Physics 927
E.Y.Tsymbal
1
Section 5: Lattice Vibrations
So far we have been discussing equilibrium properties of crystal lattices. When the lattice is at
equilibrium each atom is positioned exactly at its lattice site. Now suppose that an atom displaced from
its equilibrium site by a small amount. Due to force acting on this atom, it will tend to return to its
equilibrium position. This results in lattice vibrations. Due to interactions between atoms, various
atoms move simultaneously, so we have to consider the motion of the entire lattice.
Onedimensional lattice
For simplicity we consider, first, a onedimensional crystal lattice and assume that the forces between
the atoms in this lattice are proportional to relative displacements from the equilibrium positions.
Fig.1
This is known as the
harmonic approximation
, which holds well provided that the displacements are
small. One might think about the atoms in the lattice as interconnected by elastic springs. Therefore,
the force exerted on
n
the atom in the lattice is given by
1
1
(
)
(
)
n
n
n
n
n
F
C u
u
C u
u
+
−
=
−
+
−
,
(5.1)
where
C
is the interatomic force (elastic) constant. Applying Newton’s second law to the motion of the
n
th atom we obtain
2
1
1
1
1
2
(
)
(
)
(2
)
n
n
n
n
n
n
n
n
n
d u
M
F
C u
u
C u
u
C
u
u
u
dt
+
−
+
−
=
=
−
+
−
=
−
−
−
,
(5.2)
where
M
is the mass of the atom. Note that we neglected here by the interaction of the
n
th atom with
all but its nearest neighbors. A similar equation should be written for each atom in the lattice, resulting
in
N
coupled differential equations, which should be solved simultaneously (
N
is the total number of
atoms in the lattice). In addition the boundary conditions applied to the end atom in the lattice should
be taken into account.
Now let us attempt a solution of the form
(
)
n
i qx
t
n
u
Ae
ω
−
=
(5.3)
where
x
n
is the equilibrium position of the
n
th atom so that
x
n
=
na
. This equation represents a traveling
wave, in which all the atoms oscillate with the same frequency
ω
and the same amplitude
A
and have
wavevector
q
. Note that a solution of the form (5.3) is only possible because of the transnational
symmetry of the lattice.
Now substituting Eq.(5.3) into Eq.(5.2) and canceling the common quantities (the amplitude and the
timedependent factor) we obtain
2
(
1)
(
1)
(
)
2
iqna
iqna
iq n
a
iq n
a
M
e
C
e
e
e
ω
+
−
−
=
−
−
−
(5.4)
This equation can be further simplified by canceling the common factor
iqna
e
, which leads to
a
n
n
+1
n
−
1
u
n
−
1
u
n
u
n
+1
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Physics 927
E.Y.Tsymbal
2
(
)
(
)
2
2
2
2
1
cos
4
sin
2
iqa
iqa
qa
M
C
e
e
C
qa
C
ω
−
=
−
−
=
−
=
.
(5.5)
We find therefore the dispersion relation for the frequency
4
sin
2
C
qa
M
ω
=
,
(5.6)
which is the relationship between the frequency of vibrations and the wavevector
q
. This dispersion
relation have a number of important properties.
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 Fall '11
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 Physics, atoms, Wavelength, monatomic Bravais lattice

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