Slide

Thermal History of
Subject chosen from CHIPP Winter School
the Universe
Anne Keune
04-04-2011

02/15
Overview
• In this presentation I will shortly discuss the following topics
(which follow the expansion of the universe):
• neutrino decoupling
• e+/e- annihilation
• big bang nucleosynthesis
• recombination
• photon decoupling

03/15
Neutrino decoupling
• The early universe was dominated by relativistic particles (e + /e - /νe/γ),
hot and dense and not opaque to even weakly interacting neutrinos.
all particles in thermodynamic equilibrium
Weak interactions:
n + νe ↔ p + e - , n + νe + e + ↔ p, e + + e - ↔ νe + νē
• But as the universe expanded and cooled down, the distance between
particles increased and the expansion rate of the universe (H)
overtook the neutrino interaction rate (Γ)
• At this point the neutrino’s departed from equilibrium and decoupled
Γ >> H: neutrino’s in equilibrium
Γ

• The weak interaction rate is given by
nν = (3gζ(3)T 3 )/(4π 2 )
back-up 1
04/15
Neutrino decoupling
• The expansion rate of the universe
• Ratio:
Γ(T) = n ⋅ 〈σν〉 ≅ 2.54 ⋅ GF 2 ⋅ T 5
≅ 3.48 ⋅ 10 -22 ⋅ (T/MeV) 5 .
H(T) = (8πG/3 ⋅ ρ) 1/2 (Friedmann eq.)
≅ 4.46 ⋅ 10 -22 ⋅ (T / MeV) 2 .
Γ(T)/H(T) ≅ 0.78 ⋅ (T/MeV) 3 .
• Neutrino decoupling thus occurred at
T ≅ 1.09 MeV, only ∼7 seconds after the
creation of the universe.
Fermi coupling constant
GF = 1.17⋅10 -11 MeV -2 .
Newton constant
G = 6.707 ⋅ 10 -45 MeV -2 .
Total energy density
ρ = T 4 /30 ⋅ π 2 ⋅ g*.
Energy degrees of freedom
g* = 2 + ⅞(3 ⋅ 2 + 2 ⋅ 2)
g* = ∑B gB ⋅ (T/Tγ) 4
+ ⅞ ⋅ ∑F gF ⋅ (T/Tγ) 4
gγ = 2 : photon spin states
ge = 2 ⋅ 2 : e + /e - spin states
gν = 3 ⋅ 2 : six neutrino’s
T = Tγ = Tν : coupled species

Nowadays...
05/15
e + /e - annihilation
• Energy dropped below me = 0.5 MeV after decoupling . The energy
released by the annihilation of electrons and positrons heated the
photons, but not the decoupled neutrinos.
• For Tνdec > T > me : g*S = 2 + 7/8(2 ⋅ 2) = 11/2 (neutrino’s decoupled)
For me > T : g*S = 2 (electrons annihilated)
So (Tν/Tγ) 3 = g*Sbefore/g*Safter = 11/4
DMR 53 GHz Maps
Tγ = 2.728 K
The temperature of background photons is higher than of
background neutrinos.
Tν expected to be ∼1.95 K
Tγ ≅ 1.4 Tν
Entropy density: s(T) ∝ g*S ⋅ T 3
g*S = ∑B gB ⋅ (T/Tγ) 3 + 7/8 ⋅ ∑F gF ⋅ (T/Tγ) 3
Adiabatic expansion: d(sa 3 ) = 0 ⇒ (aT) 3 g*S = const.

06/15
Cosmic neutrino
background (CNB)
• But... current neutrino detectors cannot detect neutrinos with an
energy as low as the neutrino background.
Measuring the CNB temperature can be done through measuring
the ratio between Hydrogen and Helium.
• The proton/neutron ratio is dependent of T.
• For this we need to look at the nucleosynthesis process...
4 He
H ?

