3.6 Recombination 121most of the helium nuclei are converted into helium ions, we assume that the
reaction
He^2 ++e−He++γ (3.176)is efficient in maintaining the chemical equilibrium between He^2 +and He+.Then,
the chemical potentials satisfy
μ 2 ++μe=μ+, (3.177)and considering the ratio (n 2 +ne)/n+,where the number densities are given by
(3.61), we obtain the Saha formula:
n 2 +ne
n+=
g 2 +ge
g+(
Tme
2 π) 3 / 2
exp(
−
B+
T
)
. (3.178)
The ratio of the statistical weights here is equal to unity. Even complete recombi-
nation of helium reduces the number of free electrons by 12% at most. Therefore,
before hydrogen recombination, the number density of free electrons is
ne(0.75 to 0.88)nN 2 × 10 −^11 η 10 T^3. (3.179)Substituting this into (3.178), we obtain
n 2 +
n+exp(
35. 6 +
3
2
ln(
B+
T
)
−
B+
T
−lnη 10)
. (3.180)
If the expression in the exponent is positive, the concentration of He+ions is small
compared to the concentration of completely ionized helium. Using the method of
iteration, we find that at the temperature
T+B+
42 −lnη 1015 000×
(
1 + 2. 3 × 10 −^2 lnη 10)
K, (3.181)
the ration 2 +/n+is of order unity. At this time, the He+ions constitute about 50% of
all helium and the rest is completely ionized. Very soon after this, nearly all helium
nuclei capture an electron and are converted into He+.Expanding the expression in
the exponent in (3.180) aboutT=T+,we find, to leading order inT≡T+−T
T+,
n 2 +
n+∼exp(
−
B+
T+
T
T+
)
∼exp(
− 42
T
T+
)
. (3.182)
When the temperature falls only 20% belowT+(going from 15 000 K to 12 000 K),
the number density of He^2 +reduces ton 2 +∼ 10 −^4 n+.We see in (3.181) that the
temperatureT+varies logarithmically with the baryon densityη 10 ; the larger the
baryon density,the earlier recombination occurs (i.e., at a higher temperature).