Big Bang Nucleosynthesis

(BBN)

Eildert Slim

Timeline of the Universe

0 sec Big Bang: Start of the expansion.

10-43 sec Planck-time: Gravity splits off.

10-36 sec Strong force splits off.

10-35 sec Inflation begins.

10-33 sec Inflation ends.

10-10 sec Weak force splits off.

10-5 sec T = 2000 MeV

Nucleon and anti-nucleon pairs annihilate: Only 1 in 109 particles remain.

1 sec T = 1 MeV

Electron and positron pairs annihilate: Only 1 in 109 particles remain.

Nuclear fusion begins.

3 min T = 0.1 MeV

Nuclear fusion ends.

380 000 year Neutral atoms are formed.

Baryon Asymmetry

Baryon abundance parameter ηB = (nB - n ̅B)/nγ

ηB very small: η10 = 1010 ηB

Matter-antimatter asymmetry:

η10 = 1010 (nB/nγ)0= 274ΩBh2

η10 ≈ 6

Neutron-Proton Reactions

Weak interactions:

n ↔ p + e- +  ̅̅ν

ν + n ↔ p + e-

e+ + n ↔ p + ̅ν

nγ >> nB

Rate of Weak Reactions: Г

- Integrate square of matrix element

- Weigh by phase-space densities

- Enforce four-momentum conservation

For T > me:

Г/H ~ (T/0.8 MeV)3

Expansion rate H at BBN

Friedman equation: H2 = (8π/3)GN ρTOT

Prior to BBN: ρTOT = ργ + ρe + 3ρν = (43/8)ργ

During SBBN: ρTOT = ργ + 3ρν = 1.68ργ

Nuclear Statistical Equilibrium I

Kinetic equilibrium:

- Particles have same temperature

Chemical equilibrium:

- Same forward and reverse reaction rates

- Applies if Г >> H

Valid until T ≈ 0.8MeV

Implies μn + μν = μp + μe

Nuclear Statistical Equilibrium II

Number density nA of non-relativistic nuclear species A(Z):

nA = gA (mAT/2π)3/2 exp((μA – mA)/T)

Chemical potential μA of A(Z):

μA = Z μp + (A – Z)μn

Nuclear Statistical Equilibrium III

Number density nA of A(Z):

nA = gAA3/22-A(2π/mNT)3(A-1)/2npZnn

A-Z exp(BA/T)

With binding energy BA:

BA = Zmp + (A – Z)mn – mA

Mass Fraction

Total nucleon density:

nN = nn + np + Σi(AnA)i

Mass fraction contributed by A(Z):

XA = nAA/nN

ΣiXi = 1

Neutron-Proton ratio in Equilibrium

Until T ≈ 0.8 MeV:

nn/np = Xn/Xp = exp [-Q/T + (μe – μν)/T]

Where Q = mn – mp = 1.293 MeV

Assume (μe – μν)/T small:

(nn/np)EQ = exp(-Q/T)

Equilibrium vs non-Equilibrium I

Neutron-Proton Ratio

T >> 0.8 MeV:

Xn = Xp

T > 0.8 MeV:

Xn/Xp calculated by NSE

(n/p)freeze-out = exp (-Q/TF) ≈ 1/6

→ TF ≈ 0.7 MeV

Basic Fusion Processes I

Fusion reaction: A + B → C + …

Possible reactions: - Need maximum of 2 particles A and B- A or B must exist in sufficient quantity

2-particle reactions:p + n → 2H2H + p → 3He3He + n → 4He

Basic Fusion Processes II

Light elements’ isotopes

Mass “Stable” isotope

1 H

2 2H

3 3H, 3He

4 4He

6 6Li

7 7Li

9 9Be

Energy vs Entropy I

Energy:

Binding Energy of 2H = 2,2 MeV

At T < 2,2 MeV 2H is energetically favoured

Entropy:

Number of photons >> number of baryons

Many photons higher than average energy

High energy photons break up 2H

Thermodynamics combines energy and entropy

Energy vs Entropy II

Estimate T at which 2H becomes thermodynamically favoured:

TNUC = (BA/(A – 1))/(ln(η-1) + 1.5ln(mN/T))

