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)


\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.


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