BIG BANG NUCLEOSYNTHESIS I. EQUILIBRIUM ABUNDANCES The binding energies of the first four light nuclei, 2H, 3H, 3He and 4He are 2.22 MeV, 6.92 MeV, 7.72 MeV and 28.3 MeV, respectively (see Table I). As the universe cools below these temperatures, one expects these bound structures to form. The abundance of light elements, which are synthesized in the early universe, can be used to obtain important constraints on the cosmological parameters. Though the energy considerations suggest that these nuclei could be formed when the temperature of the universe is in the range (1 30) MeV, the actual synthesis takes place − only at a much lower temperature, T 0.1 MeV. The main reason for this delay is the BBN ≃ ‘high entropy’ of our universe, i.e., the high value for the photon-to-baryon ratio, η−1. Let us provisionally assume that the nuclear (and other) reactions are fast enough to maintain thermal equilibrium between various species of particles and nuclei. In thermal A equilibrium, the number density of a nuclear species NZ with atomic mass A and charge Z will be 3/2 mAT µA mA nA = gA exp − . (1) 2π T In particular, the equilibrium number densities of protons and neutrons are 3/2 mBT µp mp np 2 exp − , (2) ≃ 2π T 3/2 mBT µn mn nn 2 exp − . (3) ≃ 2π T The mass difference between the proton and the neutron Q m m = 1.293 MeV has ≡ n − p 3/2 to be retained in the exponent but can be ignored in the prefactor mA . We have set in the prefactor m m m , an average value. n ≃ p ≃ B A Since the chemical potential is conserved in the reactions producing NZ out of Z protons and (A Z) neutrons, µ for any species can be expressed in terms of µ and µ : − A p n µ = Zµ +(A Z)µ . (4) A p − n Writing µ m Zµ +(A Z)µ m exp A − A = exp p − n exp A T " T # − T 1 − = [exp(µ /T )]Z [exp(µ /T )](A Z) exp( m /T ) (5) p n − A and substituting for exp(µp/T ) and exp(µn/T ) from (2) and (3) we get 3A/2 µA mA −A Z (A−Z) 2π Zmp +(A Z)mn mA exp − = 2 np nn exp − − T mBT " T # 3A/2 −A Z (A−Z) 2π = 2 np nn exp(BA/T ), (6) mBT where B Zm +(A Z)m m (7) A ≡ p − n − A is the binding energy of the nucleus. Therefore, the number density (1) becomes 3(A−1)/2 −A 3/2 2π Z (A−Z) nA = gA2 A np nn exp(BA/T ). (8) mBT Since particle number densities in the expanding universe decrease as a−3 (for constant number per comoving volume), it is useful to use the total baryon density, nb = nn + np + i(AnA)i, as a fiducial quantity and to consider the ‘mass fraction’ contributed by nuclear speciesP A(Z), AnA XA . (9) ≡ nb −1 Note that i Xi = 1. Substituting for nA, np and nn in (8) by nA = nb(A XA) = −1 ηnγ(A XAP), np = ηnγ Xp and nn = ηnγXn, where nb −8 2 η =2.68 10 (Ωbh ) (10) ≡ nγ × 2 3 is the baryon-to-photon ratio and nγ = [2ζ(3)/π ]T0 is the number density of photons, we get 3(A−1)/2 A−1 Z A−Z XA = F (A)(T/mB) η Xp Xn exp(BA/T ) (11) where 5/2 A−1 (1−A)/2 (3A−5)/2 F (A)= gAA [ζ(3) π 2 ]. (12) Eq. (11) shows why the high entropy of the universe, i.e. small value of η, hinders the formation of nuclei. (For purposes of comparison, in a star like our sun, n /n 10−2; γ b ∼ even in the post-collapse core of a supernova, nγ /nb is only a few. Indeed, the entropy of the univesre is enormous.) To get X 1, it is not enough that the univesre cools to the A ≃ temperature T < BA; it is necessary that is cools still further so as to offset the small value ∼ 2 A−1 of the η factor. The temperature TA at which the mass fraction of a particular species A will be of order unity (X 1) is given by A ≃ BA/(A 1) TA − . (13) ≃ ln(1/η)+1.5 ln(mB/T ) 2 3 4 This temperature is much smaller than BA; for H, He and He the value of TA is 0.07 MeV, 0.11 MeV and 0.28 MeV, respectively. Comparison with the binding energy of these nuclei shows that these values are lower than BA by a factor of about 10, at least. II. FALLING OUT OF EQUILIBRIUM Thus, even when the thermal equilibrium is maintained, significant synthesis of nuclei can occur only at T < 0.3 MeV and not at higher temperatures. If such is the case, then we would ∼ expect significant production (XA 1) of nuclear