Summary Lectures 1 to 7 - hu-berlin.defjeger/Cosmolect1-8.pdfCosmology Summary Lectures 1 to 7...

Cosmology

Summary Lectures 1 to 7

1. Cosmology as a observational science

Olbers Paradox, Cosmological Principle, Hubble Expansion, Age of the Uni-verse, Nuclear Age Dating

2. Spectral fingerprints, Cosmic Distance Scales, Variable stars, Birth and Deadof stars, Hertzsprung Russell Sequence (Stellar Evolution). The solar Neutrinoproblem and how stars produce energy

3. Modern Cosmology, Cosmological principle, Gravitation and Geometry, Hub-ble’s law and the universal expansion and the Big Bang, observational Horizons,Sketch of cosmological evolution

4. Special relativity in a Nutshell, Summary and concepts of general relativity, Sta-tus of experimental tests, Gravitational Lensing, Black Holes

c© 2009, F. Jegerlehner ≪x Lect. 8 x≫ 47

http://www-com.physik.hu-berlin.de/~fjeger/Cosmolect1-3.pdf


Cosmology

5. Cosmological Models, Robertson-Walker Universes, Weyl’s postulate and thecosmic time, the cosmic scale factor S (t), Measuring distances in curved space,Cosmological observables: Hubble constant H(t) and deceleration parameterq(t)

6. Relativistic Hydrodynamics, The Energy-Momentum Tensor of the universes, dy-namics of the universe: Solving Einstein’s Equation for the isotropic Universe:Friedmann’s Equations, Equations of state and energy balance. Simplest exam-ples: radiation and matter denominated Universes.

7. Detailed discussion of the solutions of Friedmann’s equations. Big Bang Cos-mology, Cosmic Black-body radiation, Einstein-de Sitter Universe as a limitingcase, critical matter density




Cosmology

Radiation and Matter

Thermal History

A more realistic consideration of the evolution of the state of the universe requiresto take into account both radiation and matter as well as other contributions to thetrue energy-momentum tensor of the universe. In this lecture we consider radiationand matter in more detail. In view of the structure of the Friedmann equation

S 2

c2 =c1S 2 +

c2S − k + Λ

3 S 2

the evolution is expected to be dominated

t small radiation ∼ c1S 2

t medium matter ∼ c2S

t large curvature or cosmological constant ∼ −k + Λ3 S 2




Cosmology

In the Big Bang case S (t)→ 0 (t → 0).

The simplest possible model is that of an ideal gas in thermal equilibrium togetherwith the radiation:[-9mm]

p = n kB T +13

a T 4

ρ/c2 = n m + (γ − 1)−1 n kB T + a T 4

n number of gas atoms ; m mass of a gas atomkB Boltzmann constant ; a Stefan-Boltzmann constantγ specific heat per gas atom ; γ = 5/3 single atom gas




Cosmology

Note: the crucial point why we can apply equilibrium thermodynamics to theexpanding universe, which is not in equilibrium itself!, is the fact that the ex-pansion is slow relative to characteristic particle interaction times. i.e. pro-cess rates are large enough to contiguously restore equilibrium in comov-ing volumes. With other words, at each stage in general there was plentyof time available for equilibrium to be reached. Interaction rates are energydependent and the energy is determined by the energy density which is de-creasing due to the expansion. Typically, below some threshold interactionrates for specific processes drop to zero, such that subsystems may followtheir own equilibrium distributions once they were decoupled (Examples:photon CMB, neutrino CB, see below)




Cosmology

Particle number conservation: n S 3 = n0 S 30

Energy balance:[-9mm]

ddS

n m S 3 + (γ − 1)−1 n kB T S 3 + a T 4 S 3

= −3 n kB T S 2 − a T 4 S 2 .

With n S 3 = constant we obtain

ST

dTdS

= −σ + 1

σ + 13 (γ − 1)−1

where σkB is the photon entropy per gas atom:

σ 4 a T 3

3 n kB n =

4 a T 3

3 kBσ




Cosmology

Boarder cases: i) little radiation: σ 1

T = T0

(SS 0

)−3 (γ−1)

is the usual form of adiabatic expansion of an ideal gas.

ii) much radiation: σ 1

T = T0

(SS 0

)−1

“hot universe” situation. This is the stable case, if σ ever is big it remains big

σ =4 a T 3

0

(SS 0

)−3

3 n0 kB

(S 0S

)3 =4 a T 3

0

3 n0 kB 1 3




Cosmology

As long as radiation and matter are in equilibrium: Tmat(t) = Trad(t) i.e.

Tmat = Trad = T0γ

(SS 0

)−1

.

As long as σ 1 we have

p = a3

(4σ

+ 1)

T 4 ' a3 T 4

ρ/c2 = a(

4 m3σ kB T + 4

3 (γ−1)σ + 1)

T 4 ' a T 4

p 'ρ

3




Cosmology

In the following we assume to have : σ 1

“Hot Universe”

v early universe: Tmat = Trad = T0γ

(SS 0

)−1

σ = 4 a T 3

3 kB n = constant 1 ; p = ρ/3

v Big-bang solution universal for small t!S (t) '

(4 C

)1/4(ct)1/2 respectively S = S 0 (t/t0)1/2

independent of k,Λ, etc.

T = T0

(S 0

S

)→ ∞ (t → 0)

ρ = ρrad = ρ0,rad

(S 0

S

)4

→ ∞ (t → 0)




Cosmology

Fireball: T, ρ extremely big

At some point (Planck Temperature) classical picture must break down!

v isotropic background radiationT0γ, ρ0γ black body spectrum at any time

ρ0γ = a T 40γ · c

2

T0γ ∼ 2.725 K , ρ0γ ' 4.64 × 10−34 gr/cm3 · c2

State of matter: Density inside nuclei is ρ∗ = ρnuclei ' 2 × 1015 gr/cm3 · c2

Corresponding radiation temperature is ρ∗ = ρ0γ

(T ∗T ∗0γ

)4 T ∗ ∼ 5.4 × 1011 K

Which means that for T > T ∗ ' 1012 K transition to quark-gluon plasma takesplace. This we will consider later.

