CONSTRAINING NEUTRINOS WITH BBN (WITH A LITTLE HELP FROM THE - PowerPoint PPT Presentation

CONSTRAINING NEUTRINOS WITH BBN (WITH A LITTLE HELP FROM THE CMB) GGI NEUTRINO WORKSHOP & SMIRNOV FEST Gary Steigman Departments of Physics and Astronomy Center for Cosmology and Astro-Particle Physics Ohio State University

BBN – Predicted Primordial Abundances Depend On Three Physical / Cosmological Parameters : Baryon Density (Asymmetry) Parameter : • η B ≡ n N / n γ ; η 10 ≡ 10 10 η B = 274 Ω B h 2 10 10 Expansion Rate (Dark Radiation) Parameter : • S 2 = ( H ′ / H) 2 = G ′ ρ′ ρ′ / G ρ ≡ 1 + 7 Δ N ν / 43 Lepton (Neutrino) Asymmetry Parameter : • ξ = ξ ν = µ ν / T ν ( ξ ν = ξ ν e = ξ ν µ = ξ ν τ )

“Standard” Big Bang Nucleosynthesis (SBBN) For An Expanding Universe Described By General Relativity, With S = 1 ( Δ N ν = 0 = ξ ) The Relic Abundances Of D, 3 He, 4 He, 7 Li Depend Only On η B = η 10

SBBN – Predicted Primordial Abundances 4 He Mass Fraction Mostly H & 4 He BBN abundance of D ( 3 He, 7 Li) provides a good baryometer 7 Li 7 Be

Post – BBN Evolution of the Relic Abundances • As gas cycles through stars, D is only DESTROYED • As gas cycles through stars, 3 He is DESTROYED, PRODUCED and, some prestellar 3 He SURVIVES • Stars burn H to 4 He (and produce heavy elements) ⇒ 4 He INCREASES (along with CNO …) • Cosmic Rays and SOME Stars PRODUCE 7 Li BUT, 7 Li is DESTROYED in most stars

* Use D to constrain η B (mainly) * Use 4 He to constrain Δ N ν or ξ (mainly) (Use η B and Δ N ν or ξ to predict BBN 7 Li)

log (D/H) vs. Metallicity Observations of Deuterium In 12 High–Redshift (z), Low–Metallicity (Z) QSOALS Where is the D – Plateau ? No correlation between D/H and Metallicity

log (D/H) vs. Redshift Observations of Deuterium In 12 High–Redshift (z), Low–Metallicity (Z) QSOALS Where is the D – Plateau ? No correlation between D/H and Redshift

log (D/H) vs. Metallicity 5 + log (D/H) P = 0.42 ± 0.02 ⇒ η 10 = 5.96 ± 0.28

Y vs. O / H 4 He Observed in Low – Z Extragalactic H ΙΙ ΙΙ Regions Izotov & Thuan 2010

Y vs. O / H Y P (IT10) = 0.2565 ± 0.0010 ± 0.0050 Adopt : Y P = 0.2565 ± 0.0060 Izotov & Thuan 2010

SBBN ( Δ N ν = 0 = ξ ) IF : 5 + log(D/H) P = 0.42 ± 0.02 ⇒ η 10 = 5.96 ± 0.28 ⇒ Y P = 0.2476 ± 0.0007 Y P (OBS) − Y P (SBBN) = 0.0089 ± 0.0060 ⇒ Y P (OBS) = Y P (SBBN) @ ~ 1.5 σ IF Y P = 0.2565 ± 0.0060 ⇒ η 10 = 11.50 ± 3.77

But ! Lithium – 7 Is A Problem Li / H vs. Fe / H A(Li) ≡ 12 + log(Li/H) SBBN Asplund et al. 2006 Boesgaard et al. 2005 Aoki et al. 2009 Lind et al. 2009 Where is the Lithium Plateau ?

