Physical Foundations of Cosmology

102 The hot universe

is the equilibrium concentration of neutrons.To obtain the second equality in (3.109)
we used the relations in (3.108), assumingTν=T,as well as the fact that the proton
concentration isXp= 1 −Xn.
The exact solution of the linear differential equation (3.109), with the initial
conditionXn→Xneqast→ 0 ,is

Xn(t)=Xeqn(t)−

∫t

0

exp

⎛

⎝−

∫t

̃t

λn→p( ̄t)

(

1 +e−

QT)

d ̄t

⎞

⎠X ̇eqn

(

̃t

)

d ̃t, (3.111)

where the dot denotes the derivative with respect to time.
The second term on the right hand side in (3.111) characterizes the deviation
from equilibrium and is negligible compared to the first term at smallt. Integrating
by parts, we can rewrite the solution (3.111) as an asymptotic series in increasing
powers of the derivatives ofXeqn :

Xn=Xeqn

(

1 −

1

λn→p( 1 +exp(−Q/T))

X ̇eqn

Xeqn

+···

)

. (3.112)

If the reaction rate is much larger than the inverse cosmological time, that is,
λn→pt−^1 ∼−X ̇neq/Xeqn,then we haveXn≈Xeqn in agreement with result (3.84).
Subsequently, after the temperature has dropped significantly,Xneq→ 0 ,but the
second term on the right hand side in (3.111) approaches a finite limit. Instead
of vanishing, therefore, the neutron concentration freezes out at some finite value
X∗n=Xn(t→∞).The freeze-out effectively occurs when the second term on the
right hand side in (3.112) is of order the first one or, in other words, when the
deviation from equilibrium becomes significant. This happens beforee±annihila-
tion and after the temperature drops belowQ 1 .29 MeV (as can be checkeda
posteriori). Consequently we can setλn→p 2 λnνand neglect exp(−Q/T)in the
equality−X ̇eqn/Xeqn λn→p,which determines the freeze-out temperature. Sub-
stituting into this equality expression (3.110) forXeqn and expression (3.106) for
λnνand using the temperature–time relation (3.88), the equation for the freeze-out
temperature reduces to
(
T∗
Q

) 2 (

T∗

Q

+ 0. 25

) 2

 0. 18 κ^1 /^2. (3.113)

In the case of three neutrino species we haveκ 3 .54 and the freeze-out tem-
perature isT∗ 0 .84 MeV.The equilibrium neutron concentration at this time is
Xeqn(T∗) 0. 18 .Of course, this number gives only a rough estimate for the ex-
pected freeze-out concentration. One should not forget that atT=T∗deviations
from equilibrium are very significant and, in fact,Xn(T∗)exceeds the equilibrium
concentration by at least a factor of 2. Nevertheless, the above estimate enables