Solving a Complex Problem: Z_N, p(r), and Averages

[Jacques_Leen]

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

 

[TSny]

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

Check this integration. The middle integral looks good, but the final result doen't have the correct powers on $a$ and $R$. What should the dimensions be for the overall integral?

I have only glanced through the calculations. But it looks like the approach is correct.

 

[Jacques_Leen]

Check this integration. The middle integral looks good, but the final result doen't have the correct powers on $a$ and $R$. What should the dimensions be for the overall integral?

I have only glanced through the calculations. But it looks like the approach is correct.

Thanks for pointing that out. I haven't run the corrected computation yet but this might as well resolve my issues with the physical intepretation of what happens for $k<<1$.

edit

So, the result of the previous integral which war wrong is
$$ \frac{\pi L^{2k+1}}{(1-k)} (R^{2(1-k)} - a^{2(1-k)}) $$

this way the average radial position is
$$\langle r \rangle = \frac{(1-2k) (R^{2(1-k)} - a^{2(1-k)})}{2(1-k)(R^{1-2k} - a^{1-2k})}$$

now the limit $R>>a$ gives $a/R = \varepsilon$, and for $k>>1$ I still have that the ions are pulled towards the center stopping at $a$. The low temperature causes the Coulombian interaction to prevail. for $k<<1$ I get:
$$\langle r \rangle \rightarrow \frac{1-2k}{2(1-k)}\frac{R^{2} (1 - \varepsilon^2)}{R(1- \varepsilon)} \rightarrow 1/2 R$$
which is consistent with the idea that as the temperature grows the ions do not feel the pull towards the center due to the rod and move in a random fashion.

The same should hold for point 5)

 

[TSny]

So, the result of the previous integral which war wrong is
$$ \frac{\pi L^{2k+1}}{(1-k)} (R^{2(1-k)} - a^{2(1-k)}) $$

I believe that's right

this way the average radial position is
$$\langle r \rangle = \frac{(1-2k) (R^{2(1-k)} - a^{2(1-k)})}{2(1-k)(R^{1-2k} - a^{1-2k})}$$

I get a different expression for $\langle r \rangle$ due to getting a different value for the constant $C$ in $p(r)$.

for $k << 1$ I get:
$$\langle r \rangle \rightarrow \frac{1-2k}{2(1-k)}\frac{R^{2} (1 - \varepsilon^2)}{R(1- \varepsilon)} \rightarrow 1/2 R$$
which is consistent with the idea that as the temperature grows the ions do not feel the pull towards the center due to the rod and move in a random fashion.

For $k << 1$ in a cylinder you might expect $\langle r \rangle$ to be greater than $R/2$ due to there being "more space" at larger values of $r$ in the cylinder. If you randomly choose a large number of points inside a cylinder of radius $R$, what would be the average value of $r$ for these points?