24
The standard model of cosmology
The equilibrium solution, found by setting Xn = 0, is
xeq t _ Apn(t) - _ _ I
+
where A ==
n ( )
A(t) - 1
Apn
eAmIT(t)
+ Anp.
(1.142)
Thus, we may rewrite (1.141) as
•
Xn =
eq
-A(Xn Xn ).
(1.143)
This shows that Xn is always between its initial value and X~. At early times
A is large compared to the rate of time variation of the individual rates and X"
quickly tracks its equilibrium value X~(t). This persists until the scattering rate
A (t) decreases until it becomes comparable with the Hubble rate H (t) == R / R =
- t / T. At this point, because of the expansion of the universe, the nucleons
become too dilute to maintain the chemical equilibrium, they decouple and the
number densities become 'frozen' at the values they have at decoupling. Thus,
Xn(tdec) ~ X~(tdec) = ----:(1.144)
As explained in section 5.2, the decoupIing (or freeze-out) occurs when the
temperature is
Tdec ~ 1 MeV.
(1.145)
It is a remarkable coincidence that these two numbers, Tdec and Il.m (given in
(1.140», are of the same order. The former derives from the interplay between
the weak and gravitational interactions, while the latter derives from the difference
between the u and d quark masses, which is of unknown origin but presumably
as a result of strong and electromagnetic effects. Because of this coincidence, a
substantial fraction (of order 20%) of the neutrons survive and this, in turn, results
in a significant amount of promordial helium formed in the early universe.
This calculation of the fractional relative abundance of neutrons when the
weak interactions decouple is only a rough estimate. For a more accurate estimate,
we must solve the balance equation (1.141). It is convenient to use the variable
y == Il.m/T instead of t. In this era, the temperature T is related to the time t by
equation (1.111) with N. = 3.36, as shown in (1.121). Using (1.140), the total
decay rate is
A(y) ~ Allp(y)(1 + e- Y )
(1.146)
neglecting the neutron decay rate compared to the scattering rates. Bemstein et
al [6] have approximated Allp(y) by
Anp(y) ~ 2A(nve ~ pe-) ~ ~(12 + 6y + yl)
(1.147)
1'IIY
where a ~ 253 and l'n ~ 887 s. is the neutron lifetime. Then the solution is
Xn(y) = X~(y) + laY dy' eY/[X~(y')]21(y, y')
(1.l48)
The standard model of cosmology
The equilibrium solution, found by setting Xn = 0, is
xeq t _ Apn(t) - _ _ I
+
where A ==
n ( )
A(t) - 1
Apn
eAmIT(t)
+ Anp.
(1.142)
Thus, we may rewrite (1.141) as
•
Xn =
eq
-A(Xn Xn ).
(1.143)
This shows that Xn is always between its initial value and X~. At early times
A is large compared to the rate of time variation of the individual rates and X"
quickly tracks its equilibrium value X~(t). This persists until the scattering rate
A (t) decreases until it becomes comparable with the Hubble rate H (t) == R / R =
- t / T. At this point, because of the expansion of the universe, the nucleons
become too dilute to maintain the chemical equilibrium, they decouple and the
number densities become 'frozen' at the values they have at decoupling. Thus,
Xn(tdec) ~ X~(tdec) = ----:(1.144)
As explained in section 5.2, the decoupIing (or freeze-out) occurs when the
temperature is
Tdec ~ 1 MeV.
(1.145)
It is a remarkable coincidence that these two numbers, Tdec and Il.m (given in
(1.140», are of the same order. The former derives from the interplay between
the weak and gravitational interactions, while the latter derives from the difference
between the u and d quark masses, which is of unknown origin but presumably
as a result of strong and electromagnetic effects. Because of this coincidence, a
substantial fraction (of order 20%) of the neutrons survive and this, in turn, results
in a significant amount of promordial helium formed in the early universe.
This calculation of the fractional relative abundance of neutrons when the
weak interactions decouple is only a rough estimate. For a more accurate estimate,
we must solve the balance equation (1.141). It is convenient to use the variable
y == Il.m/T instead of t. In this era, the temperature T is related to the time t by
equation (1.111) with N. = 3.36, as shown in (1.121). Using (1.140), the total
decay rate is
A(y) ~ Allp(y)(1 + e- Y )
(1.146)
neglecting the neutron decay rate compared to the scattering rates. Bemstein et
al [6] have approximated Allp(y) by
Anp(y) ~ 2A(nve ~ pe-) ~ ~(12 + 6y + yl)
(1.147)
1'IIY
where a ~ 253 and l'n ~ 887 s. is the neutron lifetime. Then the solution is
Xn(y) = X~(y) + laY dy' eY/[X~(y')]21(y, y')
(1.l48)