Physical Foundations of Cosmology

3.5 Nucleosynthesis 115

and henceXn 0 .12. Neglecting the temperature dependence ofα,we then find
the approximate solution of (3.155):

Xn(T) 0 .12 exp

(

1

2 αη^10 R^1

(

T−TMeV(N)

))

, (3.156)

whereTMeV(N)is given in (3.140). It follows that the neutron concentration becomes
comparable to the deuterium concentration,Xn∼XD∼R 1 ,at the temperature

TMeV∗ ∼ 0. 07 + 0 .002 lnη 10 − 0. 02 K−^1 η− 101. (3.157)

In a universe with very low baryon density,Kη 10 < 0. 3 ,the abundance of free
neutrons (neglecting their decay) does not decrease belowXDand freezes out at
the value

Xnf  0 .12 exp

(

−^12 αη 10 R 1 TMeV(N)

)

∼ 0 .12 exp(− 10 Kη 10 ). (3.158)

The remaining free neutrons then decay. This explains why, for instance, the^4 He
abundance is less than 1% in a universe withη 10  10 −^2 (Figure 3.8).

Problem 3.20At which value does the deuterium concentration freeze out in low
baryon density universe? How does it depend onη 10?

In the derivation of the^4 He abundance presented above, we tacitly assumed that
the reactions converting the neutrons into^4 He are very efficient in transferring most
of the available neutrons into heavier elements. This means that (3.143) is valid
only forKη 10 > 0. 3 .Observations suggest that 10>η 10 >1 and therefore we will
assume below thatη 10 >1.
When the neutron concentration becomes of order the deuterium concentration,
(3.154) fails and the system quickly reaches another attractor. Afterwards, the
neutron concentration satisfies the quasi-equilibrium condition,dXn/dT≈ 0 ,and
it follows from (3.151) that

Xn=

1

R 1

X^2 D

[

1 +O

(

Xn

XD

)]

. (3.159)

Equation (3.152) then becomes

dXD

dTMeV

= 2 αη 10

(

X^2 D+ 2 R 2 XD

)

. (3.160)

Now the deuterium determines its own fate and regulates the quasi-equilibrium
concentrations of the neutrons and the other light elements. SinceR 2 changes
insignificantly within the relevant temperature interval (see (3.153)), it can be taken