Page 19 - 83 basic knowledge of astronomy

1. Stationary conditions:
The number of transitions Z m → Z n must be equal to the number of
opposite transitions Z n → Z m . Requiring this stationarity in equation
(15), we have:
m
m
n
n m β u ν (s) = n n [α + β u ν (s)]. (16)
n
n
m
2. Boltzmann distribution:
The probability P n for a particle to be in a state Z n , with energy level
E n , is given by the formula:
kT ,
P n = g n e − E n (17)
where k = 1.381 × 10 −23 J K −1 is the Boltzmann constant, T is the
absolute temperature of the medium in Kelvin (K), and g n is the sta-
tistical weight (reflecting in particular the degree of degeneracy) of the
state Z n .
If we denote the number density of all particles as n, the ratio n n /n must
be equal to the probability P n (at least in the statistical sense). Therefore,
we have
n m − E m
= g m e kT , (18)
n
and
n n − E n
= g n e kT . (19)
n
Inserting equations (18) and (19) into equation (16), we obtain

m
n
kT g m β u ν (s) = e
kT g n [α + β u ν (s)].
e − E m m − E n m n (20)
n
Einstein discussed the implications of this equation, as follows:
1. The energy density per unit solid angle of the thermal radiation must
tend to infinity (u ν (s) → ∞) when the temperature of the medium
tends to infinity (T → ∞) in equation (20). Hence, we obtain

m
n
g m β = g n β . (21)
n
m
Consequently, equation (20) becomes
E n −E m m m
(e kT − 1)β u ν (s) = α . (22)
n
n
If we take into account the relation hν = E n −E m (hereafter, we denote
ν mn = ν for simplicity), the energy density per unit solid angle can be
expressed as
α m 1
u ν = n . (23)
hν
m
β n e kT − 1
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