Chapter 2 Stellar Structure Equations 1 Mass conservation equation

Chapter 2
Stellar Structure Equations
To arrive at equations that determine ρ, P , T etc., we invoke the following
conservation laws.
1 Mass conservation equation
For a spherically symmetric star, let r be the distance from the star center
and m = m(r) be the mass enclosed by the sphere of radius r centered at
the center of the star. (Note: Some authors use the notation M(r) instead of
m(r).) Clearly, m(0) = 0 and m(R) = M where M and R denote the mass
and the radius of the star. More importantly, due to spherical symmetry,
∫ r
0
4πr 2 ρ(r)dr = m(r) , (1)
where ρ(r) denotes the density of the star at radius r. (Obviously, ρ(r) > 0
for 0 ≤ r < R and ρ(r) = 0 for r > R.)
Alternatively, Eq. (1) can be rewritten as the following differential form
dm
dr = 4πr2 ρ(r) . (2)
To summarize, conservation of mass relates ρ and m. So, we have introduced
one equation and yet one new unknown. So, we have to find out more
equations relating ρ, P , T , m, etc. before we can solve the problem of stellar
structure.
1

2 Momentum conservation and the equation
of hydrostatic equilibrium
The next equation we introduce comes from the conservation of momentum;
or more precisely, in LTE the net force acting on any microscopically
large but macroscopically small region is zero. Inside a star in LTE, the forces
involve are gravity and pressure gradient due to gas or photon. Specifically,
dP
dr = −Gm(r)ρ(r) r 2 . (3)
We denote the central pressure P (0) by P c . And clearly, P (R) can be
approximated by 0.
Eq. (3) can be rewritten as
dP
dm = − Gm
4πr 4 (m) . (4)
Hence,
Since r < R inside the star, we have
P (R) − P (0) =
< −
∫ M
0
∫ M
0
dP
dm dm
Gm
4πR 4 dm
= − GM 2
8πR 4 . (5)
P c > GM 2
8πR 4 (6)
in any star. In particular, this inequality tells us that the central pressure of
our Sun is at least 4.4 × 10 14 dyn/cm 2 ≈ 4 × 10 8 atm.
Multiplying Eq. (4) by the volume V (r) = 4πr 3 /3, we have
∫ P (R)
∫ M
P (0)
V (r)dP = − 1 3
2
0
Gm
r
dm . (7)

Note that the R.H.S. above is Ω/3 where Ω is the gravitational potential
energy of the entire star. Using integration by parts, the L.H.S. above equals
− ∫ V (R)
P dV = − ∫ M
P/ρ dm. Hence, we arrive at
0 0
Ω = −3
∫ M
0
P
ρ
dm . (8)
This equation is one form of the virial theorem (sometimes known as the
global form of the virial theorem).
We can use the non-relativistic classical ideal gas to obtain the usual form
of virial theorem. The kinetic energy density of an ideal gas ε is simply given
by ε = N 3
kT and the pressure of an ideal gas is P= N kT= 2 ε. Substituting
V 2 V 3
this relation into above equation we get
∫ M
∫
P
V (R) ∫ V (R)
3
0 ρ dm = 3 P dV = 2 εdV = 2U = −Ω (9)
0
Thus, virial theorem tells us that 2U + Ω = 0 where U is the thermal energy
of a star making up of non-relativistic classical ideal gas. Consequently, the
total energy of the star is U + Ω < 0. In other words, the star is in a bound
state.
One consequence of the virial theorem is that upon gravitational contraction,
the star becomes hotter, more tightly bound and have to radiate some
energy to space.
In contrast, if the star is made up of extremely relativistic particles, the
pressure equals one-third the energy density. Hence, in this case, virial theorem
implies that U + Ω = 0. In other words, a star making up of extremely
relativistic particles can be in LTE only when its total energy is 0. So, such
a star is not bounded.
Another consequence of the virial theorem concerns the conversation law.
