Download 22 Lecture 22: Degeneracy of Matter

PHYS 652: Astrophysics
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Lecture 22: Degeneracy of Matter
“Physics is very muddled again at the moment; it is much too hard for me anyway, and I wish I
were a movie comedian or something like that and had never heard anything about physics!”
Wolfgang Pauli
The Big Picture: Last time we derived the Schwarzschild metric corresponding to an isolated
mass, which led to the the introduction of black holes and even horizons. Today we introduce
degenerate matter, such as the matter in white dwarfs and neutron stars. We also introduce
polytropes as simple equilibrium stellar models.
Degeneracy
According to Pauli’s Exclusion Principle, no two fermions (particles with spin of one half) can
occupy the same quantum state. This is equivalent to requiring that the volume per fermion be
proportional to λ3c ∼ (~/mc)3 , where m is the fermion’s mass and λc is its Compton wavelength.
The average number density of the fermions is therefore nf ∼ λ−3
c . In white dwarfs the density is
nf times the mass per electron, and in neutron stars it is the nucleon mass times nf .
We can use this argument to compute the relative densities of white dwarfs, which are supported
by electron degeneracy, and neutron stars, supported by neutron degeneracy to obtain (with approximation that the mass per electron is on the order of magnitude of the mass of the nucleon):
λ−3
mn 3
m3n
ρns
n,e
=
≈ (2000)3 = 8 × 109 .
(426)
= 3 =
ρwd
m
m
λ−3
e
c,e
e
In a gas of very high fermion density, the lower momentum states are filled, so fermions must
then occupy states of higher momentum. These high-momentum fermions make a large contribution
to the pressure, and the gas is said to be (partially) “degenerate”.
Complete Degeneracy
If the fermion density is large enough, then essentially all available states having energies E <
ǫf (where ǫf is the Fermi energy, defined as the energy of the highest occupied quantum state
in a system of fermions at absolute zero temperature). As the gas temperature is lowered, the
distribution function
1
,
(427)
f (p) = [E(p)−µ]/T
e
+1
approaches unity for particle energies E . µ, and zero for E & µ, where µ is the chemical potential.
For T = 0, µ ≡ ǫf , so the distribution function becomes a step function:
1 if ǫf ≥ E(p),
f (p) = θ(ǫf − E(p)) =
(428)
0 if ǫf < E(p).
The number density of fermions corresponding to the distribution function above is
Z pf
Z ∞
Z
8π
d3 p
4πp2 dp
8π pf 2
8π 1
f (p)
=
2
=
n=g
p dp = 3 p3f = 3 p3f ,
3
3
3
(2π~)
(2π~)
h 0
h 3
3h
0
0
so
pf =
3h3
n
8π
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1/3
.
(429)
(430)
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pf is the Fermi momentum corresponding to the Fermi energy:
ǫf =
The energy density is given by
Z
ρe = g
p2f
2m
=⇒
∞
E(p)f (p)
0
pf =
d3 p
8π
= 3
3
h
h
Z
p
2mǫf .
pf
(431)
E(p)p2 dp,
(432)
0
where E(p) is the kinetic energy per fermion.
Nonrelativistic (complete) degeneracy. When the fermions are nonrelativistic, so p = mv
and E(p) = p2 /2m. The energy density then is
ρ =
=⇒
ρ =
Z
Z pf
2
8π
8π 3 pf
8π 3h3
8π 1 5
8π pf p2 2
4
p dp =
p = 3 pf
= 3
n ǫf
p dp =
h3 0 2m
2mh3 0
2mh3 5 f
5h
2m
5h
8π
3
nǫf .
(433)
5
For nonrelativistic particles
2
2
23
2 pf
n 2
n
P = ρ=
nǫf = n
=
pf =
3
35
5 2m
5m
5m
3h3
n
8π
2/3
h2
=
20m
2/3
3
n5/3 .
π
(434)
The equation above is the equation of state for a nonrelativistic, completely degenerate fermion
gas.
