Classical and Quantum Gases

Fundamental Ideas
– Density of States
– Internal Energy
– Fermi-Dirac and Bose-Einstein Statistics
– Chemical potential
– Quantum concentration
Density of States

Derived by considering the gas particles
as wave-like and confined in a certain
volume, V.
– Density of states as a function of
momentum, g(p), between p and p + dp:
V
2
g  p dp  g s 3 4p dp
h
– gs = number of polarisations

2 for protons, neutrons, electrons and photons
Internal Energy

The energy of a particle with
momentum p is given by:
Ep  p c m c
2

2
2
2
4
Hence the total energy is:
 
E  0 E p f E p g  p dp

Average no. of particles
in state with energy Ep
No. of quantum states in
p to p +dp
Total Number of Particles
 
N  0 f E p g  p dp

Average no. of particles
in state with energy Ep
No. of quantum states in
p to p +dp
Fermi-Dirac Statistics

For fermions, no more than one particle
can occupy a given quantum state
– Pauli exclusion principle

Hence:
 
f Ep 
1
    1
exp
E p 
kT
Bose-Einstein Statistics
For Bosons, any number of particles can
occupy a given quantum state
 Hence:

 
f Ep 
1
    1
exp
E p 
kT
F-D vs. B-E Statistics
100
Fermi-Dirac
Bose-Einstein
1
0.01
0.1
1
E/kT
10
0.1
0.01
Occuapncy
10
0.001
0.0001
The Maxwellian Limit

Note that Fermi-Dirac and Bose-Einstein
statistics coincide for large E/kT and
small occupancy
– Maxwellian limit
 
  
f E p  exp 
E p 
kT
Ideal Classical Gases

Classical occupancy of any one
quantum state is small
– I.e., Maxwellian

Equation of State:
N
P  kT
V

Valid for both non- and ultra-relativistic
gases
Ideal Classical Gases

Recall:
– Non-relativistic:


Pressure = 2/3 kinetic energy density
Hence average KE = 2/3 kT
– Ultra-relativistic


Pressure = 1/3 kinetic energy density
Hence average KE = 1/3 kT
Ideal Classical Gases

Total number of particles N in a volume
V is given by:
N 


0
 
exp
 E p
kT
V
g s 3 4 p 2dp
h
 
V
 N  g s 3 2mkT  exp
h
3
2
 mc 2
kT
Ideal Classical Gases

Rearranging, we obtain an expression
for , the chemical potential
g
n

s
Q 
2
  mc  kT ln

 n 
3
2
2

mkT


where nQ  
2

h


(the quantum concentration)
Ideal Classical Gases

Interpretation of 
– From statistical mechanics, the change of
energy of a system brought about by a
change in the number of particles is:
dE  dN
Ideal Classical Gases

Interpretation of nQ (non-relativistic)
– Consider the de Broglie Wavelength
h
h

 
 nQ
p mkT 
1
1
3
2
– Hence, since the average separation of
particles in a gas of density n is ~n-1/3
– If n << nQ , the average separation is
greater than  and the gas is classical
rather than quantum
Ideal Classical Gases

A similar calculation is possible for a gas
of ultra-relativistic particles:
 g s nQ 
  kT ln

 n 
 kT 
where nQ  8 

 hc 
3
Quantum Gases
Low concentration/high temperature
electron gases behave classically
 Quantum effects large for high electron
concentration/”low” temperature

– Electrons obey Fermi-Dirac statistics
– All states occupied up to an energy Ef , the
Fermi Energy with a momentum pf
– Described as a degenerate gas
Quantum Gases

Equations of State:
– (See Physics of Stars secn 2.2)
– Non-relativistic:
h
P 
5m
2
2
 3  3 53
 8  n
 
– Ultra-relativistic:
P 
hc  3 
2
3
n


4  8 
4
3
Quantum Gases

Note:
– Pressure rises more slowly with density for
an ultra-relativistic degenerate gas
compared to non-relativistic
– Consequences for the upper mass of
degenerate stellar cores and white dwarfs
The Saha Equation

Atoms within a star are ionised via
interaction with photons
– We have a dynamic equilibrium between
photons and atoms on one hand and
electrons and ions on the other
– Considering the case of hydrogen:
H + g  e- + p
The Saha Equation

