A thermodynamics polymer chain problem

  • #1
A one-dimensional polymer molecule (rubber) is chain of N links of the same length a, the links can go either forward or backward but always stay parallel to the x axis. If one denotes the coordinates of the joints are ${x_0, x_1, . . . , x_N}$ , then $|x_n − x_{n+1}| = a$. The energy of the polymer does not depend on the angles at the joints.

Two electric charges $-q$ and $q$ are attached to the two ends of the molecule. An electric field $\mathcal{E}$ is turned on. The total energy of the molecule is the electrostatic energy,
$$E=−q\mathcal{E} (x_N −x_0)$$
Find the dependence of the temperature on the distance $L$ between the two ends of the molecule. Imagine the system interacts with a thermal bath, show that the rubber molecule contracts when the temperature increases.

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So I tried this problem with first setting up energy states, with $E_n = -qE(x_N - x_0) = -qE(-N+2n)a$, and getting the partition function $$Z = \frac{e^{-\beta q \mathcal{E} N a} - e^{\beta q \mathcal{E} N a}}{1-e^{2\beta q \mathcal{E} a}}$$
but I don't know how to proceed from here. In principle I should take $\langle L \rangle = \frac{1}{Z}\sum_{n=0}^N e^{-\beta E_n} |x_N - x_0|$, but the expression is extremely complicated and there is no way to answer the second question (that the polymer contracts when temperature increases).
Could anyone help me?
 
Last edited:

Answers and Replies

  • #2
marcusl
Science Advisor
Gold Member
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Finding the mean length of a polymer chain in the absence of charges involves maximizing the entropy. I imagine that your problem involves some combination of energy and entropy (Gibbs?).
 

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