Big Bang Nucleosynthesis (BBN) Eildert Slim

Timeline of the Universe 0 sec Big Bang: Start of the expansion secPlanck-time: Gravity splits off secStrong force splits off secInflation begins secInflation ends secWeak force splits off secT = 2000 MeV Nucleon and anti-nucleon pairs annihilate: Only 1 in 10 9 particles remain. 1 secT = 1 MeV Electron and positron pairs annihilate: Only 1 in 10 9 particles remain. Nuclear fusion begins. 3 minT = 0.1 MeV Nuclear fusion ends yearNeutral atoms are formed.

Baryon Asymmetry Baryon abundance parameter η B = (n B - n ̅ B )/n γ η B very small: η 10 = η B Matter-antimatter asymmetry: η 10 = (n B /n γ ) 0 = 274Ω B h 2 η 10 ≈ 6

Neutron-Proton Reactions Weak interactions: n ↔ p + e - + ̅ ν ν + n ↔ p + e - e + + n ↔ p + ̅ ν n γ >> n B

Rate of Weak Reactions: Г -Integrate square of matrix element -Weigh by phase-space densities -Enforce four-momentum conservation For T > m e : Г/H ~ (T/0.8 MeV) 3

Expansion rate H at BBN Friedman equation: H 2 = (8π/3)G N ρ TOT Prior to BBN: ρ TOT = ρ γ + ρ e + 3ρ ν = (43/8)ρ γ During SBBN: ρ TOT = ρ γ + 3ρ ν = 1.68ρ γ

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 μ n + μ ν = μ p + μ e

Nuclear Statistical Equilibrium II Number density n A of non-relativistic nuclear species A(Z): n A = g A (m A T/2π) 3/2 exp((μ A – m A )/T) Chemical potential μ A of A(Z): μ A = Z μ p + (A – Z)μ n

Nuclear Statistical Equilibrium III Number density n A of A(Z): n A = g A A 3/2 2 -A (2π/m N T) 3(A-1)/2 n p Z n n A-Z exp(B A /T) With binding energy B A : B A = Zm p + (A – Z)m n – m A

Mass Fraction Total nucleon density: n N = n n + n p + Σ i (An A ) i Mass fraction contributed by A(Z): X A = n A A/n N Σ i X i = 1

Neutron-Proton ratio in Equilibrium Until T ≈ 0.8 MeV: n n /n p = X n /X p = exp [-Q/T + (μ e – μ ν )/T] Where Q = m n – m p = MeV Assume (μ e – μ ν )/T small: (n n /n p ) EQ = exp(-Q/T)

Equilibrium vs non-Equilibrium I

Neutron-Proton Ratio T >> 0.8 MeV: X n = X p T > 0.8 MeV: X n /X p calculated by NSE (n/p) freeze-out = exp (-Q/T F ) ≈ 1/6 → T F ≈ 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 → 2 H 2 H + p → 3 He 3 He + n → 4 He

Basic Fusion Processes II

Light elements’ isotopes Mass“Stable” isotope 1H 2 2H2H 3 3 H, 3 He 4 4 He 6 6 Li 7 7 Li 9 9 Be

Energy vs Entropy I Energy: Binding Energy of 2 H = 2,2 MeV At T < 2,2 MeV 2 H is energetically favoured Entropy: Number of photons >> number of baryons Many photons higher than average energy High energy photons break up 2 H Thermodynamics combines energy and entropy

Energy vs Entropy II Estimate T at which 2 H becomes thermodynamically favoured: T NUC = (B A /(A – 1))/(ln(η -1 ) + 1.5ln(m N /T)) 2 H: T NUC = 0.07MeV 3 He: T NUC = 0.11MeV 4 He: T NUC = 0.28 MeV

Production of Light Elements I t = sec, T = 10 MeV - Very small abundance of light nuclei -High energy photons destroy light nuclei immediately X n, X p = 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/T F ) ≈ 1/6 -(n/p) continues to decrease due to neutron decay X n ≈ 1/7, X p ≈ 6/7

Production of Light Elements III t ≈ 1 min, T ≈ 0.3 MeV - 4 He becomes thermodynamically favoured -High energy photons destroy 2 H, 3 H and 3 He quickly -Coulomb-barrier suppression becomes significant → 4 He production is slowed down

Production of Light Elements IV t ≈ 3 min, T ≈ 0.1 MeV - 2 H, 3 H and 3 He become thermodynamically favoured -More 4 He is produced -All free neutrons are bound into 4 He.

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, 3 He and 7 Li is synthesized: 7 Li/H ~ to D, 3 He/H ~ to Remaining neutrons are bound into 4 He: X 4 ≈ 2(n/p) NUC /(1 + (n/p) NUC = (2/7)/(1 + 1/7) = 1/4

Important Parameters Higher τ 1/2 (n): -Decreases weak rates -Causes freeze out at higher T -Larger 4 He abundance Higher g*: -Faster expansion rate -Causes freeze out at higher T Higher η: -Fewer photons - 2 H, 3 H, 3 He build up earlier -Less 2 H, 3 H, 3 He 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 2 H Properties: -Easy to destroy, hard to produce: -Observations provide lower bound Abundance has been measured: -In solar system studies -In studies of deuterated molecules -In UV absorption studies of local interstellar medium Results: - 2 H/H = (1.5 to 2.9) x : -η <

Observations of 3 He Properties: -Produced from 2 H in stars -Difficult to destroy without producing heavier elements Abundance has been measured: -In solar system studies: -In meteorites corresponds with pre-solar 3 He -In solar wind corresponds with pre-solar 2 H + 3 He -In galactic HII regions Results: - 3 He+/H ≈ (1.2 to 15) x [( 2 H + 3 He)/H] ≈ (3.6 ± 0.6) x Can be astrated by a factor ≤ 2: -[( 2 H + 3 He)/H] P ≤ 8 x 10 -5

Observations of 7 Li Properties: -Produced by cosmic rays and in stars -Easily destroyed Abundance has been measured: -In unevolved halo stars with low metal abundances -Plateau in 7 Li abundance was found in heavy stars: -Lower mass stars astrate 7 Li Results: -Primordial 7 Li abundance follows from plateau: - 7 Li/H ≈ (1.1 ± 0.4) x : -η = (1 to 7) x

Observations of 4 He -Produced in stars -Stars produce metals too -Correlation between 4 He and metals shows primordial 4 He -Y P ≈ 0.22 to 0.26

Observations: Conclusion -Nucleosynthesis produces only 2 H, 3 He, 4 He and 7 Li. -Good agreement between predicted and observed abundances -Corresponding parameter-values: min ≤ τ 1/2 (n) ≤ 10.7 min -4 x ≤ η ≤ 7 x

Caveats -Theorized generation of heavy stars -Could destroy 3 He and 7 Li -Present agreement would disappear -X n /X p = exp [-Q/T + (μ e – μ ν )/T] -μ ν unknown -Is assumed small or equal to μ e -Would affect 4 He abundance if not

Dark matter Assuming standard model is valid: 4 x ≤ η ≤ 7 x ≤ Ω B ≤ Measurements of Ω 0 suggest Ω 0 = 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.