Extremeprelativistic (complete) degeneracy. When the fermions are relativistic, p ≫ mc and
E(p) = c p2 + m2 c2 ≈ cp. The energy density then is
Z
Z
2π 3
2π 3h3
8πc pf 3
8πc 1 4
8π pf
2
n ǫf .
cpp dp = 3
p dp = 3 pf = 3 pf (cpf ) = 3
ρ =
h3 0
h
h 4
h
h
8π
0
3
=⇒ ρ =
nǫf .
(435)
4
For relativistic particles (recall Homework set #1):
3 1/3
hc 3 1/3 4/3
3h
13
1
1
1
=
nǫf = ncpf = nc
n
n .
P = ρ=
3
34
4
4
8π
8 π
(436)
The equation above is the equation of state for an extreme relativistic, completely degenerate
fermion gas.
Important point: For complete or nearly complete degeneracy, the pressure P is independent of
the temperature T .
Onset of Degeneracy
We now estimate the thresholds for the onset of the complete nonrelativistic degeneracy and
complete relativistic degeneracy.
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• From nondegeneracy to complete nonrelativistic degeneracy.
Let us first see under which conditions will a star end up in complete nonrelativistic degeneracy. This will happen when the pressure due to the thermal equilibrium of the particles is
balanced by the pressure due to the nonrelativistic degeneracy of electrons.
Combining the equation of state for the ideal gas
P =
ρkT
µ̄mH
(437)
and the eq. (434), we obtain
ρkT
h2
=
µ̄mH
20me
2/3
3
n5/3
e
π
(438)
where µ̄ is the mean molecular weight, defined as
1 X n̄i mH
=
,
µ̄
mi
(439)
i
mH is the mass of the hydrogen atom, and n̄i = ρi /ρ is the abundance of species by weight.
The number density ne of electrons is given in terms of the density as
ρ
ne =
.
(440)
mH µ̄e
Taking µ̄ = µ̄e ≈ 1, the eq. (438) becomes
ρ 5/3
h2
ρkT
≈
mH
20me mH
20me k 3/2 3/2
T
=⇒ ρ = mH
h2
ρ =
1.67 × 10−24 g
ρ ≈ 10
−8
T
3/2
.
20 9.11 × 10−28 g
1.38 × 10−16
erg 2
6.63 × 10−27
s
erg K
!3/2
T 3/2
(441)
Therefore
ρ > 10−8 T 3/2 ,
(442)
is the requirement for the electron gas to be completely degenerate.
• From nonrelativistic degeneracy to extreme relativistic degeneracy.
In the case of relativistic particles pf ≫ me c, but the “transition” occurs at, say, pf = 2me c:
3 1/3 3
1/3
3h
3h
ρ
pf =
=
= 2me c
take µ̄e = 1
n
8π
8π mH µ̄e
64πmH (me c)3
=⇒ ρ ≈
3h3
3
9.11 × 10−28 g 3 × 1010 cm
64π 1.67 × 10−24 g
s
=
3
(6.63 × 10−27 erg s)3
g
=⇒ ρ ≈ 107
.
(443)
cm3
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Therefore
g
.
cm3
is the requirement for the gas of electrons to reach extreme relativistic degeneracy.
ρ > 107
(444)
These degenerate forms of matter describe brown dwarfs, white dwarfs (electron degeneracy)
and neutron stars (neutron degeneracy), which we discussed in Lecture 11.
Figure 40: Simple model of a star: a sphere of gas in hydrostatic equilibrium.
Hydrostatic Equilibrium
We now present a simple model for a star in hydrostatic equilibrium.
Consider a think shell within a star in equilibrium. There are inward force acting on the shell
due to its gravitating mass and the outward force of gas pressure:
M (r) ρ(r)4πr 2 dr
Fg = −G
r2
2
Fp = 4πr [P (r + dr) − P (r)] = 4πr 2 dP
(445)
where M (r) is mass interior to the shell:
M (r) = 4π
Z
r
ρ(r̃)r̃ 2 dr̃.