For T ≤ T ∗ ' 1012 K one can understand the thermal history of the universerather well, under the following conditions




Cosmology

v isotropic background radiation is truly the relict of the Big-Bang fireball

v universe is hot: σ 1

which we take for granted (check for self-consistency).

By the expansion the universe cools down. As long as the photon energyEγ = h νγ = kB Tγ is bigger than the rest mass mi c2 of an elementary particle ofspecies i a permanent creation and annihilation of particle-antiparticle pairs takesplace:

γ + γ ↔ e+e−, µ+µ−, pp, nn, · · · ,

such that a plasma in thermal equilibrium exists. We assume here that charge Q,baryon-number B and lepton-number L are conserved and that in average there is




Cosmology

“neutrality”

nQ = nB = nL = 0 .

This means that in the thermal equilibrium the chemical potentials vanish: µi = 0.

Chemical potential is conjugate to particle number (concentration),like temperature is conjugate to entropy and pressure to the volume.

The equilibrium distributions for Fermions/Bosons are given by

ni(q) = 4 π h−3 gi q2 1(exp

(Ei(q)kBT

)± 1

) ; ±fermionsbosons

q : modulus of momentumgi : number of spin states

Ei(q) =(m2

i + q2)1/2

: for relativistic particles




Cosmology

FFFFFdigressionFFFFF

Comments concerning the chemical potentials µ:Thermodynamics: the change in internal energy U due to a change in particlenumber (species i)

dU =∑

i

µi dNi .

Thusµi =

(∂U∂Ni

)S ,V

= −T(∂S

∂Ni

)U,V

at constant entropy S or constant internal energy U, respectively, and constantvolume V. Note: internal energy U = U(S ,V,N1, · · · ,Nr)

dU = T dS − p dV +∑

i

µi dNi ,

for thermal, mechanical [work] and chemical reaction energy.




Cosmology

v Thermal equilibrium: T =(∂U∂S

)V,N

constant,

v Mechanical equilibrium: p = −(∂U∂V

)S ,N

constant,

v Chemical equilibrium for species i: µi =(∂U∂Ni

)S ,V,N j,i

constant

In the early universe the different species of elementary particles made up the“chemical composition” of the state of matter and radiation.




Cosmology

1) Photons: can be emitted and absorbed without constraint, i.e., nγ is arbitrary,such that µγ ≡ 0 ; nγ = ργ/hνγ ; q = Eγ = hνγ Planck distribution

2) Particle-antiparticle creation and annihilation:

γ + γ ↔ X + X ;

in thermal equilibrium µγ = 0 = nX + nX

hence we always haveµantiparticle = −µparticle

3) In weak processes like

e− + µ+ ↔ νe + νµ , e− + p↔ νe + n etc.




Cosmology

we have

µe− − µνe = µµ− − µνµ = µn − µp etc.

4) charge, baryon number and lepton number densities

nQ = np − np − ne− + ne+ + · · ·

nB = np + nn − np − nn essentially nucleons only

nM = nµ− + nνµ − nµ+ − nνµnE = ne− + nνe − ne+ − nνe .

In the early universe nγ nB, nM, nE, equivalent to σ 1. First approximation

nB = nM = nE = · · · = 0

In the approximation of separate lepton-number L` (` = e, µ, τ) conservation.




Cosmology

Distributions:[-13mm]

ni(q) ∝1

exp(

Ei−µikB T

)+ 1

; µi = −µi particle↔ antiparticle

Hence: nQ, nB, nM, nE odd function of four independent chemical potentialsµp, µe−, µνe, µνµ, such that

nQ = nB = nM = nE = 0 µi = 0∀i .

In more details: in the very early universe (energies above GUT scales)particle-antiparticle symmetry (pair creation via energetic photons) nQ = 0 forindividual particle species like

nQe = ne+ − ne− = 0




Cosmology

in terms of distributions:

ne+(E) =1

π2 ~3

E2

eE+µe−

kBT + 1; ne−(E) =

1π2 ~3

E2

eE−µe−

kBT + 1

nQe =

∫dE (ne+(E) − ne−(E)) =

1π2

(kBT~

)3

h(µe−

kBT

)with

h(x) =∫ ∞

0 dy y2[(ey+x + 1)−1

− (ey−x + 1)−1].

Thus, nQe = 0 requires x = 0 i.e.

µe− = −µe+ = 0 .




Cosmology

Furthermore, nQe = 0 implies nE = nνe − nνe and nE = 0 requires

µνe = −µνe = 0 .

Analogous conclusions follow for nQµ and nM. Note that the conservation of nX

implies

nX = nX0

(S 0S

)3

FFFFFFFFFF




Cosmology

Back to the early universe: basic dynamical equationsS 2

c2 = κ3 ρ S 2

universal for early universe.

Equilibrium thermodynamics:

ρ(T ) =∑

i

∫Ei(q) ni(q,T ) dq

p(T ) =∑

i

∫ (q2

3 Ei(q)

)ni(q,T ) dq




Cosmology

Applying 2nd law of thermodynamics: S (V,T ) entropy

T dS (V,T ) = dU + pdV ; U = ρ · V∂S (V,T )

∂V=

1T

(ρ(T ) + p(T ))

∂S (V,T )∂T

=VT

dρ(T )dT

with integrability condition (Maxwell relations)

∂S (V,T )∂V ∂T

=∂S (V,T )∂T ∂V

dp(T )dT

=1T

(ρ(T ) + p(T )) .




