A simple model for a rubber molecule consists of a one-dimensional chain of $n$ links each of fixed length $b$ and each of which is oriented in either the positive or negative direction. A unique state $i$ of the molecule is designated by giving the orientation $\pm 1$ of each link. If there are $n_{+}$links oriented in the positive direction and $n_{-}$links oriented in the negative direction then $n=n_{+}+n_{-}$and the length of the molecule is $l=\left(n_{+}-n_{-}\right) b$. The length of the molecule associated with state $i$ is $l_{i}$.

What is the range of $l$ ?

What is the number of states with $n, n_{+}, n_{-}$fixed?

Consider an ensemble of $A$ copies of the molecule in which $a_{i}$ members are in state $i$ and write down the expression for the mean length $L$.

By introducing a Lagrange multiplier $\tau$ for $L$ show that the most probable configuration for the $\left\{a_{i}\right\}$ with given length $L$ is found by maximizing

$$\log \left(\frac{A !}{\prod_{i} a_{i} !}\right)+\tau \sum_{i} a_{i} l_{i}-\alpha \sum_{i} a_{i} .$$

Hence show that the most probable configuration is given by

$$p_{i}=\frac{e^{\tau l_{i}}}{Z},$$

where $p_{i}$ is the probability for finding an ensemble member in the state $i$ and $Z$ is the partition function which should be defined.

Show that $Z$ can be expressed as

$$Z=\sum_{l} g(l) e^{\tau l}$$

where the meaning of $g(l)$ should be explained.

Hence show that $Z$ is given by

$$Z=\sum_{n_{+}=0}^{n} \frac{n !}{n_{+} ! n_{-} !}\left(e^{\tau b}\right)^{n_{+}}\left(e^{-\tau b}\right)^{n_{-}}, \quad n_{+}+n_{-}=n$$

and therefore that the free energy $G$ for the system is

$$G=-n k T \log (2 \cosh \tau b) .$$

Show that $\tau$ is determined by

$$L=-\frac{1}{k T}\left(\frac{\partial G}{\partial \tau}\right)_{n}$$

and hence that the equation of state is

$$\tanh \tau b=\frac{L}{n b} .$$

What are the independent variables on which $G$ depends?

Explain why the tension in the rubber molecule is $k T \tau$.