SBBN Predictions Agree With Observations Of D, 3 He, 4 He, But NOT With 7 Li When η 10 , Δ N ν , ξ are free parameters BBN abundances are functions of η 10 , Δ N ν , ξ Explore the constraints provided by D (D/H) and 4 He (Y P ) and use them to predict 7 Li (Li/H)

BBN – Predicted Y P vs. (D/H) P Δ N ν = 2 η 10 = 7.0 6.5 6.0 1 5.5 0

BBN – Predicted Y P vs. (D/H) P Δ N ν = 2 η 10 = 7.0 6.5 6.0 1 5.5 0

68 % & 95 % Contours of Δ N ν vs. η 10 BBN D & 4 He η 10 = 6.27 ± 0.34 & Δ N ν = 0.66 ± 0.46

For BBN ( Δ N ν ≠ 0 , ξ = 0) ⇒ η 10 = 6.27 ± 0.34 & Δ N ν = 0.66 ± 0.46 ⇒ Δ N ν = 0 @ ~ 1.4 σ ( Or ⇒ G BBN / G 0 = 1.11 ± 0.07 ) But, what about Lithium ? ⇒ A(Li) = 2.70 ± 0.06 (Too High !)

Chronology of Primordial Helium Abundance Determinations

Chronology Of The BBN – Inferred Values Of Δ N ν WMAP 7 Only recently is Δ N ν > 0 “favored”

The recent BBN support for Δ N ν > 0 is driven by the recent (uncertain) estimates of Y P Avoid the uncertainties in Y P by replacing BBN 4 He with CMB – determined η 10

68 % & 95 % Contours of Δ N ν vs. η 10 BBN D & CMB η 10 η 10 = 6.190 ± 0.115 & Δ N ν = 0.48 ± 0.64

68 % & 95 % Contours of Δ N ν vs. η 10

Comparing The BBN & CMB Constraints N eff = 3.046 + Δ N ν ACT WMAP 7 SPT BBN (D & 4 He) BBN (D) & CMB ( η 10 ) SPT + Cl BBN and the CMB agree , hinting at Dark Radiation (a Sterile Neutrino ?)

BBN (D & 4He) Allowing For Lepton Asymmetry (No Dark Radiation : Δ N ν = 0)

BBN – Predicted Y P vs. (D/H) P ξ = − 0.10 η 10 = 6.5 6.0 5.5 ξ = − 0.05 ξ = 0

BBN – Predicted Y P vs. (D/H) P ξ = − 0.10 η 10 = 6.5 6.0 5.5 ξ = − 0.05 ξ = 0

68 % & 95 % Contours of ξ vs. η 10 BBN D & 4 He η 10 = 6.01 ± 0.28 & ξ = − 0.038 ± 0.026

For BBN ( Δ N ν = 0 , ξ ≠ 0) ⇒ η 10 = 6.01 ± 0.28 & ξ = − 0.038 ± 0.026 ⇒ ξ = 0 @ ~ 1.5 σ But, what about Lithium ? ⇒ A(Li) = 2.69 ± 0.05 (Too High !)

BBN (D & 4 He) Allowing For Lepton Asymmetry And Dark Radiation Supplemented By A CMB Constraint On Δ N ν

ξ vs. Δ N ν (BBN D & 4 He) And CMB Δ N ν BBN CMB

For BBN ( Δ N ν ≠ 0 , ξ ≠ 0) And CMB ( Δ N ν = 0.82 ± 0.64) ⇒ η 10 = 6.34 ± 0.32 & ξ = 0.009 ± 0.035 But, what about Lithium ? ⇒ A(Li) = 2.70 ± 0.06 (Still Too High !)

CONCLUSIONS For Δ N ν ≈ 0 & ξ = 0, BBN (D, 3 He, 4 He) Agrees With The CMB + LSS (But , Lithium Is A Problem !) BBN + CMB + LSS Constrain Cosmology & Particle Physics