Cosmology

Taking into account the energy balance equation

S 3(t)dpdt

=ddt

[S 3(t) (ρ(t) + p(t))

]altogether we obtain[3mm]

ddt

[S 3(t)T (t) (ρ(t) + p(t))

]= 0

and up to a constant

S (V,T ) =VT

(ρ(T ) + p(T ))

is the total entropy of the system.




Cosmology

Result: the entropy in a volume S 3(t),s ≡ S

(S 3(t),T

)=

S 3(t)T (t) (ρ(t) + p(t)) = constant

is a constant

This provides an important test for the applicability of the thermal equilibriumproperties.

For highly relativistic particles, we have Ei = q, such that for the case hat allparticles are highly relativistic indeed

p(T ) = 13ρ(T )

and from

TdpdT

= ρ + p ρ(T ) ∝ T 4 and T ∝ S −1(t) . 3

In the highly relativistic case we get for the densities of the different particle




Cosmology

species (Fermions) because of Ei = q:

ρi/c2 =

∞∫0

dq q ni(q,T )/c2 =4 π

h3 c2 gi

∞∫0

dq q3 1

exp(

qkB T

)+ 1

=7 π5

30 h3c2 gi (kB T )4 =7

16a T 4 · gi

where a =π2 k4

B15 ~3 c2 is the Stefan-Boltzmann constant.

The contributions thus distinguish from each other by the spin multiplicity:ρν = 7

16 a T 4c2 ; ν = νe, νe, νµ, νµ, ντ, ντ, more ?ρe = 2 ρν etc.

whereasργ = a T 4c2 .

Note: ρfermion = 78 ρboson per spin degree of freedom. This is the effect of the Pauli




Cosmology

exclusion principle, which blocks part of the state space for fermions.

Formally, it is the ratio of the integrals for massless fermions and bosons∫ ∞0 dy y3 /(exp y ± 1) = π4

15 ·

781 , respectively.

The entropy is then given by: (p = ρ/3)

s =43

S 3(t)T (t)

ρ ; ρ = ργ + 6ρν + ρe− + ρe+ + · · · =g∗

2ργ .

Furthermore, because of ρ ∝ S −4 it follows

ρ

ρ= −4

SS

= −4(κ ρ

3

)1/2

which is Friedmann’s equation for k = 0 (Λ = 0),




Cosmology

and therefore

t = 1

(4 κ3 ρ)

1/2 + constant .

In the following we discuss in more detail the evolution of the universe attemperatures T <

∼T ∗ ' 1012 K , after the annihilation of the nucleons and afterµ-pair annihilation had taken place:

N + N → γ + γ (N = p, n) : mpc2 ' 938 MeV = 1.1 × 1013 Kπ± → µ± + νµ, π

0 → 2γ : mπc2 ' 139 MeV = 1.6 × 1012 Kµ+ + µ− → γ + γ : mµc2 ' 106 MeV = 1.2 × 1012 Ke+ + e− → γ + γ : mec2 ' 511 keV = 5.93 × 109 K

At this stage the relativistic particles in thermal equilibrium areγ, e±, νe, νe, νµ, νµ, ντ, ντ .




Cosmology

The density of non-relativistic particles, in particular relict nucleons now can beregarded as negligible, such that

ρ =g∗

2ργ ; ργ = ρ0γ

(TT0

)4

andt =

1(2 κ

3 ρ0γ

)1/2

1√

g∗

(T0

T

)2

+ constant

s =23

S 30

T0ρ0γ

(S (t)S 0

)3 (TT0

)3

· g∗(T )

Because the number of relativistic degrees of freedom g∗ is a function of T and atthe same time the entropy must be conserved the relation

S (t) · T (t) = constantcannot generally be true if not g∗ = constant.




Cosmology

In generalg∗(T (t)) S 3(t) T 3(t) = constant

holds and expresses the conservation of entropy.

In general, thermal equilibrium is obtained as a result of a large number ofcollisions between the particles in interaction. In other words, the average time τi

between two collisions satisfies t/τi 1. Highly relativistic particles, in particularthe massless photon and the quasi massless neutrinos, however, don’t changetheir equilibrium distribution also if no interaction is left: i.e., highly relativisticparticles behave as if they would be in thermal equilibrium with the rest of thesystem.




Cosmology

The reason may be easily explained by checking the form of the distributions:

ni =gi

2 π2~3

∞∫0

(E2 − m2

i

)1/2E dE

eE

kB T ± 1

particle density




Cosmology

ρi =gi

2 π2~3

∞∫0

(E2 − m2

i

)1/2E2 dE

eE

kB T ± 1

energy density

pi =gi

2 π2~3

∞∫0

(E2 − m2

i

)3/2dE

eE

kB T ± 1pressure

si =1T

(ρi + pi)

entropy density .

For mi ' 0 in fact pi = ρi/3 ; ρi =gi2 a T 4 , follows, such that




Cosmology

under redshift of the energy

E ∝1

S (t)./ kBT ∝

1S (t)

and

ρi ∝1

S (t)4 ; ni ∝1

S (t)3

i.e., the distributions exactly follow the expansion law in equilibrium!

At T ' T ∗ ' 1012 K the neutrinos still couple and the thermal equilibrium withγ, e± and ν′s is kept by a sufficiently high number of collision processese+ + e− ↔ γ + γ, e± + νµ ↔ µ± + νe, e± + νe ↔ e± + νe ,such that

Tγ = Te± = Tν ∝1

S (t)(@)




Cosmology

Concerning the neutrinos: below the τ and µ thresholds charged current (CC)processes like e± + ντ ↔ τ± + νe and e± + νµ ↔ µ± + νe are not any longer able tokeep neutrinos in thermal equilibrium. However, the neutral current (NC)processes

e+e− νiνi

still have sufficiently high rate to keep neutrinos in equilibrium. As neutrinomasses are far below me the NC processes are not phase space suppressed asthe CC ones are.

