6 Big Bang Nucleosynthesis One quarter (by mass) of baryonic matter in the universe is helium, heavier elements make up a few percent, and the rest is hydrogen. The building blocks of atomic nuclei, the nucleons, or protons and neutrons, are formed in the QCD phase transition at T 100 MeV and t 30 µs. Elements ∼ ∼ heavier than lithium, up to iron, cobalt, and nickel have been made from lighter elements by fusion reactions in stars. These reactions provide the energy source for the stars. Elements heavier than these have been formed in supernova explosions. However, the amount of helium and some other light isotopes in the universe cannot be understood by these mechanisms. It turns out that 2H, 3He, 4He, and 7Li were mainly produced already during the first hour of the universe, in a process called Big Bang Nucleosynthesis (BBN). Nucleons and antinucleons annihilated each other soon after the QCD phase transition, and the small excess of nucleons left over from annihilation did not have a significant effect on the expansion and thermodynamics of the universe until much later, when the universe became matter-dominated (at t 1000ω−2 years 50 000 eq ≈ m ≈ years). The ordinary matter in the present universe comes from this small excess of nucleons. Let us now consider what happened to it in the early universe. We will focus on the period when the temperature fell from T 10 MeV to T 10 keV ∼ ∼ (from t 0.01 s to a few hours). ∼ 6.1 Equilibrium The total number of nucleons minus antinucleons stays constant due to baryon number conservation. In the temperature range under consideration, the number density of antinucleons is negligible. This baryon number can be in the form of protons and neutrons or atomic nuclei. Weak nuclear reactions convert neutrons and protons into each other and strong nuclear reactions build nuclei from them. During the period of interest the nucleons and nuclei are nonrelativistic (T ≪ mp). Assuming thermal equilibrium we have 3/2 µi−mi miT n = g e T (6.1) i i 2π for the number density of nucleus type i. If the nuclear reactions needed to build nucleus i (with mass number Ai and charge Zi) from the nucleons, (A Z )n + Z p i i − i i ↔ occur at sufficiently high rate to maintain chemical equilibrium, we have µ =(A Z )µ + Z µ (6.2) i i − i n i p for the chemical potentials. Since for free nucleons we have 3/2 µ −m mpT p p n = 2 e T p 2π 3/2 µ −m mnT n n n = 2 e T , (6.3) n 2π 86 6 BIG BANG NUCLEOSYNTHESIS 87 A Z i Bi gi 2H 2.22 MeV 3 3H 8.48 MeV 2 3He 7.72 MeV 2 4He 28.3 MeV 1 12C 92.2 MeV 1 Table 1. Some of the lightest nuclei and their binding energies. we can express ni in terms of the neutron and proton densities, 3 i− 3 2 (A 1) 2 −Ai 2π Zi Ai−Zi Bi/T ni = giAi 2 np nn e , (6.4) mNT where B Z m +(A Z )m m (6.5) i ≡ i p i − i n − i is the binding energy of the nucleus. Here we have approximated m m m /A p ≈ n ≈ i outside the exponent, and denoted it by mN (“nucleon mass”). The different number densities add up to the total baryon number density Aini = nB . (6.6) X The baryon number density nB we get from the photon density 2 n = ζ(3)T 3 (6.7) γ π2 and the baryon/photon -ratio n g∗ (T ) B = s η (6.8) nγ g∗s(T0) as g∗s(T ) 2 3 nB = η 2 ζ(3)T . (6.9) g∗s(T0) π After electron-positron annihilation g∗s(T ) = g∗s(T0) and nB = ηnγ. Here η is the present baryon/photon ratio. It can be estimated from various observations, and it is about 6 10−10. × For temperatures m T & B we have N ≫ i (m T )3/2 T 3 n >n ,n N ≫ ≫ B p n and thus (6.4) implies that n n ,n i ≪ p n for Ai > 1. Thus initially there are only free neutrons and protons in large numbers. 