Finding partition functions of statistical system

[CAF123]

Homework Statement

Consider a zipper of N links, each of which can either be open or closed with associated energy 0 if closed and $\epsilon$ if open.
a) Suppose the N links are independent, compute the partition function of the system and the average number of open links
b)Now assume that the zipper opens up from one of its ends. To open the ith link, all links between i+1 and N therefore have to be open. Write down the partition function for a zipper with N links.

Homework Equations

boltzmann distribution

The Attempt at a Solution

a) For single link, $Z = e^{-\beta \epsilon} + 1$ and since links independent, $Z_N = (1+e^{-\beta \epsilon})^N$. Then
$$\langle n \rangle = \frac{1}{Z_N} \sum_{i=1}^N n_i P(n_i) = \frac{N e^{-\beta \epsilon}}{(1+e^{-\beta \epsilon})^N}$$ is it ok?

b)Just looking for a hint on how to go about starting this part, it says the geometric summation may come in helpful. Sequences are like, say for N=4 links, 0000, 0001, 0011,0111 or 1111. Could write the hamiltonian like $\mathcal H = \sum_{j=i}^{N} \epsilon$ where $j \geq i$ are open and $i \leq N$.

 

[TSny]

Does your answer for part (a) reduce to what you would expect for very high temperatures?

For part (b) you should be able to express the energy of the zipper when k links are open in a simple form. Use the energy expression to help construct the partition function.

 

[CAF123]

Hi TSny,

Does your answer for part (a) reduce to what you would expect for very high temperatures?

I suppose at high enough temperatures, there is more energy available to system so expect N links to be open which appears to be the case for my expression. Or did I misinterpret something?

For part (b) you should be able to express the energy of the zipper when k links are open in a simple form. Use the energy expression to help construct the partition function.

Ok, so $H = (N-j)\epsilon$, for j links closed then $Z$ becomes a sum over the number of closed links. So $$Z = \sum_{i=0}^N e^{-\beta(N-i)\epsilon}$$ and I think I could use the geometric summation here.

 

[TSny]

For very high temperature (small $\beta$) the thermal energy per link of the zipper is much larger than $\epsilon$. The difference in energy of a link being open or closed is negligible compared to the thermal energy. So, there shouldn't be much preference (energetically speaking) for a link to be closed rather than open. The links are behaving independently of one another in part (a). So, for the system of N independent links you should expect that at any time about half would be open and half would be closed. But your answer does not produce this in the limit of high temperature.

For part (a) I think it is helpful to first write down the probability that a single link is open. Then it should be easy to use that probability to find <n> because the links are independent.

Your expression for $Z$ in part (b) looks correct. It might simplify things to let $n = N-i$ = number of open links, and sum over $n$. Yes, you will get a sum of a geometric series.

 

[CAF123]

For very high temperature (small $\beta$) the thermal energy per link of the zipper is much larger than $\epsilon$. The difference in energy of a link being open or closed is negligible compared to the thermal energy. So, there shouldn't be much preference (energetically speaking) for a link to be closed rather than open. The links are behaving independently of one another in part (a). So, for the system of N independent links you should expect that at any time about half would be open and half would be closed. But your answer does not produce this in the limit of high temperature.

I see, so as $T \rightarrow \infty$ we expect the system to tend to a state of maximal disorder so at any given time half of the links would be open and half would be closed.

For part (a) I think it is helpful to first write down the probability that a single link is open. Then it should be easy to use that probability to find <n> because the links are independent.

$$P(\text{single link open}) = \frac{e^{-\beta \epsilon}}{(1+e^{-\beta \epsilon})^N}$$ and so $$P(\text{n are open}) = \frac{e^{-n\beta \epsilon}}{(1+e^{-\beta \epsilon})^N}$$ Then $$\langle n \rangle = \sum_{i=0}^N n P(\text{n are open})$$ Is that better?

 

[TSny]

I see, so as $T \rightarrow \infty$ we expect the system to tend to a state of maximal disorder so at any given time half of the links would be open and half would be closed.

Yes.

$$P(\text{single link open}) = \frac{e^{-\beta \epsilon}}{(1+e^{-\beta \epsilon})^N}$$

Why is the denominator raised to the power of $N$? We want the probability that a single link is open. What is the partition function for a single link?

 

[CAF123]

Why is the denominator raised to the power of $N$? We want the probability that a single link is open. What is the partition function for a single link?

Ah I see what you mean, ok so $Z = e^{-\beta \epsilon} + 1$ and so the probability that a single link is open is $P = e^{-\beta \epsilon}/Z$. For n links open it is $P(n)= P^n$ and I use this in my sum $\sum_{i=0}^N n P(n)$

 

[TSny]

Ah I see what you mean, ok so $Z = e^{-\beta \epsilon} + 1$ and so the probability that a single link is open is $P = e^{-\beta \epsilon}/Z$.

Yes, that's the probability that if you picked a link at random you would find that link to be open.

For n links open it is $P(n)= P^n$ and I use this in my sum $\sum_{i=0}^N n P(n)$

The expression $P(n)= P^n = e^{-n \beta \epsilon}/Z^n$ does not give the probability that the system of $N$ links has $n$ links open. This expression would give the probability that $n$ specifically chosen links are all open and we say nothing about the other $N-n$ links. For example it would give the probability that links 1 through $n$ are all open while we don't care about the other links. But we want the probability that the system has $n$ links open (where we don't care which particular links are open) and $N-n$ links closed. So, you would need to multiply your expression by the number of ways to select $n$ out of $N$ links and also include a factor for the probability that $N-n$ links are closed.

However, there is a more direct way to get the $P(n)$ that we want. If there were a crowd of $N$ people and you know the probability, $p$, that a person chosen at random is wearing a red shirt, then you can easily write down the expected number of people who are wearing red shirts in terms of $p$ and $N$.