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Normal mode analysis

We shall now solve the equations (6.3) to (6.6) . We first solve them leaving out the fluctuating forces. The equations of motion then form a linear set of (3N+3) first order differential equations, whose general solutions are sums of (3N+3)independent specific solutions.

As a specific solution we try

 \begin{displaymath}\vec{R}_n(t) = \vec{X}(t) \cos (an+b).
\end{displaymath} (6.10)

The equations of motion then read
$\displaystyle \frac{d \vec{X}}{dt} \cos b$ = $\displaystyle -\frac{3kT}{\gamma b^2} \{
\cos b -\cos (a+b)\} \vec{X}$ (6.11)
$\displaystyle \frac{d \vec{X}}{dt} \cos (na+b)$ = $\displaystyle -\frac{3kT}{\gamma b^2
} 4 \sin^2 \left( \frac{1}{2}a \right) \cos (na+b) \vec{X}$ (6.12)
$\displaystyle \frac{d \vec{X}}{dt} \cos (Na+b)$ = $\displaystyle -\frac{3kT}{\gamma b^2
} \{ \cos (Na+b) -\cos ((N-1)a+b)\} \vec{X}$ (6.13)

where we have used
$\displaystyle { 2\cos(na+b)-\cos((n-1)a+b)-\cos((n+1)a+b)}$
  = $\displaystyle \cos(na+b) \{ 2-2\cos a \} = \cos (na+b) 4 \sin^2 \left( \frac{1}{2}a
\right) .$ (6.14)

In order for equations (6.11) to (6.13) to be consistent we need
$\displaystyle \cos b - \cos (a+b)$ = $\displaystyle 4 \sin^2 \left( \frac{1}{2}a \right) \cos b$ (6.15)
$\displaystyle \cos (Na+b) - \cos ((N-1)a+b)$ = $\displaystyle 4 \sin^2 \left( \frac{1}{2}a \right)
\cos (Na +b)$ (6.16)

This may be rewritten as
$\displaystyle \cos (a-b)$ = $\displaystyle \cos b$ (6.17)
$\displaystyle \cos ((N+1)a+b)$ = $\displaystyle \cos (Na+b) .$ (6.18)

We find independent solutions from
a-b = b (6.19)
(N+1)a+b = $\displaystyle k2\pi - Na -b .$ (6.20)

So, finally

 \begin{displaymath}a = \frac{k\pi}{N+1}, b = \frac{1}{2} a = \frac{k\pi}{2(N+1)} .
\end{displaymath} (6.21)

The specific solution Eq. (6.10), with a and b from Eq. ( 6.21) decouple the set of differential equations.

In the following we shall frequently make use of

 \begin{displaymath}\frac{1}{N+1} \sum_{n=0}^N \cos \left( \frac{k\pi}{N+1}(n+\frac{1}{2})
\right) = \delta_{k,0} 0 \le k < 2(N+1) .
\end{displaymath} (6.22)

This is evident when k=0 or k=N+1. In the remaining cases the sum may be evaluated by using $\cos (na+b) = \frac{1}{2} (e^{ib} e^{ina} + e^{-ib}
e^{-ina})$. The result then is

\begin{displaymath}\frac{1}{N+1} \sum_{n=0}^N \cos \left( \frac{k\pi}{N+1} (n+\f...
...} \frac{\sin (k\pi)}{\sin \left( \frac{k\pi}{2(N+1)}
\right) }
\end{displaymath} (6.23)

which is consistent with Eq. (6.22).

We now turn to the solution of Eqs. (6.3) to (6.6 ). To this end we write

 \begin{displaymath}\vec{R}_{n}=\vec{X}_{0}+2\sum_{k=1}^{N}\vec{X}_{k}\cos \left( \frac{k\pi }{
N+1}(n+\frac{1}{2})\right) .
\end{displaymath} (6.24)

The factor of two in front of the summation is only for reasons of convenience. Using Eq. (6.22) we may invert this to

 \begin{displaymath}\vec{X}_{k}=\frac{1}{N+1}\sum_{n=0}^{N}\vec{R}_{n}\cos \left( \frac{k\pi }{
N+1}(n+\frac{1}{2})\right) .
\end{displaymath} (6.25)

The equations of motion then read
$\displaystyle \frac{d\vec{X}_{k}}{dt}$ = $\displaystyle -\frac{3k_{B}T}{\gamma b^{2}}4\sin ^{2}\left(
\frac{k\pi }{2(N+1)}\right) \vec{X}_{k}+\vec{F}_{k}$ (6.26)
$\displaystyle \langle \vec{F}_{0}(t)\cdot \vec{F}_{0}(t^{\prime })\rangle$ = $\displaystyle \frac{6D}{N+1}
\delta (t-t^{\prime })$ (6.27)
$\displaystyle \langle \vec{F}_{k}(t)\cdot \vec{F}_{k^{\prime }}(t^{\prime })\rangle$ = $\displaystyle \frac{3D}{N+1}\delta _{k,k^{\prime }}\delta (t-t^{\prime })k\neq 0$ (6.28)

Eqs. (6.27) and (6.28) were obtained from Eq. (6.6) and Eq. (6.22) by using

 \begin{displaymath}\vec{F}_{k}=\frac{1}{N+1}\sum_{n=0}^{N}\vec{f}_{n}\cos \left( \frac{k\pi }{
N+1}(n+\frac{1}{2})\right) .
\end{displaymath} (6.29)

We have finally arrived at a decoupled set of 3(N+1) stochastic differential equations. In fact all of them are similar to our first stochastic differential equation (4.2). They only differ in their friction coefficients , and in the characteristics of the random forces.

Using Eq. (6.22) we easily see that $
\vec{R}_{G}=\vec{X}_{0}$. Eqs. (6.26) and (6.27) for k=0 read

$\displaystyle \frac{d\vec{X}_{0}}{dt}$ = $\displaystyle \vec{F}_{0}$ (6.30)
$\displaystyle \langle \vec{F}_{0}(t)\cdot \vec{F}_{0}(t^{\prime })\rangle$ = $\displaystyle 6\frac{D}{N+1}
\delta (t-t^{\prime })$ (6.31)

from which we get again the diffusion coefficient  of the centre of mass as given in Eq. (6.9).

The specific solutions that we have found are called the normal modes  of the chain. $
\vec{X}_{0}$ describes the motion of the centre of gravity. The other modes describe vibrations of the chain leaving the centre of mass unchanged.

In the applications ahead of us, our results will always be expressed as sums over normal modes, which sums will always be dominated by the contributions from the modes with small k, i.e. those with large wavelength. We therefore approximate Eq. (6.26) by

$\displaystyle \frac{d\vec{X}_{k}}{dt}$ = $\displaystyle -\frac{1}{\tau_k} \vec{X}_k + \vec{F}_k$ (6.32)
$\displaystyle \langle \vec{F}_k(t)\cdot\vec{F}_k(t^{\prime}) \rangle$ = $\displaystyle \frac{3D}{N+1}
\delta_{kk^{\prime}} \delta (t-t^{\prime}) .$ (6.33)

These equations apply when $k \neq 0$. The characteristic time $\tau_k$ is given by

 \begin{displaymath}\tau_k = \frac{\gamma b^2(N+1)^2}{3\pi^2k_BT} \frac{1}{k^2} = \frac{
b^2(N+1)^2}{3\pi^2D} \frac{1}{k^2} .
\end{displaymath} (6.34)

next up previous contents index
Next: Correlation of the end-to-end Up: The Rouse chain Previous: The Rouse chain
W.J. Briels