecan always be chosen where v˜ = 0. Now assume that %(t, x˜) ≈ %+ δ% where δ% is some small fluctuationand the velocity is v˜(t, x˜) ≈ v˜+ δv˜ ≈ δv˜. Define the density constant δ = δ%/%. If δ > 0 then there is aconcentration of particles and if δ < 0 then there are fewer particles that average. Plugging these termsinto the fluid equations and ignoring any terms quadratic or higher gives:

0 =∂δv˜∂t

+c2s%∇δ%+∇φ

=∂δv˜∂t

+ c2s∇δ +∇φ

0 =∂δ%

∂t+ %∇ · δv˜

= δ +∇ · δv˜23

4πG% (1 + δ) = ∇2φ

Only the small term of this equation is needed since the first term is large. Eliminating the velocity bytaking the time derivative of the continuity equation and substituting in ∂v˜/∂t gives:

δ +∂

∂t(∇ · v˜)

= δ +∇ ·−c2s∇δ −∇φ

= δ − c2s∇2δ −∇2φ

= δ − c2s∇2δ − 4πG%δ

The first two terms are the components of a wave equation with speed cs. The last term is the gravitationalterm. Oscillations are expected. If the pressure dominates then there is no collapse but if gravitydominates then the fluid collapse into bound structures.

Fourier transforms are given by:

A(t, x˜) =

∫d3k

(2π)3eik˜·x˜A(t, k˜)

A(t, k˜) =

∫d3xe−ik˜·x˜A(t, x˜)

The Dirac δ-Function can be defined using the Fourier transforms:

A(t, x˜) =

∫d3k

(2π)3eik˜·x˜A(t, k˜)

=

∫d3k

(2π)3eik˜·x˜∫ d3x′e−ik˜·x˜′A(t, x˜′)

=

∫d3x′

∫d3k

(2π)3eik˜·(x˜−x˜′)A(t, x˜′)

= d3x′δ3(x˜− x˜′)A(t, x˜′)= A(t, x˜)

The Fourier transforms of the δ-Function is:

δ3(x˜− x˜′) =

∫d3k

(2π)3eik˜·(x˜−x˜′)

1 =

∫d3xe−ik˜·x˜δ3(x˜)

The Fourier transforms can be used to make the partial differential equations for the density contrastinto an ordinary differential equation. This is done by finding what ∇2δ is in Fourier space:

∇2δ(t, x˜) =

∫d3k

(2π)3∇2[eik˜·x˜] δ(t, k˜)

=

∫d3

(2π)3eik˜·x˜(−k2)δ(t, k˜)

The density contrast equation can be written in Fourier space as:

%+ (k2c2s − 4πG%)δ = 0

24

This is simply harmonic motion with frequency:

ω = k2c2s − 4πG%

When ω2 > 0 then there are two oscillatory solutions δ0 = cos(ωt) and δ1 = sin(ωt). When ω2 < 0 thenthe solutions are exponential, δ0 = eωt and δ1 = e−ωt. δ1 decays away quickly but δ0 leads to exponentialcollapse. When ω = 0 then c2skJ = 4πG% so:

kJ =1

cs

√4πG%

k is a frequency and so a wavelength can be defined using λ = 2π/k:

λJ =2πcs√4πG%

= cs

√π

G%

This is the Jeans length. If a collection of particles are in an area of side length L then when L > λJthen there is collapse and when L < λJ then the system is stable.

3.2.2 Expanding Universe

These structure formation conditions are only true for a static universe. In an expanding universe thenthe background velocity cannot be zero. Now:

v˜ = H(t)x˜Here H(t) is some function of time which describes the expansion of the universe. Now the fluid equationscan be found is exactly the same way but using:

v˜ = Hx˜+Hx˜x˜ = 0 since A˜ · ∇x˜ = A˜ for any vector A˜(t, x˜). To see this take:

A˜ =

AxAyAz

and x˜(xyz)

Now carry out the calculation:

