Sunday, September 9, 2012

Microstates, macrostates, and the Boltzmann distribution

This is Figure 10.1 from Dill and Bromberg's Molecular Driving Forces, which is my favorite book on statistical mechanics. It is a beautiful example from a beautiful book.

The Boltzmann distribution says that, at equilibrium, the probability of being in a microstate (also sometimes referred to as an energy state) is $$p_i=\frac{e^{-\varepsilon_i/kT}}{q}\text{ where }q=\sum_i^{\begin{array}{ c }\text{micro} \\
\text{states} \\
\end{array}} e^{-\varepsilon_i/kT}$$ Let's apply this formula to the simple four-bead polymer example shown in the figure above. Each of the five conformations represents a microstate.  There is one low-energy conformation with a bead-bead contact and four higher-energy conformations where the bead-bead contact is lost.  Because higher energy conformations have a common feature (the lack of bead-bead contact) and the same energies we group them into one common macrostate (a state comprised of more than one microstate).  I'll refer to this macrostate at the "unfolded" state and the to ground state the "folded" state.

According to the Boltzmann distribution  the probability of the folded state is $$p_f=\frac{1}{q}\text{ where }q=1+4e^{-\varepsilon_0/kT}$$ The probability is the unfolded state is the sum of the (equal) probabilities of being in one of the four unfolded microstates $$p_{uf}=p_{uf1}+p_{uf2}+p_{uf3}+p_{uf4}=\frac{4e^{-\varepsilon_0/kT}}{q}$$  
This figure shows a plot of the probabilities of the folded and unfolded state, together with the probability of being in one of the four unfolded microstates, as a function of temperature and for $N_A\varepsilon_0=2.5$ kJ/mol.  The polymer can be said to unfold at 217 Kelvin where the unfolded state starts to become more probable than the folded state. 

Why does the polymer unfold beyond 217 K?
Answer 1. The plot shows that if there is only one unfolded microstate then the polymer would never unfold (i.e. unfolded state would never be more probable than the folded state).  Therefore, the polymer unfolds because there are more unfolded microstates than folded microstates (4 compared to 1).

Answer 2. The number of microstates with the same energy is known as the degeneracy, so one can also say that the polymer unfolds because the unfolded state has a higher degeneracy than the folded state (4 compared to 1).

Answer 3.  The ratio of probabilities can be written as $$\frac{p_{uf}}{p_f}=4e^{-\varepsilon_0/kT}=e^{-( \varepsilon_0-kT\ln(4))/kT}$$The latter form clearly shows that the unfolded state will become more probable for temperatures were $T(k\ln(4))\gt\varepsilon_0$.  The similarly of $k\ln(4)$ to the entropy $S=k\ln(W)$ is no accident.  The entropy of the unfolded state is given by $$S_{uf}=k\ln\left(\frac{N_{uf}!}{N_{uf1}!N_{uf2}!N_{uf3}!N_{uf4}!}\right)$$where $N_{uf1}=N_{uf2}=N_{uf3}=N_{uf4}=\frac{1}{4}N_{uf}$ is the number of polymers in each of the unfolded microstates.  For large $N_{uf}$, where Stirling's approximation [$x!\approx(x/e)^x$] holds, it is easy to show that this expression reduces to $$S_{uf}=N_{uf}k\ln(4)$$So one can also say that the polymer unfolds because the unfolded state has a higher entropy than the folded state.

A simulation of the unfolding of a slightly longer polymer can be found here

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This work (except the first figure which is © by Garland Science) is licensed under a Creative Commons Attribution 3.0 Unported License.  

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