07/15
back-up 1
Nucleosynthesis
• Before neutrino decoupling all particles were in thermodynamic
equilibrium, including the non-relativistic baryons.
The baryon number density is:
n ≅ g ⋅ (mT/(2π)) 3/2 ⋅ exp(-m/T)
• At ∼0.7 MeV the ratio “freezes in” at
The proton/neutron ratio gives:
nn/np ≅ exp(-(mn-mp)/T)
nn/np ≅ 1/6
• The only reaction that hereafter changes the number of neutrons is
neutron decay, reduces the ratio to
nn/np ≅ 1/7
• Without further reactions to preserve neutrons in stable nuclei, the
universe would be pure hydrogen.
mp ≅ 938.3 MeV, mn ≅ 939.6 MeV
mn-mp ≅ 1.3 MeV

08/15
Nucleosynthesis
• The reaction that preserves neutrons is deuterium ( 2 H) formation.
The number density can be written in terms of B:
ni ≅ gi ⋅ A 3/2 ⋅ exp(Bi/T) ⋅ 2 -A ⋅ np Z ⋅ nn (A-Z) ⋅ (mnT/(2π)) 3/2⋅(1-A)
• The mass fraction of a nucleus is Xi = Aini/nB = Aini/∑ Aini (∑Xi = 1)
with nB the total Baryon density.
Xi ≅ Ci ⋅ η (A-1) ⋅ exp(Bi/T) ⋅ Xp Z ⋅ Xn (A-Z) ⋅ (T/mn) 3/2⋅(A-1)
(present) ratio
Baryon/photon η = nB/nγ
= nB⋅π 2 ⋅(2ζ(3)T 3 ) -1 .
nB = nn + np.
Binding energy (B): Zmp + (Z-A)mn - m
back-up 2
back-up 3
2 H
p + n → D + γ

09/15
Nucleosynthesis
• Xi shows that equilibrium abundances come later (lower T) for nuclei
with small binding energies.
• He nuclei which has a larger binding energy and whose abundance
would become large earlier, cannot be formed.
Deuterium bottleneck
B (MeV) T (MeV)
2 H 2.22 0.07
3 H 6.92 0.10
3 He 7.72 0.11
4 He 28.3 0.28

10/15
Nucleosynthesis
• Once D nuclei are created, the one way process of the creation of
helium nuclei can begin.
• To good approximation all neutrons are now `captured’ in 4 He nuclei,
with a fraction of:
•
Fraction of 4 He nuclei: r = # 4 He/(#H + # 4 He)
r = ½⋅nn/(np - nn + ½⋅nn) = (nn/np)/(2 - nn/np) = (1/7)/(2-1/7) = 7.7%
Mass fraction of:
m4He/mB
= 2nnmp/(npmp + nnmp)
= 2nn/(np + nn)
= 2(nn/np)/(1+ (nn/np))
= 2(1/7)/(1+1/7)
= 25%
2 H 3 H/ 3 He 4 He
D + n → 3 H + γ ⇒ 3 H + p → 4 He + γ
D + p → 3 He + γ ⇒ 3 He + n → 4 He + γ

11/15
Helium/Hydrogen
• Measurements show the Helium/
Hydrogen ratio is in the region of
25%.
• This number is consistent with
our `back of the envelope’
calculation.
• In this calculation Tν ∼ 0.7 MeV
was used.
smaller boxes: ±2σ statistical error
larger boxes: ±2σ statistical and
systematic error

12/15
Recombination
• Radiation and matter in thermal equilibrium as long as there were lots
of free electrons.
• The density of free electrons reduces as T is low enough for electrons
and protons form neutral hydrogen.
before
after
nH in terms of binding energy (B = mp+me-mH = 13.6 eV):
nH ≅ gH/(gpge) ⋅ np ⋅ ne⋅ exp(B/T) ⋅ (meT/(2π)) -3/2
p + e → H + γ

13/15
Photon decoupling
• As the number of free electrons drops, photons decouple at Γ ∼ H
• We introduce fractional ionisation Xe=ne/nB:
assume:
nH = nB-np (only protons)
ne = np (neutral)
(1-Xe)/Xe = nHnB/(nenp) = 4√2ζ(3)/√π ⋅ η ⋅ (T/me) 3/2 ⋅ exp(B/T)
• When we assume recombination at Xe ∼ 0.1 and η ∼ 10 -10 ,
then Trec ∼ 0.31 eV
• Go back to Γγ and compare to matter dominated expansion
then Tdec ∼ 0.26 eV
Origin of CMB!