Consider a star making up of non-relativistic classical ideal gas. Suppose
further that the timescale concerned is small enough that the total energy of
the star is roughly a constant. Since 2U + Ω = 0, the total energy of the star
can be expressed as a function of U (or Ω) alone. Thus, the thermal energy
and the gravitational potential energy of the star are also conserved. In turns
out that this consequence of the virial theorem is useful to understand several
properties of late stage stellar evolution.
3
0

Example. Derive the behavior of m(r),P (r), the luminosity F (r), defined
as dF/dr = 4πr 2 ρq, where q is the energy production rate, and T (r) near
the centre of a star by Taylor expansion for given composition and physical
properties at r = 0: ρ c , P c and T c . The temperature is given by dT/dr =
− (3/4ac) (kρ/T 3 ) (F/4πr 2 ), where k is the opacity and a is the radiation
constant.
If we adopt r as independent space variable, the Taylor expansion near
r = 0 for any function f(r) is
f = f c +
( ) df
r + 1 dr
c
2
( d 2 f
r
dr
)c 2 + 1 ( ) d 3 f
r 3 + ..., (10)
2 6 dr 3
and we retain only the first non-vanishing term besides f c . For the mass
m(r) we have m c = 0 (boundary condition), and from the mass continuity
equation we have
( d 3 m
( d 2 m
dr 3 )c
( ) dm
= 4π ( r 2 ρ ) = 0, (11)
dr
c
c
dr 2 )c
(
= 4π 2rρ + r 2 dρ )
= 0, (12)
dr
c
(
= 4π 2ρ + 4r dρ
dr + d2 ρ
r2 = 8πρ
dr
)c
2 c . (13)
Therefore near the centre
m(r) = 4 3 πρ cr 3 , (14)
as if the density were uniform and equal to the central value. For the pressure
P (r) we have
( ) dP
(
= − ρ gm ( ) 4πGρ
= −
dr
c
r
)c
2 r
= 0, (15)
2 3
c
4

( d 2 P
dr 2 )c
= −
( dρ Gm
+ ρ G dm
dr r 2 r 2 dr − 2ρGm r
)c
3
= − 4πGρ2 c
, (16)
3
Therefore near the centre
P (r) = P c − 2 3 πGρ2 cr 2 . (17)
For the luminosity F (r) we have F c = 0 (boundary condition), and
( d 2 F
dr 2 )c
( ) dF
= 4π ( r 2 ρq ) = 0, (18)
dr
c
c
(
= 4π 2rρq + r 2 dρ
dr q + r2 ρ dq )
= 0, (19)
dr
c
( d 3 F
dr 3 )c
= 4π [ 2ρq + r (...) + r 2 (..) ] c = 8πρ cq c . (20)
Therefore near the centre
F (r) = 4 3 πρ cq c r 3 . (21)
For the temperature T (r) we obtain
( ) dT
= − 3 ( kρ F
= −
dr
c
16πac T 3 r
)c
1 ( ) kρ 2 q
2 4ac T r = 0, (22)
3
( d 2 T
dr 2 )c
= − 3 [ ( )
kρ d F
+ F (
d kρ
= −
16πac T 3 dr r 2 r 2 dr T
)]c
1
3 4ac
k c ρ 2 cq c
. (23)
Tc
3
5

Therefore near the centre
T (r) = T c − 1
8ac
k c ρ 2 cq c
r 2 . (24)
Tc
3
Note that these relations hold regardless of the functional dependence
P (ρ, T ), q (ρ, T ), and k (ρ, T ).
3 Simple stellar models
So far, we have 3 unknowns, namely, m(r), ρ(r) and P (r) but only two
equations, namely, the mass conservation equation and equation of hydrostatic
equilibrium. So, we need another equation to relate m, ρ and P .
Astronomers have proposed a number of extremely simplified equations to
fulfill this task. Since all these proposed equations are not derived from first
principle, they are nothing more than crude approximations of the realistic
situation. Yet, these approximations are sometimes quite useful to investigate
the approximate structure of a star.
