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/*T*^{2} (MeV).

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

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

(11) |

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

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

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
*T*_{} becomes
slightly lower than the photon *T*_{g}.

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*
*g*^{*}*T*^{3}. Here *i* and
*f* will stand for the initial state
(before annihilation) and the final state (after annihilation);

(12) | |

thus *T*_{f} / *T*_{i} = (11/4)^{1/3} =
1.31.

Since the neutrinos did not receive their share of this annihilation
phase they remained at *T*_{i}. Today we measure a photon
temperature
*T*_{} of 2.7 K. Thus we expect that the neutrino radiation
is at *T*_{} = 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'* >> *G*_{F}. 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*
+
*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*
^{3}*He* +
;*D* +
*n*
*T* +
. The
population of these mass-3 nuclei increases in turn as
the *D* fall down by many orders of magnitude.

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