Big Bang Nucleosynthesis Grazia Luparello PhD Course Physics of the early Universe Grazia Luparello 1 / 24
Summary 1 Introduction 2 Neutron - proton ratio (at T<1MeV, T 1MeV, T>1MeV) 3 Reactions for the synthesis of light nuclei 4 Light elements abundances 5 Baryon content of the Universe Grazia Luparello 2 / 24
Introduction (I) Big Bang Nucleosynthesis refers to the production of light nuclei (D, 3 He, 4 He and 7 Li) during the early phase of the universe. It took place in the early Universe when: T (10-0.1) MeV t (10 2-10 2 ) s Grazia Luparello 3 / 24
Introduction (II) BBN is a good probe of the standard hot big bang cosmology. The teoretical predicted quantities are in good agreement with the primordial aboundances inferred from observational data. This is particularly impressive given that these abundances span nine orders of magnitude: from 4 He/H 0.08 down to 7 Li/H 10 10 BBN also provides powerful constraints on possible deviations from the standard cosmology and on new physics beyond the Standard Model. Grazia Luparello 4 / 24
Neutron proton ratio The baryons (non relativistic particles) have a number density n given by the equation: n p(n) g ( ) 3/2 mp(n) T e 2π m p(n) T Neutron proton ratio n n n p ( ) 3/2 mn e Q k B T m p where: Q = (m n m p )c 2 1.3MeV is the neutron-proton mass dierence. Grazia Luparello 5 / 24
T>1 MeV At T>1 MeV the plasma contains the following particles at equilibrium: relativistic particles e +, e, ν i, ν i, γ non-relativistic particles n,p Neutrons and protons are maintained both in kinetic and chemical equilibrium by charged current weak interactions: n p + e + ν e n + ν e p + e n + e + p + ν e At T>1 MeV nn n p ( ) 3/2 m n e m p Q k B T = 1 Grazia Luparello 6 / 24
T 1 MeV: neutron freeze-out (I) To decide if a specie of particles is in equilibrium with the plasma we have to compare: the expanction rate of the Universe H the interaction rate Γ When Γ/H=1 the particles go out of the equilibrium. The temperature at which Γ/H=1 is called freeze-out temperature and for neutrons it is 1 MeV Grazia Luparello 7 / 24
T 1 MeV: neutron freeze-out (II) Interaction rate: Expantion rate: Γ G F T 5 with G F = 10 5 m 2 p H 2 = 8πG 3 ρ(t ) = 4 45 π3 g (T )GT 4 with G = 1 (where ρ(t ) = π2 30 g (T )T 4 and g (T ) = 43/4) m 2 Pl Grazia Luparello 8 / 24
T 1 MeV: neutron freeze-out (III) Freeze-out temperature: ( ) 3 Γ T H 0.8MeV T fr 1MeV We can calculate the neutron - proton ratio: n n n p = e Q/T fr 1/6 Grazia Luparello 9 / 24
T<1MeV After freeze-out the only reaction that changes the number of neutrons is neutron dacay: The n n /n p ratio goes down to 1/7. n p + e + ν e Without further reactions to preserve neutrons within stable nuclei, the Universe would be pure hydrogen. The reaction that preserves the neutrons is deuterium formation. Grazia Luparello 10 / 24
Deuterium synthesis The reaction that produce D is: p + n D + γ The reaction is exothermic with a bound energy of 2.2 MeV. But photo-dissociation by the high number density of photons delays production of deuterium until T 0.1MeV, about 100s after Big Bang. Grazia Luparello 11 / 24
4 He Synthesis Once deuteron formation has occurred, further reactions proceed to make helium nuclei. 3 H, 3 He and 4 Heare made. Example of sequence of reactions: D + n 3 H + γ 3 H + p 4 He+ γ D + p 3 He+ γ 3 He+n 4 He+ γ The creation of deuterium is crucial. It has to be formed in appreciable amount, before the other reactions can start. Many body interactions such as 2p + 2n 4 He are in general ineective, due to the small number densities of nucleons. These reactions only go one way. Grazia Luparello 12 / 24
