Boson and Fermion “Gases”
• If free/quasifree gases mass > 0 non-relativistic
P(E) = D(E) x n(E) --- do Bosons first
• let N(E) = total number of particles. A fixed
number
(E&R use script N for this)


0
0
N   n( E ) D( E )dE  
D( E )
dE
 E / kT
e e
1
• D(E)=density (~same as in Plank except no 2 for
spin states)
(E&R call N)
N 

0
4V (2m3 )1/ 2 E1/ 2 dE
h3
e e E / kT  1
• If know density N/V can integrate to get
normalization. Expand the denominator….
(2mkT)3 / 2V 
1 
N
e (1  3 / 2 e ....)
3
h
2
P461 - Quan. Stats. II
1
Boson Gas
• Solve for e by going to the classical region (very
good approximation if m and T both large)
e 
N
h3

V (2mkT )3 / 2
• this is “small”. For helium liquid (guess) T=1 K,
kT=.0001 eV, N/V=.1 g/cm3
e
•
3
(
1240
eVnm
)
 1022 / cm3
 0.5
3/ 2
(2  4GeV  .0001eV )
work out average energy


0
0
E   En( E ) D( E )dE /  n( E ) D( E )dE
E  kT (1  25 / 2
3
2
1
N
h3
V ( 2mkT ) 3 / 2
......)
• average energy of Boson gas at given T smaller
than classical gas (from BE distribution ftn). See
liquid He discussion
P461 - Quan. Stats. II
2
Fermi Gas
• Repeat for a Fermi gas. Add factor of 2 for S=1/2.
Define Fermi Energy EF = -kT
change “-” to “+” in distribution function
8V (2m3 )1/ 2 E1/ 2
n( E ) D( E )  3 ( E  EF ) / kT
h (e
 1)
• again work out average energy


0
0
E   En( E ) D( E )dE /  n( E ) D( E )dE
E  kT (1  25 / 2
3
2
1
N
h3
V ( 2mkT )3 / 2
......)
• average energy of Fermion gas at given T larger
than classical gas (from FD distribution ftn). Pauli
exclusion forces to higher energy and often much
larger
P461 - Quan. Stats. II
3
Fermi Gas
• Distinguishable <---> Indistinguishable
Classical
<----> degenerate
• depend on density. If the wavelength similar to the
separation than degenerate Fermi gas
particle   separation  n
h
p
1 / 3
• larger temperatures have smaller wavelength 
need tighter packing for degeneracy to occur
• degenerate electron examples
- conductors and semiconductors
- pressure at Earth’s core (at least some of it)
-aids in initiating transition from Main Sequence
stars to Red Giants (allows T to increase as electron
pressure is independent of T)
- white dwarves and Iron core of massive stars
• Neutron and proton examples
- nuclei with Fermi momentum = 250 MeV/c
- neutron stars
P461 - Quan. Stats. II
4
Degenerate vs non-degenerate
P461 - Quan. Stats. II
5
Conduction electrons
• Most electrons in a metal are attached to individual
atoms.
• But 1-2 are “free” to move through the lattice. Can
treat them as a “gas” (in a 3D box)
• more like a finite well but energy levels (and
density of states) similar (not bound states but
“vibrational” states of electrons in box)
• depth of well V = W (energy needed for
electron to be removed from metal’s surface photoelectric effect) + Fermi Energy
• at T = 0 all states up to EF are filled
W
V
EF
Filled levels
P461 - Quan. Stats. II
6
Conduction electrons
• Can then calculate the Fermi energy for T=0 (and it
doesn’t usually change much for higher T)
T=0
n

N   n( E ) D( E )dE
0

EF
0
2  8V (2m 3 )1/ 2
D( E )dE 
2h 3
 EF 
•
h  3N 


8m  V 
2

EF
0
E 1/ 2 dE
E
2/3
Ex. Silver 1 free electron/atom
N 6.02  10 23 atoms / mole 10.5 g 1e
 
V
108 g / mole
cm 3 atom
 5.9  10 22 free electrons / cm 3
EF 
34
(6.6  10 joule  sec)
8  9.1 10 31 kg
or E F 
2
 3  5.9  10

3

m

(1240eVnm)  3  5.9  10
 
8  .511MeV   m 3
2
P461 - Quan. Stats. II
28



28



2/3
1eV
1.6  10 19 J
2/3
 5.5eV
7
Conduction electrons
• Can determine the average energy at T = 0

