Chapter 12: Solid state physics
63
12.3
Crystal vibrations
12.3.1
A lattice with one type of atoms
In this model for crystal vibrations only nearestneighbour interactions are taken into account. The force on
atom
s
with mass
M
can then be written as:
F
s
=
M
d
2
u
s
dt
2
=
C
(
u
s
+1

u
s
) +
C
(
u
s

1

u
s
)
Assuming that all solutions have the same timedependence
exp(

i
ω
t
)
this results in:

M
ω
2
u
s
=
C
(
u
s
+1
+
u
s

1

2
u
s
)
Further it is postulated that:
u
s
±
1
=
u
exp(
isKa
) exp(
±
iKa
)
.
This gives:
u
s
= exp(
iKsa
)
. Substituting the later two equations in the fist results in a system of linear
equations, which has only a solution if their determinant is 0. This gives:
ω
2
=
4
C
M
sin
2
(
1
2
Ka
)
Only vibrations with a wavelength within the first Brillouin Zone have a physical significance. This requires
that

π
< Ka
≤
π
.
The group velocity of these vibrations is given by:
v
g
=
d
ω
dK
=
Ca
2
M
cos(
1
2
Ka
)
.
and is 0 on the edge of a Brillouin Zone. Here, there is a standing wave.
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 Spring '10
 Ye
 Force, Mass, Photon, Solid State Physics, Brillouin zone, M1 M2 ±C, crystal vibrations, ﬁrst Brillouin Zone

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