# Rubber bands and entropy

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.34.

We can model a rubber band as a one-dimensional chain of ${N}$ polymer links each of length ${\ell}$, where each link can point either to the left or right. The total length ${L}$ of the rubber band is therefore

$\displaystyle L=\left(N_{R}-N_{L}\right)\ell=\left(2N_{R}-N\right)\ell \ \ \ \ \ (1)$

where ${N_{R,L}}$ is the number of links pointing to the right or left, so that ${N=N_{R}+N_{L}}$. We’ll assume that ${N_{R}>N_{L}}$, although the converse is just the mirror image and gives the same behaviour.

The entropy can be found by noting that the system is equivalent to a coin-flipping experiment, so for a given ${N_{R}}$, the multiplicity is

$\displaystyle \Omega=\binom{N}{N_{R}}=\frac{N!}{N_{R}!\left(N-N_{R}\right)!} \ \ \ \ \ (2)$

Using Stirling’s approximation, the entropy is

$\displaystyle S=k\left[N\ln N-N_{R}\ln N_{R}-\left(N-N_{R}\right)\ln\left(N-N_{R}\right)\right] \ \ \ \ \ (3)$

Since this is a one-dimensional system, the role of the pressure in a 3-d system is taken here by the tension force ${F}$ generated by stretching the rubber band. If ${F>0}$ when the band is pulling inward, then work ${F\cdot dL}$ is done on the band when it is stretched through a distance ${dL>0}$.

The thermodynamic identity for this system is therefore

$\displaystyle dU=TdS+FdL \ \ \ \ \ (4)$

If the band is stretched in such a way that its energy remains constant (e.g. by losing heat), then ${dU=0}$ and the force is given by

$\displaystyle F=-T\left(\frac{\partial S}{\partial L}\right)_{U} \ \ \ \ \ (5)$

From 1

$\displaystyle dL=2\ell dN_{R} \ \ \ \ \ (6)$

so from 3

 $\displaystyle \frac{\partial S}{\partial L}$ $\displaystyle =$ $\displaystyle \frac{1}{2\ell}\frac{\partial S}{\partial N_{R}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k}{2\ell}\left[-\ln N_{R}-1+\ln\left(N-N_{R}\right)+1\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k}{2\ell}\ln\frac{N-N_{R}}{N_{R}} \ \ \ \ \ (9)$

From 1

$\displaystyle N_{R}=\frac{1}{2}\left(\frac{L}{\ell}+N\right) \ \ \ \ \ (10)$

so

 $\displaystyle \frac{\partial S}{\partial L}$ $\displaystyle =$ $\displaystyle \frac{k}{2\ell}\ln\frac{\frac{N}{2}-\frac{L}{2\ell}}{\frac{N}{2}+\frac{L}{2\ell}}=\frac{k}{2\ell}\ln\frac{N\ell-L}{N\ell+L}\ \ \ \ \ (11)$ $\displaystyle F$ $\displaystyle =$ $\displaystyle -\frac{kT}{2\ell}\ln\frac{N\ell-L}{N\ell+L} \ \ \ \ \ (12)$

If ${L\ll N\ell}$, the band is almost fully contracted since its length is much less than the maximum length. In this case we can get an estimate of the force:

 $\displaystyle F$ $\displaystyle =$ $\displaystyle -\frac{kT}{2\ell}\ln\frac{1-L/N\ell}{1+L/N\ell}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle -\frac{kT}{2\ell}\ln\left[\left(1-L/N\ell\right)^{2}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{kT}{\ell}\ln\left[1-L/N\ell\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{kT}{N\ell^{2}}L \ \ \ \ \ (16)$

That is, ${F\propto L}$ which is Hooke’s law for a spring, with spring constant ${kT/N\ell^{2}}$.

The formula 12 for ${F}$ says that ${F}$ should increase as the temperature is increased, that is, a rubber band should contract when heated (up to a point, obviously; after a while it will just melt). Although this relation was derived in the special case of constant energy ${U}$, it does seem to make sense. A higher temperature would increase the entropy, meaning that ${N_{R}}$ gets closer to its maximum entropy value of ${\frac{N}{2}}$, which means that ${L}$ gets smaller. By the same token, we’d expect a rubber band to get warmer if it is suddenly stretched. I did try the experiment suggested, although I couldn’t find a heavy rubber band, and it did seem to warm up a bit when it was stretched.

## 2 thoughts on “Rubber bands and entropy”

1. Mr 4

>a rubber band should contract when heated (up to a point, obviously; after a while it will just melt)

I think the term should be expand instead of contract.

1. gwrowe Post author

No, it should contract, since if the tension force ${F}$ increases with temperature, there is more force pulling the rubber together.