A˜ · ∇A˜ =

Ax∂xAy∂yAz∂z

xyz

=

AxAyAz

= A˜

So if A˜ = v˜ it can be seen that v˜ · ∇x˜ = v˜ and as such:

v˜ =dx

dt

=∂x˜∂t

+ v˜ · ∇x˜= (0) + v˜= v˜

25

Now plugging the background velocity, v˜ into the the continuity equation gives:

∂%

∂t+ %∇ · v˜ = %+ %∇ · (Hx˜)

= %+ %H∇ · x˜= %+ 3H%

= 0

The Euler equation is now:

∂v˜∂t

+ v˜ · ∇v˜+∇φ = Hx˜+H2x˜+∇φ

= 0

∇φ = −(H +H2)x˜This can now be substituted into Poisson’s equation:

∇2φ = −3(H +H2)

= 4πG%

The rate of change of H is therefore:

H +H2 +4πG%

3= 0

Now looking at small fluctuations again %(t, x˜) = %(t)+δ% where δ = δ%/% with % = %(1+δ) and v˜ = v˜+u˜where u˜ is small and v˜ = H(t)x˜. The potential is also φ = φ+ δφ. This reveals the equations:

Continuity Equation δ +∇ · u˜+Hx˜ · ∇δ = 0

Euler Equation u˜+H(u˜+ x˜ · ∇u˜) = −c2sδ −∇δφ

Poisson’s Equation ∇2δφ = 4πG%δ

Again going to Fourier space makes analysis simpler, but because everything has time dependence a newcomoving coordinate, r˜ is introduced such that:

dr

dt=

∂r˜∂t

+ v˜ · ∇r˜=

∂r˜∂t

+Hx˜ · ∇r˜= 0

An ansatz for r˜ would be r˜ = x˜/a(t) so that:

dr

dt= − a

a2x˜+Hx˜ · ∇(xa)

=x

a

(− aa

+H

)= 0

This means that H = a/a. The Fourier transforms are made using r˜:δ(t, r˜) =

∫d3k

(2π)3eik˜·r˜δ(t, k˜)

=

∫d3k

(2π)3e(ik˜·x˜)/aδ(t, k˜)

26

Now introducing a new coordinate for conveinience, ϑ˜ = u˜/a the liearised equations are:

Continuity Equation δ + ik˜ · ϑ˜ = 0

Euler Equation ϑ˜ + 2Hϑ˜ = − ika2

[c2sδ − δφ

]Poisson’s Equation −k2δφ = 4πG%a2δ

The %˜ variable can be eliminated by multiplying the Euler equation by ik˜ and substituted into the partial

derivative with respect to time of the continuity equation:

δ + 2Hδ +

(c2sk

2

a2− 4πG%

)δ = 0

If H = 0 and a = 1 then the static case is recovered. In general then this equation is a damped harmonicoscillator equation where 2Hδ is the damping term. The frequency is:

ω2 =c2sk

2

a2− 4πG%

ω2 > 0 Stable System, Oscillatesω2 < 0 Instability, Collapses

Setting ω = 0 means a Jeans length can be defined using:

kJ =

√4πGa2%

c2s

=2π

λJ

λJ =csa

√π

G%

When L > λJ then there is collapse under gravity. This Jeans length is time dependent and so a statewhich is initially stable can eventually begin to collapse, depending on a.

For the collapsing case where ω2 < 0 then c2sk2/a2 is negligible and so the equation is:

δ + 2Hδ − 4πG%δ = 0

In matter domination then % ∝ a−3, a ∝ t2/3 and H = 2/(3t) so that:

δ +4

3tδ − 2

3t2δ = 0

The solution to this equation is of the form δ = δ0tn so using n(n − 1)tn−2 + 4/(3n)tn−2 − 2/3t−2 = 0

gives 3n2 + n− 2 = 0 which can be solved and gives solutions n = −1 and n = 2/3. When n = −1 thenδ ∝ t−1 and so decays and is unimportant. When n = 2/3 then δ ∝ t2/3. During the matter era thenfluctuations grow as t2/3 ∼ a. The collapse has slowed down in an expanding universe but collapse stilloccurs.

