– 1 –1. Big Bang Nucleosynthesis and the Case of Li, B e, and BEarly in the Big Ba ng, T so high that electrons, protons, positrons, and neutrons werein thermal equilibrium. The fact that there are baryons today means that initially in theearly universe there were more baryons than anti-baryons. All the anti-baryons annihilatedwith baryons very early, leaving the excess baryons we see today. The ratio o f baryons tophotons, denoted ηb, does not change with time as both numerator and denominator declinewith a(t)−3. Since nγcan be calculated from the CMB radiation we see today, ηbcan alsobe measured.After the annihilation was over, with T ∼ 1 Mev (1010K), we hadηb≡nbnγ= 5.5 × 10−10(Ωbh20.020)Thus in the early universe with T ∼ 1 MeV there were many more photons thanbaryons.The mass of a proton is 938.3 MeV/c2. A neutron is slightly heavier than a proton,(mn− mp)c2= 1 .3 Mev; the difference is ∼0.14% of the proton mass.Neutrons and protons can interconvert via weak interactions. When only the rightarrow holds the second reaction listed below is electron capture, while the third is β-decay.p +ν⇀↽n + e+, p + e−⇀↽n + ν, n⇀↽p + e−+ν.For T > 1 MeV (1010K), the reactions listed above can proceed in both directions.Statistical equilibrium then determines the ratio between n and p,– 2 –NnNp= (mnmp)3/2exp[−(mn− mp)c2kT]Note that under these circumstances, once T is specified, the ratio of neutronsto protons is determined.At very high T , there are no atomic nuclei, just free neutrons, protons and electrons.Any nuclei beyond p that might be formed are destroyed immediately by high energyphotons. And the ratio of neutrons to protons is almost one.As T decreases, the ratio Nn/Npdecreases, and protons outnumber neutrons. AtT ∼ 0 .8 MeV, the mean time for the second reaction becomes longer than the age of theuniverse at that epoch, ∼2 sec. At this time, the neutrons freezeout; no more neutronscan be created, and the neutron-proton ratio is set t o exp[−1.3/0.8] = 0.20. A detailedcalculation finds that the fraction of neutrons at freezeout Xn= Nn/(Nn+ Np) is 0.15.Production of deuterium and other light elements begins at about T ∼ 0.07 MeV. Butfree neutrons have a finite lifetime, only 15 min. By the time the reactions to make nucleiare at their peak, only 74% of the neutrons ar e left, the remainder having become protonsvia β-decay.Because of the low binding energy of the isotopes after4He (see the appended fig ure),only elements up to He and slightly beyond can be produced. Lets consider deuteriumfirst. We are looking at the equilibrium of the reaction n + p⇀↽D + γ. Nuclear statisticalequilibrium (the equivalent of the Saha equation for ionization equilibrium between variousions of a specific atom) requiresNdNnNp=34(2πmDmnmpT)3/2e[mn+mp−md]c2/kT≈34(4πmpT)3/2eBD/T,where BDis the binding energy of deuterium. Since η ∝ T3/Nb, this in turn becomes– 3 –(approximately)NdNb∝ ηb(Tmp)3/2eBD/T.The exponential f actor increases rapidly as T becomes less than BD, but η, whichis (nb/nγ), is a very small number as shown above. The binding energy for deuterium is2.2 MeV. But the small baryon fraction (very small η) inhibits the production o f nucleisuch as D or He until the temperature drops well below the nuclear binding energy, whichis where one would naively think nuclear reactions would begin.The equilibrium deuterium abundance (ignoring He) is when all the neutrons are indeuterium, and the rest of the protons ar e free1H nuclei without any electrons. To findit, all we need to know is the value of T ; given that we can find Np/Nnand hence t hedeuterium fraction.But the binding energy of He is much larger than for deuterium, so very little deuteriumgets produced; almost all of the neutrons go into4He nuclei, each of which requires 2neutrons.Some neutrons β-decay into +e−+ν before making it into a He nucleus, and a verysmall fraction end up in isootopes other than4He, i.e2H,3He, Li and Be. That plus thedelay from the high photon/baryon density reducing Tnucchanges Nn/Npfrom about 1/5to 1/7. The predicted ratio of4He/H is then 1/ 12 (i.e. for every4He nucleus, there are 12free protons). The mass fraction of He is then predicted to be 4/(12 + 4) = 1/4, which isextremely close to that observed. Obviously this arguement can be worked out in detail topredict the4He/H more carefully.We expect that X4= 4 × n(4He)/nb= 2 Xn(Tnuc), a nd the crucial thing is establishingthe appropriate Tnuc. Using the the value of Tnucwe found f r om the D/H ra t io , we then– 4 –find a He mass fraction from Big Bang nucleosynthesis of Yp= 0.22. Observations fromHII regions, where emission lines of both He and H can be detected, extrapolated to zerometallicity, suggest Yp= 0.238, in good agreement with this rather crude prediction.A more careful detailed treatment yields ratios o f D/H (10−5),3He/H (10−5),7Li/H(10−9), and almost nothing else. Clearly H and4He dominate. These predictions dependon the baryon density at the time of neutron freezeout, nb.Measurements of the CMB photon density today can be used to infer η. The CMBfluctuation spectrum also depends on the baryon density, and observations give a consistentvalue of nbtoday, usually expressed as a fraction of the critical closure density. Thisprecision measurement of the value of η can be compared to the value required for thepredicted primordial abundances of the elements and their isotopes listed above to a greewith those observed in the most metal-poor stars, extrapolated to zero metallicity.The WMAP measurement of baryon/photon ra t io η is 6.23±0.17 × 10−10(Cyburt etal 2008). We stress that this determines the primordial abundances of the stable isotopesof He, Li, Be and B, with the predicted value of log[7Li/H] = 2.72±0.06 dex. (Here theconventional notation is used, abundances are calculated with respect n(H) set to 1 012and are given as logs in base 10, so the above converts to n(Li)/n(H) = 5 × 10−10.) Onemust assume for these calculations a value for the number of flavors of neutrinos; almost allcalculations adopt that as three.Agreement between the observed and predicted big bang nucleosynthesis for theisotopes of He and the lightest elements is a key test of big bang cosmology. The elementof choice for this test is the D /H ratio, as D is only destroyed in stars. So (D/H)(obs) ≤(D/H)(big bang), and in stars of very low metallicity, the observed value