Chapter 2 Stellar Structure Equations 1 Mass conservation equation

Note
∑
that G is minimized in chemical equilibrium. Since G = E − T S −
s µ sN s , we have
N s = − ∂G
∂µ s
∣
∣∣∣T,V,{µi
} i≠s
= g sV
λ 3 s
( )
µs − E 0,s
exp
. (63)
kT
By rearranging terms, the above equation becomes
( )
( )
Ns λ 3 s
ns λ 3 s
µ s = kT ln + E 0,s = kT ln + E 0,s , (64)
g s V
g s
where n s is the number density of species s.
Now consider a specific atomic species and use the label s, 0 to denote its
sth ionized state in its ground level. For example, the label 0, 0 of H refers
to the state of hydrogen atom with its electron in the lowest energy level.
And I use the label e to denote electron. Clearly, at non-zero temperature,
reactions such as
X s+ ⇋ X (s+1)+ + e − (65)
may occur, where X denotes the atomic species. Hence, at thermodynamic
equilibrium,
µ s,0 = µ s+1,0 + µ e . (66)
Since E 0,(s+1,0) + E 0,e − E 0,(s,0) = χ s+1 is the (s + 1)th ionization potential
of the atomic species, substitute this and Eq. (64) into above equation, we
have
n s+1,0 n e λ 3 s+1,0λ 3 eg s,0
n s,0 λ 3 s,0g s+1,0 g e
= e −χ s+1/kT . (67)
From Eq. (57), g e = 2 and the approximation that the masses of the sth and
the (s + 1)th ionized atom are about the same, the above equation becomes
where
n s+1,0
n s,0
f s+1 (T ) = 2(2πm ekT ) 3/2
h 3
= g s+1,0
g s,0 n e
f s+1 (T ) , (68)
(
exp − χ )
s+1
, (69)
kT
In general, not all the sth ionized atoms in a thermalized system are in
the ground state. From Eq. (55), Eq. (68) can be re-written as
n s+1
n s
= Z s+1
Z s n e
f s+1 (T ) , (70)
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