Exercise: estimate the collision rate for the above NC process at T ∼ 1011 K andcompare it with the CC muon capture (inverse muon decay) process rate.

In the next stage, in the range 1011 K >∼T >

∼ 1010 K all neutrinos decouple, asneutrino processes get very rare. This happens still before e+e−-annihilationbegins.




Cosmology

As argued above (@) hold until e+e− pair-annihilation sets in, thus effectively

[9mm]

g∗ = 2 +78 ( 4 + 6 ) = 43

4 = constantγ e± ν’s

2 spins 2 spins 1 spineach each

In order to describe in detail what happens when the electrons becomenon-relativistic, especially in the range where pair-annihilation takes place, wemay use the following facts

1. charged particles (here e±) are in thermal equilibrium all the time with the pho-tons (Compton scattering)

2. for relativistic particles (massless particles) what we said above holds

3. entropy remains conserved




Cosmology

s =S 3(t)T (t)

43

a T 4γ + 6

43

716

a Tν + ρe− + ρe+ + pe− + pe+

If one assumes that the chemical potentials of e± are negligible, we haveρe+ = ρe− , pe+ = pe− and

ρe =1

π2~3

∞∫me

(E2 − m2

e

)1/2E2 dE

eE

kB T ± 1

pe =1

3 π2~3

∞∫me

(E2 − m2

e

)3/2dE

eE

kB T ± 1




Cosmology

A transformation of variables yields

ρe− + ρe+ + pe− + pe+ = 2(kBT )4

π2~3 ·7π4

90f(

me

kBT

)= 4 ·

43

716

a T 4 f(

me

kBT

)

f (z) =907π4

∞∫0

dy y2

√y2 + z2 +y2

3√

y2 + z2

1

exp√

y2 + z2 + 1; f (0) = 1 .

Herewith, provided all particles are relativistic and in thermal equilibrium,

s = S 3(t) · T 3(t) ·43

a

1︸︷︷︸γ

+218︸︷︷︸

6ν′s

+74

f(

me

kBT

)︸︷︷︸

e±

= constant

When electrons get non-relativistic, such that f , 1, only Te± = Tγ remains equal,while Tν must go its own way.




Cosmology

Since the neutrinos by themselves satisfy an equilibrium distribution withtemperature Tν and pν = ρν/3 their contribution to the entropy is always

sν = S 3(t) · T 3(t) ·43

a ·218.

As mentioned before, the cosmic redshift lets the energy scale like E ∝ 1S (t) which

implies sν = constant as the distribution follows Tν ∝ 1S (t).




Cosmology

We thus can work out the neutrino temperature as follows:

s = sν︸︷︷︸const.

+43

a(S Tγ

)3

1 +74

f(

me

kBT

)= constant

s − sν = S 3(t) · T 3ν (t) ·

43

a ·(1 +

74

)valid at first for t small and is independent of t

= S 3(t) · T 3γ(t) ·

43

a ·(1 +

74

f(

me

kBT

))

which yields for Tν

Tν =(

411

)1/3Tγ ·

(1 + 7

4 f(

mekBT

))1/3




Cosmology

The time dependence is obtained with the Friedmann equation

dSS

=

(κρ

3

)1/2dt

with

ρ = ργ + ρe+ + ρe− + 6 ρν

= a T 4γ︸︷︷︸

photon

+218

a T 4ν︸︷︷︸

neutrinos

+74

a T 4γ g

(me

kBT

)︸︷︷︸

electron−positron

where

g(z) =47

30π4

∞∫0

dy y2√

y2 + z2 1

exp√

y2 + z2 + 1; g(0) = 1 .




Cosmology

We are now able to calculate the relationship between the elapsed time and thedensity, given earlier to be t = 1

(4 κ3 ρ)

1/2 up to a constant, with the exact

pair-annihilation taken into account. The pair-annihilation heats up the plasma,such that during this era S (t) · T (t) , constant.

The precise relationship is obtained as follows: first introduce the auxiliaryfunctions F and G

F(z) 1 +74

f (z) ; G(z) 1 + Nν78

(411

)4/3

F4/3(z) +74

g(z)

where f (z) and g(z) were given above and Nν = 3 the number of neutrino species.Then from

Tν = T0νS 0

S; dTν = −Tν

dSS




Cosmology

together with

Tν =

(411

)1/3

F(

me

kBT

)1/3

Tγ ; dTν = TνdTγTγ

+13

TνdFF

,

we obtain

dSS

= −dTνTν

= −

(dTγTγ

+13

TνdFF

)=

(κρ

3

)1/2dt ; ρ = a T 4

γ G(

me

kBT

).

The integration yields the following important relation between the elapsed time tand Tγ:[3mm]

t = −

∫ (κ

3a T 4

γ G(

me

kBT

))−1/2dTγ

Tγ+

13

TνdF

(me

kBT

)F

(me

kBT

) c© 2009, F. Jegerlehner ≪x Lect. 8 x≫ 86



Cosmology

Denoting z mekBT we have

dF(z)dT

=dF(z)

dzdzdT

= −zT

F(z) = −74

zT

f (z) ,

and with dTT = −dz

z we may write

t = te∫

dz z√

G(z)

(3 − zF

F (z))

wherete = (3 κ a c2 (me/kB)4)−1/2 = 4.3694 sec.

