In statistical mechanics, for a system of $N$ particles $x_1, \ldots, x_N$ in three dimensions, the Gibbs free energy is defined in terms of the Hamiltonian $H$ as $$ G = -k_\mathrm B T \log \int_{\mathbb{R}^{3N}} \mathrm e^{-H(x_1, \dots, x_N)}\,\mathrm dx_1 \dots \mathrm dx_N.$$ People talk about the "free energy difference between conformations". For instance, for some region $A \subset \mathbb{R}^N$, the particles could be in a certain conformation, and for some other region $B \subset \mathbb{R}^N$, the particles could be in another conformation.


In this paper, equation (3), what is the (implied) definition of $F_A$ or $F_B$ in terms of equation (2)?

Equivalent Question

Is the free energy difference between the two conformations defined as $$G_B - G_A = -k_\mathrm B T\log \frac{\int_B \mathrm e^{-H(x)}\,\mathrm dx}{\int_A \mathrm e^{-H(x)}\,\mathrm dx}?$$

Auxiliary details

The above definition made sense to me, until I was told that "it does not make sense to talk about the free energy of a conformation". An alternative I see is to think of two Hamiltonians $H_A$ and $H_B$, and instead define the free energy difference as

$$G_B - G_A = -k_\mathrm B T \log \frac{\int_{\mathbb{R}^{3N}} \mathrm e^{-H_B(x)}\,\mathrm dx}{\int_{\mathbb{R}^{3N}}\mathrm e^{-H_A(x)}\,\mathrm dx}. $$

Are either/both/neither of these definitions correct?

I apologize if the question is exceedingly obvious. I am unable to find a straightforward answer by Googling. Any references are welcome.


Your question implies that you define the Gibbs free energy as $G=-RT\ln(Z)$ where $Z$ is the partition function, in which case there will be a difference in $G$ only if there is a difference in the partition function for your two configurations, i.e differences in energy levels and/or in the population in these levels. That will depend entirely on your system (and also the temperature).

  • $\begingroup$ I'm not sure I understand. In the original post, the boxed-in question assumes a common temperature $T$ and Hamiltonian $H$, which together determine all energy levels across $\mathbb{R}^{3N}$. In particular, $H$ is common to both configuration regions $A$ and $B$. The restatement in terms of a partition function is a mathematical tautology for me, but I understand there may be physical intuition behind it (which I do not have). The question is whether it makes sense or is common to define the "partial partition function" $Z_A = \int_A e^{-H(x)} dx$, and if this is what is implicitly done. $\endgroup$
    – Anne Onyme
    Mar 7 '19 at 15:09
  • $\begingroup$ The partition function is normally described as $\sum_i g_i \exp(-E_i/k_BT)$ for energy levels $E_i$ with degeneracy $g_i$. In your case you use the Hamiltonian (energy operator) to give the energy levels, and presumably the sum can be replaced by an integral without error as the energy levels must be closely spaced, as in the translational $Z$; (See Sakur-Tetrode eqn for example). Unless your $H$ and hence $E_i$ are different then there can be no change in $G$. ps thanks for edit. $\endgroup$
    – porphyrin
    Mar 7 '19 at 15:35
  • $\begingroup$ I am not accepting your answer because it does not address the question of free energy difference between conformational states and you have not addressed whether it makes sense to define $Z_A$. The passage from an integral to a sum is mathematically inconsequential. $\endgroup$
    – Anne Onyme
    Mar 7 '19 at 15:43
  • $\begingroup$ I have added a reference in the OP to make the question foolproof. What I want is an expression for $F_A$ or $F_B$. $\endgroup$
    – Anne Onyme
    Mar 7 '19 at 17:01
  • $\begingroup$ The direct counting in 3.1 in the paper gives an answer, but it boils down to knowing what the energy levels are and if they have different populations, which I pointed out in my answer. If you don't know that you are stuck no matter how fancy the model is that you use. $\endgroup$
    – porphyrin
    Mar 7 '19 at 18:29

I'd say that either way is correct as long as you are using the same potential function for the Hamiltonian in both cases. If you aren't, then you would need to indicate distinct Hamiltonians, but then the free energy difference that is calculated is pretty meaningless. Given that, my preference would be to write it the way you originally did, with the same Hamiltonian integrated over two conformations (or ensembles of conformations).

I don't have a reference on-hand to dispute the assertion that "it does not make sense to talk about the free energy of a conformation", but I'll make an argument against it. First, I don't think it's controversial to say that it is possible to determine an equilibrium constant for interconversion of two conformations and that the interconversion can be treated thermodynamically like any other transition between two states. The distribution between the two states is determined by the difference in free energy.

We can also use the paper you cited as an example. The authors look at an ensemble of states of a given molecule and divide it into two sub-ensembles, one of "bound" molecules and one of "unbound" molecules. They then calculate an ensemble average free energy for each state. There is no reason why the same procedure could not be used to divide a large ensemble into ensembles which each represent a different conformation (even if your definition of conformation is so restrictive that each individual molecule is a unique conformation) and thereby to calculate a difference in free energy between two conformations. That one can calculate a difference in free energy between two conformations suggests quite strongly (to me at least) that there is such a thing as "the free energy of a conformation."


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.