The first such model is called constant density model, namely, we assume
ρ is a constant inside the entire star. The second one is called the linear
density model, namely, ρ(r) = ρ c (1 − r/R) where ρ c is a constant. You may
solve m(r) and P (r) in the above two models yourself.
4 The polytrope model
The last such model I am going to introduce, which is the most important
model of this kind, is called the polytrope model. We assume that
P = Kρ γ (25)
for some constants K and γ. Note that this model is based on the observation
that P and ρ follow the above equation for ideal gas in various situations such
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as adiabatic expansion. With this in mind, we expect that γ varies from 4/3
to 5/3.
Combining the mass conservation and hydrostatic equilibrium equations,
we have
( )
1 d r
2
dP
= −4Gπρ . (26)
r 2 dr ρ dr
Write γ = 1 + 1/n, then the above equation can be written as
[
]
(n + 1)K d
r 2 (1−n)/n dρ
ρ = −ρ . (27)
4πGnr 2 dr
dr
The boundary conditions for this differential equations are ρ(R) = 0 and
dρ(0)/dr = 0. (Do you know why?)
Let ρ = ρ c θ n where ρ c is the central density of the star, we have
(
1 d
ξ 2 dθ )
= −θ n , (28)
ξ 2 dξ dξ
where r = [(n + 1)Kρ (1−n)/n
c /4πG] 1/2 ξ. The corresponding boundary conditions
are θ = 1 and dθ/dξ = 0 at ξ = 0. This equation is called Lane-Emden
equation and can be solved numerically.
The significance of the polytrope model is that Lane-Emden equation is
independent of the mass M, radius R and central density ρ c of a star. So,
once you have numerically solved the Lane-Emden equation for a given value
of n, the numerical solution can be used to deduce the solution of any star
with the same polytropic index n (or γ).
For the special cases of n = 0, n = 1 and n = 5, Lane-Emden equation
can be solved exactly. The case of n = 0 is easy. You may work out by
yourself that θ(ξ) = 1 − ξ 2 /6; and hence
ρ = ρ c whenever ξ ≠ √ 6 . (29)
The case of n = 1, Lane-Emden equation can be rewritten as
d 2
(ξθ) = −ξθ (30)
dξ2 7

whose solution that matches the correct boundary conditions above is θ(ξ) =
sin ξ/ξ. Hence,
ρ = ρ c
sin ξ
ξ
. (31)
For the case of n = 5, Lane-Emden equation can be rewritten as
d 2 z
dt = z(1 − z4 )
2 4
, (32)
where ξ = e −t and θ = z/(2ξ) 1/2 . Multiplying Eq. (32) by dz/dt, we obtain
1
2
( ) 2 dz
= 1 dt 8 z2 − 1 24 z6 + C . (33)
Boundary conditions demand that C = 0 and hence
dz
dt = −z 2
) 1/2 (1 − z4
. (34)
3
After making the substitution z 4 /3 = sin 2 ζ and upon integration, we have
e −t = C ′ tan(ζ/2) where C ′ is a constant of integration. Applying the boundary
conditions and after simplification, we obtain
( ) −5/2
ρ = ρ c 1 + ξ2
. (35)
3
(Could you express ρ c as a function of M and R? Besides, could you
express ρ and P as a function of r?)
It is straight-forward to check that the smallest positive root for the
solution θ(ξ) of the Lane-Emden equation equals ξ = ξ 1 ≡ √ 6 and ξ = ξ 1 ≡ π
when n = 0 and 1 respectively. This value of ξ 1 corresponds to the boundary
of a star. On the other hand, θ ≠ 0 for any real-valued ξ for n = 5. In
other words, the solution of the Lane-Emden equation for n = 5 does not
correspond to a physical star. In fact only n < 5 and n ≠ 1 can correspond
to a physical star (cf. table 1 in supplementary notes).
