Page 149 - Numerical methods for chemical engineering

Normal mode analysis 135

Let us assume for the moment that each degree of freedom has the same effective mass
m eff , so that the kinetic energy is
F
m eff 2
K(q, ˙q) = ˙ q k (3.178)
2
k=1
The dynamics of q are governed by the equations of motion
2 2
d q j d δ j ∂
m eff 2 = m eff 2 =− U(ˆ q 1 + δ 1 ,..., ˆ q F + δ F ) (3.179)
dt dt ∂δ j
which, in the vicinity of the local minimum, reduce to

2 F F
d δ j ∂ 1
m eff =− U(ˆ q 1 ,..., ˆ q F ) + δ m H mn δ n
dt 2 ∂δ j 2
m=1 n=1
2 F F F
d δ j 1 1
m eff 2 =− H jn δ n − δ m H mj =− H jn δ n (3.180)
dt 2 2
n=1 m=1 n=1
This system of second-order ODEs is written more compactly as
d 2
m eff 2 q =−Hδ (3.181)
dt
The eigenvalues and eigenvectors of the Hessian matrix satisfy
Hw [ j] = λ j w [ j] j = 1, 2,..., F (3.182)

Since the Hessian is real symmetric, the eigenvectors are mutually orthogonal, and so form
a convenient basis set for representing any vector. We therefore write a trial form of δ(t)as
the linear combination
d 2 [1] [F]
[1]
[F]
δ(t) = c 1 (t)w +· · · + c F (t)w q = ¨ c 1 w +· · · + ¨ c F w (3.183)
dt 2
The dynamical equations then become
[1] [F] [1] [F]
m eff ¨ c 1 w +· · · + ¨ c F w =−H c 1 w +· · · + c F w

m eff ¨ c 1 w [1] +· · · + m eff ¨ c F w [F] =− c 1 Hw [1] +· · · + c F Hw [F]
[1] [F]
=− c 1 λ 1 w +· · · + c F λ F w (3.184)
Equating the left- and right-hand sides separately in each eigenvector direction yields the
following uncoupled set of equations for each c j (t),
2
d c j
m eff =−λ j c j (3.185)
dt 2
We propose a trial form of the solution
c j (t) = a j sin(ω j t) (3.186)
and substitute it into the differential equation
2
d c j 2 −1 −1
= a j ω sin(ω j t) =−m eff λ j a j sin(ω j t) =−m eff λ j c j (3.187)
j
dt 2