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.


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

| cite | improve this answer | |
  • $\begingroup$ 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 '19 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.