# Free Energy Landscape - Construction and meaning?

I struggle to understand the concept of free-energy landscape.

It seems to me the concept makes perfect sense for energies, but not for (canonical) free energies.

In my actual, hopefully to be bettered, understanding, a free energy landscape is built by fixing certain internal coordinates and computing the free energy under such constraints.

One should then be able to get information on the system behavior (for example, locating the equilibrium minima).

Let me clarify my doubts with a simple example: a particle in a harmonic well. There are two coordinates of interest: position $x$ (with related potential energy via a constant $\alpha$) and a fictitious internal coordinate $s$ (just for explanatory purposes, let us say the size of the particle): further we assume there is an energy related to the size, also quadratic w.r.t to the latter, via a constant $\gamma$.

For simplicity I completely neglect kinetic energy. So, the partition function for such a system would read

$$Z = \int_0^{\infty} \int_{-\infty}^{\infty} \exp{[-\beta({\gamma s^2 + \alpha x^2})]} \mathrm{d}x\mathrm{d}s = \frac{1}{2} \sqrt {\frac{\pi}{\beta \alpha}} \sqrt{\frac{\pi}{\beta \gamma}}$$

From which the free energy at equilibrium follows. At equilibrium, both mean $x$ and $s$ will be different from zero.

Now I could be tempted to describe the free energy landscape by fixing the size $s$, and calculate the constrained free energy.

The “constrained” partition function reads $$\bar{Z} (s) = \exp^{-\beta \gamma s^2} \sqrt {\frac{\pi}{\beta \alpha}}$$

The free energy landscape with respect to $s$ is a decreasing function of $s$, so one would conclude the system will try to minimize the size $s$ (to minimise its free energy), which is not correct.

I must be making quite some confusion between variables that can be constrained and variables for which constraining goes exactly against the meaning of canonical ensemble. Could somebody please help me to understand? Thanks in advance

LATER EDIT As a matter of fact, the more I think about I start to think the overall picture could be saved. Indeed the probability of having a size $s$ is proportional to $\exp{-\beta \gamma s^2}$ but it is possibly incorrect to state this will result in the system selecting the smallest possible size $s$: indeed is the expected value that should matters. Any further hint would be most welcome.

• I think you picked a poor example. Free energy landscapes are good for describing conformations of macromolecules (e.g., protein folding). Feb 16, 2018 at 12:19
• I though I would pick an example in the simplest setting, just to expose "the flesh" of the matter. Feb 16, 2018 at 12:38
• @Smerdjakov "At equilibrium, both mean $x$ and $s$ will be different from zero"... It seems mean $x$ will be zero from the symmetry of the model Feb 16, 2018 at 15:28
• @AlekseyDruggist, thanks for your comment. As a matter of fact you mention an aspect I might well completely misunderstand. For an oscillator in contact with thermal bath: indeed the symmetry of the potential dictates that $x=0$ is the equilibrium. Yet the average energy will be different from $0$, so does it make sense to talk about the equilibrium being $0$? Feb 20, 2018 at 16:40
• @Smerdjakov $\langle x^2\rangle \neq \langle x\rangle^2$. With your model, the average position (and size) is 0, even though their fluctuation is not. To have non-zero equilibrium size, you should use $(s-S)^2$, with $S=\langle s\rangle$.
Feb 22, 2018 at 7:52

Here is how to understand the minimization of free energy. In the canonical ensemble $$(\beta,V,N)$$ there are $$\Omega(E,V,N)$$ microstates with energy $$E$$ and each of them has probability $$e^{-\beta E}/Q(\beta,V,N)$$. The probability to find a microstate with energy $$E$$ is $$p(E) = \Omega(E,V,N) \frac{e^{-\beta E}}{Q(\beta,V,N)}$$ Here, $$\Omega$$ is the microcanonical partition function and $$Q$$ is the canonical partition function.

The equilibrium state is the most probable state –all other states are fluctuations from equilibrium. Increasing $$E$$ increases $$\Omega$$ but decreases $$e^{-\beta E}$$, which means that the most probable state is not the lowest or largest energy, but somewhere in between . We obtain the exact value by maximizing $$\ln p$$:

$$\max_E\Big\{\ln p\Big\} = \max_E\Big\{\ln\Omega -\beta E - \ln Q\Big\}$$

But $$\ln\Omega = S/k$$, therefore the maximization of $$\ln p$$ is equivalent to

$$\max_E\Big\{\ln p\Big\} = \max_E\Big\{\frac{S}{k} -\beta E - \ln Q\Big\} = \max_E\Big\{\frac{T S-E}{kT} - \ln Q\Big\} = \max_E\Big\{\frac{-A}{kT} - \ln Q\Big\} = \min_E\Big\{\frac{A}{kT} + \ln Q\Big\}$$ and finally $$\boxed{ \max_E\Big\{\ln p\Big\} = \min_E\Big\{\frac{A}{kT} + \ln Q\Big\} }$$

Therefore, the most probable state is the state that maximizes the free energy $$A$$ ($$Q$$ is constant in this maximization).

The most probable state is overwhelmingly more probable than all others. This means that $$p\to 1$$, and the abaove result becomes $$\boxed{ \ln Q \to -\frac{A}{kT} }$$ which is another important result and relates the partition function to the free energy at equilibrium.

The bottom line is that the system does not go to the minimum energy $$E$$, as one might assume because the canonical probability increases as $$e^{-\beta E}$$. We must also take into consideration the number of microstates with energy –that's entropy. Indeed, free energy contains the opposing effects of energy and entropy.

• I understand what you say, I see that the energy is not minimised in the canonical ensemble, but I fail to see how your explanation answers my post. My issue is that if I consider the whole system I get equilibrium for $x,s \neq 0$, while I get a different conclusion if I consider the constrained free energy, was to $s$. Sep 9, 2022 at 6:02
• @Smerdjakov The free energy is not minimized with respect to any internal coordinate. The rule is this: minimize $A(T,V,N)$ with respect to different partitions of $V$ and $N$ while keeping $T$ constant. In general, minimize a potential function ($A(T,V,N)$, $G(T,P,N)$,$U(S,V,N$, $H(S,P,N)$ with respect to partitioning the extensive variables while keeping all intensive variables constant. Sep 9, 2022 at 14:19