Homework #7
1. Singularity in density of states. (a) From the dispersion derived in Chapter 4 for a
monatomic linear lattice of N atoms with nearest neighbor interactions, show that the
density of modes is
2N
1
,
D( ) 
 2
 (m   2 )1 / 2
where m is the maximum frequency. (b) Suppose that an optical phonon branch has
the form  ( K )  0  AK 2 , near K  0 in three dimension.
 L  2
1/ 2
for   0 and D( )  0 for   0 .
D( )  
  3 / 2 (0   )
2

A


density of modes is discontinuous.
Show that
3
Here the
Solution:
(a)  
4C
ka
ka
sin
 m sin
m
2
2
k  0 , with  m 
4C
being the maximum
m
frequency.
With positive k,
L
L dk
L dk
.
D( )d  2
dk 
d . Thus D( ) 
 d
2
 d
d  m a
ka  m a
ka  m a
2 a

cos

1  sin 2

1 2 
 m2   2 .
dk
2
2
2
2
2
m 2
dk
2

.
d a  m2   2
L dk
2L / a
2N
D( ) 


 d   m2   2   m2   2
(b) For  ( K )  0  AK 2 ,
  0
d
  2 AK   2 A
 2 A(0   ) .
dk
A
3
3
dk
 L 
 L 
In 3D, D( )d    4k 2 dk    4k 2
d
d
 2 
 2 
Then
 
1
 L 
 L 
 L  4
2 dk
D( )  



 4k
 4 0
 3 / 2 0   .
d  2 
A
2 A(0   )  2  A
 2 
3
3
3
For   0 , there is no normal modes, so that D( )  0 .
2. Zero point lattice displacement and strain. (a) In the Debye approximation, show
that the mean square displacement of an atom at absolute zero is
R 2  3D2 / 8 2 v3 , where v is the velocity of sound.
u02  4(n  1 / 2) / V
summed
over
the
Start from the result
independent
lattice
modes:
R 2  ( / 2 V ) 1 . We have included a factor of ½ to go from mean square
amplitude to mean square displacement. (b) Show that

1
and R 2 diverge for
a one-dimensional lattice, but that the mean square strain is finite. Consider
(R / x)2  (1/ 2) K 2u02 as the mean square strain, and show that it is equal to
D2 L / 4MNv3 for a line of N atoms each of mass M, counting longitudinal modes
only. The divergence of R 2 is not significant for any physical measurement.
Solution:
(a) For the motion of frequency  , the average kinetic energy should be half of the
E
m 2

1 
total energy: N
or v2 
.
v  0  
2
2
2 Nm
2 2
On the other hand, v2   2 R2 and Nm  M  V .
Therefore, we have
v2

.

2 V

1
 D D( )d
Then R 2 
.

2 V 
2 V 0

For 3D, there are three modes (two transverse and one longitudinal) for each  , thus
the density of states with the dispersion of   vk is
V 4k 2 dk
V 4 2 d 3V 2 d
D( )d  3 

3


.
(2 ) 3
(2 ) 3 v 3
2 2 v 3
R 
2
2
Then R 2 


2 V
D

D( )d
0



3V
 2 3
2 V 2 v
D

0
d 
3 D2
.
8 2 v 3
2 Ldk 3Ld
.

2
v
  D D( )d
 3L  D d
 3L

R2 




 ln  0 D   .


2 L 0

2 L v 0

2 L v
C 2


R 
For 1D longitudinal wave, N
or R2 
.
2
4
2CN

 D
R2 
 
D( )d .
2CN
2CN 0
2 Ldk Ld
D( )d  1 

.
2
v
D
L D2
 D
L

D
(

)
d



d


Then R 2 
.
2CN 0
2CNv 0
4CNv
(b) For 1D, D( )d  3 
C
 a , where a is the lattice constant, then we have
m
L D2 a 2
.

