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CHAPTER 19
FERMI-DIRAC GASES
1
Okay, why should we want to discuss a Fermi-Dirac gas?
Gosh, its probably because it is necessary for the
understanding of some important problems.
OK, what problems?
Some major problems that can be tackled with this
formalism are free electrons in conductors, free electrons in
white dwarf stars and neutrons in a neutron star.
Fine, lets press on.
Before we do, I would like to mention the Third Law of
Thermodynamics.
T0 S0
One expression of this law is that as
(There are exceptions to this statement, such as with glasses.)
Another expression is: It is impossible to reduce the temperature
of a system to absolute zero with a finite number of processes.
2
We now consider particles with half-integer spin (s=1/2, 3/2, ….)
called fermions. These particles obey the Pauli Exclusion Principle.
They are indistinguishable in a gas and so obey Fermi-Dirac Statistics.
N ( )
1
We have
f ( ) 
 (   ) / kT
g( ) e
1
f ( ) is called the fermi function. 0  f ( )  1
f ( ) gives the probability that a single state  will be occupied.
At any T, if   
f ( )  1 / 2
  ( T)
We define the fermi energy by  F  (T  0)
At T=0 [   (0)] / kT  

Therefore
(not fermi level!)
if    (0)
if    (0)
f ( )  1
if    (0)
0
if    (0)
[The ground state is
taken to be at 0eV
and so μ(0)>0] 3
The PEP permits only one particle per state so the N particles are
crowded into the N lowest energy states. At T=0 only one
microstate is possible, so w=1 and S=kln(w)=0. Hence S=0 at T=0.
This is in accordance with the 3rd law of thermodynamics.
T=0
1
f ( )
0
F

The density of states is the same as before except that there are two
possible values for the quantum number of sz : (1 / 2 ,  1 / 2)
Hence we have 2 particles per spatial state.Taking this into
consideration gives, from our previous calculation of the density of
states:
4
 2m 
g( )d  4 V  2 
h 

N 
3/2
 d
 f ( )g( )d
0
At T=0 the fermi function is unity up to the fermi energy
 2m 
N  4 V  2 
h 
3/2 F
 2m 
8
 d   V  2 
3
h 

0
3/2
 F3 / 2
2/3
h  3N 


F 
2m  8 V 
2
The energy of an ordinary gas molecule is of order kT. We define a
fermi temperature by  F  kTF
2/3
h  3N 


TF 
2mk  8 V 
2
or
2/3


N


TF 
2 mk  1.504 V 
2
h
5
This is analogous to the bose temperature defined earlier. The fermi
temperature is a reference temperature, it is not the temperature of
the gas. We will calculate the fermi temperature a little later: it is in
the range 104  105 K
The more difficult situation in when the temperature is
greater than zero. Now states above the fermi energy are occupied.
3/2 
 2m 

N  4 V  2 
d

(   ) / kT
1
h 
0 e
The 1 in the denominator makes the integration difficult. Numerical
calculations can be made to determine the chemical potential as a
function of the temperature. However approximations can be made,
2
valid for T  TF ,giving
2


 T


   (0) 1 
 12  TF



 (1)

T  TF
Notice that   0 for T  TF in contrast to bosons of which   0
6
A knowledge of the chemical potential permits the calculation of
the fermi distribution function.
T=0K
f ( )
  0.97(0)
T=0.2 TF

F
7
Next we are going to treat the electrons in a solid as a gas. This
may seem unreasonable given the strong coulomb interaction
between electrons and the presence of the positively charged
lattice sites. The many-body theory of solids shows that these
effects can be reasonably ignored, at least to a good 1st
approximation.
8
Free electrons in a metal. (This is relevant to conduction in materials,
white dwarfs, He-3, and nuclear matter.)
To a good first approximation, the electrons confined in the interior
of a metal are similar to molecules in a gas, so we speak of an electron
gas. Of course this is a simplification because it ignores, among other
things, the periodic ionic potential, which leads to band structure so
important in semiconductor physics. (An effective mass is often
introduced to partially compensate for the simplifications.)
The metal will be at a fixed temperature and there will be a
corresponding chemical potential, called the fermi level. The potential
energy for the electron gas is as follows:
!