8

By substituting θ and ξ in the Lane-Emden equation back our equation of
mass conservation (also known as equation of (mass) continuity) and equation
of hydrostatic equilibrium, we obtain
[ ] 1/2 (n + 1)K
R =
ρ (1−n)/2n
c ξ 1 , (36)
4πG
where ξ 1 is the smallest positive value of ξ having θ(ξ) = 0,
[ ] 3/2 (n + 1)K
M = 4π
ρ (3−n)/2n
c
4πG
[
−ξ 2 dθ ]∣
∣∣∣ξ=ξ1
, (37)
dξ
and
[
P c = GM ( ) ] ∣ 2
2 −1∣∣∣∣∣ξ=ξ1 dθ
4π(n + 1)
, (38)
R 4 dξ
¯ρ =
M [
4πR 3 /3 = ρ c − 3 ]∣
dθ ∣∣∣ξ=ξ1
, (39)
ξ dξ
Ω = − 3 GM 2
5 − n R . (40)
Note that R is independent of ρ c for n = 1, Ω > 0 for n > 5, Ω is
undefined for n = 5. Therefore, the polytropic models for n = 1 and n ≥ 5
are not physical. (Can you verify that M is finite, P c is infinite and ¯ρ is 0
for case of n = 5?)
Interestingly, for n = 3, the mass of the star is independent of its central
density. This particular polytropic index is sometimes also called the Eddington
standard model. (We shall say more about the Eddington standard
model later in the next chapter.)
Example. For a given mass M and central pressure P c , which polytrope
yields a bigger star: that of index 1.5 or that of index 3?
The central pressure of the star can be written as
P c = (4π) 1/3 B n GM 2/3 ρ 4/3
c . (41)
9

(Remark : You can check above equation yourself and B n contains all n-
dependent coefficients. In table 2 of Supplementary notes, you can see that
B n varies very slowly with n. )
For a given M and P c we obtain
( ) 3/4
ρ c,1.5 B3
= . (42)
ρ c,3 B 1.5
The central density of the polytropic star is given in terms of the mean
density as
M
ρ c = D n ¯ρ = D n
4πR 3 /3 , (43)
where
[ ( ) ] −1
3 dθ
D n = −
, (44)
ξ 1 dξ
ξ 1
are some constants. For a given M we obtain the ratio of radii R(n) as
R (1.5)
R (3)
=
=
(
D1.5
(
D1.5
) 1/3 ) 1/3 ( ) 1/4
ρ c,3
B1.5
=
D 3 ρ c,1.5 D 3 B 3
( ) 1/3 ( ) 1/4 5.991 0.206
< 1, (45)
54.81 0.157
and therefore
R(3) > R (1.5) . (46)
This result makes sense because for a large polytropic index, the star is stiffer
and hence its pressure can resist a bigger gravity so the star should be larger
for the same mass.
Example. Capella is a binary star discovered in 1899, with a known
orbital period, which enables the determination of the mass and radius of the
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ighter component: M = 8.30 × 10 33 g and R = 9.55 × 10 11 cm. Assuming
that the star can be described by a polytrope of index 3, find the central
pressure and the central density. Check wether the central pressure satisfies
the inequality P c > GM 2 /8πR 4 .
The central density is given by
M
ρ c = D n ¯ρ = D n
4πR 3 /3 , (47)
and we obtain ρ c = 1.2 × 10 −1 g/cm 3 . In order to obtain the central pressure
as a function of M and R we eliminate ρ c between the above equation and
thus obtaining
P c = (4π) 1/3 B n GM 2/3 ρ 4/3
c , (48)
P c = GM 2
4πR 4 [
(3D n ) 4/3 B n
]
. (49)
The term in square brackets exceeds unity for all n and hence
P c > GM 2
4πR 4 > GM 2
8πR 4 . (50)
Thus the inequality is generally satisfied by polytropic models. For Capella,
with n = 3, P c = 6.1 × 10 13 dyn/cm 2 .
5 Energy conservation and the energy production
equation
Now, we want to go further than simple stellar models.
introducing a few more equations.
We do so by
11

With our LTE assumption, matter in the star is in thermal equilibrium.