In radiation domination then the density fluctuations in matter is given by:

δ +δ

t= 0

This is because 4πG%δ H2 and a ∝√t. A trivial solution to this is when δ is constant. Another

solution is δ = δ0 + δ1 ln t and so collapse happens but extremely slowly. This is called the Meszaroseffect. It means that any structure formation in the universe must have happen during matter dominationreally.

27

3.3 Relativistic Perturbation Theory

Newtonian perturbation theory is good for pressureless matter for scales much smaller than the cosmo-logical horizon. This means that it can be used to calculate structure on sub-horizon scales.

Relativity is needed to study the CMB, inflation and dark energy. To start working out the relativisticperturbation theory then the metric is needed:

gµν = a2(η)[−dη2 + γijdx

idxj]

Here γij are the background variables which satisfy γajγjb = δab . The overline is use to denote that it isin conformal time, η. The metric can be placed into the Einstein equation:

Gµ

ν = 8πGTµ

ν

This then gives the two Friedmann equations where H = a′/a and a′ = ∂a/∂η. Small fluctuations canbe added to gµν to give:

gµν = gµν + δgµν

=

(−1 ·· δij

)+

(δg00 δg0i

δgij

)δg00 is a scalar, δg0i is a three-vector and δgij is a tensor. This is called the 3+1 decomposition. Ingeneral relativity then the physics cannot depend on the coordinates system and so four components ofδgµν can be fixed. The metric has 10 components since it is symmetric. These can be reduced by fixingsome of the modes.

3.3.1 Scalar-Vector-Tensor Decomposition

Since δgoi is a vector then in can be denoted vi. This can then be written in terms of a scalar mode anda vector mode:

∇i = ∇iv + vi

∇iv is the scalar mode and vi is the pure vector mode. The tensor δgij can be decomposed into:

dgij = hγij +

(∇i∇j −

1

3∇2γij

)A+∇ifi +∇j fi + hTij

γij are the background variables and h is a scalar mode and A is another scalar mode where the thirdcomes from making this term cancel out when contracting with γij . The vector modes have the conditionthat ∇ifi = 0. The vector mode has two degrees of freedom since one is removed by this condition.The scalar fields have one degree of freedom each. The tensor has six degrees of freedom but this canbe reduced by making it traceless, γijhTij = 0 and stating that its divergence is zero, ∇ihTij = 0 and sothe degrees of freedom are reduced to two meaning δgij has the expected six degrees of freedom. Tensormodes are pure gravitational waves.

In total δgµν has four scalar modes:

δ00 , δ0i = ∇iB , δgij = hγij +

(∇i∇j −

1

3∇2γij

)A

There are also four vectors and two tensors. Two of the scalars and two of the vectors can be removedby covariance and so that there are now six degrees of freedom. The scalars which are fixed are δ0i and(∇i∇j − 1/3∇2γij

)A. The effect of the vectors is so small that they are negligible so they can be ignored

and so there are now four degrees of freedom. The tensors cannot be removed. The remaining scalarmodes are:

δgscalarµν =

(−2ψ(t, x˜) ·

−2φ(t, x˜)γij

)

28

Here φ and ψ are gravitational potentials. The fluctuations in the energy momentum tensor can also befound in terms of Scalar-Vector-Tensor decomposition and gives:

δTµν =

−%δ −(%− p)∇iu

%

[1

3Πδij + (1 + w)Di

jσ

] −(% − p)∇iu is the velocity decomposed into a scalar mode and a vector mode where only the scalarmode is considered in δTµν . Π is the pressure contrast given by δp = %Π. σ is the shear and Di

j =

∇i∇j − 1/3∇2δij . The four scalars are therefore δ, u,Π and σ and none of these can be fixed.

3.3.2 Perturbative Einstein Equations

The four perturbative Einstein equations (with %δ the sum of all species, i.e. baryons, cold dark matter(CDM), photons, γ and neutrinos, ν) are:

δG00 : 2∇2φ− 6H(φ′ +Hψ) = 8πGa2%δ

δG0i : 2(φ′ +Hψ) = 8πGa2(%+ p)u

δGij :

φ′′ +Hψ′ + 2Hφ′ + (2H′ +H2 + 1/3∇2)ψ − 1/3∇2φ

φ− ψ

=

=

4πGa2%Π i = j

8πGa2(%+ p)σ i 6= j

These equations replace Poisson’s equation in Newtonian gravity. They can be simplified slightly bysubstituting in three times the δG0

i equation into the δG00 term to get:

∇2φ = 4πGa2% [δ + 3H(1 + w)u]

This is very similar to the Newtonian gravity Poisson’s equation but it is also sourced by the velocity.

