# 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.

## 5 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.

2. Nikos

Also in part (c) F is defined as positive when the rubber band is pulling INWARD. Increasing T would make it pull inward more, hence heating it would contract.

2. Nikos

One part I do not fully understand is the last bit, namely “By the same token, we’d expect a rubber band to get warmer if it is suddenly stretched.”. If the band is stretched L should increase, if I am not mistaken. Increasing L increases NR in (1) assuming that little l does not change. If this is the case why would temperature change higher or lower? According to (3.6) in his book T is the change of internal energy with S at constant N and in this case at constant L (corresponding to V in 3.6). However, stretching the rubber it would change L and this is the confusing part to me. I would very much appreciate it if this part is clarified further.

1. gwrowe Post author

I think the key word here is ‘suddenly’ – if we suddenly stretch the rubber band, this is analogous to an adiabatic compression of an ideal gas, where the temperature does increase. If we stretch the rubber band slowly, allowing it to remain in thermal equilibrium with its environment, then the temperature won’t change.