Thermal History of the Universe and the CosmicMicrowave Background.

I. Thermodynamics of the Hot Big Bang

Matthias BartelmannMax Planck Institut fur Astrophysik

IMPRS Lecture, March 2003

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Part 1: Thermodynamics of the Hot Big Bang

1. assumptions

2. properties of ideal quantum gases

3. adiabatic expansion of ideal gases

4. particle freeze-out

5. neutrino background

6. photons and baryons

7. physics of recombination

8. nucleosynthesis

9. the isotropic microwave background

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1. Assumptions

1. adiabatic expansion

2. thermal equilibrium

3. ideal gases

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1.1. Adiabatic Expansion

The Universe expands adiabatically.

• isotropy requires it to expand adiathermally: no heat can flow because anyflow would define a preferred direction

• an adiathermal expansion is adiabatic if it is reversible, but irreversibleprocesses may occur

• the entropy of the Universe is dominated by far by the cosmic microwavebackground, so entropy generation by irreversible processes is negligible

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1.2. Thermal Equilibrium

Thermal equilibrium can be maintained despite the expansion.

• thermal equilibrium can only be maintained if the interaction rate of particlesis higher than the expansion rate of the Universe

• the expansion rate of the Universe is highest at early times, so thermalequilibrium may be difficult to maintain as t → 0

• nonetheless, for t → 0, particle densities grow so fast that interaction rates areindeed higher than the expansion rate

• as the Universe expands, particle species drop out of equilibrium

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1.3. Ideal Gases

Cosmic “fluids” can be treated as ideal gases.

• ideal gas: no long-range interactions between particles, interact only by directcollisions

• obviously good approximation for weakly interacting particles like neutrinos

• even valid for charged particles because oppositely charged particles shieldeach other

• consequence: internal energy of ideal gas does not depend on volumeoccupied

• cosmic “fluids” can be treated as possibly relativistic quantum gases

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2. Properties of Ideal Quantum Gases

1. occupation number

2. number density

3. energy density

4. grand canonical potential

5. pressure

6. entropy density

7. results for bosons and fermions

8. numbers

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2.1. Occupation Number

• in thermal equilibrium with a heat bath of temperature T, the phase-spacedensity for a quantum state of energy ε is

〈 f 〉=

exp

[−ε−µ

kT

]±1

−1

(“+” for fermions, “−” for bosons)

• in thermal equilibrium with a radiation background, the chemical potential µ=0: Helmholtz free energy F(T,V,N) = E−TS is minimised in equilibrium for asystem at constant T and V, so from dF =−SdT−PdV +µdN:

∂F∂N

= 0 = µ

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2.2. Number Density

• particles in volume V, number of states in k-space element:

dN = gV

(2π)3d3k

• g is statistical weight, e.g. spin degeneracy factor

• momentum ~p = ~~k, related to energy by

ε(p) =√

c2p2+m2c4

• spatial number density in thermal equilibrium:

n =g

(2π~)3

∫ ∞

0

4πp2dpexp[ε(p)/kT]±1

/.

2.3. Energy Density

• mean energy density: number of states per phase-space cell, timesoccupation number, times energy per state, integrated over momentumspace:

u =g

(2π~)3

∫ ∞

0

4πp2ε(p)dpexp[ε(p)/kT]±1

• integrals can most easily be carried out by substituting a geometrical series:

∫ ∞

0

xmdxex−1

=∫ ∞

0

xme−xdx1−e−x

=∫ ∞

0dxxme−x

∞

∑n=0

e−nx =∞

∑n=1

∫ ∞

0dxxme−nx = m!ζ(m+1) ;

for fermions, use1

ex+1=

1ex−1

− 2e2x−1

/.

2.4. Grand Canonical Potential

• extremal principles yield: (thermodynamical potential)=−kT ln(partition sum)

• adequate potential here: grand canonical potential Φ(T,V,µ) (particlenumbers can change)

• grand-canonical partition sum:

Z =∞

∑N=0

eµN/kT ∑Nα|N

e−∑εαNα/kT

• grand-canonical potential:

Φ(T,V,µ) =∓kTgV

(2π~)3

∫ ∞

0dp4πp2 ln

[1±eµ/kTe−ε(p)/kT

] (FermionsBosons

)

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2.5. Pressure

• Helmholtz free energy:

F(T,V,N) = U−TS

• grand-canonical potential:

Φ(T,V,µ) = F−µN= U−TS−µN

• thermodynamical Euler relation:

U = TS−PV +µN

• it follows:

Φ =−PV ⇒ P =−ΦV

• example: relativistic boson gas inthermal equilibrium; ε = cp, µ= 0,

PB = gπ2

90(kT)4

(~c)3

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2.6. Entropy Density

• Helmholtz free energy:

dF(T,V,N) =−SdT−PdV +µdN

• grand-canonical potential:

dΦ(T,V,µ) =−SdT−PdV−Ndµ

• thus, the entropy is:

S=−∂Φ∂T

• example: entropy density forrelativistic boson gas in thermalequilibrium

s=SV

= gk2π2

45

(kT~c

)3

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2.7. Results for Bosons and Fermions

relativistic non-Bosons Fermions relativistic

n gBζ(3)π2

(kT~c

)3 34

gF

gBnB g

(kT2π~

)3/2

e−kT/mc2

u gBπ2

30(kT)4

(~c)3

78

gF

gBuB

32

nkT

P gBπ2

90(kT)4

(~c)3=

uB

378

gF

gBPB nkT

s gBk2π2

45

(kT~c

)3 78

gF

gBsB

/.

2.8. Numbers

note: 1eV = 1.6×10−12erg correspond to kT = 1.16×104K

•nB = 10gB

(TK

)3

cm−3 = 1.6×1013gB

(kTeV

)3

cm−3

•uB = 3.8×10−15gB

(TK

)4 ergcm3

= 2.35×10−3gB

(kTeV

)4 ergcm3

•sB

k= 36gB

(TK

)3

cm−3 = 5.7×1013gB

(kTeV

)3

cm−3

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3. Adiabatic Expansion of Ideal Gases

1. temperature

2. density

3. matter-radiation equality

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3.1. Temperature

• for relativistic boson or fermiongases:

P =u3

=E3V

• first law of thermodynamics,dE +PdV = 0,then implies:

dE =−PdV = 3d(PV) ⇒ P ∝ V−4/3

(adiabatic index is γ = 4/3)

• for non-relativistic ideal gas: γ = 5/3

• since P ∝ T4 for relativistic gases:

T ∝ V−1/3 ∝ a−1

(a is cosmological scale factor)

• for non-relativistic ideal gas:

T ∝ PV ∝ V−5/3+1 ∝ a−2

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3.2. Density

• non-relativistic gases:rest mass energy kinetic energymass density:

⇒ ρ ∝ V−1 ∝ a−3

• relativistic gases: mass density

ρ =uc2

∝ T4 ∝ a−4

• density of relativistic particles dropsfaster than that of non-relativisticparticles:

• dilution due to volume expansion,plus energy loss due to redshift

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3.3. Matter-Radiation Equality

• today’s matter density in the Universe:

ρ0 = Ω03H2

0

8πG= 1.9×10−29Ω0h

2gcm−3

• today’s radiation density in the Universe:

T = 2.73K ⇒ ρR,0 =uc2

= 4.7×10−34gcm−3

• radiation domination:

ρR(aeq) = ρR,0a−4eq = ρ0a

−3eq = ρ(aeq) ⇒ aeq =

ρR,0

ρ0= 2.5×10−5(Ω0h

2)−1

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4. Particle Freeze-Out

1. interaction rates

2. freeze-out conditions

3. particle distributions after freeze-out

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4.1. Interaction Rates

• expansion timescale:

texp ∼ (Gρ)−1/2

ρ is density of dominant fluid

• in early Universe, radiationdominates, thus

ρ ∝ a−4 ⇒ texp ∝ a2

expansion timescale is shortest atearly times and increases as a2

• thermal equilibrium maintained bytwo-body interactions, collision ratefor one particle species is

Γ ∝ n ∝ T3 ∝ a−3

• thus, the collision time scale is

tcoll ∝ Γ−1 ∝ a3

• for a → 0, tcoll decreases fasterthan texp, so equilibrium can bemaintained

/.

4.2. Freeze-Out Conditions

• continuity equation in absence ofcollisions:

n+3Hn = 0

• collision rate:

Γ = 〈σv〉n

• source term from thermal particlecreation:

S= 〈σv〉n2T

• continuity equation now:

n+3Hn =−Γn+S

• introduce comoving number densityN = a3n:

d lnNd lna

=−ΓH

(1−NT

N

)

• freeze-out can occur if Γ H:particle number N cannot adapt

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4.3. Particle Distributions After Freeze-Out

• relativistic particles:

n ∝ T3 ⇒ N = a3n = const.

• from freeze-out equation:

d lnNd lna

= 0 ⇒ N = NT

independent of Γ!