fermi level (T )
(T )

=work function (3-4eV)
9
An electron near the surface of the metal feels a strong attractive force
due to the positive metal ions in its vicinity. To free an electron from
a metal one must give an electron at the fermi level a certain energy ,
called the work function. In the photoelectric effect, this energy is
supplied by absorbing a photon.
Note that the fermi level is not the same as the fermi energy
but, for T  TF
(T )  (0)   F
As shown in your textbook, for Ag, the fermi energy is 5.54eV. This
is high and the most striking property of a fermi gas. Contrast this
with a gas molecule at room temperature, which has an energy of
about 0.025eV. The energy of the electrons at the fermi energy, at
absolute zero, is about that of a gas molecule at a temperature of
about 64,000K.
The energy of an electron gas at T=0K is called the zero point
energy.
T
300 K
 0.0047 { F  kTF }10
At room T (Ag) T 
4
F
6.4  10 K
When T  TF the gas is said to be in the degenerate region. This
is not to be confused with the degeneracy of an energy level.
 2
2
0.0047    0.99998  F
   (0)1 
 12

Using equation (1)
(300 K )  0.99998  F This is why
We now plot f ( ) and g( )
Fermi level

g( )
f ( )
0
Fermi energy

T=0
1
0
(T ) is often confused with  F
T
F


11
N( )  f ( )g( )
T=0
N ( )

F
It is the electrons in the tail of this distribution that can be most
easily extracted from a metal by various processes.

Internal energy of the gas: U   N ( ) d


0
 2m 
U    f ( )g ( )d  4 V  2 
h 
0
3/2 
e
 3/2
(   ) / kT
0
1
d
Again we can make approximations to obtain a series solution
to this equation.
12

3
5 2
U  N F 1 
5
12

At T=0
3
U  N F
5
T

 TF
2

4  T
 

16  TF

4


    (2)


This is a large energy. It comes about by
the PEP.
For example, for Ag at T=0,
U (0) 3
3
  
  F  (5.54eV )  3.33eV
N
5
5
For an ordinary gas at T=300K
   0.025eV
The electrons in an electron gas have much more energy at 0K than
the molecules of a gas at room temperature.
13
Specific heat of metals (one of the great accomplishments of the
theory).
The law of Dulong and Petit: cV  3R for all elementary solids
at room temperature. (This is an experimental result.)
This law has a simple explanation based on the principle of
equipartition of energy. Each atom of the solid is considered to be a
linear oscillator with 6 degrees of freedom (vibrating in 3 dimensions
and the oscillator has both kinetic and potential energy).
1 
U  N6 kT  3NkT  3nRT
2 
 u 
u  3RT c V  
  3R c V  3 R
 T  V
At lower temperatures the specific heat decreases and , at these
temperatures, a more sophisticated quantum-mechanical approach
14
is required.
Continuing with the electron gas, we associate a specific heat with
 dU 
the free electrons by defining:

Ce  
 dT V
2
4
2
4


3
5  T 
 T 
  
     (2)
U  N F 1 
Using equation (2)
5
12  TF 
16  TF 


3
2
4





3
5
T

T 
  
   (approx)
C e  N F 
5
 6TF  TF  4TF  TF  
3
2
2

 N F  T  3  T  
  
   (approx)
Ce 
2 TF  TF  10  TF  


3
2
2






T
3
T 
  
Ce 
Nk   
Since  F  kTF
2
 TF  10  TF  
Since we are considering low temperatures it is certainly true that
T is very much smaller than the fermi temperature and so the second
15
term is negligible.
T
Ce 
Nk 
2
 TF
2
 2
 kT 
 
