1549380323-Statistical Mechanics Theory and Molecular Simulation

Polymer model 27

ω^2 kak=Aak. (1.7.9)

Here,Ais a matrix given by

A=ω^2











1 −1 0 0 0 ··· 0 0

−1 2 −1 0 0 ··· 0 0

0 −1 2 −1 0 ··· 0 0

···

0 0 0 0 0 ··· −1 1











, (1.7.10)

and theω^2 kandakare the eigenvalues and eigenvectors, respectively. The square roots
of the eigenvalues are frequencies that correspond to a set of special modes of the chain
known as thenormal modes. By diagonalizing the matrixA, the frequencies can be
shown to be

ω^2 k= 2ω^2

[

1 −cos

(

(k−1)π

N

)]

. (1.7.11)

Moreover, the orthogonal matrix U whose columns are the eigenvectorsakofAdefines
a transformation from the original displacement variablesηito a new set of variables
ζivia

ζi=

∑

k

ηkUki, (1.7.12)

known asnormal mode variables. By applying this transformation to the Hamiltonian
in eqn. (1.7.4), it can be shown that the transformed Hamiltonian is given by

H=

∑N

k=1

p^2 ζk

2 m

+

1

2

∑N

k=1

mωk^2 ζ^2 k. (1.7.13)

(The easiest way to derive this result is to start with the Lagrangianin terms of
η 1 ,...,ηN,η ̇ 1 ,...,η ̇N, apply eqn. (1.7.12) to it, and then perform the Legendre transform
to obtain the Hamiltonian. Alternatively, one can directly compute the inverse of the
mass-metric tensor and substitute it directly into eqn. (1.6.10).) Ineqn. (1.7.13), the
normal modes are decoupled from each other and represent a setof independent modes
with frequenciesωk.
Note that independent ofN, there is always one normal mode, thek= 1 mode,
whose frequency isω 1 = 0. This zero-frequency mode corresponds to overall transla-
tions of the entire chain in space. In the absence of an external potential, this transla-
tional motion is free, with no associated frequency. Considering this fact, the solution
of the equations of motion for each normal mode

̈ζk+ωk^2 ζk= 0 (1.7.14)

can now be solved analytically:

ζ 1 (t) =ζ 1 (0) +

pζ 1 (0)

m

t