• relativistic particles maintainthermal distribution even afterfreeze-out

• non-relativistic particles, comovingnumber density:

NT ∝ T−3/2exp

(−mc2

kT

)

• once T mc2, NT drops rapidly,hence NT N, and

d lnNd lna

=−ΓH→ 0

particle production stops: freeze-out

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5. Neutrino Background

1. neutrino decoupling

2. electron-positron annihilation

3. photon heating

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5.1. Neutrino Decoupling

• neutrinos are kept in thermal equilibrium by the weak interaction

ν+ ν↔ e+ +e−

• this interaction freezes out when the temperature drops to

Tν ∼ 1010.5K∼ 2.7MeV

• neutrinos are ultrarelativistic at that “time”, i.e. their comoving number densityis that of an ideal, relativistic fermion gas

/.

5.2. Electron-Positron Annihilation

• the reaction

e+ +e−↔ 2γ

is suppressed once temperaturedrops below

T ∼ 2mec2≈ 1MeV≈ 1010K

• thus, electrons and positronsannihilate shortly after neutrinosfreeze out

• the entropy of the annihilatingelectron-positron gas heats thephotons, but not the neutrinos

• the microwave backgroundis heated above the neutrinobackground by the annihilation

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5.3. Photon Heating

• entropy before annihilation equalsentropy after annihilation:

s′e+ +s′e− +s′γ = sγ

• before annihilation,

Te+ = Te− = Tγ ≡ T ′

thuss′e+ = s′e− =

78

s′γ

(ge+ = ge− = gγ = 2)

• since s∝ T3,(2· 7

8+1

)(T ′)3 = T3

• temperature after annihilation:

T =(

114

)1/3

T ′ ≈ 1.4T ′

• temperature of ν background today:

Tν,0 = 1.95K

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6. Photons and Baryons

1. baryon number

2. photon-baryon ratio

/.

6.1. Baryon Number

• number density of baryons today:

nB ≈ρB

mp=

ΩB

mp

3H20

8πG= 1.1×10−5ΩBh2cm−3

• measurements: abundance ratios of light elements, microwave background(see later), gas in galaxy clusters:

ΩBh2≈ 0.025

• only ∼ 10%−20%of the matter in the Universe is contributed by baryons

• baryon number density ∝ a−3 ∝ T3

/.

6.2. Photon-Baryon Ratio

• number density of photons today:

nγ = 407cm−3

• scales with temperature like T3 likebaryon number density, hence theirratio is independent of time

• constant baryon-photon-ratio:

η =nB

nγ= 2.7×10−8ΩBh2

• there is one baryon for about abillion photons in the Universe

• photon entropy completelydominates entropy of the Universe

• very important for process ofrecombination

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7. Physics of Recombination

1. recombination process

2. Saha equation

3. recombination temperature

4. two-photon recombination

5. thickness of the recombination shell

6. expectation on radiation spectrum

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7.1. Recombination Process

• as the temperature drops, electronsand protons can combine to formatoms:

e−+ p+ ↔ H + γ

freezes out

• How does recombination proceed,i.e. how does the electron densitychange with T?

• solution: minimisation of Helmholtzfree energy F(T,V,N)

• canonical partition function Z,

F =−kT lnZ

with

Z =ZNe

e

Ne!Z

Npp

Np!ZNH

H

NH!

• baryon number NB = Np + NH,electron number Ne = Np, thusNH = NB−Ne

• Stirling: lnN! ≈ N lnN−N

/.

7.2. Saha Equation

• from ∂F/∂Ne = 0:

N2e

NB−Ne=

ZeZp

ZH

• partition function for single species:

Z =4πgV(2π~)3

∫ ∞

0dpp2e(µ−ε)/kT

• nonrelativistic limit:

ε = mc2+p2

2m

• chemical potential:

µ= 0 ⇒ µe +µp = µH

• ionisation potential:

(me +mp−mH)c2 = χ = 13.6eV

• x = Ne/NB, nB = NB/V; Sahaequation:

x2

1−x=

(2πmekT)3/2

(2π~)3nBe−χ/kT

/.

7.3. Recombination Temperature

• nB is the baryon number density; wehad

nB = ηnγ = η2ζ(3)

π2

(kT~c

)3

• insert this into Saha equation:

x2

1−x≈ 0.26

η

(mec2

kT

)3/2

e−χ/kT

• want x 1, hence x2/(1−x)≈ x2

• 1/η is huge number, so kT χrequired for x to be small

• putting x = 0.1 yields kTrec ≈ 0.3eV,or

Trec ≈ 3500K

• high photon number density delaysrecombination considerably

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7.3. Recombination Temperature

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7.4. Two-Photon Recombination

• direct hydrogen recombinationproduces energetic photons; laststep to ground state is Lyman-α(2P→ 1S);

hν≥ ELyα =34

χ = 10.2eV

• abundant Ly-α photons can reionisethe cosmic gas

• photons cannot be lost as inclouds; energy loss due to cosmicexpansion is slow

• How can recombination proceed atall?