species A at temperatures T < TA. It ≃ ∼ turns out, however, that the rate of nuclear reactions is not high enough to maintain thermal equilibrium between various species. We have to determine the temperatures up to which thermal equilibrium can be maintained and redo the calculations to find non-equilibrium mass fractions. In particular, we used the equilibrium densities for np and nn in the above analysis. In thermal equilibrium, the interconversion between n and p is possible through weak interac- tion processes: n p + e− +¯ν , ↔ e ν + n p + e−, e ↔ e+ + n p +¯ν . (14) ↔ e When the rates for these interactions are rapid enough compared to the expansion rate H, chemical equilibrium obtains, µ µ = µ µ , (15) n − p e − ν from which it follows that in chemical equilibrium nn Xn = = exp[ Q/T +(µe µν)/T ]. (16) np Xp − − Based upon the charge neutrality of the universe, we can infer that µ /T n /n = n /n e ∼ e γ p γ ∼ η, from which it follows that µ /T 10−10. The electron-neutrino number of the universe is e ∼ 3 similarly related to µν/T ; however, since this relic background has not been detected, none of the neutrino numbers is known. We will assume that the lepton numbers, like the baryon numbers, are small, so that µ /T 1. With this assumption, ν ≪ n n = exp( Q/T ). (17) np !EQ − The ratio will be maintained as long as the n p reactions are rapid enough. But when the − reaction rate Γ falls below the expansion rate, H =1.66g1/2T 2/m 5.5(T 2/m ) at some Pl ≃ Pl temperature T , the ratio (n /n ) will get frozen at the value exp( Q/T ). The only process D n p − D which can continue to change this ratio thereafter will be the beta decay, n p + e +¯ν, of → the free neutron. The neutron decay will continue to decrease the ratio until all neutrons are used up in forming bound nuclei. To determine TD we have to estimate Γ and use the condition Γ = H. The rate Γ can be calculated from the theory of the weak interactions. The rates are found by integrating the square of the matrix element for a given process, weighted by the available phase- space densities of particles (other than the initial nucleon), while enforcing four-momentum conservation. One obtains ∞ − 2 Γ(nνe pe ) = A dqν qν qe Ee (1 fe) fν , Ee = Eν + Q, → 0 − Z ∞ + 2 Γ(ne pν¯e) = A dqe qe qν Eν (1 fν ) fe , Eν = Ee + Q, → 0 − Z q0 − 2 2 2 Γ(n pe ν¯e) = A dqe qe qν Eν (1 fν) (1 fe) , q0 = Q me. (18) → 0 − − − Z q Here A is an effective coupling while fν and fe are the distribution functions for electrons and neutrinos. Although the weak interaction coupling GF is known quite accurately from muon decay, the value of A or, equivalently, the neutron lifetime, cannot be directly determined from this alone because neutrons and protons also interact strongly, hence the ratio of nucleonic axial vector (GA) and vector (GV ) couplings is altered from unity. Moreover, relating these couplings to the corresponding experimentally measured couplings of the u and d quarks is complicated by weak isospin violating effects. If we assume conservation of the weak vector current (CVC), then G = G cos θ where sin θ =0.22 = V . However, V F c c | us| the weak axial current is not conserved and GA for nucleons differs from that for the first generation quarks. These non-perturbative effects cannot be calculated reliably, hence GA (in practice, GA/GV ) must be measured experimentally. The neutron lifetime is then given 4 by 5 2 −1 me 2 GA τn = 3 GV 1+ 2 f, (19) 2π GV ! where f = 1.715 is the integral over the final state phase space (including Coulomb cor- rections) and G is usually determined from superallowed 0+ 0+ pure Fermi decays of V → suitable light nuclei. It is thus more reliable to measure the neutron lifetime directly, and then relate it to the coupling A in (18) in order to obtain the other reaction rates. The experimental value is τ = 885.7 0.8 sec. (In the literature one often finds the neutron half n ± life τ1/2(n)= τn ln 2.) Thus we obtain, for example, τ −1(T/m )3 exp( Q/T ) T Q, m , n e − ≪ e Γ(pe νn) 7π(1+3g2 ) (20) → ≃ A G2 T 5 T Q, m .