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Suppose we have a 2D polymer model described by a set of 2D vectors {$\mathbf{t}_i$} ($i=1,2,\dots N$) of length $a$.

The energy of the polymer is given by: $$ \mathcal{H}~=~-k\sum^N_{i=1}\mathbf{t}_i\cdot\mathbf{t}_{i+1} ~=~-ka^2\sum^N_{i=1}\cos\phi_i. $$

The constant $k$ is a measure of bending rigidity so that a probability of finding any configuration of the polymer is proportional to $e^{-\frac{\mathcal{H}}{k_BT}}$.

Essentially, this is a discrete version of the wormlike-chain model by Kratky and Porod.

  1. What is the expression for propability density $p(\mathbf{R})$, where $\mathbf{R}$ is an end-to-end vector $\sum_i\mathbf{t_i}$, for this model? Let's assume that $N \rightarrow \infty$.

  2. How do I find linear force extension relation,when a force $\mathbf{F}$ is applied to the ends of the chain? Let's assume that the force is weak.

  3. Can anyone point out useful literature?

KBriggs
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molkee
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1 Answers1

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This question has been worked out in the first chapter of "Statistical Physics of DNA: An Introduction to Melting, Unzipping and Flexibility of the Double Helix", by Nikos Theodorakopoulos. The first chapter is freely available and can be found here: https://www.worldscientific.com/doi/pdf/10.1142/9789811209543_0001. As can be seen, there is no closed form for this problem. Nevertheless, since you are pointing out the limit of $N \rightarrow \infty$, a good approximation is to rescale your average intermonomer distance $a$ to the value of the persistence length $l_p = \kappa/(k_BT)$ and then use it as a freely jointed chain. The bending stiffness will then not be important anymore making this a valid approximation. For the freely jointed chain, the exact expression is also given in the link. Notice also that the freely jointed chain for $N \rightarrow \infty$ becomes a Gaussian.

So to wrap up and give a short answer: For your problem, a Gaussian distribution with an intermonomer distance of $a = \kappa/(k_BT)$ would be a good approximation for $N \rightarrow \infty$. Hope this answers your question :)