Π = 0 for pressureless matter such as baryons, dark matter or cosmological constant and Π = 1/3δfor radiation like photons or neutrinos. The shear σ = 0 for baryons, cold dark matter or cosmologicalconstant but is not for photons and neutrinos. It is still very small and so when ignoring it then φ = ψ.

The conservation of mass from Newtonian perturbation theory now becomes ∇µTµν . When ν = 0 thenthe relativistic analogue of the continuity equation is:

δ′ = 3H(wδ −Π) + (1 + w)(∇2u+ 3φ′)

Taking ν = i then the relativistic analogue of Euler equation is:

u′ = −H(1− 3w)u+1

1 + w+

2

3∇2σ + ψ

There is additional sourcing from the shear and time dependence of the potentials. Solutions can be foundfor the potential and the matter variables. Assuming that there is only one fluid then the Friedmannequation in conformal times is:

3H2 = 8πGa2%

δ can be eliminated using the δG00 equation and the δGi i equation to get:

φ′′ + 3H(1 + w)φ′ + wk˜2φ = 0

This is valid for a single fluid with parameter w = 0 on all scales.

Starting with a given mode with wavenumber k˜ then it can be determined whether it is a superhorizonor subhorizon mode. The horizon is given by 1/H and so a superhorizon mode occurs when k Hand so the wavelength is the much greater than the length of the universe and subhorizon mode occurswhen k H such that the wavelength is much less than the length of the universe. This mode is timedependent in an expanding universe and so at early times the mode may be superhorizon and become

29

subhorizon after H ∼ k˜.

For superhorizon scales then the k˜2 term is negligible so that:

φ′′ + 3H(1 + w)φ′ = 0

There are two solutions, φ = constant and φ′ = A/a3(1+w). This can be integrated to get φ =

∫dη

a3(1+w).

This is a decaying mode and so only the φ = constant needs to be considered. This means on superhorizonscales φsup = φsup(k˜). The superhorizon scale total density becomes:

−6H(φ′ +Hφ) = 8πGa2%δ

Substituting in φsup gives:

−6H2φsup =8πG2%

3H2δ

Therefore δsup = −2φsup. The density contrast is constant on superhorizon scales. The total velocity onsuperhorizon scales is the same and so after substituting in φsup and δsup then:

u˜sup =2

3(1 + w)

φsup

H

Since H = 2/(1 + 3w)η then:

u˜sup =1 + 3w

3(1 + w)φsupη

These results give the conditions of curvature and adiabatic initial conditions. When η → 0 then usup → 0so all the particles at the start of the universe are comoving with the expanding universe. The fluctuationsare regular as η → 0.

If w = 0 (pressureless matter domination) then the solutions are the same as the superhorizon calculationseven on sub horizon scales. On small scales (subhorizon) then the Newtonian calculations are good.During the matter era then φ ∼ constant so that:

δ′ = −k˜2u˜ and u˜′ = −Hu˜+ φ

These are exactly the Newtonian equations of motion.

For two fluids, radiation and matter, then the total density, total velocity and total pressure become:

2∇2φ− 6H(φ′ +Hψ) = 8πGa2[%rδr + %mδm]

= 3H2 (Ωrδr + Ωmδm)

2(φ′ +Hψ) = 8πGa2(%r + %m + pr + pm)u

= 3H2

(4

3Ωrδr + Ωmδm

)φ′′ +Hψ′ + 2Hφ′ + (2H′ +H2 + 1/3∇2)ψ − 1/3∇2φ = 4πGa2%rδrΠ

Only one fluid dominates at a given time and so the solution for φ for a single fluid is still valid. Theadiabatic condition states that δm = 3δr/4. In the radiation era then δr = −2φsup and so δm = −3φsup/2and in the matter era then δm = −2φsup and so δr = −8φsup/3.