The integral can be evaluated numerically; see the Table




Cosmology

T (K ) T/Tν t(sec)3 × 1011 1.0000 02 × 1011 1.0000 1.3807 × 10−3

1 × 1011 1.0001 8.8375 × 10−3

6 × 1010 1.0002 2.6519 × 10−2

3 × 1010 1.0009 1.0952 × 10−1

2 × 1010 1.0020 2.4829 × 10−1

1 × 1010 1.0080 1.00706 × 109 1.0219 2.86753 × 109 1.0804 12.6712 × 109 1.1587 32.3011 × 109 1.3454 168.113 × 108 1.4010 1979.61 × 108 1.4010 1.7779 × 104

1 × 107 1.4010 1.7774 × 106

1 × 106 1.4010 1.7774 × 108




Cosmology

Until pair-annihilation sets in one has t∼∝ 1/T 2. After pair-annihilation we again

have t∼∝ 1/T 2, however, with Tν , Tγ since

Tν =

(4

11

)1/3

Tγ ·(1 +

74

f(

me

kBT

))→ Tν =

(411

)1/3

Tγ

as f(

mekBT

)' 1 ; kBT me while f

(me

kBT

)' 0 ; kBT me. As a result

[3mm]|| Tγ/Tν = 1.401 ⇒ Tν = 1.945 K




Cosmology

10−2 100 102 104 106 108 1010106

107

108

109

1010

1011

1012

e+e−

annihilation

γ, νe, νe, νµ, νµντ , ντ , e

+, e−

γ, νe, νe, νµ, νµ, , ντ , ντ

T K

t sec.

↑kBT = mec

2 (plot: 2 log T vs. log t)

today: T0ν = T0γ/1.401 ≃ 1.9 K

Neutrino CBR Tν

T = Tγ

Era of decoupling (freeze out) of neutrinos, e+e−-annihilation reheating the photonsea




Cosmology

Comment on the evaluation of the integrals f and g:by a change of integration variables y→ x =

√y2 + z2 and x→ u : x = zu we

obtain

f =907π4 z4 (I1(z) + I2(z))

g =1207π4 z4 (I1(z))

I1(z) = =

∞∫1

du u2√

u2 − 11

ezu + 1

I2(z) = =

∞∫1

du

(√u2 − 1

)3

31

ezu + 1.




Cosmology

Furthermore, we need the derivative of f :

f =907π4

[4 z3 (I1(z) + I2(z)) + z4

(I1(z) + I2(z)

)],

where the integrals Ii are obtained from Ii by

1ezu + 1

→ddz

[1

ezu + 1

]= −u

ezu

(ezu + 1)2 .

For the numerical evaluation one has to write

1ezu + 1

=e−zu

1 + e−zu .




Cosmology

FFFFFdigressionFFFFF

More on the evaluation of the integrals:Electron density:

ρe =74

a T 4 g(

me

kBT

); g(z) =

47

30π4 I(z)

We want to study the basic integral I(z) defined by

I(z) =

∞∫0

dy y2√

y2 + z2 1

e√

y2+z2+ 1

.




Cosmology

We perform two successive transformationsy→ x : x =

√y2 + z2, dx =

yx dy, y =

√x2 − z2 and x→ u : uz = x, dx = z du, thus

I(z) =

∞∫z

dxx2√

x2 − z2 1ex + 1

= z4

∞∫1

duu2√

u2 − 11

ezu + 1

= z4

∞∫1

duu2√

u2 − 1 e−zu 11 + e−zu .

We note that the following bound apply:

12≤

11 + e−zu ≤ 1 ; 0 < e−zu ≤ 1




Cosmology

and we expand

11 + e−zu ' 1 − e−zu + · · ·

Then the leading term reads

∞∫1

du u2√

u1 − 1 e−zu

=

∞∫1

du(u2 − 1

)3/2e−zu +

∞∫1

du(u2 − 1

)1/2e−zu

=Γ(2 + 1

2)√π(

12 z

)2 K2(z) +Γ(1 + 1

2)√π(

12 z

)1 K1(z)

= 3K2(z)

z2 +K1(z)

z=

(6 K2(z) + z2 K1(z)

)/z4 ,




Cosmology

where Kν(z) are modified Bessel functions

Kν(z) =

√π

(12z

)νΓ(ν + 1

2)

∞∫1

du(u2 − 1

)ν−12 e−zu .

We have introduced modified functions K normalized to unity at zero:

Kν(z) '12

Γ(ν)(z2

)−ν Kν(z)

2 K(z)

Γ(ν)(

z2

)ν ; Kν(0) = 1 .

Furthermore, we define

M(z) K2(z) +z2

6K1(z)

such that

I(z) = 6 M(z) + · · ·




Cosmology

The next term in the expansion we obtain by replacing e−zu by −e−2zu. If we replacethe prefactor z4 →

(2z)4

16 we see that the correction term is − 116 M(2z), thus

I(z) = 6(M(z) −

116

M(2z))

+ · · ·

I(0) = 6(1 −

116

)+ · · · ' 5.625

The exact value

I(0) =

∞∫0

dx x3 1ex + 1

=74π4

30' 5.682...

For larger z values the expansion converges better, since 11+e−zu gets better

approximated by 1.




Cosmology

Utilizing these results, for the electron number density we find

ne =2 (kBT )3

2π2~3

∞∫z

dx x√

x2 − z2 1ex + 1

=2 (kBT )3

2π2~3 2

K2(z) −18

K2(2z)

+ · · ·

The expanded integral at z = 0 yields 2(K2(z) − 1

8 K2(z))→ 2

(1 − 1

8

)= 1.75 while

the exact result is

∞∫0

dxx2

ex + 1=

34· 2 · ζ(3) = 1.803 . . . ; ζ(3) = 1.202 . . .

As mentioned before convergence is worst at z = 0 and improves for larger zvalues.