Thermodynamic equilibrium is reached
when the chemical potentials on both
sides of the equation are equal
– I.e, changes in numbers of particles
doesn’t affect the energy, hence:
(H) + (g) = (e-) + (p)
The Saha Equation
– Chemical potential of a photon: (g) = 0
– Also have to allow the hydrogen atom to
be in any electronic quantum state, q, with
energy:
Eq = -13.6/q2 eV
– Then:
(Hq) = (e-) + (p)
(1)
The Saha Equation
– Assuming the density is low and energies are nonrelativistic:

See Workshop 3, Question 1
– Evaluate the chemical potentials in terms of the
quantum concentrations using functions derived in
Lecture 5:
The Saha Equation
– For electrons:
 g se nQe 
(e )  me c  kT ln

 ne 
– For protons:
 g sp nQp 
( p )  m p c  kT ln

 n p 
– For atoms:


2
2
 g sH nQp 
(H q )  mH c  kT ln

 nH 
2
(Note nQ depends on mass and is almost
identical for protons and hydrogen atoms)
The Saha Equation

Note that the total energy of a hydrogen
atom in state q is:
m (H q )c 2  m (e  )c 2  m (p )c 2  E q

Also,
gse = gsp = 2 and
gsH = gq gse gsp with gq =q2
The Saha Equation

Combining these relationships with the
condition for equilibrium (equation (1)),
we obtain:
n (H q ) g n
 Eq 

exp 

ne n p
nQe
 kT 
The Saha Equation
Consequences

Degree of ionisation
– Sum over all q levels to obtain the ratio of
protons to all neutral states of H
n (H )
1

ne n p nQe
 Eq 
g q exp 


q 0
 kT 

Degree of Ionisation
– We may re-write this in the form:
Ei


exp



kT
  (E q  E i ) 
n(H )



g q exp 


ne np
nQe
kT
q 0


– The summation is truncated at a value of q
such that the spatial extent of the atom is
comparable to the separation of the atoms
– In practice, the summation ~1
Degree of Ionisation
– The ratio of ionised to neutral hydrogen (or
indeed, any atom) can now be written as:
n (H ) nQe
 E i


exp 
(2)

kT
n (H )
ne



Degree of Ionisation

Degree of ionisation in the sun:
– Average density, r~1.4x103 kg/m-3
– Typical temperature T~ 6x106 K
– nQe ~ 1021T3/2
– Assume electron density ~ proton density
Degree of Ionisation
– Denote the fraction of hydrogen ionised as
x(H+). Then, we can write:
– ne = n(H+) = x(H+).r/mH
– n(H) = (1-x(H+)).r/mH
– We can now re-write (2) as:
x (H )
10 mHT


r
1  x (H )
 2
21
3
2
 E i

exp 
kT 

Degree of Ionisation
– Substituting the values for the sun and for
hydrogen, we obtain
x(H+) ~ 95%
– I.e., the interior of the sun is almost
completely ionised

For further discussion, see Phillips, ch. 2 secn
2.5
Balmer Absorption
– To find the degree of Balmer absorption in
a stellar atmosphere, we require:
n(H(2))/(n(H)+n(H+))
(3)
– Saha gives us n(H+)/n(H)
– Boltzmann gives us n(H(2))/n(H(1))

Assume n(H(1))~ n(H)
Balmer Absorption
From Boltzmann
– We can rewrite (3) as:
n (H (2 ) )
 
n (H )  n (H )
n (H (2) )
1
n (H (1) )
n (H )

n (H )
From Saha
Balmer Absorption
– Hence:
n( H ( 2 ) )

n( H )  n( H )

 E2 

exp

kT 
21
10 T
1
ne
– Typically, ne ~ 1019 m-3
3
2

E


i
exp

kT 
Balmer Absorption
Temperature (K)
100000
10000
1000
Relative Population
1.E-05
B
1.E-06
1.E-07
1.E-08
1.E-09
1.E-10
O
A
F
G
K
M
Reminder

Assignment 1 available today on
module website
Next Week

Private Study Week - Suggestions
– Assessment Worksheet
– Review Lectures 1-3
– Photons in Stars (Phillips ch. 2 secn 2.3)


The Photon Gas
Radiation Pressure
– Reactions at High Temperatures (Phillips ch. 2 secn
2.6)


Pair Production
Photodisintegration of Nuclei