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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.

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  • $\begingroup$ You appear to have a pretty solid background on this, so forgive me if this is insultingly oversimplified, but could this just be a question of kinematics? Many things stay in a higher energy state for a very long time before their eventual collapse. $\endgroup$ – Cort Ammon Jun 9 at 15:36
  • $\begingroup$ I'm not sure I understand the contradiction. The Helmholtz free energy predicts that you should have an open configuration above a critical temperature. The Boltzmann distribution predicts you should have an open configuration as $T$ gets large. Both say that open configurations should happen at high temperatures, no? Can you explain where you think the results disagree? $\endgroup$ – Jahan Claes Jun 9 at 15:38
  • $\begingroup$ @Cort Ammon. Right cinematics certainly play a role when doing a real life experiment. However, here we make statements about the thermodynamic equilibrium. At no point are we concerned how fast this equilibrium will be achieved.. $\endgroup$ – Lenus Stueli Jun 9 at 15:39
  • $\begingroup$ @Jahan Claes. I do not understand why not ALL polymers would collapse, if the process of collapsing clearly lowers the Helmholtz free energy of the overall system. The two methods both predict that the open configuration is favoured at high temperatures, however in my mind they disagree on the exact distribution between open and closed states. $\endgroup$ – Lenus Stueli Jun 9 at 15:42
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    $\begingroup$ OK, I think I see two points of confusion. 1. The closed form has a lower Helmholtz function below a threshold temperature, not (as you just said) above. This is because the closed form has a lower energy, which dominates the entropy term at low $T$. This is the collapsed form. 2. Above the threshold temperature, you see predominantly the expanded form. At very high $T$, the different energies become almost unimportant, and the situation simply reflects the number of available states. This is dominated by entropy. $\endgroup$ – user197851 Jun 9 at 17:44
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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).

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  • $\begingroup$ Thank you for your elaborative answer! The last few sentences of your second paragraph made it clear that the remaining 1/5 of the molecules indeed should not collapse. My confusion comes from the change in Helmholtz free energy upon collapse that Dill and Blundell use (p. 136): $$\Delta F_{collapse}=F_{closed}-F_{open}=-\epsilon +k_{B}T \ln{4}$$ Which somehow suggests that this collapse would still be favourable. I suppose this arises due to them considering the collapsed and closed conformations as two distinct systems? $\endgroup$ – Lenus Stueli Jun 9 at 18:58

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