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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.
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