Superhorizon Superhorizon Subhorizon SubhorizonRadiation Era Matter Era Radiation Era Matter Era

φ constant constant oscillatory decay constantδm constant constant constant+log term grow ∝ η2

u˜m grow ∝ grow ∝ η decay grow ∝ η

30

η

δmφ

Equality

ηhor1

ηhor1

ηhor2

ηhor2k˜ 2

k˜ 1

k˜ 2

k˜ 1

This is only true for truly pressureless matter. The universe contains baryons, 75% H and 25% He. Athigh temperatures, i.e. in the early universe, then these are ionised and so are charged. This means thatthe baryons interact with photons via Compton scattering on the scale 1/m2

p as do the electrons with scale1/m2

e. Baryons also interact with electrons via Rutherford scattering. This is called the tight-couplingregime.

When the temperature drops then neutral atoms begin to form which do not interact with photons andso the Compton scattering switches off and so photons can travel without much interaction. This is calledthe free-streaming regime.

In tight coupling then baryons have pressure because they are just a constituent of a photon-baryon fluid.Since δp ∼ c2sδ% then there is a speed of sound c2s 6= 0.

In free-streaming then the baryons decouple and so there are two separated fluids and the speed of soundc2s → 0. This means that the baryons become pressureless. The Jeans scale is:

λJ =csa

√π

G%b(a)

During tight-coupling then λJ ∼ 1/H and so is of the horizon scale. The baryons form waves and cannotform structures. During free-streaming then the Jeans length λJ → 0 since cs → 0 and as such structurescan form.

Where the horizon crossing occurs sets the phase of the oscillations:

31

η

δb

Equality

ηhor1

ηdec

ηhor2

Oscillations should be imprinted in the galaxy distribution but this is not seen on the predicted scale.This is because in the real universe there is cold dark matter which is completely pressureless since itdoes not interact electromagnetically.

η

δb

Equality

Cold Dark Matter

ηdec

Baryons

The baryons fall in to potential wells, φ and so the oscillation pattern is mostly erased. Of course darkmatter feels some back reaction so there can still be seen some small oscillations.

3.3.3 Observing Structure

Using the galaxy distribution across the sky can give an idea of the structure.

Galaxies in a volume formed of the some box of area dA and some thickness given by the redshift, z, can

32

be counted to find a number density contrast:

δn(z, n) =N(z, n)−N(z)

N(z)

n is the direction of of the galaxies, N(z, n) is the number of galaxies in that direction and N(z) isthe mean number of galaxies at redshift z. The galaxy mass density contrast in this region is givenby δs(z, n) = b1δn where b1 is a phenomenological number which depends on the type of galaxy beingobserved. What is really needed is δg(η, r, n) where r is the position of the galaxies. Since galaxies are inthe past lightcone then η ≈ r. Now as the galaxies are being observed by redshift then a redshift spacedistortion needs to be taken into account. Galaxies have velocities which cause doppler shift. This meansthat the galaxies which are moving further away have a greater redshift and as such are closer than theyappear and galaxies moving towards are blue shifted and so are really further away than they appear.The redshift is therefore given by:

z = z + δz

≈ H0r + v˜ · n≈ H0

Here s is the redshift vector:

s = r +v˜ · nH0

As a vector then:

s˜ =

(1 +

v˜ · nH0r

)r˜

The volume should be the same whether in redshift space or normal space:

nsd3s = nrd

3r

What is found is that:

ns = nr

[1− 1

H0

∂

∂r(v˜ · n)

]The last term is small. The number density is therefore nr(r˜) = n(z)

(1 + δg(r˜)) and so the mass density

contrast of the galaxies is:

δs = δg(r˜)− 1

H0

∂

∂r

(v˜ · n)

Fourier transforming and looking in k-space gives:

δs(k˜) = δg(k˜) +µ2

H0k2v(k˜)