E
EF
E  D( E )n( E )dE
0

EF
0


EF
E  E 1/ 2 dE
0

D( E )n( E )dE
EF
0
E 1/ 2 dE
EF5 / 2  52 3
 3 / 2 2  EF
EF  3 5
• for silver  3.3 eV
• can compare to classical statistics
3.3eV
3.3eV
o
T


40
,
000
K
5
k
8.6 10 eV / K
• Pauli exclusion forces electrons to much higher
energy levels at “low” temperatures. (why e’s not
involved in specific heat which is a lattice
vibration/phonons)
P461 - Quan. Stats. II
8
Conduction electrons
P461 - Quan. Stats. II
9
Conduction electrons
• Similarly, from T-dependent

3 
1
Nh3
E  kT 1  5 / 3
....
3/ 2
2  2 V (2mkT)

• the terms after the 1 are the degeneracy
terms….large if degenerate. For silver atoms at
T=300 K
3
E  kT 1  820
2
•
not until the degeneracy term is small will the
electron act classically. Happens at high T
• The Fermi energy varies slowly with T and at
T=300 K is almost the same as at T=0
EF (300)  EF (0)
• You obtain the Fermi energy by normalization.
Quark-gluon plasma (covered later) is an example
of a high T Fermi gas
P461 - Quan. Stats. II
10
Fermi Gases in Stars
• Equilibrium: balance between gravitational
pressure and “gas” (either normal or degenerate)
pressure
• total gravitational Energy:
GM ( r)  Mass ( r  r)
E 

r
R
r
 ( r)
2
E G
4r dr 4r 2  ( r )dr
0
0
r
E
3 M 2G

if   cons tan t pressure   V
5 R
• density varies in normal stars (in Sun: average is 1
g/cm3 but at r=0 is 100 g/cm3). More of a constant
in white dwarves or neutron stars
• will have either “normal” gas pressure of P=nkT
(P=n<E>) or pressure due to degenerate particles.
Normal depends on T, degenerate (mostly) doesn’t
• n = particle density in this case
P461 - Quan. Stats. II
11
Degenerate Fermi Gas Pressure
• Start with p = n<E>
• non-relativistic
relativistic
density states : D ( p )  p 2 both cases
D( E )  E1/ 2
N 
EF
0
 E2
1/ 2
AE dE
 EF  n 2 / 3
E 

EF
0
AE 2dE
EF  n1 / 3
1/ 2
E

E
dE

1/ 2
E
 dE
E  53 EF  K1n 2 / 3
P  K1n  n 2 / 3  n 5 / 3
2
E

E
dE

2
E
 dE
 43 EF  K 2n1 / 3
 K 2n  n1 / 3  n 4 / 3
• P depends ONLY on density
P461 - Quan. Stats. II
12
Degenerate Fermi Gas Pressure
non-relativistic
relativistic
EF  n 2 / 3
EF  n1 / 3
E  53 EF  K1n 2 / 3
P  K1n  n 2 / 3  n5 / 3
 43 EF  K2n1 / 3
 K2n  n1 / 3  n 4 / 3
• P depends ONLY on density
• Pressure decreases if, for a given density, particles
become relativistic
• if shrink star’s radius by 2
 density increases by 8
 gravitational E increases by 2
• if non-relativistic. <E> increases by (N/V)2/3 = 4
• if relativistic <E> increases by (N/V)1/3 = 2
•  non-relativistic stable but relativistic is not. can
aid in collapse of white dwarf
P461 - Quan. Stats. II
13
Older Sun-like Stars
• Density of core increases as HHe. He inert (no
fusion yet). Core contracts
• electrons become degenerate. 4 e per He nuclei.
Electrons have longer wavelength than He
Ee  EHe thermal equilibriu m
 pe 
me
mHe
pHe  e 
mHe
me
He
• electrons move to higher energy due to Pauli
exclusion/degeneracy. No longer in thermal
equilibrium with p, He nuclei
• pressure becomes dominated by electrons. No
longer depends on T
Ptotal  Pe  PH  PHe
Pe  PH  PHe
• allows T of p,He to increase rapidly without
“normal” increase in pressure and change in star’s
equilibrium.
• Onset of 3HeC fusion and Red Giant phase
(helium flash when T = 100,000,000 K)
P461 - Quan. Stats. II
14
White Dwarves
• Leftover cores of Red Giants made (usually) from
C + O nuclei and degenerate electrons
• cores of very massive stars are Fe nuclei plus
degenerate electrons and have similar properties
• gravitational pressure balanced by electrons’
pressure which increases as radius decreases 
radius depends on Mass of star
• Determine approximate Fermi Energy. Assume
electron density = 0.5(p+n) density
N M Sun 1
1

