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Homework #6
1. Continuum wave equation. Show that for long wavelengths the
equation of motion,
)2( 112
2
ssss uuuC
dt
udM , reduces to the continuum elastic wave equation
2
22
2
2
x
uv
t
u
where v is the velocity of sound.
Solution:
For a , ss uu 1 is small. Replacing su with )(xu , we have
)(sauus ,
))1((1 asuus , and 2
2
2
2 ),(
t
txu
dt
ud s
. Thus
2
2
2
12
1)())1(( a
x
ua
x
uauasuus
or 22
2
)(2))1(())1(( ax
usauasuasu
Then )2( 112
2
ssss uuuC
dt
udM becomes
2
2
2
2
2 ),(a
x
uC
t
txuM
or
2
22
2
2 ),(
x
uv
t
txu
with M
Cav
2
.
Note the mass density is 3/aM and the Yongs modulus is aCY / .
Then
Y
M
Cav
2
.
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 0K and
aK / . Sketch in the dispersion relation by eye. This problem
simulates a crystal of diatomic molecules such as H2.
Solution:
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 )exp( tiikxAu .
C C C 10C 10C
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The second group has spring 10C at the left and spring C at
right, and its motion can
be described by )exp( tiikxBv .
Then the equations of motion for these two groups are:
)(10)(
)()(10
112
2
112
2
sssss
sssss
vuCvuCdt
vdm
uvCuvCdt
udm
Substitute )exp( tiikxAu and )exp( tiikxBv into these two
equations,
)(10)(
)()(102/2/2
2/2/2
BAeCBAeCBm
ABeCABeCAmikaika
ikaika
;
or
0)11()10(
0)10()11(22/2/
2/2/2
BmCAeeC
BeeCAmCikaika
ikaika
.
To have non-zero solution, we have
00)11()10(
0)10()11(det
22/2/
2/2/2
BmCAeeC
BeeCAmCikaika
ikaika
.
0)cos201100()11( 222 kaCmC , kaCmC cos20101)11( 2
m
kaC )cos2010111(2
At k=0, m
C2221 , BA ; and 0
2
2 , BA .
At k=/a, m
C2021 , iBA ; and
m
C222 , iBA .
k
a a
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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 )4/(3 3R ,
which defines R. (a) Show
that the frequency of a single ion set into oscillation is 2/132
)/( MRe . (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
within the sphere is rrr
rE3
4
3
41)( 3
2
. Now
33 4
3
3/4 R
e
R
e
, so
3)(
R
errE .
The force acting on the ion is KrR
reeEF
3
2
with 3
2
R
eK .
Then 3
2
MR
e
M
K .
(b) For Sodium, ggM 2324 108.31067.123 . Taking R=1, we
have
)/1(10~10108.3
)108.4( 142423
210
3
2
sMR
e
.
(c) scmav /101010~~ 6814 .
4. Soft phonon modes. Consider a line of ions of equal mass but
alternating in charge,
with p
p ee )1( as the charge on the pth ion. The interatomic potential
is the sum of
two contributions: (1) a short-range interaction of force
constant RC1 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 332 /)1(2 apeC ppC , where a is the
equilibrium nearest-neighbor distance. (b)
From )cos1()/2(0
2 pKaCM pp
, here C includes both nearest neighbor and
other neighbors, show that the dispersion relation may be
written as
3
1
22
0
2 )cos1()1(2
1sin/
ppKaKa p
p
,
where M/420 and 32 / ae . (c) Show that 2 is negative (unstable
mode) at
the zone boundary Ka if 475.0 or )3(7/4 , where is a Riemann
zeta
function. Show further that the speed of sound at small Ka is
imaginary if
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721.0)2ln2( 1 . Thus 2 goes to zero and the lattice is unstable
for some
value of Ka in the interval ),0( if 721.0475.0 . 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 22 / re .
For two ions at sth and (s+p)th sites, their separation distance
is pa without ion
displacement, and is sps uupa after the sth and (s+p)th ions are
displaced by su
and psu , respectively. Then the force between the two ions will
change by an
amount of )()(
2)1(
)(
)1(
)(
)1(3
2
2
2
2
2
sps
pp
sps
p
uupa
e
pa
e
uupa
e
. Here p)1(
accounts for the sign change of the two charges.
(b) The equation of motion for the sth ion is
)()(
2)1()()(
3
2
1
112
2
spssps
p
p
sssss uuuu
pa
euuuu
dt
udm
.
Let )exp( tiikxAu ss , and substitute it into the above
equation.
)1(cos)(
4)1()1(cos2
)11()(
2)1()1()1(
3
2
1
3
2
1
2
pkapa
eka
eepa
eeem
p
p
ikpaikpap
p
ikaika
)cos1()1(4
2sin
4
)cos1()1(4
)cos1(2
31
3
22
31
3
22
pkapma
eka
m
pkapma
eka
m
p
p
p
p
Define m
40 and 3
2
a
e
, then
)cos1()1(
2sin
31
2
2
0
2
pkap
ka p
p
us+p us
pa
pa+us+p-us
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(c) At ka , ppka )1()cos( and 12
sin
ka. Then
106.21
...5
1
3
1121...
5
2
3
2
1
21
))1(1()1(
1
33333
31
2
0
2
pp
p p
If 475.0106.2/1 , 02 . For 1ka ,
)2ln21(2
)1(21
2
2
)1(2
2
2
1
2
2
31
2
2
0
2
ka
p
ka
kpa
p
ka
p
p
p
p
If 721.02ln2/1 , 02 . Strong Coulomb interaction makes the
lattice unstable.
5. For a 1D lattice, if ankk /221 ,
(a) Show that 1k and 2k describe the same elastic wave.
(b) For a special case of ak 3/1 and ak 3/72 , make a plot of
)cos( 1xk and
)cos( 2xk versus ax / . Confirm the conclusion of (a) from the
plot.
Solution:
(a) For )exp( 11 tixikAu , )exp( 22 tixikAu , ])exp[(/ 1212
xkikuu . For
ankk /221 and pax , 1]2exp(/ 12 npiuu .
(b)
0 1 2 3 4 5 6 7 8 9 10
-1.0
-0.5
0.0
0.5
1.0
x/a
Note the two curves cross at x/a=integer.
