Homework #6
1. Continuum wave equation. Show that for long wavelengths the equation of motion,
d 2 us
M 2  C (us 1  us 1  2us ) , reduces to the continuum elastic wave equation
dt
2
 2u
2  u

v
t 2
x 2
where v is the velocity of sound.
Solution:
For   a , us 1  us is small.
Replacing us with u(x ) , we have us  u(sa) ,
d 2 us  2 u ( x , t )

. Thus
dt 2
t 2
u
1  2u 2
us 1  u(( s  1)a )  u(a )  a 
a
x
2 x 2
 2u
or u(( s  1)a )  u(( s  1)a )  2u( sa)  2 a 2
x
2
d u
Then M 2 s  C (us 1  us 1  2us ) becomes
dt
2
2
 2 u ( x, t )
 u ( x, t )
 2u 2
2  u
M


C
v
a
or
t 2
t 2
x 2
x 2
us 1  u(( s  1)a ) , and
Ca 2
with v 
.
M
Note the mass density is   M / a 3 and the Yong’s modulus is Y  C / a . Then
v
Ca 2
Y

.
M

2. Diatomic chain. Consider the normal modes of a linear chain in which the force
constants between nearest-neighbor atoms are alternately C and 10C. Let the masses
be equal, and let the nearest-neighbor separation be a/2. Find  (K ) at K  0 and
K   / a . Sketch in the dispersion relation by eye. This problem simulates a crystal
of diatomic molecules such as H2.
Solution:
C
10C
C
10C
C
The atoms can be separated into two groups:
The first group has spring C at the left and spring 10C at right, and its motion can be
described by u  A exp(ikx  it ) .
The second group has spring 10C at the left and spring C at right, and its motion can
be described by v  B exp(ikx  it ) .
Then the equations of motion for these two groups are:
 d 2us
m dt 2  10C ( v s 1  u s )  C ( v s 1  u s )

2
m d v s  C (u s 1  v s )  10C (u s 1  v s )
 dt 2
Substitute u  A exp(ikx  it ) and v  B exp(ikx  it ) into these two equations,
  m 2 A  10C ( Be ika / 2  A)  C ( Be ika / 2  A)
;

2
ika / 2
 B)  10C ( Ae ika / 2  B)
 m B  C ( Ae
 (11C  m 2 ) A  C (10e ika / 2  e ika / 2 ) B  0
or 
.
ika / 2
ika / 2
2

C
(
10
e

e
)
A

(
11
C

m

)
B

0

To have non-zero solution, we have
 (11C  m 2 ) A  C (10e ika / 2  e ika / 2 ) B  0 
  0.
det 
ika / 2
 e ika / 2 ) A  (11C  m 2 ) B  0 
  C (10e
(11C  m 2 ) 2  C 2 (100  1  20 cos ka)  0 , (11C  m 2 )  C 101  20 cos ka
C (11  101  20 cos ka )
m
22C
At k=0, 12 
, A   B ; and  22  0 , A  B .
m
20C
2C
At k=/a, 12 
, A  iB ; and  22 
, A  iB .
m
m
2 



k
a

a
3. Atomic vibrations in a metal. Consider point ions of mass M and charge e immersed
in a uniform sea of conduction electrons. The ions are imagined to be in stable
equilibrium when at regular lattice points. If one ion is displaced a small distance r
from its equilibrium position, the restoring force is largely due to the electric charge
within the sphere of radius r centered at the equilibrium position. Take the number
density of ions (or of conduction electrons ) as 3 /( 4R 3 ) , which defines R. (a) Show
that the frequency of a single ion set into oscillation is   (e2 / MR3 )1 / 2 . (b)
Estimate the value of this frequency for sodium, roughly. (c) From (a), (b), and some
common sense, estimate the order of magnitude of the velocity of sound in metal.
Solution:
(a) For a uniformly charged sphere of charge density  , the electrical field at r
1 4 3 4
e
3e
r 

r . Now   
within the sphere is E ( r )  2 
, so
3
4R / 3
r
3
4R 3
3
er
E (r)   3 .
R
e2r
e2
The force acting on the ion is F  eE   3   Kr with K  3 .
R
R
K
e2

.
M
MR 3
(b) For Sodium, M  23  1.67  10 24 g  3.8  10 23 g .
have
Then  
Taking R=1Å, we
e2
(4.8  10 10 ) 2