14/15
Summary
Energy (γ) Time Event
1 MeV 7 sec neutrino decoupling
0.5 MeV 10 sec e + /e - annihilation
70 keV 3 min nucleosynthesis
0.31 eV 300,000 yr recombination
0.26 eV 380,000 yr photon decoupling
0.2 meV 14 Gyr today

15/15
The End

ack-up 1
back-up: number density
• Bose-Einstein/Fermi-Dirac distribution function:
• fFB(k)d 3 k = g/(2π) 3 ⋅ [exp(E/T) ± 1] -1 d 3 k
• Number density: n = ∫ fFB(k) d 3 k
• T >> m (E ∼ k):
• nB ≅ gζ(3)/π 2 ⋅ T 3 ,
• nF ≅ 3gζ(3)/(4π 2 ) ⋅ T 3 ,
• T

ack-up 2
back-up: number density
• nB ≅ g ⋅ (mT/(2π)) 3/2 ⋅ exp(-m/T)
≅ g ⋅ (mT/(2π)) 3/2 ⋅ exp(B/T) ⋅ exp(-Zmp/T) ⋅ exp(-(A-Z)mn/T)
≅ g ⋅ (mT/(2π)) 3/2 ⋅ exp(B/T) ⋅ exp(-mp /T) Z ⋅ exp(-mn /T) (A-Z)
≅ g ⋅ (mT/(2π)) 3/2 ⋅ exp(B/T) ⋅ (2 -1 ⋅ np ⋅ (mpT/(2π)) -3/2 ) Z ⋅ (2 -1 ⋅ nn ⋅
(mnT/(2π)) -3/2 ) (A-Z)
≅ g ⋅ (mT/(2π)) 3/2 ⋅ exp(B/T) ⋅ 2 -A ⋅ np Z ⋅ nn (A-Z) ⋅ (mpT/(2π)) -3/2⋅Z ⋅
(mnT/(2π)) -3/2⋅(A-Z)
back-up 1
Binding energy (B): Zmp + (Z-A)mn - m
np/n ≅ 2 -1 np/n(mp/nT/(2π)) -3/2
m/A ∼ mp ∼ mn outside exponent
≅ g ⋅ (AmnT/(2π)) 3/2 ⋅ exp(B/T) ⋅ 2-A ⋅ np Z ⋅ nn (A-Z) ⋅ (mnT/(2π)) -3/2⋅A
≅ g ⋅ A3/2 ⋅ exp(B/T) ⋅ 2-A ⋅ np Z ⋅ nn (A-Z) ⋅ (mnT/(2π)) 3/2⋅(1-A)
Slide 8

ack-up 3
back-up: mass fraction
• ni = C0 ⋅ exp(Bi/T) ⋅ np Z ⋅ nn (A-Z) ⋅ (mnT) 3/2⋅(1-A)
Xi = Aini/nB
• Xi = C1 ⋅ nB -1 ⋅ exp(Bi/T) ⋅ np Z ⋅ nn (A-Z) ⋅ (mnT) 3/2⋅(1-A)
= C1 ⋅ nB (A-Z)-1 ⋅ exp(Bi/T) ⋅ np Z ⋅ Xn (A-Z) ⋅ (mnT) 3/2⋅(1-A)
= C1 ⋅ nB A-1 ⋅ exp(Bi/T) ⋅ Xp Z ⋅ Xn (A-Z) ⋅ (mnT) 3/2⋅(1-A)
η = nB/nγ = nB⋅π 2 ⋅(2ζ(3)T 3 ) -1
= C2 ⋅ ηA-1 ⋅ T3(A-1) ⋅ exp(Bi/T) ⋅ Xp Z ⋅ Xn (A-Z) ⋅ (mnT) 3/2⋅(1-A)
= C2 ⋅ ηA-1 ⋅ exp(Bi/T) ⋅ Xp Z ⋅ Xn (A-Z) ⋅ (T/mn) -3/2⋅(1-A)
C0, C1, C2 constants
back-up 2
Slide 8