3.5 Nucleosynthesis 113
If the universe were not expanding, then nearly all free neutrons would end up
in^4 He nuclei and there would be negligible abundances of the other light elements.
However, in an expanding universe, expansion acts as a “shut-off valve” for the
pipes. At the moment the expansion rate becomes larger than a particular reaction
rate, the corresponding pipe closes. When all pipes entering a reservoir have closed,
the abundance of that light element freezes out. The final abundances of^3 He and
T are determined by the freeze-out concentration of deuterium, which we now
calculate.
Let us derive the system of kinetic equations for the abundances by weightXn,
XD,XTandX (^3) He.The concentration of free neutrons decreases due to the reactions
pn→Dγand^3 Hen→Tpbut increases in the processes DD→^3 Henand DT→
(^4) Hen.Therefore, taking into account (3.133) and (3.134), we obtain
dXn
dt
=−λpnXpXn−^13 λ (^3) HenX (^3) HeXn+^14 λDD1X^2 D+^16 λDTXDXT. (3.144)
Deuterium is produced only in the reactionpn→Dγand destroyed in the reactions
DD→^3 Hen,DD→Tp,Dp→^3 Heγ,^3 HeD→^4 Hep,DT→^4 Hen. Hence,
dXD
dt
= 2 λpnXpXn−^12 λDDX^2 D−λDpXDXp−^13 λDTXDXT−^13 λ (^3) HeDX (^3) HeXD,
(3.145)
whereλDD=λDD1+λDD2.The equation for tritium is obtained similarly:
dXT
dt
=^34 λDD2X^2 D+λ (^3) HenX (^3) HeXn−^12 λDTXDXT. (3.146)
We assume that the tritium concentration satisfies the quasi-equilibrium condition,
that is, the rate of its overall change is much smaller than the rates of the individual
reactions on the right hand side of (3.146). Therefore, we setdXT/dt≈0 and
(3.146) reduces to
3
4 λDD2X
2
D+λ^3 HenX^3 HeXn≈
1
2 λDTXDXT. (3.147)
The quasi-equilibrium condition for helium-3 takes the form
3
4 λDD1X
2
D+
3
2 λDpXDXp≈
1
2 λ^3 HeDX^3 HeXD+λ^3 HenX^3 HeXn. (3.148)
Using (3.147) and (3.148) to expressX^3 HeandXTthrough the neutron and deu-
terium concentrations, (3.144) and (3.145) become
dXn
dt
=^14 λDDX^2 D−λpnXpXn, (3.149)
dXD
dt
= 2 λpnXpXn−λDDX^2 D− 2 λDpXDXp. (3.150)