 5  1035 e / m 3
V
m p 2 volume Earth
E F (non  rel ) 
E F (rel )  hc
 
h2 N 3 2 / 3
8m V 

N 3 1/ 3
V 8
 0.3 MeV
 0.8 MeV
• electrons are in this range and often not completely
relativistic or non-relativistic  need to use the
correct E2 = p2 + m2 relationship
P461 - Quan. Stats. II
15
White Dwarves + Collapse
• If the electron energy is > about 1.4 MeV can have:
e  p    n EThreshold  1.4MeV
• any electrons > ET “disappear”. The electron
energy distribution depends on T (average E)
#e’s
EF
ET
• the “lost” electrons cause the pressure from the
degenerate electrons to decrease
• the energy of the neutrinos is also lost as they
escape  “cools” the star
• as the mass increases, radius decreases, and
number of electrons above threshold increases
P461 - Quan. Stats. II
16
White Dwarves+Supernovas
• another process - photodisentegration - also absorbs
energy “cooling” star. Similar energy loss as e+p
combination
  56Fe  134 He  4n
  He  2 p  2n
4
• At some point the not very stable equilibrium
between gravity and (mostly) electron pressure
doesn’t hold
• White Dwarf collapses and some fraction (20-50%
??) of the protons convert to neutrons during the
collapse
• gives Supernovas
LSN (light )  109  LSun (light )
LSN (neutrinos )  100(1000 ?)  LSN (light )
P461 - Quan. Stats. II
17
White Dwarves+Supernovas
P461 - Quan. Stats. II
18
Neutron Stars-approx. numbers
• Supernovas can produce neutron stars
- radius ~ 10 km
- mass about that of Sun. always < 3 mass Sun
- relative n:p:e ~ 99:1:1
• gravity supported by degenerate neutrons
N 2 M Sun

V
mn
1
44
3
3

6

10
/
m

.
6
/
F
4  (10 km) 3
3
separation  1.2 F
•
plug into non-relativistic formula for Fermi Energy
 140 MeV (as mass =940 MeV, non-rel OK)
• look at wavelength
h
h
1240MeVF
 

 2F
p
2mE
2  940MeV 140MeV
• can determine radius vs mass (like WD)
• can collapse into black hole
P461 - Quan. Stats. II
19
Neutron Stars
• 3 separate Fermi gases: n:p:e p+n are in the same
potential well due to strong nuclear force
• assume independent and that p/n = 0.01 (depends
on star’s mass)
EF ( p)  EF (n) VN 
2/3
EF (e)  EF (n) VN 
 14525MeV  6MeV
2 / 3 mn
me
 14525MeV
940
.5
 10,000MeV ( wrong )
• so need to use relativistic for electrons
EF (e)  

N 3 1/ 3
V 8
hc  500MeV
• but not independent as p <---> n
n  p  e 
p  n  e  
p  e  n 
n  e  p  
• plus reactions with virtual particles
• free neutrons decay. But in a neutron star they can
only do so if there is an available unfilled electron
state. So suppresses decay
P461 - Quan. Stats. II
20
Neutron Stars
• Will end up with an equilibrium between n-p-e
which can best be seen by matching up the Fermi
energy of the neutrons with the e-p system
• neutrons with E > EF can then decay to p-e-nu
(which raises electron density and its Fermi energy
thus the balance)
E F (e)  E F ( p )  E F ( n )
 3n p 
 3ne 
2


 hc  m p c  
 8 
 8 
1/ 3
 3nn 


 8 
2/3
2/3
h2
2m p
h2
 mn c 2
2mn
• need to include rest mass energies. Also density of
electrons is equal to that of protons
• can then solve for p/n ratio (we’ll skip algebra)
• gives for typical neutron star:
  2 1017 kg / m 3 
nn  110 44 / m 3
ne  n p  nn / 200
P461 - Quan. Stats. II
21