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Next: Appendix A Up: The Gaussian chain Previous: Green's function method

One Gaussian chain in a box

As an example of the use of the Green's function method, we calculate the pressure exerted by a Gaussian chain on the walls of a confining box .

The potential is zero everywhere inside the box, and infinite everywhere outside the box. Then $\Psi_n(
\vec{R})=0$ at the walls of the box. Solving Eq. (3.18) yields

\begin{displaymath}\Psi_{n_1,n_2,n_3}(\vec{R}) = \varphi_{n_1}^{(1)} (R_1) \varphi_{n_2}^{(2)}
(R_2) \varphi_{n_3}^{(3)} (R_3)
\end{displaymath} (3.21)


\begin{displaymath}E_{n_1,n_2,n_3} = \epsilon_{n_1}^{(1)} + \epsilon_{n_2}^{(2)} +
\epsilon_{n_3}^{(3)}
\end{displaymath} (3.22)

where

 \begin{displaymath}\varphi_n^{(i)}(R) = \sqrt{2/L_i} \sin (\pi nR/L_i)
\end{displaymath} (3.23)


 \begin{displaymath}\epsilon_n^{(i)} = \frac{\pi^2 b^2}{6L_i^2} n^2 .
\end{displaymath} (3.24)

The partition function then reads
Z = $\displaystyle \left\{ \frac{3}{2\pi b^2} \right\}^{-\frac{3}{2}N} Z_1 Z_2 Z_3$ (3.25)
Zi = $\displaystyle \int dR_i \int dR^{\prime}_i \sum_n \exp\{ -\epsilon_n^{(i)} N
\} \varphi_n (i)(R_i) \varphi_n (i)(R^{\prime}_i)$ (3.26)

Introducing the eigenfunctions and eigenvalues from Eqs. (3.23) and (3.24) we get

\begin{displaymath}Z_i = \frac{8}{\pi ^2} L_i \sum_{n=1}^{\infty}{\,}^{\prime}\frac{1}{n^2} \exp
\{ - \frac{\pi ^2 b^2}{6L_i^2} n^2 N \}
\end{displaymath} (3.27)

where the prime at the summation sign indicates that only odd n should occur in the sum.

Now look at two limits

i.
$\sqrt{N}b\ll L_{i}$
The polymer is much smaller than the box
Zi = $\displaystyle \frac{8}{\pi ^{2}}L_{i}\sum_{n=1}^{\infty }{\,}^{\prime }\frac{1}{
n^{2}}=L_{i}$ (3.28)
Z = $\displaystyle \left\{ \frac{3}{2\pi b^{2}}\right\} ^{-\frac{3}{2}N}L_{1}L_{2}L_{3}.$ (3.29)

The pressure on wall one is

\begin{displaymath}P_{1}=-\frac{1}{L_{2}L_{3}}\frac{\partial }{\partial L_{1}}A=...
...
L_{2}L_{3}}\frac{\partial }{\partial L_{1}}\ln Z=\frac{kT}{V}
\end{displaymath} (3.30)

i.e. independent of the wall number, and equal to the ideal gas result.

ii.
$\sqrt{N}b\gg L_{i}$
The polymer is very much constrained by the box

\begin{displaymath}Z_{i}\approx \frac{8}{\pi ^{2}}L_{i}\exp \{-\frac{\pi ^{2}b^{2}}{6L_{i}^{2}}
N\}
\end{displaymath} (3.31)


P1 = $\displaystyle \frac{kT}{L_{2}L_{3}}\frac{\partial }{\partial L_{1}}\{\ln L_{1}-
\frac{\pi ^{2}b^{2}}{6L_{1}^{2}}N\}$ (3.32)
  = $\displaystyle \frac{kT}{L_{1}L_{2}L_{3}}\{1+\frac{\pi ^{2}b^{2}}{3L_{1}^{2}}N\}\approx
\frac{\pi ^{2}b^{2}}{3L_{1}^{2}}\frac{NkT}{V}.$ (3.33)

In this case the pressure on the different walls depends on the size of the box orthogonal to the wall.


next up previous contents index
Next: Appendix A Up: The Gaussian chain Previous: Green's function method
W.J. Briels