Thus, the gas inside a star will not expand or contract; the work done by the
gas is zero. Let L(r) denotes the luminosity at the spherical shell with radius
r centered at the core of the star. (That is, L(r) is the energy per unit time
moves out of the spherical shell of radius r.) Energy conservation demands
that
dL
dr = 4πr2 ρ(r)ϵ(r) , (51)
where ϵ is the rate of energy production per unit mass of material at the
radius r from the stellar center. Note that ϵ is an implicit function of density
ρ, temperature T and chemical composition. (Note: a few authors use the
notation q instead of ϵ.)
Using Eq. (2), the above energy production equation can be written as
dL
dm
= ϵ(r) . (52)
Note that L(0) = 0 and L(R) = L where L is the luminosity of the star.
Besides, the region with ϵ(r) > 0 is the region where nuclear fusion takes
place.
6 Some important timescales
• The free-fall / dynamical timescale: For a star with mass M and
radius R, the escape velocity or free fall velocity is of order of √ GM/R.
The free-fall timescale is, therefore, of the order of τ ff ≈ R/ √ GM/R =
√
R3 /GM ≈ 1/ √ G¯ρ where ¯ρ denotes the mean density of the star.
For our Sun, the free-fall timescale is of order of an hour. In short, any
force that is unbalanced inside a star occurs at the free-fall timescale.
Thus, if we are interested in the processes of a stable star over a time
span which is much longer than the free-fall timescale, then we are safe
to assume that the star is in mechanical equilibrium.
• The Kelvin-Helmholtz timescale: This is the timescale τ KH for a
star to radiate its thermal energy away at constant luminosity. By
virial theorem,i.e. KE=PE/2 or KE (the thermal energy) ∼ GM 2 /R,
therefore τ KH ≈ GM 2 /RL. For our Sun, τ KH is about 3 × 10 7 yr. For a
time much longer than the Kelvin-Helmholtz timescale, we may safely
assume that the star is in thermal equilibrium.
12

• The nuclear burning timescale: This is the timescale for burning
all the nuclear fuel of a star at constant luminosity and is therefore
equal to τ nucl ≈ Mc 2 ∆/L, where ∆ ≈ 10 −3 is the typical binding
energy of a nucleon divided by the rest energy of the nucleon. For our
Sun, τ nucl ≈ 10 11 yr. The nuclear burning timescale is a rough estimate
of the lifespan of a star. (You may notice that the nuclear burning
timescale overestimates the lifespan of a star. Do you know why?)
Since τ ff ≪ τ KH ≪ τ nucl , we know once again that our assumption
of LTE is a good approximation to study the interior structure and
evolution of a star.
7 Equation of state
So far, we have used the conservation of mass, energy and momentum
to construct three equations for stellar structure. Yet, we have P , T , m, ρ,
L, ϵ, as well as the chemical composition as a function of r for variables.
To further our analysis, we have to look for equations that tell us specific
properties of the material in a star. One such equation is the equation of
state (EOS), namely, an equation expressing the pressure P as a function of
T , ρ and chemical compositions.
To model a star, astrophysicists may use some extremely accurate and
hence complicated EOSs. I shall not do so in this introductory course for the
following reason. Although those EOSs more accurately model the material
behavior, their predictions on the interior structure of a star are in most cases
not significantly different from the simple EOS we will use in this course. The
simplest EOS, which at the same time turns out to be the one we will use in
this course, is the ideal gas law that you have learned in high school. That
is, P = nkT where n is the number density of gas particle.
Ideal gas law is a good approximation to describe the properties of matter
in a star. To see why, I first have to introduce the Saha equation discovered
by M. Saha in 1920.
