20
The standard model of cosmology
the only relativistic particles are the photon and three neutrinos. Neutrinos drop
out of thennal equilibrium below about I MeV when the (weak) interaction rate
that keeps them in thennal equilibrium becomes less than the Hubble parameter.
(See section 5.2.) When the temperature drops below the electron mass (about
0.5 MeV), electrons and positrons annihilate via e+e- -+ yy, and the entropy of
the electron-positron pairs is transferred to the photons. However, no entropy is
transferred to the neutrinos, which are now decoupled. Before electron-positron
annihilation, we have
7
11
N. = 2+ ~ = T
(1.117)
but afterwards we should take
N. =2.
(1.118)
Note that we are not keeping any contribution from the neutrinos because they
have now dropped out of thennal equilibrium and no longer contribute to the
entropy. If Ty; and Tyf are the photon temperatures before and after electronpositron annihilation, conservation of entropy requires that
T 11 T3. = 2T 3
(1.119)
yl
yf
so that
Tyf _ (11)1/3
(1.120)
Ty; -
4'
~ 1.4.
However, the neutrinos do not share in this temperature increase. Thus, there is
an effective N. for the purpose of calculating the energy density of radiation
N •• eff = 2 + t x 6 x (rt )4/3 ~ 3.36.
(1.121)
Note that the relativistic neutrinos still have an energy density that varies as R- 4
as the universe expands, because their number density varies as R- 3 and they
undergo redshifting of their energy as R- 1 • Thus, the neutrino energy density
also varies as T 4 , where T is the photon temperature, i.e. the temperature in the
usual sense. Then. from (1.103). in the radiation-dominated universe below the
temperature at which e+e- annihilation occurs,
11'2
p(T) = 30 N.,eff T4 .
(1.122)
If the matter energy density of (1.114) and the radiation energy density
(1.103) are equal at a temperature Teq. we find that
9OM2H, 2 0o
T. -
P 0
(1.123)
eq- 2
3'
11' N •• effTO
Using (1.65), (1.121) and ( 1.40) gives
Teq = 5.6800h 2 eV.
(1.124)
The standard model of cosmology
the only relativistic particles are the photon and three neutrinos. Neutrinos drop
out of thennal equilibrium below about I MeV when the (weak) interaction rate
that keeps them in thennal equilibrium becomes less than the Hubble parameter.
(See section 5.2.) When the temperature drops below the electron mass (about
0.5 MeV), electrons and positrons annihilate via e+e- -+ yy, and the entropy of
the electron-positron pairs is transferred to the photons. However, no entropy is
transferred to the neutrinos, which are now decoupled. Before electron-positron
annihilation, we have
7
11
N. = 2+ ~ = T
(1.117)
but afterwards we should take
N. =2.
(1.118)
Note that we are not keeping any contribution from the neutrinos because they
have now dropped out of thennal equilibrium and no longer contribute to the
entropy. If Ty; and Tyf are the photon temperatures before and after electronpositron annihilation, conservation of entropy requires that
T 11 T3. = 2T 3
(1.119)
yl
yf
so that
Tyf _ (11)1/3
(1.120)
Ty; -
4'
~ 1.4.
However, the neutrinos do not share in this temperature increase. Thus, there is
an effective N. for the purpose of calculating the energy density of radiation
N •• eff = 2 + t x 6 x (rt )4/3 ~ 3.36.
(1.121)
Note that the relativistic neutrinos still have an energy density that varies as R- 4
as the universe expands, because their number density varies as R- 3 and they
undergo redshifting of their energy as R- 1 • Thus, the neutrino energy density
also varies as T 4 , where T is the photon temperature, i.e. the temperature in the
usual sense. Then. from (1.103). in the radiation-dominated universe below the
temperature at which e+e- annihilation occurs,
11'2
p(T) = 30 N.,eff T4 .
(1.122)
If the matter energy density of (1.114) and the radiation energy density
(1.103) are equal at a temperature Teq. we find that
9OM2H, 2 0o
T. -
P 0
(1.123)
eq- 2
3'
11' N •• effTO
Using (1.65), (1.121) and ( 1.40) gives
Teq = 5.6800h 2 eV.
(1.124)