3.4 Lepton era 89corresponding formulae are the standard ones found in any book on statistical
physics. Having completed our brief review of relativistic statistical mechanics, we
now apply the results derived to the early universe.
3.4 Lepton era
When the temperature in the universe drops below a hundred MeV(att> 10 −^4 s),
the quarks and gluons are confined and form color-singlet bound states−baryons
and mesons. We recall that baryons are made out of three quarks, each of which
has baryon number 1/3, while the mesons are bound states of one quark and one
antiquark, so that their resulting baryon number is zero.
The main ingredients of ordinary matter at temperatures below 100 MeV are pri-
mordial radiation(γ), neutrons(n), protons(p), electrons and positrons
(
e−,e+)
,
and three neutrino species.Mesons, heavy baryons,μ- andτ-leptons are also
present, but their number densities are very small and become increasingly negli-
gible as the temperature decreases.
At energies of order a few MeV, the most important processes involve the weak
interactions in which leptons, such as neutrinos, participate. Therefore, one calls
this epoch thelepton era. At low energies, the baryon number and the lepton
numbers are each conserved. The total electric charge is obviously also conserved.
To enforce these conservation laws, a chemical potential is introduced for each
particle species. The number of the independent potentials, however, is equal to the
number of conserved quantities; any remaining potentials are expressed through
these independent potentials using the chemical equilibrium conditions (3.24).
To demonstrate this, let us consider a medium containing the following ingredi-
ents: photons, leptonse,μ,τ, neutrinosνe,νμ,ντ, the lightest baryonsp,n,,
and mesonsπ^0 ,π±.The corresponding antiparticles are also present in the state of
equilibrium. To enforce the conservation laws for electric charge, baryon number
and the three different lepton numbers, we take as independent the following five
chemical potentials:μe−,μn,μνe,μνμ,μντ.All other potentials will be written in
terms of the members of this set. To start with,
μπ 0 = 0 (3.64)because, as a result of electromagnetic interaction, theπ^0 meson quickly decays
(tπ^0 8. 7 × 10 −^17 s) into photons (π^0 →γγ) which haveμγ= 0 .From→
nπ^0
(
t 2. 6 × 10 −^10 s)
,we find
μ=μn+μπ 0 =μn. (3.65)The muon is unstable
(
tμ 2. 2 × 10 −^6 s)
and decays into an electron, an antineu-
trino and a neutrino,μ−→e−ν ̄eνμ,and hence
μμ=μe−−μνe+μνμ. (3.66)