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Jun 10, 2019 at 11:54 vote accept Confinement
Jun 10, 2019 at 12:44
Jun 9, 2019 at 19:01 vote accept Confinement
Jun 10, 2019 at 9:53
Jun 9, 2019 at 18:28 answer added user197851 timeline score: 0
Jun 9, 2019 at 17:50 comment added Confinement @LonelyProf. (To your first point: Ah yes, of course the collapse occurs below the theshold temperature and not above!) To your second point: But why exactly do the other 1/5 of the polymers at very high temperatures not also collapse if this would still lower the overall Helmholtz free energy? Even if their favoured only by a small margin, this shouldn't this eventually be the equilibrium configuration?
Jun 9, 2019 at 17:44 comment added user197851 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.
Jun 9, 2019 at 17:35 comment added Confinement @LonelyProf. Thanks for pointing out my mistake! My question is: If the closed form of the polymers if favoured in terms of having a lower Helmholtz free energy above a certain threshold temperature, why do not ALL polymers collapse? When considering the Boltzmann distribution one finds that only 4/5 of the polymers seem to collapse. However, since the thermodynamic equilibrium under the conditions of constant temperature and volume corresponds to a minimum of F, I do not understand why this is the case.
Jun 9, 2019 at 17:31 history edited Confinement CC BY-SA 4.0
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Jun 9, 2019 at 17:29 comment added user197851 Your question contains some incorrect statements (you say the open configuration has lower entropy and the closed configuration has higher entropy, when the opposite is true) and does not make clear what the precise problem is. Polymer collapse (on lowering temperature or solvent quality) is a quite general phenomenon. Can you clarify exactly what you are not understanding?
Jun 9, 2019 at 16:53 history edited Confinement
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Jun 9, 2019 at 15:42 comment added Confinement @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.
Jun 9, 2019 at 15:39 comment added Confinement @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..
Jun 9, 2019 at 15:38 comment added Jahan Claes 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?
Jun 9, 2019 at 15:36 comment added Cort Ammon 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.
Jun 9, 2019 at 15:29 history edited Confinement CC BY-SA 4.0
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Jun 9, 2019 at 15:25 review First posts
Jun 9, 2019 at 15:28
Jun 9, 2019 at 15:21 history asked Confinement CC BY-SA 4.0