110 The hot universe
deuterium to produce^4 He:
TD→^4 Hen. (3.136)In this sequence, two of the three neutrons end up in the newly formed^4 He nucleus
and one neutron returns to thenpreservoir. The^3 He nucleus can interact either
with a free neutron and proceed to the T reservoir,
(^3) Hen→Tp, (3.137)
or with deuterium and go directly to the^4 He reservoir,
(^3) HeD→ (^4) Hep. (3.138)
The ratio of the rates for these reactions is
λ (^3) HenX (^3) HeXn
λ (^3) HeDX (^3) HeXD
∼ 6
Xn
XD. (3.139)
Hence, until the concentration of free neutrons,Xn,drops belowXD(which never
exceeds 10−^2 ),(3.137) is more efficient than (3.138). Therefore, most of the neu-
trons are fused into^4 He through the reaction chainsnp→D→T→^4 He and
np→D→^3 He→T→^4 He.Within a short interval around the time when the deu-
terium concentration reaches its maximal valueXD 10 −^2 ,nearly all neutrons,
except a very small fraction∼ 10 −^4 ,end up in^4 He nuclei. Therefore, the final^4 He
abundance is completely determined by the available free neutrons at this time.
According to (3.130),XDis of order 10−^2 at temperature
TMeV(N) 0. 07 ( 1 + 0 .03 lnη 10 ), (3.140)or, equivalently, at time
tsec(N) 269 ( 1 − 0. 07 (Nν− 3 )− 0 .06 lnη 10 ), (3.141)where we have used (3.132) forκin (3.88). Because half of the total weight of^4 He
is due to protons, its final abundance by weight is
X 4 fHe= 2 Xn(
t(N))
= 2 Xn∗exp(
−
t(N)
τn)
. (3.142)
Substituting hereXn∗from (3.115) andt(N)from (3.141), one finally obtains
X 4 fHe 0. 23 + 0. 012 (Nν− 3 )+ 0 .005 lnη 10 (3.143)This result is in good agreement with the numerical calculations shown in Figure 3.8.
The^4 He abundance depends on the number of ultra-relativistic speciesNνand
the baryon density characterized byη 10 .The presence of an additional massless
neutrino increases the final abundance by about 1.2%.This increase comes from two