Homework 8.
1. (Kittel 7.1) Density of orbitals in one and two dimensions.
(a) Show that the density of orbitals of a free electron in one dimension is
L  2m 
D1 (ε ) =  2 
π  ε 
12
where L is the length of the line.
(b). Show that in two dimensions, for a square of area A,
D2 (ε ) =
Am
π 2
independent of ε.
Solution.
(a) p = ± ( 2mε )1 2 . Hence
2L
2L
dε
L ( 2 m )1 2
L  2m 
( 2 m )1 2 1 2 2 =
dε 2 =
dp =
D(ε )dε =

 dε
12
2ε
2π{ε
π{  ε 
h
h
12
and so
D(ε ) =
L  2m 


π{  ε 
12
(b)
D(ε )dε =
2A
2 Aπ
2 Aπ 2m
2πpdp = 2 dp 2 =
dε
2
h2
h
h
Hence
D(ε ) =
4 Aπm Am
= 2
h2
π{
2. (Kittel 7.2) Energy of a relativistic fermi gas. For electrons with an energy
ε >> mc 2 , where m is the rest mass of the electron, the energy is given by ε ≈ pc ,
where p is the momentum. For electrons in a cube of volume V = L3 the momentum
(
)
is of the form (πh L) multiplied by n x + n y + n z
, exactly as for the
nonrelativistic limit. (a) Show that in this extreme relativistic limit the Fermi energy
of a gas of N electrons is given by
2
 3n 
ε F = πc 
π 
2
2 12
13
where n=N/V. (b) Show that the total energy of the ground state of the gas is
U0 =
3
Nε F
4
The general problem is treated by F. Juttner, Zeitschrift fur Physik 47, 542 (1928).
Solution.
(a) ε ≈ pc . Hence
D (ε )dε =
2V
8πV
V
4πp 2 dp = 3 3 ε 2 dε = 2 2 3 ε 2 dε
3
h
hc
π c
N =∫
εF
0
εF
3
V
V
ε 2 dε = 2 2 3 ε F 3
2 3
π c
3π c
2
N
3n
 3n 
= 3π c
= ( πc )1 3
⇒ ε F = ( πc ) 
V
π
π 
2
3 3
(b)
εF
U 0 = ∫ εD(ε )dε and D(ε ) = Aε 2 . Hence
0
εF
U 0 = A∫ ε 3 d ε =
0
εF
εF
0
0
A 4
ε F (Eqn 1)
4
N = ∫ D (ε )dε = A∫ ε 2 dε =
A 3
ε F (Eqn 2)
3
13
Eqn 1 and Eqn 2 give
U0 =
3
Nε F
4
3. (Kittel 7.3) Pressure and entropy of degenerate Fermi gas.
(a) Show that a fermi electron gas in the ground state exerts a pressure
(3π 2 ) 2 3 2  N 
p=
 
m V 
5
53
In a uniform decrease of the volume of a cube every orbital has its energy raised: The
energy of an orbital is proportional to 1 L2 or 1 V 2 3 .
(b) Find an expression for the entropy of a fermi electron gas in the region τ << ε F .
Notice that σ → 0 as τ → 0 .
Solution.
(a)
2  3π 2 N 


εF =
2m  V 
23
3
3 2  3π 2 N 


U = Nε F = N
5
5 2m  V 
23
T =0K. Hence
∂U 3 N 2
2
(3π 2 ) 2 3 2  N 
(3π 2 ) 2 3 N 2 3 V 5 3 =
p=−
=
 
