(^00) omega_max
1
2
3
4
5
6
ω
D(ω
)
(a)
(^00) pi/(2a) pi/a
0.2
0.4
0.6
0.8
1
k
ω
(b)
Figure 13: a) Dispersion relation of a 1-dimensional chain of atoms; b) Density of states for a 1-
dimensional chain of atoms
that for phonons, there are as many normal modes as there are degrees of freedom in the crystal,
therefore our 1-dimensional chain can only have 1 normal mode.
Thedensity of statesD(ω)can be obtained the same way as for photons, by computingD(ω) =
D(k)dkdωwhich leads us to fig. 13(a).
Linear chain of two different types of atoms
Now we generalize our assumption a bit, as we take two different massesM 1 andM 2 into account, so
every atom of massM 1 , to which we assign the position vectorsus, is coupled to two atoms of masses
M 2 with position vectorsvsand vice versa.
M 1 M 2 M 1 M 2
Figure 14: Linear chain of two different types of atoms
Hence, Newton’s law has to be altered as follows:
M 1
∂^2 us(x,t)
∂t^2=C(vs− 1 − 2 us+vs)M 2
∂^2 vs(x,t)
∂t^2=C(us− 2 vs+us+1)As before, we assume harmonic solutions forusandvs
us=uei(ksa−ωt)vs=vei(ksa−ωt)to obtain the equations:
−ω^2 M 1 u=C(v(1 +e−ika)− 2 u)