Number density of D The number density of D is given by the equation (non relativistic particles): n d = g ( md T 2π ) 3/2 e m d T n d = 3 ( ) 3/2 md 2π e m (mn +mp ) d T n n n p 4 m n m p T Grazia Luparello 13 / 24
Mass fraction We can dene the Mass Fraction of D: where: X d = n d = 3 ( ) 3/2 md n tot 4 X 2π nx p n tot e B/T m n m p T n tot : n n + n p + 2n d ; X n, X p : mass fraction of n and p; B = m d + (m n + m p ). Grazia Luparello 14 / 24
Relation between X and η We can dene the barion to photon ratio η as: η = n tot n γ n tot = ηn γ The number density of γ is given by: X d = 3 n γ = 2ζ(3) π 2 T 3 ( ) 1/2 ( ) 3/2 2 T ζ(3)x n X p η e B/T π m d Grazia Luparello 15 / 24
We can extend the formula to nucleos with Z and A: X A = n AA n tot where: ( ) 3(A 1)/2 n A = g(a)a 3 2 2 A 2π n Z m N T p na Z n e B A/T n tot = n n + n p + (An A ) i B A = Zm p + (A Z)m n m A g(a) = 2S A + 1 (S:spin) «3(A 1)/2 X A = g(a)(ζ(3)π (1 A)/2 2 (3A 5)/2 )A 3/2 T η A 1 X Z p X A Z n e B A/T m N The D, 3 He, 4 He and 7 Li abundances depend only on the parameter η. Grazia Luparello 16 / 24
Mass fraction of 4 He Nearly all neutrons surviving down to freeze out are captured in 4 He because of its large binding energy per nucleon. Its mass fraction is then approximately given by: X4 He 4n/2 n + p = 2n/p 1 + n/p = 2 1/7 1 + 1/7 = 1 4 The 4 He abundance is quite insensitive to η and almost only depends on the weak reaction rates determining neutron freeze out. Grazia Luparello 17 / 24
End of nuclosynthesis For t>10 3 s BBN is over since the low temperature and low density in the Universe suppress nuclear reactions. Nuclei heavier than 4 He do not form in any signicant quantity both because of absence of stable nuclei with A = 5,6 and the large Coulomb barriers for reactions. The most abundant nuclides with which we are left are hydrogen (p) and 4 He (incorporating almost all neutrons), followed by trace amounts of D, 3 He and 7 Li. Grazia Luparello 18 / 24
Light nuclei abundances vs. η Aboundances of 4 He, D, 3 He and 7 Li as predicted by the standard model of the hot BBN. Boxes indicated the observed light element aboudances. The narrow vertical band indicates the CMB measure of the cosmic baryon density. (from Fields and Sarkar, astro-ph/0601514) Grazia Luparello 19 / 24
Barion density The relation between the entropy density and n γ is given by: s = π4 ζ(3) g s (T )n γ = 1.8g s (T )n γ 45 η = n B s 1.8g s(t ) So to evaluate η we have to know the n B /s ratio. Grazia Luparello 20 / 24
Barion density If there isn't any interaction that change the baryon number this ratio is costant: n B s = n B,0 s 0 n B,0 = ρ B,0 = ρ B,0 1 ρ c = (Ω B h 2 ) ρ c/h 2 = 1.07 10 5 (Ω B h 2 )cm 3 m ρ c m m where: ρ c = 1.05 10 4 h 2 evcm 3 : m is the eective mass of the barion into the nucleon = m p B 0.931MeV n B,0 = 1.07 10 5 cm 3 (Ω s 0 2926cm 3 B h 2 ) = 0.37 10 8 (Ω B h 2 ) At present time: η = 1.8 43 11 0.37 10 8 = 2.6 10 8 (Ω B h 2 ) Grazia Luparello 21 / 24
Baryon density 4.7 10 10 η 6.5 10 10 0.017 Ω B h 2 0.024 We note that: Ω B 1 baryons cannot close the universe. Ω B Ω lum most baryons are optically dark Ω M 0.3 most matter in the universe is not only dark but also takes some non-baryonic form. Grazia Luparello 22 / 24
Beyond the Standard Model BBN can be used to test the Big Bang model or, alternatively one can consider possible physics beyond the SM and then use all the abundances to test such model. For example: The 4 He is proportional to the n/p ratio which is determinated when the weak interaction rates fall behind the Hubble expantion rate. The presence of other neutrino avors (or any other relativistic species) increase g ast leading to a large value of T fr, n/p and X4 He. We can use measured helium abundance to place limit on g Grazia Luparello 23 / 24
That's all. Thank you Grazia Luparello 24 / 24