Physical Foundations of Cosmology

3.5 Nucleosynthesis 107

Problem 3.18Verify that after electron–positron annihilation, the value ofκin
(3.85) becomes

κ 1. 11 + 0. 15 (Nν− 3 ), (3.132)

whereNνis the number of neutrino species. (HintRecall that the neutrino and
radiation temperatures are different aftere±annihilation.)

After the deuterium concentration reaches X(bn)D , everything proceeds very
quickly. According to (3.130) the equilibrium concentrationXDincreases from
10 −^4 to 10−^2 as the temperature drops from 0.08 MeV to 0.07 MeV. As a result, the
rate of deuterium conversion into heavier elements, proportional toXD,becomes
100 times larger than the expansion rate. Such a system is far from equilibrium and
nucleosynthesis is described by a complicated system of kinetic equations which
are usually solved numerically. In Figures 3.7 and 3.8 below we present the results
of highly precise numerical calculations for the time evolution of the light element
concentrations and for their final abundances, respectively. We will now show how
these results can be reproduced analytically with good accuracy. The system of ki-
netic equations will be solved using the quasi-equilibrium approximation. This will
provide us with a solid physical understanding of primordial nucleosynthesis and
will reveal the reasons for the dependence of final abundances on the cosmological
parameters. To simplify our task we consider only the most abundant isotopes up
to^7 Be, among which are^4 He,D,^3 He,T,lithium-7 (^7 Li) and beryllium (^7 Be) itself.
Other elements such as^6 Li,^8 B etc. are produced in much smaller amounts and will
be ignored.
The most important nuclear reactions are shown schematically in Figure 3.6.
The reader is encouraged to keep a copy of this figure at hand throughout the
rest of this section. Every element corresponds to a “reservoir.” The reservoirs are
connected by “one-way pipes”, one for each nuclear reaction converting an element
into another. To simplify the diagram, we include only the initial elements involved
in the reaction; the outcome can easily be inferred from the diagram. The efficiency
of the pipe is determined by the reaction rate. For example, for the rate of escape
from the reservoir A due to the reaction AB→CD, we find

X ̇A=−A−B^1 λABXAXB, (3.133)

and the rate of increase of the element C is

X ̇C=ACA−A^1 A−B^1 λABXAXB. (3.134)

HereXA≡AAnA/nN,etc. are the concentrations by weight of the corresponding
elements,Aare their mass numbers (for example,AD=2 andAT= 3 ,etc.) and
λAB=〈σv〉ABnN.The reaction is efficient only ifX ̇A/XA>t−^1.