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## Kittel Zipper problem

Posted by peeterjoot on March 20, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

## Question: Zipper problem ([1] pr 3.7)

A zipper has $N$ links; each link has a state in which it is closed with energy $0$ and a state in which it is open with energy $\epsilon$. we require, however, that the zipper can only unzip from the left end, and that the link number $s$ can only open if all links to the left $(1, 2, \cdots, s - 1)$ are already open. Find (and sum) the partition function. In the low temperature limit $k_{\mathrm{B}} T \ll \epsilon$, find the average number of open links. The model is a very simplified model of the unwinding of two-stranded DNA molecules.

The system is depicted in fig. 1.1, in the $E = 0$ and $E = \epsilon$ states.

Fig 1.1: Zipper molecule model in first two states

The left opening only constraint simplifies the combinatorics, since this restricts the available energies for the complete molecule to $0, \epsilon, 2 \epsilon, \cdots, N \epsilon$.

The probability of finding the molecule with $s$ links open is then

\begin{aligned}P_s =\frac{e^{- \beta s \epsilon}}{Z},\end{aligned} \hspace{\stretch{1}}(1.0.1)

with

\begin{aligned}Z = \sum_{s = 0}^N \frac{e^{- \beta s \epsilon}}{Z}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

We can sum this geometric series immediately

\begin{aligned}\boxed{Z =\frac{e^{-\beta (N+1) \epsilon} - 1}{e^{-\beta \epsilon } - 1}.}\end{aligned} \hspace{\stretch{1}}(1.0.3)

The expectation value for the number of links is

\begin{aligned}\left\langle{{s}}\right\rangle &= \sum_{s = 0}^N s P_s \\ &= \frac{1}{{Z}} \sum_{s = 1}^N s e^{- \beta s \epsilon} \\ &= -\frac{1}{{Z}} \frac{\partial {}}{\partial {(\beta \epsilon)}} \sum_{s = 1}^N e^{- \beta s \epsilon}.\end{aligned} \hspace{\stretch{1}}(1.0.4)

Let’s write

\begin{aligned}a = e^{-\beta \epsilon},\end{aligned} \hspace{\stretch{1}}(1.0.5)

and make a change of variables

\begin{aligned}-\frac{\partial {}}{\partial {(\beta \epsilon)}} &= \frac{\partial {}}{\partial {\ln a}} \\ &= \frac{\partial {a}}{\partial {\ln a}}\frac{\partial {}}{\partial {a}} \\ &= \frac{\partial {e^{-\beta \epsilon}}}{\partial {(-\beta \epsilon)}}\frac{\partial {}}{\partial {a}} \\ &= a\frac{\partial {}}{\partial {a}}\end{aligned} \hspace{\stretch{1}}(1.0.6)

so that

\begin{aligned}-\frac{\partial {}}{\partial {\ln a}} \sum_{s = 1}^N a^s &= a \frac{d}{da} \left( \frac{a^{N+1} - a}{a - 1} \right) \\ &= a\left( \frac{(N+1) a^N - 1}{a - 1} - \frac{a^{N+1} - a} { (a - 1)^2 } \right) \\ &= \frac{a}{(a-1)^2}\left( \left( (N+1) a^N - 1 \right) (a - 1) - a^{N+1} + a \right) \\ &= \frac{a}{(a-1)^2}\left( N a^{N+1} -(N+1) a^N + 1 \right) \\ &= \frac{a}{(a-1)^2}\left( a^N ( N (a - 1) - 1 ) + 1 \right).\end{aligned} \hspace{\stretch{1}}(1.0.7)

The average number of links is thus

\begin{aligned}\left\langle{{k}}\right\rangle = \frac{a - 1}{a^{N+1} - 1}\frac{a}{(a-1)^2}\left( a^N ( N (a - 1) - 1 ) + 1 \right),\end{aligned} \hspace{\stretch{1}}(1.0.8)

or

\begin{aligned}\boxed{\left\langle{{k}}\right\rangle = \frac{1}{1 - e^{-\beta \epsilon(N+1)} }\frac{1}{e^{\beta \epsilon} - 1}\left( e^{-\beta \epsilon N} ( N (e^{-\beta \epsilon} - 1) - 1 ) + 1 \right).}\end{aligned} \hspace{\stretch{1}}(1.0.9)

In the very low temperature limit where $\beta \epsilon \gg 1$ (small $T$, big $\beta$), we have

\begin{aligned}\left\langle{{k}}\right\rangle \approx\frac{1}{e^{\beta \epsilon}}= e^{-\beta \epsilon},\end{aligned} \hspace{\stretch{1}}(1.0.10)

showing that on average no links are open at such low temperatures. An exact plot of $\left\langle{{s}}\right\rangle$ for a few small $N$ values is in fig. 1.2.

Fig 1.2: Average number of open links

# References

[1] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.

1. ### Mikesaid

I think it would be easier to just make the approximation $Z = 1 + \exp(-\beta \epsilon)$ since beta is very large, higher terms can be neglected. Then $U = -\partial \ln Z/\partial \beta = \epsilon \exp(-\epsilon/kT) = \epsilon$ so $\langle k \rangle = \exp(-\epsilon/kT)$

• ### peeterjootsaid

You mean since $N$ is very large, and then use $\langle k \rangle = U/\epsilon$? Yes, that’s a much easier way to do it!