(446)
0
In hydrostatic equilibrium, these two forces are balanced, so
Fp = Fg
4πr 2 dP
=⇒
dP
dr
M (r) ρ(r)4πr 2 dr
= −G
r2
GM (r)
= −ρ(r)
.
r2
The equation above is the equation of hydrostatic equilibrium.
121
(447)
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Isothermal Atmospheres in Hydrostatic Equilibrium
Stellar atmospheres are usually thin when compared to the stellar radius, which allows us to
approximate the force due to gravity as a constant throughout the atmosphere:
g≡
GM
≈ const.
R2
(448)
Let h be the height of the atmosphere (r-derivative can be replaced with an h-derivative). Then
the equation of hydrostatic equilibrium [eq. (447)] then becomes
dP
= −ρg.
dh
(449)
But from the equation of state for ideal gas [eq. 437]:
P =
ρkT
µ̄mH
=⇒
ρ=
µ̄mH
P,
kT
(450)
so the eq. (449) becomes
µ̄mH g
dP
=−
P.
dh
kT
If we define the “e-folding height” (“scale height”) of the atmosphere as
H≡
(451)
kT
,
µ̄mH g
(452)
and define the initial condition P (0) = P0 , we can rewrite the eq. (449) and integrate it to obtain
dP
dh
=⇒
P
dP
dh
=⇒
=−
H
P
H
h
=⇒
P (h) = Ce−h/H
log P = − + c
H
P (h) = P0 e−h/H .
= −
but P (0) = P0
(453)
Important point: the equation of hydrostatic equilibrium must be accompanied by an equation
of state.
Polytropes
Polytropes are a family of equations of state for which the pressure P is given as a power of
density ρ. A gas governed by a polytropic process has the equation of state
P V γ = const.
Since ρ = M/V , where M is the mass of gas contained in volume V , we have
−γ
M
−γ
,
P ∝ V
∝
ρ
=⇒
P = κργ ,
κ = const.
(454)
(455)
Gas obeying an equation of state of this form is called a polytrope. Examples of polytropes are
given in Table 7.
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Table 7: Examples of polytropic gases.
Type of polytropic gas
nonrelativistic, completely degenerate gas
extreme relativistic completely degenerate gas
isothermal gas
gas and radiation pressure
γ
5/3
4/3
1
4/3
Eddington standard model. The polytrope with γ = 4/3 is a simple model of a star supported
by both radiation pressure
Pr =
1 π2 4 π2 4 1 4
1
ργ =
T =
T ≡ aT ,
3
3 15
45
3
(456)
ρkT
.
µ̄mH
(457)
and ideal gas pressure:
Pg =
Now introduce the constant β quantifying the relative contribution of gassy pressure to the total
pressure (both gas and radiation) (P = Pr + Pg ):
Pg = βP,
=⇒
=⇒
β=
Pg
,
P
Pr = (1 − β)P,
(458)
so that
3(1 − β)
1
=⇒
T4 =
P,
Pr = (1 − β)P = aT 4
3
a
Next, we eliminate the temperature T in from the equation of state:
4
β P
4
P3
=⇒
=⇒
P
ρk 4 4
ρk 4 3(1 − β)
=
=
P
T =
µ̄mH
µ̄mH
a
4
3(1 − β) 4
k
ρ
=
µ̄mH
aβ 4
4/3 k
3(1 − β) 1/3 4/3
=
ρ .
µ̄mH
aβ 4
Pg4
(459)
(460)
The term multiplying ρ4/3 in the equation above is constant if β is constant (the relative breakdown
of radiation and gas pressure remains unchanged) and µ̄ is constant (composition of gas does not
change). If this is indeed the case, then we have the Eddington standard model
P = κρ
4/3
,
κ≡
k
µ̄mH
4/3 3(1 − β)
aβ 4
1/3
.
(461)
This model is a special case of Lane-Emden equations governing the polytropes in hydrostatic
equilibrium which we will discuss next time.
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