2H: TNUC = 0.07MeV3He: TNUC = 0.11MeV4He: TNUC = 0.28 MeV

Production of Light Elements I

t = 10-2 sec, T = 10 MeV

- Very small abundance of light nuclei

- High energy photons destroy light nuclei immediately

Xn, Xp = 0.5

Production of Light Elements II

t ≈ 1 sec, T ≈ 1 MeV

- Very small abundance of light nuclei

- Weak interactions freeze out at Г < H

- (n/p)freeze-out = exp (-Q/TF) ≈ 1/6

- (n/p) continues to decrease due to neutron decay

Xn ≈ 1/7, Xp ≈ 6/7

Production of Light Elements III

t ≈ 1 min, T ≈ 0.3 MeV- 4He becomes thermodynamically favoured

- High energy photons destroy 2H, 3H and 3He quickly

- Coulomb-barrier suppression becomes significant

→ 4He production is slowed down

Production of Light Elements IV

t ≈ 3 min, T ≈ 0.1 MeV

- 2H, 3H and 3He become thermodynamically favoured

- More 4He is produced

- All free neutrons are bound into 4He.

Production of Light Elements V

Heavy Elements

- Stable at high T

- Need lighter elements to form

- Lighter elements have too low abundance at high T

- At low T Coulomb-barrier suppression too strong

Elements Produced

Some D, 3He and 7Li is synthesized: 7Li/H ~ 10-10 to 10-9

D, 3He/H ~ 10-5 to 10-4

Remaining neutrons are bound into 4He:

X4 ≈ 2(n/p)NUC/(1 + (n/p)NUC = (2/7)/(1 + 1/7) = 1/4

Important Parameters

Higher τ1/2(n):- Decreases weak rates- Causes freeze out at higher T- Larger 4He abundance

Higher g*:- Faster expansion rate- Causes freeze out at higher T

Higher η: - Fewer photons- 2H, 3H, 3He build up earlier- Less 2H, 3H, 3He remains unburnt

Dependence on η

Observations

We would like to measure primordial, cosmic abundances.

We can only measure present-day abundances in selected sites.

Observations of 2H

Properties:- Easy to destroy, hard to produce:

- Observations provide lower boundAbundance has been measured:- In solar system studies- In studies of deuterated molecules- In UV absorption studies of local interstellar

mediumResults:- 2H/H = (1.5 to 2.9) x 10-5:

- η < 10-9.

Observations of 3He

Properties:- Produced from 2H in stars- Difficult to destroy without producing heavier elementsAbundance has been measured:- In solar system studies:

- In meteorites corresponds with pre-solar 3He- In solar wind corresponds with pre-solar 2H + 3He

- In galactic HII regionsResults:- 3He+/H ≈ (1.2 to 15) x 10-5

- [(2H + 3He)/H] ≈ (3.6 ± 0.6) x 10-5

- Can be astrated by a factor ≤ 2:- [(2H + 3He)/H]P ≤ 8 x 10-5

Observations of 7Li

Properties:- Produced by cosmic rays and in stars- Easily destroyedAbundance has been measured:- In unevolved halo stars with low metal abundances- Plateau in 7Li abundance was found in heavy stars:

- Lower mass stars astrate 7LiResults:- Primordial 7Li abundance follows from plateau:

- 7Li/H ≈ (1.1 ± 0.4) x 10-10:- η = (1 to 7) x 10-10

Observations of 4He

- Produced in stars

- Stars produce metals too

- Correlation between 4He and metals shows primordial 4He

- YP ≈ 0.22 to 0.26

Observations: Conclusion

- Nucleosynthesis produces only 2H, 3He, 4He and 7Li.

- Good agreement between predicted and observed abundances

- Corresponding parameter-values:- 10.3 min ≤ τ1/2(n) ≤ 10.7 min

- 4 x 10-10 ≤ η ≤ 7 x 10-10

Caveats

- Theorized generation of heavy stars

- Could destroy 3He and 7Li

- Present agreement would disappear

- Xn/Xp = exp [-Q/T + (μe – μν)/T]

- μν unknown

- Is assumed small or equal to μe

- Would affect 4He abundance if not

Dark matter

Assuming standard model is valid:

4 x 10-10 ≤ η ≤ 7 x 10-10

0.015 ≤ ΩB ≤ 0.016

Measurements of Ω0 suggest Ω0 = 0.2 ± 0.1

Hence a non-baryonic form of matter must account for difference.

References:

Graphs from “The Early Universe”, by Edward W. Kolb and Michael S.Turner

Fusion image from ned.ipac.caltech.edu.