6 BIG BANG NUCLEOSYNTHESIS 88 6.2 Neutron-proton ratio What can we say about np and nn? Protons and neutrons are converted into each other by the weak interaction in the reactions − n+ νe p+e + ↔ n+e p +ν ¯e (6.10) ↔ − n p+e +ν ¯ . ↔ e If these reactions are in equilibrium, we have µn + µνe = µp + µe, and the neu- tron/proton ratio is nn n − − = e Q/T +(µe µνe )/T , (6.11) np ≡ p where Q m m = 1.293 MeV. ≡ n − p We need now some estimate of the chemical potentials of electrons and electron neutrinos. The universe is electrically neutral, so the number of electrons (or n − e − ne+ ) equals the number of protons, and µe can be calculated exactly in terms of η and T . We leave the exact calculation as an exercise, but give below a rough estimate for the ultrarelativistic limit (T m ): ≫ e 3 3 2T 2 µe µe ∗ 2 3 − + ne ne = 2 π + = np <nB ηnγ = η 2 ζ(3)T . (6.12) − 6π T T ≈ π Here n∗ includes the protons inside nuclei. Since η is small, µ T , and we can drop p ≪ the (µ/T )3 term to get µ 6 e . ζ(3)η. (6.13) T π2 Thus µ /T η 10−9. The nonrelativistic limit can be dealt with in a similar e ∼ ∼ manner. It turns out that µe rises as T falls, and somewhere between T = 30 keV and T = 10 keV µe becomes larger than T , and, in fact, comparable to me. For T & 30 keV, µ T , and we can drop µ in (6.11). e ≪ e Since we have not measured the cosmic neutrino background (and probably will not do so in the foreseeable future, as neutrinos interact so weakly), we don’t know the neutrino chemical potentials. Usually it is assumed that the neutrino asymmetry is small, like the baryon asymmetry, so that µ e T . The observational upper | ν | ≪ limit from BBN is µ e /T . 0.1; if neutrinos are their own antiparticles, their | ν | chemical potentials are exactly zero. Thus, we ignore both µe and µνe and get the equilibrium neutron/proton ratio n − = e Q/T . (6.14) p (This is not valid for T . 30 keV, since then µe is no longer small, but we will use this formula only at higher temperatures.) We can thus express the number densities of all nuclei in terms of the free proton number density np, as long as chemical equilibrium holds. 6 BIG BANG NUCLEOSYNTHESIS 89 6.3 Bottlenecks We define the mass fraction of nucleus i as Aini Xi . (6.15) ≡ nB Since nB = Aini, (6.16) Xi (where the sum includes protons and neutrons), we have Xi = 1. 1 Using also the normalisation condition (6.16)), or i PXi = 1, we get all equilib- rium abundances as a function of T (they also depend onP the value of the parameter η). There are two items to note: 1. The normalisation condition (6.16) includes all nuclei up to uranium and be- yond. Thus we would get a huge polynomial equation from which to solve Xp. 2. In practice we don’t have to care about the first item, since as the tempera- ture falls the nuclei no longer follow their equilibrium abundances. The reac- tions are in equilibrium only at high temperatures, when the other equilibrium abundances except Xp and Xn are small, and we can use the approximation Xn + Xp = 1. In the early universe the baryon density is too low and the time available is too short for reactions involving three or more incoming nuclei to occur at any appreciable rate. The heavier nuclei have to be built sequentially from lighter nuclei in two-particle reactions, so deuterium is formed first in the reaction n+p d+ γ. → Only when deuterium nuclei are available can helium nuclei be formed, and so on. This process has “bottlenecks”: the lack of sufficient densities of lighter nuclei hin- ders the production of heavier nuclei, and prevents them from following their equi- librium abundances. As the temperature falls, the equilibrium abundances rise fast. They become large later for nuclei with small binding energies. Since deuterium is formed directly from neutrons and protons it can follow its equilibrium abundance as long as there are large numbers of free neutrons available. Since the deuterium binding energy is rather small, the deuterium abundance becomes large rather late (at T < 100 keV). Therefore heavier nuclei with larger binding energies, whose equilibrium abundances would become large earlier, cannot be formed. This is the deuterium bottleneck. Only when there is enough deuterium (X 10−3) can helium be produced in large d ∼ numbers. The nuclei are positively charged so they repel each other electromagnetically. The nuclei need large kinetic energies to overcome this Coulomb barrier and get within the range of the strong interaction. Thus the cross sections for these fusion 1 For np and nn we know just their ratio, since we do not know µp and µn, only that µp = µn. Therefore this extra equation is needed to solve all ni. 6 BIG BANG NUCLEOSYNTHESIS 90 reactions fall rapidly with energy and the nuclear reactions are “shut off” when the temperature falls below T 30 keV.