Cosmology

For the pressure we have

pe =2 (kBT )4

6π2~3

∞∫z

dx(x2 − z2

)3/2 1ex + 1

=2 (kBT )3

6π2~3 6

K2(z) −1

16K2(2z)

+ · · ·

and similarly for the energy density

ρe =2 (kBT )4

2π2~3

∞∫z

dx x2√

x2 − z2 1ex + 1

=2 (kBT )3

2π2~3 6

M(z) −116

M(2z)

+ · · ·




Cosmology

So far we assumed the electron gas to be non-degenerate i.e. µe ' 0. Thisapproximation can only be good as long as

ne−, ne+ |ne− − ne+|

In the pair-annihilation regime, however, ne− and ne+ decrease exponentially

(∝ e−mec2kBT ). We have

Kν(z) '

√π

2ze−z

1 +

µ − 18z

+(µ − 1)(µ − 9)

2!(8z)2 +(µ − 1)(µ − 9)(µ − 25)

3!(8z)3

+ · · ·

µ 4ν2 ; z =

mec2

kBT.

FFFFFend digressionFFFFF

The exponential suppression is absent when the electron gas is degenerate:




Cosmology

1

eE

kBT +1⇒ 1

eE−µkBT +1

||1

ezu+1 ⇒ 1ezu−m+1

E =(p2 − m2

)1/2⇒ E − µ =

(p2 − m2

)1/2− µ ; m =

µ

kBT.

When µe , 0 , one must determine µe via ne. One may assume that matter inaverage is neutral and consists of p, n and e− (n p, nn, ne+ ' 0), typical for ourgalaxy. After the pair-annihilation one expects a certain density of non-relativisticelectrons, where now

ne ∝ S −3(t)

The charge neutrality of cosmic matter requires[3mm]

ne = np




Cosmology

The proton number density is known from the density of baryons (nucleons),which is usually expressed normalized to the relic black-body photon density as

η ≡ nB/nγ = (4.7 − 6.5) × 10−10 (95% CL))Here nγ is fixed from the 2.725 K CMB radiation. The allowed range for thebaryon mass density today is[3mm]

ρ0B = (3.2 − 4.5) × 10−31 gr/cm3 or ΩB = ρ0B/ρ0,crit ' (0.017 − 0.024) h−2

where h ≡ H0/100 kms−1 Mpc−1 = 0.72 ± 0.08 is the present Hubble parameter.

As we will learn later baryonic matter consists to 98% of Hydrogen and Helium-4,observed in a ratio H : 4He ∼ 3 : 1 such that about 5/7 are protons and 2/7 areneutrons. We thus assume ne ∼ 5/7nB such that

ρ0e ' 1.5 × 10−34 gr/cm3

We thus find the following general picture about the development of the densitiesafter about 1011 K :




Cosmology

110−12 10−10 10−8 10−6 10−4 10−2

1010

1

10−30

10−20

10−10all

B=0constituents

inth

erm

alequilibrium

radiation era matter era

e+e−

annihitation

density

gr/

cm

3

todayexpansion factor S/S0

ρν = 32ρe

ρe =74ργ

ρe−

ρB

ργ

ρν = 218

(411

)4/3ργ

Possible Neutrino Degeneracy




Cosmology

So far we always have assumed that lepton numbers are essentially zero:nM ' nE ' 0. In the hot universe scenario we assumed

nM, nE, · · · nγ ; nγ ∝(kBTγ~

)3

.

To high accuracy individual lepton numbers nT , nM and nE are conserved. Chargeneutrality nQ = 0 is always assumed. For temperatures T <

∼T ∗ ' 1012 K only e−

and e+ show up as charged particles (not yet annihilated), the density of relictbaryons is negligible. Therefore

nQ = ne+ − ne− = 0

and the chemical potentials must vanish

µe+ = µe− = 0 .




Cosmology

Hence,

nE = ne+ − ne− + nνe − nνe = nνe − nνe ,

and similarly for the muon and tau leptons. With,

nν(E) =1

2π2~3

E2

eE−µνkBT + 1

nν(E) =1

2π2~3

E2

eE+µνkBT + 1

we obtain

nE =

∫dE

(nνe(E) − nνe(E)

)=

12π2

(kBT~

)3

h(µνe

kBT) ,




Cosmology

and corresponding expressions for nM and nT , with

h(x) =

∞∫0

dy y2[(

ey−x + 1)−1−

(ey+x + 1

)−1].

Because of the conservation of nE, nM, nT , on the one hand, we must have

nL = n0L

(S 0

S

)3

, L = T,M, E .

On the other hand we know that

Tν = T0ν

(S 0

S

),

which requires for all temperatures T <∼T ∗ ' 1012 K up to date:




Cosmology

µνkBT = constant ↔ µν = µ0ν

(S 0S

).

Our previous calculation concerning the e+e−-annihilation era changes only slightlyif µν , 0: it amounts to slightly increase the neutrino energy density: by replacing

ρν,all = 6ρν =218

a T 4ν

by

ρν,all = ρνe + ρνe + ρνµ + ρνµ + ρντ + ρντ

=78

a T 4ν

(r(µνe

kBTν

)+ r

(µνµ

kBTν

)+ r

(µντ

kBTν

))︸︷︷︸

constant in t

,




Cosmology

where

r(x) =60

7 π4

∞∫0

dy y3[(

ey−x + 1)−1−

(ey+x + 1

)−1]

; r(0) = 1 .

Thus a partial neutrino degeneracy would lead to smaller t values. In the limitingcase of complete degeneracy

∣∣∣∣ µνkBT

∣∣∣∣ 1, for example, for µνkBT 1, we get

nν(E) '

12π~3 E2 E < µν

0 E > µν

and

nν(E) ' 0 ,




Cosmology

and thus

r(x) =60

7 π4

∞∫0

dy y3 =157π4 x4 =

157π4

(µν

kBT

)4

.

With a =π2k4

B15~3 =

8π5k4B

15h3 we find

ρν,all = π h−3(µ4νe

+ µ4νµ

+ µ4ντ

).