~ 1014 (1 / s ) .
3
23
24
MR
3.8  10  10
14
(c) v ~ a ~ 10  108  106 cm / s .
4. Soft phonon modes. Consider a line of ions of equal mass but alternating in charge,
with e p  e(1) p as the charge on the pth ion. The interatomic potential is the sum of
two contributions: (1) a short-range interaction of force constant C1R   that acts
between nearest-neighbors only, and (2) a coulomb interaction between all ions. (a)
Show that the contribution of the coulomb interaction to the atomic force constants is
C pC  2( 1) p e2 / p 3a 3 , where a is the equilibrium nearest-neighbor distance. (b)
From  2  (2 / M )  C p (1  cos pKa ) , here C includes both nearest neighbor and
p 0
other neighbors, show that the dispersion relation may be written as

1
 2 / 02  sin 2  Ka     ( 1) p (1  cos pKa ) p  3 ,
2 
p 1
2
2
3
where 0  4 / M and   e / a . (c) Show that  2 is negative (unstable mode) at
the zone boundary Ka   if   0.475 or 4 / 7 (3) , where  is a Riemann zeta
function. Show further that the speed of sound at small Ka is imaginary if
  (2 ln 2)1  0.721 . Thus  2 goes to zero and the lattice is unstable for some
value of Ka in the interval (0, ) if 0.475    0.721 . Notice that the phonon
spectrum is not that of a diatomic lattice because the interaction of any ion with its
neighbors is the same as that of any other ion.
Solution:
(a) The force between two ions is  e 2 / r 2 .
us
us+p
pa
pa+us+p-us
For two ions at sth and (s+p)th sites, their separation distance is pa without ion
displacement, and is pa  us  p  us after the sth and (s+p)th ions are displaced by u s
and us  p , respectively. Then the force between the two ions will change by an
( 1) p e 2
( 1) p e 2 ( 1) p 2e 2


(u s  p  u s ) .
( pa  us  p  us ) 2
( pa ) 2
( pa ) 3
accounts for the sign change of the two charges.
amount of
Here (1) p
(b) The equation of motion for the sth ion is

d 2 us
( 1) p 2e 2
m 2   (us 1  us )   (us 1  us )  
(u s  p  u s  u s  p  u s ) .
dt
( pa ) 3
p 1
Let us  A exp(ikxs  it ) , and substitute it into the above equation.

 m 2   (e ika  1)   ( e ika  1)  
p 1

 2 (cos ka  1)  
p 1
( 1) 4e
(cos pka  1)
( pa ) 3
p
2
4e 2
 
(1  cos ka) 
m
ma 3
2

4
 ka  4e
sin 2   
3
m
 2  ma
Define  0 
2
( 1) p 2e 2 ikpa
( e  1  e ikpa  1)
3
( pa )


p 1
2


p 1
( 1) p
(1  cos pka)
p3
( 1) p
(1  cos pka)
p3
4
e2
and   3 , then
m
a

2
( 1) p
2  ka 

sin


(1  cos pka)
 

 02
p3
 2
p 1
 ka 
(c) At ka   , cos( pka)  (1) p and sin   1 . Then
 2
2
p


( 1)
 1 
(1  ( 1) p )
2
3
0
p
p 1
1 1
 2 2 2



 1     3  3  3  ...   1  2 1  3  3  ... 
 1 3 5

 3 5

 1  2.106
If   1 / 2.106  0.475 ,  2  0 .
For ka  1 ,

 2  ka 
( 1) p  kpa 
    2 


 02  2 
p3  2 
p 1
2
2
2


( 1) p   ka 
1

2




   (1  2 ln 2)
p
p

1

  2
2
If   1 / 2 ln 2  0.721 ,   0 .
Strong Coulomb interaction makes the lattice unstable.
 ka 
 
 2
2
5. For a 1D lattice, if k1  k2  2n / a ,
(a) Show that k1 and k 2 describe the same elastic wave.
(b) For a special case of k1   / 3a and k2  7 / 3a , make a plot of cos(k1 x ) and
cos(k2 x ) versus x / a . Confirm the conclusion of (a) from the plot.
Solution:
(a) For u1  A exp(ik1 x  it ) , u2  A exp(ik 2 x  it ) , u2 / u1  exp[(ik 2  k1 ) x] . For
k1  k2  2n / a and x  pa , u2 / u1  exp(i 2np]  1 .
(b)
1.0
0.5
0.0
-0.5
-1.0
0
1
2
3
4
5
x/a
Note the two curves cross at x/a=integer.
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