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7.1 Derivation of Saha Equation
Recall from the course Statistical Mechanics and Thermodynamics that
for a particle (or system of particles), the probability of a particle is in state
s is proportional to g s e −E s/kT where g s is the degeneracy of state s and E s is
the energy of state s relative to some fixed reference energy level (sometimes
taken to be the ground state). More importantly, the proportionality constant
is the same for all the states. Thus, the ratio of the probability that
the particle is in state (s + 1) to the probability that it is in state s is given
by the Boltzmann formula
g s+1
g s
e −(E s+1−E s)/kT . (53)
The Statistical Mechanics and Thermodynamics course also tells us that the
thermodynamic properties of a system with fixed number of particles N and
fixed temperature T and fixed volume V is encoded in the so-called partition
function
Z ≡ ∑ ( ) −Ei
g i exp , (54)
kT
i
where the sum is over all possible energy states of the system. Furthermore,
the ratio between the mean number of particles N s in state s and the total
number of particles N ≡ ∑ N j in a sample is given by
N s
N =
g se −E s/kT
∑
j g je −E j/kT
≡ g se −E s/kT
Z
. (55)
For an ideal classical non-relativistic gas particle with ground state energy
E 0 and degeneracy g, its partition function is given by
∫ ∫
g
Z(1, V, T ) =
e −(E 0+p 2 /2m)/kT d 3 ⃗x d 3 ⃗p
(2π) 3
=
gV
(2π) 3 e−E 0/kT
= gV
λ 3 e−E 0/kT
∫ +∞
0
4πp 2 e −p2 /2mkT dp
(56)
where m is the mass of the particle and
√
2π
λ ≡
mkT . (57)
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Note that we have assumed that the momentum of the gas particle is isotropically
distributed in order to obtain the above expression for Z(1, V, T ). The
partition function of a system of N indistinguishable classical non-relativistic
ideal gas particles each with ground state energy E 0 and degeneracy g is then
given by
Z(1, V, T )N
Z(N, V, T ) = . (58)
N!
Recall again from the Statistical Mechanics and Thermodynamics course
that the most convenient method to study the situation in which the particle
number in the system may change is to use the so-called grand partition
function
+∞∑
( ) µN
Q(µ, V, T ) ≡ Z(N, V, T ) exp , (59)
kT
N=0
where µ is the chemical potential. In the thermodynamic limit, µ can be
interpreted as the work needed to add one extra particle to the system at
constant V and T . The grand partition function encodes the thermodynamical
information of a system with fixed chemical potential, volume and
temperature. For N → ∞, equation (59) can be evaluated exactly as
[ ]
gV
Q(µ, V, T ) = exp
λ 3 e(µ−E 0)/kT
. (60)
You can refer to the detail of any standard Statistical Mechanics and Thermodynamics
textbook for detail.
More generally, the grand partition function of a system of interacting
particles of various different species such that each species can be approximated
by a classical non-relativistic ideal gas is given by
Q({µ s }, V, T ) ≡ ∏ s
Q(µ s , V, T ) = exp
[ ∑
s
g s V
λ 3 s
e (µ s−E 0,s )/kT
]
, (61)
where the s is the species label. Hence, the corresponding grand potential
equals
G({µ s }, V, T ) ≡ −kT ln Q = −kT ∑ ( )
g s V µs − E 0,s
exp
. (62)
λ 3 s s kT
15

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)
16

where
Z s = ∑ i
[ ]
−(Es,i − E s,0 )
g s,i exp
kT
is the partition function of those sth ionized atoms in various energy levels.
(71)
Eqs. (68) and (70) are known as the Saha equation. Saha equation is
useful in astronomy. For instance, if we know the electron partial pressure P e ,
then the electron number density n e can be well approximated by P e /kT in
many cases. We now apply Saha equation to estimate the degree of ionization
of our Sun. For simplicity, we consider only the form of Saha equation
in Eq. (68). The mean density of our Sun is about 1.4 g/cm 3 . By virial
theorem, the mean temperature of our Sun is about 6 × 10 6 K. We simplify
our analysis by assuming that the Sun is made up of entirely hydrogen. Thus,
χ 1 = 13.6 eV. Let us define the degree of ionization by n 1 /(n 0 + n 1 ) ≡ x.
