5.3 Crystals of Rigid Molecules
95
i αβ =
ρ(r)
γ=x,y,z
(r
2
γ δ αβ − r α r β )dV,
(5.52)
where r = R − R 0 . Using these quantities, the kinetic energy of a rigid body is given
by
T =
1
2
m
d r
dt
2
+
1
2
t
ωiω.
(5.53)
Here, ω is an angular momentum around the center of mass and is related to angular
coordinates Θ by
ω =
dΘ
dt
.
(5.54)
If the potential energy is written as U (Θ), thus, the equation of angular motion
around the center of mass reads
i
d
2
Θ
dt 2 = −∇ Θ U (Θ).
(5.55)
For the molecular crystal, a unit cell of which contains n molecules consisting
of p atoms, the potential energy can be expanded in terms of displacements from
equilibrium structure,
r p = R − R 0 p
(5.56)
θ p = Θ − Θ 0 p
(5.57)
where r p is the coordinate of the center of mass of the pth molecule with p that symbolically indicates all indexes necessary for distinguishing molecules. The expansion
of potential energy (lattice energy) up to the second-order is
V ≈ V 0 +
1
2
p
p
α,β=x,y,z
φ
TT
αβ ( p, p
)r α, p r β, p
+
1
2
p
p
α,β=x,y,z
φ
TR
αβ ( p, p
)r α, p θ β, p
+
1
2
p
p
α,β=x,y,z
φ
RT
αβ ( p, p
)θ α, p r β, p
+
1
2
p
p
α,β=x,y,z
φ
RR
αβ ( p, p
)θ α, p θ β, p
.
(5.58)
Here, the force constants are defined as