Where µ = k · z. The velocities appear to add to the density of the galaxies. Defining δ = k2v allows agrowth rate to be written:

f =d ln δ

d ln a

Now δs = b(1 + µ2β)δ(k˜) where β = f/b is the bias which is included since galaxies do not trace thedensity field exactly. The density contrast is then given by:

δ(k˜) = δ0(k˜ 0)T (k˜)

Here δ0(k˜) is an initial condition and T (k˜) is a transfer function which describes how to get from theinitial conditions to todays density contrast. The transfer function contains the density of baryons andcold dark matter:

δ(k˜) = δ0(k˜)(Ωbδb(k˜) + ΩCDMδCDM (k˜)

)δ(k˜) is a random variable and so 〈δ(k˜)〉 = 0. The power spectrum is what is actually observed:

P (k˜) = 〈δ(k˜)δ(k˜)〉

= P0(k˜)T 2(k˜)

33

Here P0(k˜) is the initial power. The power spectrum is sketched as:

k˜

P (k˜)Subhorizon

Superhorizon

Baryon Oscillations

3.4 Cosmic Microwave Background

The cosmic microwave background is made of photons and was first observed (accidentally) by Penziasand Wilson for which they won a Nobel prize. The CMB has a current temperature T0 ∼ 2.725K asmeasured by the COBE satellite in 1992. There are 10× 1012s−1cm−2 photons in the CMB. It is almostisotropic since it is the same temperature in all directions. The anisotropies are very small with a dipoleon the scale of ∼ 10−3 and higher multipoles ∼ 10−5. The CMB has a Planck spectrum which impliesequislibrium. The intensity is:

I(ν) =4πν3

e2πν/T − 1

The power per unit area can be calculated:

P = σT 4

σ is a constant. The CMB is the best example of a Plank spectrum in the universe. The fact that the pho-tons are in equilibrium means that they can be analysed using statistical mechanics. The Bose-Einsteindistribution is a function of phase space F (t, x˜, k˜). Since the universe is isotropic and homogeneous thenthe background is:

f(t, k˜) =2

(2π)3(e k˜/T − 1

)The energy-momentum tensor can be calculated using:

Tµν =

∫d3kf(t, x˜, k˜)

kµkνE

For the Friedmann-Robertson-Walker universe then this is:

Tµν =2

(2π)3

∫kµkνE

1

e k˜/T − 1

34

The energy density is given by %γ = T 00:

%γ =2

(2π)3

∫d3k

k

e k˜/T − 1

=8π

(2π)3

∫ ∞0

dkk3

e k˜/T − 1

=T 4

π2

∫ ∞0

dxx3

ex − 1

=π2

15T 4

=π2

15

T 40

a4

The pressure is obtained from T ij and is:

pγ =1

3%γ

Energy conservation is given by ∇µTµν = 0 and in FRW is given by %γ + 4H%γ = 0. This can be seento be the result of the Boltzmann equation, which also gives the evolution of the temperature. For FRWthen the Boltzmann equation is:

∂f

∂t−Hk∂f

∂k= 0

So the partial derivatives are:

∂f

∂t=

ek/T

(ek/T − 1)2

k

T 2

∂T

∂t

∂f

∂k= − ek/T

(ek/T−1)21T

Substituting these into the Boltzmann equation gives:

∂T

∂t+HT = 0

This implies that T = T0/a.

3.4.1 Anisotropies

The dipole is apparent on a scale of ∼ 10−3 and corresponds to v = (627 ± 22)km s−1 which is thevelocity the Earth is moving with respect to the CMB. Removing this dipole allows the higher multipoleanisotropies to be seen. These multipoles are seen by observing in direction n and then constructing T (n)for all directions. The average is then T = 〈T (n)〉 and the temperature anisotropy is then given by:

Θ(n) =T (n)− T

T

Θ(n) is position dependent so 〈Θ(n)〉 = 0. Spherical harmonics can be used to describe the temperatureanisotropy:

Θ(n) =∑lm

almYlm(n)

Where Ylm(n) is the spherical harmonic and l = 0, 1, 2, 3, · · · and m = −l, · · · , l. alm can be found using:

alm =

∫dΩnY

∗lm(n)Θ(n)

