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

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

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

where

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

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

The partition function then reads

$$\left\{ \frac{3}{2\pi b^2} \right\}^{-\frac{3}{2}N} Z_1 Z_2 Z_3$$ Z = (3.25)

$$\int dR_i \int dR^{\prime}_i \sum_n \exp\{ -\epsilon_n^{(i)} N \} \varphi_n (i)(R_i) \varphi_n (i)(R^{\prime}_i)$$ Z_i = (3.26)

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

$$Z_i = \frac{8}{\pi ^2} L_i \sum_{n=1}^{\infty}{\,}^{\prime}\frac{1}{n^2} \exp \{ - \frac{\pi ^2 b^2}{6L_i^2} n^2 N \}$$ (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

$$\frac{8}{\pi ^{2}}L_{i}\sum_{n=1}^{\infty }{\,}^{\prime }\frac{1}{ n^{2}}=L_{i}$$ Z_i = (3.28)

$$\left\{ \frac{3}{2\pi b^{2}}\right\} ^{-\frac{3}{2}N}L_{1}L_{2}L_{3}.$$ Z = (3.29)

The pressure on wall one is

$$P_{1}=-\frac{1}{L_{2}L_{3}}\frac{\partial }{\partial L_{1}}A=... ... L_{2}L_{3}}\frac{\partial }{\partial L_{1}}\ln Z=\frac{kT}{V}$$ (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

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

$$\frac{kT}{L_{2}L_{3}}\frac{\partial }{\partial L_{1}}\{\ln L_{1}- \frac{\pi ^{2}b^{2}}{6L_{1}^{2}}N\}$$ P₁ = (3.32)

$$\frac{kT}{L_{1}L_{2}L_{3}}\{1+\frac{\pi ^{2}b^{2}}{3L_{1}^{2}}N\}\approx \frac{\pi ^{2}b^{2}}{3L_{1}^{2}}\frac{NkT}{V}.$$ = (3.33)

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