- #1

Jacques_Leen

- 11

- 0

- Homework Statement
- A solution of negatively (-e) charged ions is inserted into a cylinder with ##(R, L)## as parameters. At the center of this cylinder there is a rod ##(r=a)## of a positively charged polymer. The Hamiltonian of the system is:

##\mathcal{H} = \sum_{i=1}^{N} \frac{p_i ^2}{2m} + 2e^2 n \ln(r_i/L)##

Where ##n=N/L## and ##N## is the number of positive charges on the polymer.

Determine:

1 ##Z_N## partition function (let ##k=e^2 n \beta##)

2 ##p(r)## and ##\langle r \rangle##

3 Discuss the behavior at 2) in the limits ##k>>1## and ##k<<1## assuming ##R>>a##

4 Helmoltz function

5 The pressure at the side of the cylinder in the limits ##k>>1## and ##k<<1## assuming ##R>>a##

- Relevant Equations
- ##Z_N = \frac{1}{N! h^{3N}} \int e^{\beta \mathcal{H}}d^3pd^3q##

##A = -1/\beta \ln{Z_N}##

##P(r=R) = -\frac{1}{2\pi R L} \frac{\partial A}{\partial R}##

Hi everyone,

I want to post this exercise and my attempt to a solution since there are a couple of points I am not entirely sure of and I might need your help. I'll address them while posting my solution

1) ##Z_N = (Z_1)^N##. I can evaluate ##Z_1## integrating with respect of both parts of the ##\mathcal{H}##.

$$\int e^{-\beta p^2/2m}d^3p = 4\pi \int_{0}^{\infty} p^2 e^{-\beta p^2/2m}dp = (\frac{2 m \pi}{\beta})^{3/2}$$

$$\int e^{-2 k \ln(r/L)} d^3q = 2\pi L \int_{a}^R r (\frac{r}{L})^{-2k} dr = 2\pi L^{2k+1} (\frac{1}{2k a^{2k}} - \frac{1}{2k R^{2k}} ) $$

which yields:

$$Z_N = \frac{1}{N!} (2\pi L^{2k+1} (\frac{1}{2k a^{2k}} - \frac{1}{2k R^{2k}} ) )^N (\frac{2 m \pi}{h^2\beta})^{3N/2}$$

2) ##p(r) = C e^{-2k \ln(r/L)}##. The constant C can be found as done here. It should be

$$ C= \frac{1-2k}{L^{2k}(R^{1-2k} - a^{1-2k})} ,$$

which substituted gives the result for ##p(r)##. As for the average radial position it gets obtained by ##\int_a^R r A e^{-2k \ln(r/L)}dr##. The result is:

$$\langle r\rangle = \frac{AL^{2k}}{2k(a^{-2k} - R^{-2k})} $$

3) That is were the issues starts. Let ## k>>1##, then in the limit ##R>>a## the term ##\frac{1}{2k(a^{-2k} - R^{-2k})} ## tends to ##0##. It seems coherent with the fact that if there is no disorder introduced in terms of kinetic energy into the composite system (rod + ions) then the negatively charged ions tend towards the positively charged rod at the center.

But if I follow the same line of reasoning what happens with the limit ## k<<1## is that ##\frac{1}{2k(a^{-2k} - R^{-2k})} ## diverges if I am not mistaken. I have been trying to wrap my head around that problem but I cannot find a physical counterpart for that. Intuitively with the temperature growing the ions should be less and less influenced by the presence of the rod at the center of the cylinder. This seems at odds with my calculations.

4) ##A## is obtained from ## -1/\beta \ln(Z_N)## the result:

$$ -N/\beta ( \ln(N) + \ln(2\pi L^{2k+1} (\frac{1}{2k a^{2k} } - \frac{1}{2k R^{2k}} ) ) +3/2 \ln(\frac{2 m \pi}{h^2\beta}) - 1)$$

5) ##P## is associated with the work that gives an expansion ##2 \pi R L dR##, hence the general formula:

$$\frac{N}{\beta 2\pi L R} \frac{2k}{R^{2k+1} (a^{-2k} - R^{-2k})}$$

In the limits discussed above I have the same issue. In one case the computation is coherent with all the ions beeing pulled towards the center. Whereas the other case is odd.

Am I missing something?

Thanks to everyone who is willing to participate

I want to post this exercise and my attempt to a solution since there are a couple of points I am not entirely sure of and I might need your help. I'll address them while posting my solution

1) ##Z_N = (Z_1)^N##. I can evaluate ##Z_1## integrating with respect of both parts of the ##\mathcal{H}##.

$$\int e^{-\beta p^2/2m}d^3p = 4\pi \int_{0}^{\infty} p^2 e^{-\beta p^2/2m}dp = (\frac{2 m \pi}{\beta})^{3/2}$$

$$\int e^{-2 k \ln(r/L)} d^3q = 2\pi L \int_{a}^R r (\frac{r}{L})^{-2k} dr = 2\pi L^{2k+1} (\frac{1}{2k a^{2k}} - \frac{1}{2k R^{2k}} ) $$

which yields:

$$Z_N = \frac{1}{N!} (2\pi L^{2k+1} (\frac{1}{2k a^{2k}} - \frac{1}{2k R^{2k}} ) )^N (\frac{2 m \pi}{h^2\beta})^{3N/2}$$

2) ##p(r) = C e^{-2k \ln(r/L)}##. The constant C can be found as done here. It should be

$$ C= \frac{1-2k}{L^{2k}(R^{1-2k} - a^{1-2k})} ,$$

which substituted gives the result for ##p(r)##. As for the average radial position it gets obtained by ##\int_a^R r A e^{-2k \ln(r/L)}dr##. The result is:

$$\langle r\rangle = \frac{AL^{2k}}{2k(a^{-2k} - R^{-2k})} $$

3) That is were the issues starts. Let ## k>>1##, then in the limit ##R>>a## the term ##\frac{1}{2k(a^{-2k} - R^{-2k})} ## tends to ##0##. It seems coherent with the fact that if there is no disorder introduced in terms of kinetic energy into the composite system (rod + ions) then the negatively charged ions tend towards the positively charged rod at the center.

But if I follow the same line of reasoning what happens with the limit ## k<<1## is that ##\frac{1}{2k(a^{-2k} - R^{-2k})} ## diverges if I am not mistaken. I have been trying to wrap my head around that problem but I cannot find a physical counterpart for that. Intuitively with the temperature growing the ions should be less and less influenced by the presence of the rod at the center of the cylinder. This seems at odds with my calculations.

4) ##A## is obtained from ## -1/\beta \ln(Z_N)## the result:

$$ -N/\beta ( \ln(N) + \ln(2\pi L^{2k+1} (\frac{1}{2k a^{2k} } - \frac{1}{2k R^{2k}} ) ) +3/2 \ln(\frac{2 m \pi}{h^2\beta}) - 1)$$

5) ##P## is associated with the work that gives an expansion ##2 \pi R L dR##, hence the general formula:

$$\frac{N}{\beta 2\pi L R} \frac{2k}{R^{2k+1} (a^{-2k} - R^{-2k})}$$

In the limits discussed above I have the same issue. In one case the computation is coherent with all the ions beeing pulled towards the center. Whereas the other case is odd.

Am I missing something?

Thanks to everyone who is willing to participate