The spherical harmonic satisfies:[1

sinϑ

∂

∂ϑ

(sinϑ

∂

∂ϑ

)+

1

sin2 ϑ

∂2

∂φ2

]Ylm = −l(l + 1)Ylm

35

Legendre polynomials can be related to the spherical harmonics. Legendre polynomials are Pl(µ) whereµ = −1, · · · , 1 and satisfy:

(1− µ2)d2Pldµ2

− 2µdPldµ

+ l(l + 1)Pl = 0

Since for two directions n and hatn′ then n · n′ = cosϑ then it can be seen that:

Pl(cosϑ) =4π

2l + 1

∑m

Y ∗lm(n)Ylm(n)

For the CMB then Θ(n) =∑lm almYlm(n) is used and the average between Θ(n) and Θ(n′) is taken:

〈Θ(n)Θ(n′)〉 = C(Θ)

=∑lm

∑l′m′

Ylm(n)Y ∗l′m′(n′)〈alma∗l′m′〉

If the universe is statistically isotropic then 〈alma∗lm〉 = Clδll′δmm′ and so the correlator becomes:

C(Θ) =∑lm

ClYlm(n)Y ∗lm(n′)

=1

4π

∑l

(2l + 1)Cl

Cl is the angular power spectrum where C0 = 0 is the monopole, when l = 1 then C1 is the dipoles andthen the anisotropies are given by the l ≥ 2 terms. Each l has 2l + 1 samples.

3.5 Recombination

In tight-coupling then there is thermal equilibrium so that e− + p↔ H + γ. This means that:

nenpnH

=n

(0)e n

(0)p

n(0)H

Here ne and np are the free electrons and protons and n(0)i are the equilibrium number density:

n(0)i = gi

(miT

2π

)3/2

e−mi/T

The Boltzmann equation for massive particles is:

1

a3

d

dt(a3ne) = n(0)

e np(0)〈σv〉

(nH

n(0)H

− nenp

n(0)e n

(0)p

)

This is zero at equilibrium. The electron ionisation fraction is:

Xe =ne

ne + nH

=np

np + nH

Xe → 1 for the completely ionised case and Xe → 0 for the completely neutral case. When including theHe then Xe = 1.15 when completely ionised. Using this it can be seen that:

nenpnH

=n2e

nH

=

(meT

2π

)3/2

e−(me+mp−mH)/T

36

Here me +mp −mH is the binding energy εH ≈ 13.6eV. By measurement ne/nH = Xe/(1−Xe) whichreveals the Saha equation:

X2e

1−Xe=

1

ne + nH

(meT

2π

)3/2

e−εH/T

ne + nH = np + nH is the baryon density, nb ≈ 109T 3. Using this it can seen that X2e/(1−Xe) diverges

so Xe → 1. Beyond equilibrium then the Boltzmann equation needs to be solved:

1

a3

d

dt(a3ne) = nb〈σv〉

[(1−Xe)

(meT

2π

)3/2

e−εH/T −X2enb

]Then use ne = Xenb to get:

dXe

dt= (1−Xe)β −X2

enbα(2)

β is the ionisation rate:

β = 〈σv〉(meT

2π

)3/2

e−εH/T

α(2) is called the recombination rate which describes how electrons and protons form a neutral hydrogen:

a

XeWith Helium

Free Electronsand Protons

Decoupling

From Hydrogen

3.5.1 Decoupling

Decoupling occurs when neσT ∼ H. The number density of baryons is:

nb =%bmb

=3H2

0 Ω0b

mba3

This means:

neσT ∼ H

=σ0Ω0bh

2Xe

a3

The Friedmann equation gives:

H = H0

√Ω0m√a3

+

√1 +

aeq

a

37

Taking the ratio gives:neσTH≈ 0.069

Ω0bhXe√Ω0ma3

(1 +

aeq

a

)−1/2

Plugging in todays values gives aeq ≈ 3× 10−4 so:

neσTH≈ 117Xe

This means that decoupling happens when Xe ∼ 1/117. This means decoupling happens during theperiod of recombination.

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