Then Saha equation demands that
x
1 − x = f 1(T )
≈ m Hf 1 (T )
n e ρx
. (72)
By solving above equation we get x ≈ 0.95. In other words, almost all
hydrogen in our Sun is ionized. By the same token, it is straight-forward to
check that most atoms in a main sequence star are ionized. On the other
hand, in the solar atmosphere where T ≈ 6000 K and n e ≈ 10 13 cm −3 . Saha
equation tells us that the degree of ionization of hydrogen is about 10 −4 .
Note however that Saha equation has its limitations. It works for the
most region of a star, where is in thermal equilibrium. On the other hand it
may not be applicable to stellar atmosphere, where thermal equilibrium may
not be applied. Also, it assumes classical non-relativistic ideal gas, so it may
not work near the core of some stars, where degenerate may occur.
7.2 Simple Equation of State for Stars
As most of the stellar material should be ionized, the dominant interaction
between particles in stellar interior (besides gravity) is electrostatic in nature.
The typical distance between particles in stellar interior d ≈ (Ām H /¯ρ) 1/3
where Ā is the average atomic weight of particles, m H is the mass of hydrogen
atom, and ¯ρ is the average density of a star. Consequently, the typical
17

Coulomb energy per particle is about ¯Z 2 e 2 /d where ¯Z is the average atomic
number of particles. In contrast, virial theorem tells us that the gravitational
potential energy and internal energy of a star are of the same order. Consequently,
the ratio of Coulombic energy ( M ) to internal energy of a star
( GM 2
R ) is about ¯Z 2 e 2
¯Z 2 e 2
Ām H d
. (73)
G(Ām H ) 4/3 M
2/3
Note that ¯Z and Ā are of order of 1. Therefore, the above ratio is about 10 −2 ,
which is much less than 1. Thus, electrostatic potential energy contribute
only a small fraction of the internal energy of a star; and most of the internal
energy are in the form of heat. More specifically, the thermal K.E. of a
typical particle inside a star is much greater than the electrostatic potential
energy it experiences.
Well, you may also ask if quantum effects may alter the ideal gas EOS
in a star. We may estimate the uncertainty in position ∆x to be of order of
d. Similarly, the uncertainty in momentum ∆p is of the order of √ kT Ām H .
Thus, for all living stars, ∆x ∆p is much larger than h. Hence, quantum
effects play little role in affecting the ideal gas EOS of a star. (Note: the
situation is completely different for dead stars such as white dwarfs and
neutron stars. Because the density of dead stars is high, ∆x ∆p for some
particle species inside these stars are of order of h and hence quantum effect
drastically changes their EOS.)
To summarize, ideal gas EOS is a good approximation in the study of
the interior of most living stars. (However, quantum mechanical effect is
important in some other “stars” such as brown dwarfs and supernovae. Radiation
pressure also plays an important role in the structure and stability
of a super-massive star. We shall come across them later on in this course.)
From now on, we model the EOS of a star by ideal gas law. More precisely,
we assume that the gas pressure of a star is given by P gas = nkT . Since the
temperature and luminosity of a star are high, we conclude that electrons,
ions, and photons are the three most important constituents of a star. The
electron pressure is given by
where n e is the electron number density.
P e = n e kT (74)
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The ion pressure is given by
where n i is the ion number density.
P i = n i kT (75)
What are the values of n e and n i ? To answer this question, we denote the
mass fraction of H, He and “heavy elements” (that is, those elements heavier
than H and He) in a star by X, Y and Z = 1 − X − Y respectively. (More
precisely, X and Y are the mass fraction of 1 H and 4 He respectively. Do
you know why we can forget about the contribution of 2 H, 3 H, 3 He?) Then,
number density of hydrogen and helium nuclei equal Xρ/m H and Y ρ/4m H ,
respectively. Therefore,
n i =
ρ (
X + Y m H 4 + Z )
, (76)
⟨A⟩
where ⟨A⟩ denotes the average mass number of heavy elements in a star.