This compares to the photon density

ργ = a T 4γ =

8 π4

15︸︷︷︸'52

π h3(kBTγ

)4 π h−3

(µ4νe

+ µ4νµ

+ µ4ντ

),

which means that the evolution would be dominated completely by ρν and not beργ. From the observed value of the deceleration parameter one can derive bounds




Cosmology

on chemical potential contribution µ4νe

+ µ4νµ

+ µ4ντ

. Such degenerate neutrinobackground would also affect laboratory experiments like Tritium β-decay3H →3 He + e− + νe which shows that |µe| can be at most a few eV. Much morestringent constraints come from cosmological nucleosynthesis.

Summary so far: we have worked out the thermal history of our universe back to astate when electrons, positrons, neutrinos and photons were in thermalequilibrium at temperature of about 3 × 1011 K .

Exactly the same reasoning allow us to go further back, the only thing we need toknow are the particles and their thresholds, which can be created and annihilatedat higher energies. In fact we can go back as much as we know the elementaryparticle content, which is what we have in the Standard Model of elementaryparticle physics. Sufficiently high above the thresholds (e.g. beyond the

electroweak scale v = 1/√√

2Gµ = 246.22 GeV) we can take all particles ashighly relativistic, and what matters then is just the number of relativistic degreesof freedom.




Cosmology

The Standard Model

Constituents of matter:

Spin 1/2 fermions

1st family

2nd family

3rd family

SU(3)ccolor

siglets triplets anti-triplets

SU(2)Ldoublet

singlet

⇑weak isospin

singlet

sterile νs

Leptons Quarks

νeL uL uL uL

e−L dL dL dL

νeR uR uR uR

e−R dR dR dR

uL uL uL

dL dL dL

uR uR uR

dR dR dR

νµL cL cL cL

µ−L

sL sL sL

νµR cR cR cR

µ−R

sR sR sR

cL cL cL

sL sL sL

cR cR cR

sR sR sR

ντL tL tL tL

τ−L bL bL bL

ντR tR tR tR

τ−R bR bR bR

tL tL tL

bL bL bL

tR tR tR

bR bR bR




Cosmology

γ

W− Z0 W+

ϕ−ϕ0

ϕ+

H

Carrier of Forces:

Spin 1 Gauge Bosons

“Cooper Pairs”:

Spin 0 Higgs Boson

Photon

Vector Bosons

Higgs Ghosts

Higgs Particle

Octetof

Gluons

⇔

Higgs mechanism: breaking weak isospin spontaneously

weak bosons W , Z and all fermions get masses

mi ∝ yi v; v = 〈H〉 >≃ 246.21 GeV; GFermi =1√2v2

(radioactivity)




Cosmology

Dynamics: strongly interacting quarks and gluons

LQCD = −14F(a)

µν F(a) µν + i∑

q ψiqγ

µ(Dµ

)i jψ

jq

−∑

q mqψiqψqi + LGF + LFP

F(a)µν = ∂µAa

ν − ∂νAaµ + gs fabcAb

µAcν(

Dµ

)i j

= δi j∂µ − igs∑

aλa

i j2 Aa

µ

v Quarks:q = u, d, s, c, b, t (Dirac fields)

v Gluons: color-octet Aa, a = 1, 2, . . . , 8 (Yang–Mills fields)




Cosmology

SM particle content:Quarks t, t 171.3 ± 1.6 GeV spin= 1

2 g = 2 · 2 · 3 = 12b, b 4.20+0.17

−0.07 GeV 3 colorsc, c 1.27+0.07

−0.11 GeVs, s 105+25

−35 MeVd, d 3.5 − 6.0 MeVu, u 1.5 − 3.3 MeV

72Gluons 8 massless bosons spin= 1 g = 2 · 8 = 16

16Leptons τ−, τ+ 1776.84 ± 0.171 MeV spin= 1

2 g = 2 · 2 = 4µ−, µ+ 105.658 MeVe−, e+ 0.510999 MeV

12Neutrinos ντ, ντ < 18.2 MeV spin= 1

2 g = 2νµ, νµ < 190 keVνe, νe < 2 eV

6




Cosmology

Weak bosons W± 80.403 ± 0.029 GeV spin= 1 g = 3Z 91.1876 ± 0.0021 GeV

Photon γ 0 (< 6 × 10−17 eV) g = 211

Higgs H > 114.4 GeV spin= 0 g = 11

g f = 72 + 12 + 6 = 90gB = 16 + 11 + 1 = 28

Thus the total energy density

ρ(T ) =∑

ρi(T ) =π2

30g∗(T ) T 4

which in general defines an effective number of degrees of freedom g∗(T ).




Cosmology

If all particles are ultra relativistic

g∗(T ) = gB(t) +78

g f (T )

with gB =∑

i gi is the sum over relativistic bosons and g f =∑

i gi the sum over therelativistic fermions.

If all SM are relativistically excited we haveg∗(T ) = 106.75

Mass effects can be taken into account as done above for electrons and positronsin the discussion of the e+e−-annihilation. Mass effects manifest themselvesslightly different in energy density ρ, pressure p and entropy density s. Therefore

ρ(T ) =π2

30g∗(T ) T 4 ; p(T ) =

π2

90g∗p(T ) T 4 ; s(T ) =

2 π2

45g∗s(T ) T 3

define slightly different effective numbers of degrees of freedom when masses




Cosmology

play a role. Numerical results are shown in the following figure.

The functions g∗(T )) (solid), g∗p(T ) (dashed), and g∗s(T ) (dotted) calculated for theSM particle content.

In the following we will often adopt the convention to express temperatures in




Cosmology

energy units, as E = kB T (hence 1 K ∼ 8.6 × 10−5 eV) is a universal relation andas particle physicist we are more familiar with energy units.