• production of lower-energyphotons: forbidden transition2S→ 1S, requires emission of twophotons; this process is slow

• recombination proceeds at asomewhat slower rate thanpredicted by the Saha equation

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7.5. Thickness of Recombination Shell

• recombination is not instantaneous;“time” interval between beginningand end?

• time: measured in terms of scalefactor a, or redshift 1+z= a−1

• optical depth throughrecombination shell:

τ =∫

nexσTdr

(σT: Thomson cross section)

• scattering probability for photonswhen travelling from z to z−dz:

p(z)dz= e−τdτdz

dz

• probability distribution e−τdτ/dzis well described by Gaussianwith mean z∼ 1100 and standarddeviation δz∼ 80

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7.5. Thickness of Recombination Shell

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7.6. Expectation on Radiation Spectrum

• last-scattering shell has finite width: CMB photons received today werereleased at different redshifts

• cosmic plasma cooled during recombination, photons were released atdifferent mean temperatures: T = T0(1+z), thus

δT = T0δz≈ 200K

• photons were redshifted after release: those released earlier, i.e. from hotterplasma, were redshifted by larger amount: E = E0(1+z)

• these effects cancel if plasma temperature depends on scale factor like T ∝a−1; then: Planck spectrum of single temperature expected

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8. Nucleosynthesis

1. formation of the light elements

2. Gamow criterion

3. prediction of the microwave background temperature

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8.1. Formation of the Light Elements

• high temperature in the earlyUniverse allows nuclear reactionslike in stars; density is muchlower, so higher temperatures arerequired:

Tnuc ∼ 109K

• before: neutrons and protonsformed; equilibrium through

p+e−↔ n+ν , p+ ν↔ n+e+

• relative abundance freezes outonce weak interaction becomes tooslow;

Tfreeze−out ∼ 1.4×1010K

afterwards, free neutrons decay

• nucleosynthesis proceeds throughstrong interactions until thesefreeze out, e.g. deuterium formationn+ p↔ D+ γ stops at

TD ∼ 7.9×108K

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8.2. Gamow Criterion

• deuterium fusion is the most important step towards the fusion of higherelements; e.g. Helium:

D+ p→ 3He+ γ , 3He+n→ 4He

• deuterium needs to be produced in sufficient abundance for higher elementsto form, but if all neutrons are immediately locked up into deuterium, no higherelements can form either

• George Gamow noticed in 1948 that deuterium formation has to proceed at“just right” rate:

nB 〈σv〉 t ∼ 1 ,

i.e. collision rates for baryons should not be too small or too large

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8.3. Prediction of CMB Temperature

• for deuterium formation, t ∼ 3min

• from a theoretical estimate for〈σv〉, Gamow estimated the baryondensity at deuterium formation:

nB ∼ 1018cm−3

• today’s baryon density is

nB,0 ∼ 1.1×10−5ΩBh2cm−3

• baryon density drops like T3; thus:

nB

nB,0=

(TD

T0

)3

• using TD = 7.9×108K:

T0 =(

nB,0

nB

)1/3

TD ∼ 4K

CMB temperature is predicted bythe Big-Bang model!

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9. The Isotropic Microwave Background

1. temperature and spectrum

2. limits on chemical potential and Compton parameter

3. the dipole

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9.1. Temperature and Spectrum

• microwave background wasdetected, but not recognised,by Penzias and Wilson in 1965

• NASA’s Cosmic BackgroundExplorer (COBE) preciselymeasured its spectrum (andother things, see below)

• its temperature isT0 = (2.726±0.002)K

• its spectrum is the best black-bodyspectrum ever measured

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9.2. Limits on µ and y

• finite width of the last-scatteringsurface: it is important that the CMBhas a black-body spectrum

• shape of the spectrum allowsconstraints on the chemicalpotential:

|µ| ≤ 9×10−5

• hot gas between the last-scatteringsurface and us can distort thespectrum through Comptonscattering

• Compton-y parameter:

y =kT

mec2

∫neσTdl

(typical energy change timesscattering probability)

• constraint from COBE-FIRASspectrum strictly contrains heatinput in young Universe:

y≤ 1.5×10−5

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9.3. Dipole

• “freely falling” observers inthe Universe define comovingcoordinates

• the CMB is at rest with respect tothis coordinate frame

• Earth moves around the Sun, Sunaround the centre of the Galaxy,Galaxy within the Local Group etc.

• motion causes temperature dipole:

T(θ) = T0

(1+

vc

cosθ)

+O(v2/c2)

• COBE has measured the dipole:

v = (371±1)km s−1

towards

l = (264.3±0.2) , b = (48.1±0.1)

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