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Let's consider a polymer that is formed by an horizontal linear chain of $N$ disc-shaped monomers. Each monomer can either adopt either a vertical alignment (with length $l_1$ and energy $E_1$) or an horizontal alignment (with length $l_2$ and energy $E_2$) (obviously $l_2>l_1$). The chain is also subject to a tension $T$.

I want to compute the average energy $<E>$ and average length $<L>$ of the chain using the Gibbs canonical ensemble, whose partition function is given by:

$$Z_G=\sum_S\sum_i exp[\beta(E_i+pV_S)]$$

I began by calculating the denegeracy of energies $g(n$) of the system having $n$ monomers aligned horizontally, which is a basic problem of combinatorics:

$$g(n)={N \choose n}$$

We can exchange the sum in the partition function to:

$$\sum_S\sum_i\rightarrow\sum_{\{ n \}}g(n)$$

So the partition function is reduced to,

$$Z_G=\sum_{\{ n \}} {N \choose n} exp[\beta(E_i+pV_S)]$$

If I can find the form of the partition function, the calculation of $<E>$ and $<L>$ is straightforward as:

$$<E>=\frac{\sum_{\{ n \}}E_i {N \choose n} exp[\beta(E_i+pV_S)]}{\sum_{\{ n \}} {N \choose n} exp[\beta(E_i+pV_S)]}$$

$$<L>=\frac{\sum_{\{ n \}}L_i {N \choose n} exp[\beta(E_i+pV_S)]}{\sum_{\{ n \}} {N \choose n} exp[\beta(E_i+pV_S)]}$$

However, what I don't understand is how to express the term $E_i+pV_S$ in terms of the energies of the monomers, their lengths, and the total tension (a general way to express this would be useful for more general problems).

For example, I thought about substituting $pV_S\rightarrow TL_S$, with $L_S=l_1+l_2$ but I'm not even sure if this is valid. I found a similar problems online but they don't have a solution.

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In my opinion, this problem gives not much freedom in choosing expressions for $E$ and $L$. Energy of a chain with $n$ monomers aligned horizontally is $E_n = E_1(N-n) + E_2n$ and length of this chain is $L_n = l_1(N-n) + l_2n$. Correspondent Hibbs exponent $\exp(-\beta(E_n+TL_n))$ in this case has the form $x^{N-n}y^n$, where $x = \exp(-\beta(E_1+Tl_1))$, $y=\exp(-\beta(E_2+Tl_2))$. Hence you can use the binomial theorem for sums. For example, the partition function is $$ Z = \sum_{n=0}^N \frac{N!}{n!(N-n)!} x^{N-n} y^n = (x+y)^N $$

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  • $\begingroup$ Thank you, I managed to find the solution and it gave me some exponential distributions. I'll post the solution during the weekend to close the question. $\endgroup$ – Charlie Mar 21 at 19:51

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