Structure and Evolution of Stars II
Lecture 4
Equations of State - Degeneracy, Ionization
3.3. Ionization Equilibrium
3.3. Ionization Equilibrium:
e.g., γ + H � H+ + e−
generally, γ + S i+ � S (i+1)+ + e−
Density of atoms in one state compared to state above given by
the Saha Equation:
�
�
3/2
n(i+1) ne
(2πme kT )
ui+1 −χi /kT
=
.2
e
(37)
ni
ui
h3
3.3. Ionization Equilibrium
Saha Equation:
n(i+1) ne
=
ni
�
kT )3/2
(2πme
h3
�
.2
ui+1 −χi /kT
e
ui
χi =ionization potential
3.3. Ionization Equilibrium
Saha Equation:
n(i+1) ne
=
ni
�
kT )3/2
(2πme
h3
�
.2
ui+1 −χi /kT
e
ui
ui =partition function for state i,
ui =
∞
�
gn e−Eg /kT
(38)
n=0
E0 = 0, En → χi
(sum of the statistical weights, or how likely an electron is to be
in a particular state i).
3.3. Ionization Equilibrium
Saha Equation
Generally, we need to simultaneously solve Saha
equations for every ionization state (bllrgghh...)
But for H, there is only one ionization state, and for
H � H + + e− , u0 = 2, u1 = 1.
So, Saha equation reduces to...
3.3. Ionization Equilibrium
Saha Equation
ln
�
nH +
nH
�
= ln
�
kT )3/2
(2πme
ne h3
�
−
χi
kT
(39)
ne depends on ionization of all elements (each ionizes at a
different T ).
First term on RHS (≡ Θ) is slowly varying during ionization.
Near surface of the the Sun, ne ≈ 1013 cm−3 , T ≈ 6000K,
Θ ≈ 18.5.
So, ionization occurs quickly when χi /kT ≈18.5.
3.3. Ionization Equilibrium
3.3. Ionization Equilibrium
3.4. Degeneracy
When a star begins to run out of nuclear fuel near the end of its
life, thermal pressure drops.
Core contracts, gas becomes extremely dense.
Heisenberg: electon’s position (x, y , z) and momentum
(px , py , pz ) can be known to a precision of h/4π.
Each electron occupies a volume of h3 in 6-D phase-space.
3.4. Degeneracy
Pauli: only 2 electrons (spin up & down) can occupy this
volume in phase space (2/h3 ).
Density of states f (�x , �p),
f (�x , �p) d 3 x d 3 p =
2 3
d x d 3p
3
h
(40)
As gas density increases, positional states fill up, electrons
are forced into higher momentum states – requires work.
Gas begins to exert degeneracy pressure.
3.4. Degeneracy
Consider a volume V containing Ne electrons,
Number density ne = Ne /V = ρ/(µe mH )
... where µe is the mean molecular weight of electrons, and
1/µe is the number of electrons per nucleon.
µe = 1 for hydrogen.
µe = 2 for He and metals.
The energy states are filled from the lowest up, the highest
energy filled being known as the Fermi energy, Ef , moving with
momentum pf .
3.4. Degeneracy
The total number of electrons is then given by,
� � pf
Ne =
f (�x , �p) d 3 x d 3 p
V
0
Because f (�x , �p) is spherically symmetric...
� � pf
2
2
3
Ne =
4πp
dp
d
x
3
h
V 0
8πpf3 V
=
3h3
Hence, from ρ we can find pf , which characterizes the
degenerate gas.
(41)
(42)
3.4. Degeneracy
Pressure
Degeneracy Pressure:
3.4. Degeneracy
Pressure
3.4. Degeneracy
Pressure
3.4. Degeneracy
Pressure
3.4. Degeneracy
Pressure
3.4. Degeneracy
Pressure
Degeneracy Pressure:
Number of electrons with momentum between p and p + dp,
passing through area dσ at an angle to the normal θ, into solid
angle dΩ, per time dt, is:
N(p) = f (�x , �p) dp . v (p) dt . dσ cos θ .
dΩ
4π
(43)
3.4. Degeneracy
Pressure
The pressure is then the momentum carried by these electrons
p cos θ per time dt per area dσ. Substituting Equ.(41), and
swapping d 3 x for v (p)dt cos θ dσ,
� pf �
2
dΩ
3
2
Pe =
4πp
dp
v
cos
θ
(44)
3
4π
h
0
sphere
3.4. Degeneracy
Pressure
We can simplify this by saying,
�
�
cos2 θdΩ = 2π
π
0
sphere
sin θ cos2 θdθ =
4π
3
(45)
And so in the end, we get:
8π
Pe =
3h3
�
0
pf
p3 v dp
(46)
3.4. Degeneracy
Pressure
3.4.1. Non-relativistic, p = me v
When the electrons are non-relativistic, we get...
� pf
8π
4
Pe =
p
dp
3me h3 0
8πpf5
=
15me h3
(47)
Eliminate pf with Ne ,
� �2/3
3
h2
1
5/3
5/3
Pe =
ρ
=
K
ρ
nr
π
20me (µe mH )5/3
(48)
That is, an n = 3/2 polytrope with fixed Knr .
3.4. Degeneracy
Pressure
So, from Eq. (15), we can say that the mass M is...
�
5Knr
M = 4π
8πG
�3/2
1/2
ρC Y3/2
(49)
where Y3/2 is found by numerical integration. Therefore, the
core density is...
ρC
�
�
5Knr −3
=
8πG
�
�2
M
= 1.32 × 108 µ5e
kg m−3
M⊙
M
4πY3/2
�2 �
(50)
3.4. Degeneracy
Pressure
Since R ∝ M −1/3 (see ‘polytropes’ section), and µe =2 (for a mix
of He, C, O),
−5/3
R = 0.04 µe
�
M
M⊙
�−1/3
R⊙
(51)
→ typical size of a white dwarf, the naked core of a ∼Solar
mass star.
Upper limit of M is when electrons become relativistic,
pf ≈ me c,
ρmax = µe mH
8π
(me c)3 ≈ 2 × 109 kg m−3
3
3h
So, M ≈ 0.7M ⊙ if µe = 2.
3.4. Degeneracy
Pressure
3.4.2. Extremely relativistic, v = c
This time...
Pe =
=
�
8πc pf 3
p dp
3h3 0
2πcpf4
3h3
(52)
Again, eliminate pf with Ne ,
� �1/3
�
�4/3
3
hc
ρ
Pe =
= Ker ρ4/3
π
8 (µe mH )
(53)
So this time, an n = 3 polytrope.
3.4. Degeneracy
Pressure
Again, from Eq. (15), we can say that the mass M is...
M = 4π
�
Ker
πG
�3/2
Y3
Note that it is independent of ρc and R.
For µe = 2, M = 1.4M ⊙ ≡ ‘Chandrasekhar Mass’.
This is the maximum mass of a white dwarf.
(54)
3.4. Degeneracy