Why don’t ALL polymers collapse even when their compact state would clearly be favoured in terms of having a lower Helmholtz free energy?

Question

This question is based on example 8.3 in Molecular Driving Forces, Dill and Bromberg (https://books.google.ch/books?id=1gYPBAAAQBAJ&pg=PA136&lpg=PA136&dq=polymer+collapse+a+toy+model&source=bl&ots=lQrGS148oi&sig=ACfU3U1-3V0Vt8tHwba9cyWAD_cYSTCTyw&hl=en&sa=X&ved=2ahUKEwjMkrfI99XiAhXwx6YKHTHjCwQQ6AEwAXoECAkQAQ#v=onepage&q=polymer%20collapse%20a%20toy%20model&f=false)

In this problem Dill and Bromberg calculate the change in Helmholtz Free energy when a polymer collapses from an open configuration (higher energy, higher entropy) into a closed configuration (lower energy, lower entropy). They find that above a critical temperature the open configuration has a lower Helmholtz free energy and therefore conclude that this configuration is favoured at temperatures higher than this value. However, they also say that the exact composition needs to be calculated by means of considering a canonical ensemble.

My question now is: One can easily calculate the Boltzmann distribution corresponding to this system and see that at high temperatures one will get a closed configuration with a probability of 1/5 and an open configuration with a probability of 4/5 when evaluating as T goes to infinity. (They assume that there are four open configurations (energy: 0) and only one closed configuration (energy: -E)). But why do no ALL polymers collapse when such a process would clearly be favourable at high temperatures in terms of the Helmholtz free energy?

Of course I agree with the result obtained from the Boltzmann distribution! I just don't seem to grasp why not ALL of the polymers collapse.

Answer 1

This question seems to arise from a misreading of the source material. The model is a toy one, intended to illustrate that typical polymer molecules have a relatively small number of low-energy collapsed states, and a much larger number of high-energy expanded states. The "collapse transition" occurs on lowering the temperature below a critical or threshold value.

The simplified model seems to have a single compact microstate, with a negative energy, and four microstates of zero energy. The occupation fractions of 1/5 and 4/5 apply to the high temperature situation. As $T\rightarrow\infty$, the different energies cease to matter, and (for this toy model) all the microstates are occupied with equal probability. So one could say that at very high temperatures 1/5 of the molecules are found in the "collapsed" states, and 4/5 of them are in the "expanded" state. For this model, that's as disordered as it gets. In real life, of course, there are many many more microstates that would be classified as "expanded", and so the expanded states would be far more likely to be seen than the very few collapsed states. This is the entropy-dominated regime, and if we can roughly count the number of accessible microstates, $\Omega$, we can set $S=k_B\ln \Omega$. This is not a precise formula, in the canonical ensemble: a better formula is $S=-k_B\sum_i p_i \ln p_i$ where $p_i$ is the probability of occupying microstate $i$, and this is maximized when all the probabilities are equal (and in this case, that means each $p_i=1/5$). Also roughly, the free energy will be $F\approx -TS$. It is not correct to say that the free energy would be lowered if the remaining 1/5 of molecules in the collapsed microstate were to be converted into expanded microstates. That would make $p_0=0$, say, and $p_1=p_2=p_3=p_4=1/4$, which would give a lower (worse) entropy.

At the other extreme, below the collapse transition temperature, the Boltzmann distribution tells us that the lowest-energy microstate will be the most populated. In fact, as $T\rightarrow 0$, for this model, all the molecules will be in the lowest energy (collapsed) state, and none of them in the expanded state. We can set $F\approx E$, where $E$ is the energy; the entropy term becomes insignificant. At $T=0$ we expect $p_0=1$ (the ground state) and $p_1=p_2=p_3=p_4=0$ (the higher energy states).