Nk  
 2
 F 
(Notice the linear dependence on T)
At room temperature and for Ag 2

 0.025eV 
Ce  nR 
 c e  0.022R
2
 5.54eV 
This small value explains a puzzle regarding the specific heat of
metals. It was expected that the free electrons, having 3 translational
degrees of freedom, would contribute an additional (3/2)R to the
specific heat (from equipartition theorem). This is obviously
not in agreement with experiment. Our calculation shows that, indeed,
the electronic contribution is small.
The reason that it is small is as follows. Even though the
kinetic energies of the free electrons are much greater than the thermal
energy of the particles in a gas, the change in energy (dU/dT) of the
electrons is small. Only the electrons near the fermi level can
increase their energies because of the availability of unoccupied
states, and only a small fraction of the free electrons are near  F 16
The entropy of the electron gas:
đ Q  đ
CV  
Q  C V dT ( V constant) TdS  C V dT

 dT  V
T
Ce
For the free electrons S   dT
T
0
 T

S 
Nk  
2
T
0 
 F
2
T
 3
 
 10
2
T

 TF
1

   dT
 T

3
3
2


T  T

S
Nk        ( 3)
2
 TF  10  TF 

2
S=0 at T=0 in accordance with the 3rd law of thermodynamics. This
is also true for a Bose gas.
17
Helmholtz thermodynamic potential. F=U-TS
Eqn(2) –T[eqn(3)]

3
5 2
F  NkTF 1 
5
12

T

 TF
2


   


4

 T
 

 16  TF
4
 T 2  T


T
Nk  
2
 TF 10  TF
3


  


2
3  2
F  NkTF  
 5 4
T

 TF
2

3  T
 

80  TF

 2
NkTF 
 2
4
T

 TF


   


4
2

4  T
 

20  TF

3  2  T

F  NkTF  
 5 4  TF
2
4


  



4  T
 

80  TF

4


  


18
The pressure of an electron gas: Now that we have the Helmholtz
function, we can calculate the pressure. (potentials yield properties)
 F 

P  
 V T ,N
Rewriting F
Our first chore is then to write F explicitly in terms
2/3
of the volume!!!
2
h  3N 

  aV 2 / 3
kTF 
2m  8 V 
3

 2T 2 kTF  4T 4 kTF
F  N  kTF 

 
2
4
4TF
80TF
 5

3

 2T 2 k 2
 4T 4k 4
F  N  kTF 

 
3
4(kTF ) 80(kTF )
5

 3 2 / 3  2T 2 k 2V 2 / 3  4T 4k 4V 2

F  N  aV


 
3
4a
80a
5

 3 a(2 / 3)  2T 2 k 2 (2 / 3)  4T 4k 42V

P  N 


 
5/3
1/ 3
3
4aV
80a
5 V

19
2 a

 2T 2 k 2V 2 / 3  4T 4k 4V 2


 

2/3
3
6a
40a
5 V


N 2
 2T 2 k 2
 4T 4k 4
P   kTF 

 
3
V 5
6kTF
40(kTF )

N
P
V
and a  kTF V 2 / 3

2 NkTF 
5 2T 2 k 2 5 4T 4k 4
P

 
1 
2
4
5 V  12(kTF )
80(kTF )

2 NkTF 
5 2
1 
P
5 V 
12

T

 TF
2

4  T
 

16  TF

4


  


Comparing with equation (2) (slide 15)
2U
P   
 3 V
N
 5.91  1028 m3
Example: For Ag
V
kg
  10.5  10 3
m
3
A  107
Since T  TF
TF  64  103 K
P
2N
 kTF
5V 
20
2  5.91  1028 
23 J 
3
10



P  
1
.
38

10
64

10
K

2
.
1

10
Pa


3

5
K
m

P  2.1  105 atm


In spite of this tremendous pressure, the potential barrier at the
surface of the metal keeps the free electrons from evaporating from
the surface.
This pressure tends to increase the volume. This is balanced
by the interaction between the electrons and the ions.
If we were to continue with a description of solids we would
then choose a more realistic potential. The regular spacing of the
ions is described by a periodic potential. This would lead to an
energy level diagram which has bands of energy states separated
by gaps in which there are no states. This structure permits us to
21
understand electrical conduction in semiconductors.
Here is how you can coax MAPLE to give you a numerical
answer to an integral.
> assume(x>=0);
> int((x^2)/(exp(x)-1),x=0..infinity);
2*Zeta(3)
> evalf(%);
2.404113806
22
White Dwarf Stars. These stars have masses comparable to the mass
of the sun and radii comparable to the radius of the Earth. Therefore
they are extremely dense. The core temperature is of the order of 107 K
Under these conditions the atoms are completely ionized so we
have nuclei and an electron gas. At the white dwarf stage the H has
been used up in thermonuclear reactions, fusion is greatly reduced, the
star cools and begins to collapse. This collapse is stopped by the
pressure of the electron gas.
In an elementary discussion of white dwarfs, several
approximations are made. Relativistic effects are not usually dominant
so it is not too egregious to make a non-relativistic calculation. All
densities are assumed uniform (perhaps the most serious
approximation).
The first white dwarf discovered was Sirius B, so we shall use
it as an example.
M  2.09  10 30 kg
R  5.57  106 m
V  7.23  1020 m3
T  10 7 K
23
A
The white dwarf consists of light elements and hence  2
Z
Let Nnuc be the number of nuclei in the star and A the average
mass number. The total mass, M, is then M=NnucAMH
Since the atoms are completely ionized the number of electrons is:
N=ZNnuc
N  ZN nuc
M
M
2.09  1030 kg
Z M
Z
 


