Physical Foundations of Cosmology

3.5 Nucleosynthesis 103

us to see how the freeze-out concentration depends on the number of relativistic
species present at the freeze-out time. BecauseT∗∝κ^1 /^8 ,additional relativistic
components increaseT∗and, hence, more neutrons survive. Subsequently, nearly
all neutrons fuse with protons to form^4 He and we anticipate, therefore, that addi-
tional relativistic species increase the primordial helium abundance. For example,
in the extreme case of a very large number of unknown light particles, the tem-
peratureT∗would exceedQand the neutron concentration at freeze-out would be
almost 50%.This would lead to an unacceptably large abundance of^4 He.Thus,
we see that primordial nucleosynthesis can help us to restrict the number of light
species.

Problem 3.16Find the freeze-out temperature using the simple criteriont 1 /λ
and verify that in this approximation one obtains the result quoted in many books
on cosmology, namely,T∗∝κ^1 /^6. What accounts for the difference between this
and the above result,T∗∝κ^1 /^8?

Now we turn to a more accurate estimate for the freeze-out concentration.
SinceXeqn →0asT→ 0 ,Xn∗is given by the integral term in (3.111) when we
take the limitt→∞.The main contribution to the integral comes fromT>me.
Therefore we setλn→p 2 λnν,whereλnνis given in (3.106). Using (3.88) to
replace the integration variabletbyy=T/Q, we obtain

Xn∗=

∫∞

0

exp

(

− 5. 42 κ−^1 /^2

∫y

0 (x+^0.^25 )

2 ( 1 +e− 1 /x)dx)

2 y^2 ( 1 +cosh( 1 /y))

dy. (3.114)

For the case of three neutrino species (κ 3 .54) one findsX∗n 0. 158 .This result is
in very good agreement with more elaborate numerical calculations. The presence
of an additional light neutrino, accompanied by the corresponding antineutrino,
increasesκby amount 2·κf 0. 58 ,and the freeze-out concentration becomes
X∗n 0. 163 .Thus, two additional fermionic degrees of freedom increaseX∗nby
about 0.5% and we conclude that

X∗n 0. 158 + 0. 005 (Nν− 3 ), (3.115)

whereNνis the number of light neutrino species.

Neutron decayUntil now we have neglected neutron decay,

n→p+e−+ν. ̄ (3.116)

This was justified because the lifetime of a free neutronτn≈886 s is large compared
to the freeze-out time,t∗∼O( 1 )s. However, after freeze-out the interactions (3.96)
and the inverse three-body reaction (3.116) become inefficient and neutron decay