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Many crucial events take place around the first seconds of the universe: a) the weak interaction decoupling at one MeV (one second or so); b) electron-positron annihilation around 0.5 MeV (four seconds or so); c) the nucleosynthesis of the light nuclides around 0.1 MeV (one hundred seconds or so). A useful approximation for the time temperature relationship is the following: t(sec) = 1/T2 (MeV).

As discussed before, at T > 1 MeV the weak interaction equilibrium is insured by the reactions:

Equation 11

In consequence, the neutron-proton ratio is given by Boltzman law (fig 1):

Equation 11 (11)

where DeltaM is the neutron-proton mass difference (1.293 MeV). As T decreases, then/p ratio goes down from one to a value close to 0.2 at decoupling. After Td the neutrino capture reactions are too slow to compensate the neutron free desintegration n longrightarrow p + e + nu.

All through this period, the neutrons also have a nuclear interaction with the protons, inducing the formation of a deuteron: n + p longleftrightarrow D + gamma.

The electron-positron annihilation at 0.5 MeV creates a flux of new photons which increases slightly the photon radiation (more exactly they slow down the cooling rate). Since the neutrino interactions are now very weak, the neutrinos are essentially decoupled from the rest of the universe and they receive no share of the energies released by the annihilation. As a result the neutrino temperature Tnu becomes slightly lower than the photon Tg.

This effect can be calculated through the conservation of entropy per co-volume during the annihilation phase. The entropy density is proportional to the number density of relativistic interacting particles s propto g*T3. Here i and f will stand for the initial state (before annihilation) and the final state (after annihilation);

Equation 12 (12)
Equation 12

thus Tf / Ti = (11/4)1/3 = 1.31.

Since the neutrinos did not receive their share of this annihilation phase they remained at Ti. Today we measure a photon temperature Tgamma of 2.7 K. Thus we expect that the neutrino radiation is at Tnu = 2 K. Such a cosmological background of neutrinos is a necessary prediction of the theory if indeed the universe has reached temperatures over one MeV in the past. Because of the low mean energy of these particles (one MeV) its detection is outside of the realm of contemporary technology.

The previous example explains why various relativistic gases can be at different temperature as expressed in eq. (7) for g*. Suppose for instance that there exist right-handed neutrinos interacting as a weak interaction particle but with a coupling constant G' >> GF. If their decoupling temperature, evaluated through eq. (10), is larger than 107 MeV, the mass of the muon, they will receive no share from the muon annihilation at the equivalent temperature. Conservation of entropy during this phase, evaluated as in the previous example, will allow a determination of the temperature of this radiation, still lower today than the neutrino temperature.

Around 0.1 MeV the gamma rays (the tail of the Bose-Einstein photon energy distribution) are no more numerous enough to keep the deuteron D population in statistical equilibrium with the nucleons. This is the onset of primordial nucleosynthesis. Through the reaction n + p longrightarrow gamma + D, the population of D increases rapidly, as shown in fig. 2. As they reach a ratio of D/H of 10-3 or so, they undergo further nuclear reactions and are transformed in mass-3 nuclei: D + p longrightarrow 3He + gamma;D + n longrightarrow T + gamma. The population of these mass-3 nuclei increases in turn as the D fall down by many orders of magnitude.

Figure 2

Figure 2. Time history of primordial nucleosynthesis. The mass fractions of the various nuclides are displayed as a function of temperature in billion degrees at the top, or as a function of time in seconds at the bottom.

The same fate happens to the mass-3 nuclei, as 4He starts its rise through the 3He + 3He longrightarrow 4He + 2p. Because there are no stable nuclide with mass 5 and 8, the 4He suffers essentially no further nuclear depletion. Only a very small fraction of its population gets transformed in 7Li through 4He + T longrightarrow longrightarrow 7Li + gamma and the 4He + 3He longrightarrow 7Be + gamma, followed after many days by 7Be + e longrightarrow 7Li + nu. Some 7Li is further destroyed but, because of its larger Coulomb charge (Z = 4), the 7Be remains almost intact. This fact will play an important role in our discussion of the quark-hadron transition.

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