4mNv 3
Note the speed of sound v 
R
2
L D2


4CNv
L D2
v 2m
4 2 Nv
a
The strain is   R / a , thus  2  R 2 / a 2 
L D2
.
4mNv 3
3. Gruneisen constant. (a) Show that the free energy of a phonon mode of frequency 
is kBT ln[2 sinh( / 2kBT )] . It is necessary to retain the zero-point energy  / 2 to
obtain this result. (b) If  is the fractional volume change, then the frequency of the
crystal may be written as
1
F (, T )  B2  k BT  ln[ 2 sinh(K / 2k BT )] ,
2
where B is the bulk modulus. Assume that the volume dependence of  K is
 /    , where  is known as the Grűneisen constant. If  is taken as
independent of the mode K, show that F is a minimum with respect to  when
B    ( / 2) coth( / 2kBT ) , and show that this may be written in terms of the
thermal energy density as   U (T ) / B . (c) Show that on the Debye model
   ln  /  ln V . Note: Many approximations are involved in this theory: the result
(a) is valid only if  is independent of temperature;  may be quite different for
different modes.
Solution:

(a) Z    e ( n 1 / 2 )  / k BT 
n 0
e   / 2 k BT
1
  / 2 k BT

  / 2 k BT
1 e
e
 e  / 2 k BT
1
.
  

2 sinh
 2k B T 

  
 .
F  k BT ln Z  k BT ln 2 sinh
2
k
T
 B 


  k 
1 2
 .
B  k BT  ln 2 sinh
2
2
k
T
k
 B 

F is minimized when F /   0 .
  k 

2 cosh
2k BT   d k

B  k BT 
 0.
  k  2k B T d
k

2 sinh
 2k B T 
From the given condition of  /    , there is d / d   .
(b) F ( , T ) 
  k   k

 0.
Then the F /   0 becomes B    coth
k
 2k B T  2
  k 
 k


 
coth
B k
2
2
k
T
 B 

k
  k 
 k
  
coth
2
2
k
T
 B  k
  k 
  k 
  exp  

exp 
 k
 2k B T 
 2k B T 
2
  k 
  k 
  exp  

exp 
 2k B T 
 2k B T 
  
exp  k   1
 k
 k
 k
 k BT 



2
2
  
  
k
k
k
exp  k   1
exp  k   1
 k BT 
 k BT 
nd
The first term is zero point energy and the 2 term is thermal energy. If the zero
energy can be ignored, we have  

B
U (T ) .
V  V0
 D
 V
 ln  D
V
, there is    D /
.
   

 
D
D V
V0
 ln V
V
 ln  D
 ln 
With  D  k B ,   
.

(c) With
 ln V
 ln V
4. For d-dimensional crystal,
(a) Show that the density of states varies as  d 1 .
(b) Deduce from this that the low-temperature specific heat vanishes as T d . Graphite
has layered structure with very weak bounding between the layers. What
temperature dependence of the specific heat would you expect for graphite at low
temperature?
(c) Show that if the dispersion were  ~ k  , the low-temperature specific heat would
vanish as T d /  .
Solution:
(a) For d-dimensional system with   vk ,
Vk d 1dk V d 1d
D( )d ~

. Thus D( )   d 1
(2 ) d
(2 ) d v d

D
(b) The energy of the system is U  
0
Let x 
e  / kBT  1
 D / k BT
xd

dx .
, U ~ T d 1 
0
ex  1
k BT
D
D( )d ~ 
0
 d
e  / kBT  1
d
At low temperature ( k BT   D ), the up limit of the integral can be replaced by  .
xd
U
dx  T d 1 , and C 
Td .
x
0 e 1
T
The layered structure of graphite behaves like a 2D system  C ~ T 2 .
Then U ~ T d 1 

1
Vk d 1dk V ( d 1) /  d 1/ 
(c) For  ~ k , D( )d ~

~   d .
d
d d
(2 )
(2 ) v
d


D
U 
0
D
e  / kBT  1
D( )d ~ 
0
 d / 
d
e  / kBT  1
1  D / k BT
xd

, U ~T 
dx .
0
ex  1
k BT
At low temperature ( k BT   D ), the up limit of the integral can be replaced by  .
d
Let x 
d
Then U ~ T

1 

0
1
xd
U

, and C 
 T d / .
dx

T
x
T
e 1
d