# One-dimensional polymer (Gibbs canonical ensemble)

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 $$$$ and average length $$$$ 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 $$$$ and $$$$ is straightforward as:

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

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

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.

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 $$Z = \sum_{n=0}^N \frac{N!}{n!(N-n)!} x^{N-n} y^n = (x+y)^N$$