Electroweak Interactions 127At this time the number density of nucleons decreases quickly because they have
become nonrelativistic. Consequently, they have a larger probability of annihilating
into lepton pairs, pion pairs or photons. Their number density is then no longer given
by the Fermi distribution [Equation (6.28)], but by the Maxwell–Boltzmann distribu-
tion, Equation (6.29). As can be seen from the latter, when푇drops below the mass,
the number density decreases rapidly because of the exponential factor. If there had
been exactly the same number of nucleons and anti-nucleons, we would not expect
many nucleons to remain to form matter. But, since we live in a matter-dominated
Universe, there must have been some excess of nucleons early on. Note that neutrons
and protons exist in equal numbers at the time under consideration.
Although the nucleons are very few, they still participate in electromagnetic reac-
tions such as the elastic scattering of electrons,
e±+p→e±+p, (6.47)and in weakcharged currentreactions in which charged leptons and nucleons change
into their neutral partners, and vice versa, as in
e−+p→휈e+n, (6.48)
휈e+p→e++n. (6.49)Other such reactions are obtained by reversing the arrows, and by replacing e±by휇±
or휈eby휈휇or휈휏. The nucleons still participate in thermal equilibrium, but they are
too few to play any role in the thermal history any more.
Below the pion mass (actually at about 70MeV) the temperature in the Universe
cools below the threshold for pion production:
(e−+e+)or(휇−+휇+)→훾virtual→휋++휋−. (6.50)The reversed reactions, pion annihilation, still operate, reducing the number of pions.
However, they disappear even faster by decay. This is always the fate when such lighter
states are available, energy and momentum, as well as quantum numbers such as
electric charge, baryon number and lepton numbers, being conserved. The pion, the
muon and the tau lepton are examples of this. The pion decays mainly by the reactions
휋−→휇−+휈휇,휋+→휇++휈휇. (6.51)Thus푔∗decreases by 3 to^574. The difference in mass between the initial pion and the
final state particles is
푚휋−푚휇−푚휈=( 139. 6 − 105. 7 − 0. 0 )MeV= 33 .9 MeV, (6.52)so 33.9MeV is available as kinetic energy to the muon and the neutrino. This makes
it very easy for the휋±to decay, and in consequence its mean life is short, only 0.026μs
(the휋^0 decays even faster). This is much less than the age of the Universe at 140MeV,
which is 23μs from Equation (6.45). Note that the appearance of a charged lepton
in the final state forces the simultaneous appearance of its anti-neutrino in order to
conserve lepton number.