Once again by virial theorem and the above analysis, we know that the
thermal energy of a star is of the order of GM 2 /R. That is,
GM 2 /R ∼ k ¯T M/m H , (77)
where ¯T ≈ 10 6 K is the average temperature of a star. Hence, for most
part of a star, the temperature is so high that most atoms are completely
ionized. (As we have already seen, this is not a valid assumption for stellar
atmospheres. Nonetheless, this assumption has little effect in determining
the overall structure of a star. Similarly, this assumption is not valid for
high atomic number atoms, but their numbers are small compared to that of
hydrogen and helium for a typical main sequence star.) With this assumption
plus the charge neutrality in mind,
n e = ρ (X + 2 Y ⟨ z
⟩ )
m H 4 + Z A
≈
ρ (X + Y m H 2 + Z )
2
ρ(1 + X)
= , (78)
2m H
where ⟨z/A⟩ ≈ 1/2 is the ratio of atomic number to mass number averaged
over all heavy elements.
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The gas pressure P gas = P i + P e . For our Sun, it is approximately equal
to 1.61ρkT/m H .
We have not finished the discussions on EOS yet since we leave out radiation
pressure. Recall from the course Statistical Mechanics and Thermodynamics
that photons are bosons. Therefore, in thermal equilibrium, photons
obey Bose-Einstein statistics. That is, the distribution of photons is isotropic
and the number density of photons with frequencies between ν and ν + ∆ν
equals
n(ν) dν = 8πν2
c 3
dν
exp(hν/kT ) − 1 . (79)
To see why photon number density follows Eq. (79), one recall that the
energy of having s photons each with momentum ⃗p equals spc. As photon
is bosonic, s can take on any natural number. Moreover, the probability
P (s) that there are exactly s photons each with momentum ⃗p follows the
constraint
P (s + 1)
= e −pc/kT = e −hν/kT , (80)
P (s)
where ν is the frequency of the photon. Consequently,
P (s) =
e −shν/kT
∑ +∞
s ′ =0 e−s′ hν/kT = e−shν/kT (1 − e −hν/kT ) (81)
provided that ⃗p is an allowed momentum of a photon. Thus, the expected
number of photons with an allowed momentum ⃗p is given by
⟨N γ (⃗p)⟩ = 2
+∞∑
s=0
sP (s) =
2
e hν/kT − 1 . (82)
(Note that the 2 above reflects the fact that a photon has two possible polarizations.)
As a result, the total expected number of photons equals
⟨N γ ⟩ = 1 ∫ ∫
⟨N
h 3 γ (⃗x)⟩ dV d 3 ⃗p
= 1 ∫ ∫
2
dV
h 3 e hν/kT − 1 d3 ⃗p
Hence, Eq. (79) is valid.
= V
∫
8πν 2
dν . (83)
c 3 (e hν/kT − 1)
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Note that the first fraction in Eq. (79) is called phase space factor while
the second fraction is the Bose-Einstein distribution factor. Moreover, this
distribution n(ν) is sometimes called the blackbody spectrum.
For a photon of momentum ⃗p hitting a wall and then reflected back elastically,
the magnitude of the change in momentum equals 2p cos θ = 2hν cos θ/c
where θ is the angle between ⃗p and the normal of the wall. The radiation
pressure due simply the momentum transfer to the wall per unit time per
unit surface area, i.e. P rad = ∫ dN γ (p,θ)
2p cos θ= ∫ c cos θ dN γ(p,θ)
2p cos θ, in
dAdt
dV
other words
P rad = 1 h 3 ∫ +∞
=
0
∫ +∞
0
∫ π/2 ∫ 2π
0
0
hν
3 n(ν)dν
c cos θ
2hν cos θ
c
2
e hν/kT − 1
( ) 2 ( )
hν
hν
sin θ dϕdθd
c
c
= 4σ
3c T 4 , (84)
where σ = 2π 5 k 4 /15c 2 h 3 is the Stefan’s constant.
Thus, our ideal gas EOS for stellar matter is
P = P i + P e + P rad . (85)
Finally, note that the energy density of a photon gas at temperature T is
given by
u rad =
Hence, u rad = 3P rad .
∫ +∞
0
hνn(ν)dν = 4σ c T 4 . (86)
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