3.5 Nucleosynthesis 97remains constant. Taking into account thatsγ∝Tγ^3 andsν∝Tν^3 ,we have
(
Tγ
Tν) 3 (
1 +
se±
sγ)
=C, (3.93)
whereCis a constant. Just after neutrino decoupling, but beforee±annihilation,
Tγ=Tνandse±/sγ= 7 /4 (see (3.59)).Therefore,C= 11 /4 and
(
Tγ
Tν)
=
(
11
4
) 1 / 3 (
1 +
se±
sγ)− 1 / 3
. (3.94)
When the electron–positron pairs begin to annihilate atT 0 .5 MeV the ratio of
entropiesse±/sγdecreases and finally becomes completely negligible (see (3.82),
where one has to substitutemeinstead ofmp). Hence, after electron–positron
annihilation we have
(
Tγ
Tν
)
=
(
11
4
) 1 / 3
= 1. 401. (3.95)
Thus, the massless primordial neutrinos should have a temperature today of
Tν 2 .73 K/ 1. 4 1 .95 K.Unfortunately it is not easy, if even possible, to detect
the primordial neutrino background and verify this very robust prediction of the
standard cosmological model.
Problem 3.13Assuming that neutrinos have a small, but nonvanishing mass, esti-
mate their temperature today.
Problem 3.14Calculate the contribution of neutrinos to the energy density
aftere±annihilation and determine at which redshiftzthe total energy density
of radiation and relativistic neutrinos is exactly equal to the energy density of cold
(nonrelativistic) matter.
3.5 Nucleosynthesis
The most widespread chemical element in the universe is hydrogen, constituting
nearly 75% of all baryonic matter. Helium-4 constitutes about 25%. The other light
elements and metals have only very small abundances.
Simple arguments lead to the conclusion that the large amount of^4 He could not
have been produced in stars. The binding energy of^4 He is 28.3 MeV, and therefore,
when one nucleus of^4 He is formed, the energy released per one baryon is about
7 .1 MeV 1. 1 × 10 −^5 erg. Assuming that one quarter of all baryons has been
fused into^4 He in stars during the last 10 billion years (3. 2 × 1017 s), we obtain the