∂V
5 2m
3
5
m V 
(b)
CV =
2
π 2 Nk B T
2 εF
 ∂S 
T   = CV
 ∂ T V
53
CV π 2 Nk B
 ∂S 
=
  =
T
2 εF
 ∂ T V
S≈
2
2
π 2 Nk B T
2 εF
4. (Kittel 7.4) Chemical potential versus temperature. Explain graphically why the
initial curvature of µ versus τ is upward for a fermion gas in one dimension and
downward in three dimensions (Figure 7.7).
Hint: The D1 (ε ) and D3 (ε ) curves are different, where D1 is given in Problem 1. It
will be found useful to set up the integral for N, the number of particles, and to
consider from the graphs the behavior of the integrand between zero temperature
and a finite temperature.
Solution.
∞
N = ∫ D(ε )
0
1
β (ε − µ )
e
µ (0)
At T=0K, N = ∫0
+1
dε
D(ε )dε . For T ≠ 0K , ε below µ will be excited to ε > µ . µ
will shift to ensure the number of particles removed from ε < µ equals the number of
particles at ε > µ . If D(ε ) = constant , the shift of µ will be approximately zero.
If D(ε ) increases as ε increases, the number of particles is weighted more for ε > µ
than ε < µ . Therefore µ decreases.
Similarly, if D(ε ) decreases as ε increases, then µ will shift upward.
1
For 1-dim, D(ε ) ≈ 1 2 , µ(T ) > µ (0) .
ε
For 3-dim, D(ε ) ≈ ε 1 2 , µ(T ) < µ (0) .
5. (Kittel 7.5) Liquid 3 He as a fermi gas. The atom 3 He has spin I =
1
and is a
2
fermion.
(a) Calculate as in Table 7.1 the fermi sphere parameters v F , ε F and TF for 3 He at
absolute zero, viewed as a gas of noninteracting fermions. The density of the liquid is
−3
0.081gcm .
(b) Calculate the heat capacity at low temperatures T << TF and compare with the
experimental value CV = 2,89NkB T as observed for T < 0.1K by A. C. Anderson, W.
Reese, and J. C. Wheatley, Phys. Rev. 130, 495 (1963); see also Figure 7.18.
Excellent surveys of the properties of liquid 3 He are given by J. Wilks, Properties of
liquid and solid helium, Oxford, 1967, and by J. C. Wheatley, "Dilute solutions of
3
He and 4 He at low temperatures ," American Journal of Physics, 36, 181-210
(1968). The principles of refrigerators based on 3 He - 4 He are reviewed in Chapter
12 on cryogenics; such refrigerators produce steady temperatures down to 0.01K in
continuously acting operation.
Solution.
(a)
m = 3 × 1.67 × 10
n=
− 27
g = 5.01 × 10
−24
g
ρ
0.081
22
−3
=
−24 = 1.62 × 10 cm
m 5.01 × 10
k F = (3π n) = (3π × 1.62 × 10 ) = 7.83 × 10 cm
2
13
2
22 1 3
7
−1
2 k F 2 (1.05 × 10 −27 × 7.83 × 10 7 )2
εF =
=
= 6.75 × 10 −16 erg
−24
2m
2
×
5.01×
10
88
TF =
ε F 6.75 × 10−16
=
= 4.9K
kB 1.38 × 10 −16
k F 1.05 × 10 −27 × 7.83 × 10 7
vF =
=
= 1.64 × 10 3 cm s
−24
m
5.01 × 10
88
(b)
CV =
π 2 NkB 2 T π 2
π2
=
Nk BT =
Nk T = 1.01Nk B T
2 εF
2TF
2 × 4.9 B
6. (Kittel 7.6) Mass-radius relationship for white dwarfs. Consider a white dwarf of
mass M and radius R. Let the electrons be degenerate but nonrelativistic.
(a) Show that the order of magnitude of the gravitational self-energy is − GM 2 R ,
where G is the gravitational constant. (If the mass density is constant within the
sphere of radius R, the exact potential energy is − 3GM 2 5R ).
(b) Show that the order of magnitude of the kinetic energy of the electrons in the
ground state is
2N 5 3
2M 5 3
≈
2
m(MH )5 3 R 2
8mR
where m is the mass of an electron and MH is the mass of a proton.
(c) Show that if the gravitational and kinetic energies are of the same order o
magnitude (as required by the virial theorem of mechanics), M1 3 R ≈ 1020 g1 3 cm .
(d) If the mass is equal to that of the Sun ( 2 × 1033 g ), what is the density of the white
dwarf?
(e) It is believed that pulsars are stars composed of a cold degenerate gas of neutrons.
Show that for a neutron star M1 3 R ≈ 1017 g1 3 cm . What is the value of the radius for a
neutron star with a mass equal to that of the Sun? Express the result in km.
Solution.
(c) g( r ) = −
dV
. Hence
dr
r < R : g( r ) = −
GM
3 r
R
r > R : g( r ) = −
GM
2
r
and so
r < R : V (r) =
GM 2 3GM
r −
2R 3
2R
r > R : V (r ) = −
Ug =
GM
r
1
4π
ρVd 3 r =
∫
2
2
∫
R
0
 GM
3GM 
M
1 M  GM R5 3GM R 3 
3GM 2
 3
 =−
r2  3 r 2 −
 dr π
=
−

 2R
5R
2R  4 R 3 2 1 R3  2R 5
2R 3 
3
3
(b)
23
2  3π 2 N 
3
3

Ue = Neε F = Ne

5
5 2m  V 
88
Ne = NH =
M
4π 3
and V =
R . Hence
mH
3

2 3
53
53
M 
2  M 
3 M 2  2
3  4π  2 3 2  M 
Ue =
3π 4π
=  
2 
 = 1.10 mR2 



3
4
mR
m
5 mH 2m 
10


 mH 
H
mH R 


3
88
(c) Let | Ug |= Ue . Then
53
2
2  M 
5 × 1.10
2
3GM2
13
20 1 3
= 1.10
⇒
M
R
=
=
1.8

53
5 3 ≈ 10 g cm
2
mR  mH 
GmmH
GmmH
3
88 5R
(d) M = 2 × 1033 g . Hence R ≈
1020
8
3
33 1 3 ≈ 8 × 10 cm = 8 × 10 km .
(2 × 10 )
Density: M1 3 R = 1020 g1 3 cm , ie R =
10 20
. Hence
M1 3
M
M
M2
4 × 1066
6
3
ρ = 4π
=
=
= 0.95 × 10 g cm
60 =
4
π
4
π
R 3 4π × 10
× 1060
× 10 60
3
3
3
3
M
(e) For neutron stars, replace the electron mass m with the proton mass m H .
53
53
GM 2
2  M 
GM 2
2  M 
≈
 ⇒ R ≈ m R2  m 
2 
m
R
mR


 H
H
H
88
and so
m
2
2
M1 3 R =
=
GmH mH 5 3 mH GmmH 5 3
88
Hence M1 3 R for a neutron star is
M R = 10
13
20
m
that for a white dwarf.
mH
m
20
−3
17 1 3
≈ 10 × 10 = 10 g cm
mH
M = 2 × 10 g . So R ≈
33
m
2
2
M1 3 R =
=
GmH mH 5 3 mH GmmH 5 3
88
1017
1017
6
≈
33 1 3
11 = 10 cm = 10km .
(2 × 10 )
10