80
4 Representations
The kinetic energy is then given by
T =
1
2
21
i=1
m i
dS i
dt
2
=
M
2
3
i=1
dS i
dt
2
+
m
2
21
i=4
dS i
dt
2
(4.92)
Here, M is the atomic mass of uranium, and m is the atomic mass of fluorine. The
kinetic energy can be reduced to a uniform scalar product by mass weighting the
coordinates, i.e., by multiplying the S coordinates with the square root of the atomic
mass of the displaced atom. We shall denote these as the vector Q. Hence, Q i =
√
m i S i :
T =
1
2
i
dQ i
dt
2
=
1
2
i
˙
Q
2
i
(4.93)
where the dot over Q denotes the time derivative. The potential energy will be approximated by second-order derivatives of the potential energy surface V(Q) in the
mass-weighted coordinates:
V ij =
∂ 2 V
∂Q i ∂Q j
(4.94)
These derivatives are the elements of the Hessian matrix, V, which is symmetric
about the diagonal. The potential minimum coincides with the octahedral geometry.
The resulting potential energy is
V =
1
2
i,j
V ij Q i Q j
(4.95)
The kinetic and potential energies are combined to form the Lagrangian, L = T −V .
The equation of motion is given by
∂L
∂Q k
=
d
dt
∂L
∂ ˙
Q k
(4.96)
The partial derivatives in this equation are given by
∂L
∂Q k
=−
∂V
∂Q k
=−V kk Q k −
1
2
i =k
(V ik + V ki )Q i
=−V kk Q k −
i =k
V ki Q i =−
i
V ki Q i
(4.97)
d
dt
∂L
∂ ˙
Q k
=
d
dt
∂T
∂ ˙
Q k
=
d
dt
˙
Q k = ¨
Q k
(4.98)