AM H  A  M H 2M H 2(1.66  10 27 kg)
h  3N

F 
2m  8 V
2



2/3
34
(6.63 10 J  s )

2(9.1110 31 kg)
2
N  6.30  1056
 3  6.30 10



20 3 
 8  7.23 10 m 
56
2/3
 F  5.34  10 14 J  0.333 MeV
The fermi temperature can now be calculated.
F
5.34  10 14 J
TF 
k

1.38  10 23
J
K
TF  3.87  109 K
24
The temperature of the star is much less than the fermi temperature
and so we have a degenerate electron gas.
Now we can calculate the pressure of this degenerate gas.
From slide 20
2N
2  6.30  1056 
14
22

P    F  
5
.
34

10
J

1
.
86

10
Pa
20
3
5V 
5  7.23  10 m 
P  1.86  1017 atm
Is this pressure, due to the degenerate electron gas, sufficient to
prevent future collapse?
Condition for stability: U=minimum
U has two terms: electron gas and gravitational U  U e  U G
25
We will write U=U( R ) and determine the radius at equilibrium
The gravitational U can be obtained by building up the star
by bringing in mass elements from infinity. We obtain
3 GM 2
UG  
5 R
a
UG  
R
3
a  GM 2
5
Since T  TF and using the expression for the energy on slide 15
2/3
3
3 h  3N 


U e  N F  N
5
5 2m  8 V 
2
2/3
3 h  3N 


 N
5 2m  8 
2
4
3
  R 
3

26
2 / 3
3 h  3N 3 


Ue  N
5 2m e  8 4 
2
b
Ue  2
R
2/ 3
1  3  Nh  9N

 
2
R  5  2m e  32 2
2
2
3h
b
10 me
 9

2
32





2/ 3
1
R2
2/3



N 5/3
a
b
dU
U   2
0
For equilibrium
R R
dR
a
2b
2b
 3 0 R
(equilibriu m)
2
a
R
R
2
3h
R 2
10 me
R
 9

2
32


2/3



34
2
N
5/3
5/3
5
h N

2
3GM
meGM 2
 9

2
32


(6.63  10 J  s) (6.30  10 )
2

31
11 N  m 
30
2

(9.11  10 kg) 6.67  10
(
2
.
09

10
kg
)
2

kg


2
56 5 / 3
2/3



 9

2
32


27
2/3



R  7.15  106 m
(measured R  5.57  106 m)
This is equal to the observed radius, to within the accuracy of the
calculation. Hence it is reasonable to assume that Sirius B is now
a stable white dwarf. However it will eventually become invisible.
In most known white dwarfs the core contains C and O with
an outer layer which consists of H or He or both. The star cools
slowly. It is estimated that white dwarfs take about 1010 years
to become invisible. Since the age of the universe is about
1.4  1010 years, most white dwarfs are still visible.
28
If the mass of a star is sufficiently large so that the electrons are
relativistic, a stable equilibrium is not possible. The largest
possible mass is called the Chandrasekhar limit. This limit is
about 1.44 solar masses. A star exceeding this mass collapses,
the density approaches that of nuclear matter and the electrons
combine with the protons (inverse β-decay) to form a neutron
star. Now a degenerate neutron gas stabilizes the star. Again there
is a limiting mass, which is about three solar masses.
More massive stars collapse to form black holes.
29