The thermal history of the universe as a scan of the SM:

T event g∗(T ) comment∼ 200 GeV all states relativistic 106.75∼ 100 GeV EW phase transition 106.75< 170 GeV top annihilation 96.25< 80 GeV W±,Z,H annihilation 86.25< 4 GeV bottom annihilation 75.75< 1 GeV charm, τ annihilation 61.75∼ 150 MeV QCD phase transition 17.25 (u, d, g→ π±,0, 37→ 3)< 100 MeV π±, π0, µ annihilate 10.75 e±, ν, ν, γ left< 500 keV e± annihilation 7.25




Cosmology

The role of the QCD phase transition

Quark-Gluon Plasma: The Stuff of the Early Universe




Cosmology

QCD under extreme conditions:

r High temperature matter: early universe

r High density matter: in neuron stars and other supernova remnants




Cosmology

r Laboratory test to come: heavy ion collisions RICH/Brookhaven, LHC/Geneva

Nuclei in collision:




Cosmology

QCD phase transition: non-perturbative regime à only known approach

lattice QCD [field theory in a heat bath at finite temperature]




Cosmology

Physics of the early universe:

à Entropy per co-moving volume is conserved

à All entropy is in relativistic speciesExpansion covers many decades in T, so typicallyeither T m (relativistic) or T m (frozen out)

à All chemical potentials are negligible

Entropy S in co-moving volume (Dc)3 preserved

g∗S effective number of relativistic species

Entropy density SV = S

D3c

1S (t)3 =

2p2

45 g∗S T 3

à T = (g∗S )−1/3 1S (t)




Cosmology




Cosmology

Broniowski, et al. 2004

How many hadrons?Density of hadron mass states dN/dM increases exponentially:

dNdM ∼ Ma exp M/TH (TH ∼ 2 × 1012 K = 170 MeV)




Cosmology

Before [QCD] we could not go back further than 200,000 years after the Big Bang.Today since QCD simplifies at high energy, we can extrapolate to very early timeswhen nucleons melted to form a quark-gluon plasma. David Gross, Nobel Lecture(RMP 05)




Cosmology

Cosmological phase transition....

...when the universe cools down below 175 MeV

10−5 seconds after the big bang...

Quarks and gluons form baryons and mesons

before: simply not enough volume per particle available

Cosmological phase transitions

QCD phase transition T=175 MeV Electroweak phase transition T= 150 GeVbaryogenesis? others earlier?

Electroweak phase transition ?

10−12 seconds after big bang

fermions, W and Z bosons get mass




Cosmology

Standard model: cross over transition baryogenesis if 1st order bubble formation“out of vacuum”

QCD at High temperature, at high T less order more symmetry (magnets crystals)

l Quark-gluon plasma

l Chiral symmetry restored ( no pions, no quark condensates)

l “Deconfinement” (no linear heavy quark potential at large distances)

l Lattice QCD simulations: both effects happen at the same temperature




Cosmology

QCD - phase transition

Quark-gluon plasma Hadron gasv Gluons: 8 × 2 = 16 v Light mesons: 8v Quarks: 9 × 7/22 = 12.5 v (Pions: 3)v DOF: 28.5 v DOF: 8Chiral symmetry Chiral symmetry brokenLarge difference in number of degrees of freedom !Strong increase of density and energy density at Tc !

quarks + gluons

〈qq〉〈qq〉pion’s

protons, neutrons etc

T

µ

QCD phases




Cosmology

quark-gluon plasma“deconfinement”

quark matter: superfluidB spontaneously broken

vacuum

nuclear matter: B,I spontaneously brokenS conserved

〈qq〉 6= 0

〈qq〉 6= 0

〈qq〉 6= 0

1st order

2nd order ?

T

µ

QCD phases: ms > mu, md

?

pion’s

protons, neutrons etc




Cosmology

Exercise pages

Estimate the collision rate for the NC process

e+e− νiνi

at T ∼ 1011 K and compare it with the charged current (CC) muon capture

e− + µ+ ↔ νe + νµ

(inverse muon decay) process rate.

Exercise: Reproduce the Table T versus t given above by numerically integratingthe corresponding integral equation.

Exercise: In calculating the cosmic neutrino background we assumed neutrinos tobe massless. Discuss what modifications we have to expect due to the fact thatobserved neutrino oscillations require neutrino masses at the meV scale.




Cosmology

Exercise: For large enough temperatures, when T |µ| and T m, we may usethe approximations µ = 0 and m = 0. Show that in this ultrarelativistic limit thefollowing results are found (here we use units such that kB = 1, i.e. temperaturesin energy units):

n =g

(2π)3

∞∫0

4πq2 dqeq/T ± 1

=

34π2 ζ(3) gT 3 fermions1π2 ζ(3) gT 3 bosons

ρ =g

(2π)3

∞∫0

4πq3 dqeq/T ± 1

=

78π2

30 gT 4 fermionsπ2

30 gT 4 bosons

p =g

(2π)3

∞∫0

43πq3 dq

eq/T ± 1=

13ρ




Cosmology

Above ζ(3) ≡∑∞

n=1(1/n3) = 1.20206... where ζ(n) is Riemann’s zeta function.Furthermore, calculate the average particle energy

〈E〉 =ρ

n

Exercise: In the non-relativistic limit T m and T m − µ, the typical kineticenergies are much below the mass m and E = m + q2/2m. The second conditionleads to small occupation numbers (dilute gas) and is usually satisfied incosmology if the first one is satisfied. We can the use the approximation

e(E−µ)/T ± 1 ' e(E−µ)/T

such that bosons and fermions have the same distributions, the non-relativistic




Cosmology

Maxwell-Boltzmann one. Show that the following relations hold in this case:

n = g(mT

2π

)3/2

e−m−µ

T

ρ = n(m +

3T2

)p = n T ρ

〈E〉 = m +3T2

n − n = 2g(mT

2π

)3/2

e−mT sinh

µ

T

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