# What is the potential of mean force?

I've come across the term potential of mean force (PMF) in polymer physics, colloidal physics etc., but have not come across a complete definition.

As far as I understand, the PMF determines the equilibrium distribution of some degree of freedom of interest (call it x) in a (usually large) system with all other degrees of freedom that we don't care about averaged out. In the textbook "Structured Fluids" by Witten and Pincus, they derive that in the canonical ensemble, the work done by the full system (with all degrees of freedom) to change x can be directly calculated from the PMF of x. Therefore, the PMF is a form of free energy.

I am interested in a more complete, statistical mechanics description of the PMF. Can it also be defined in the same way for other ensembles (e.g. grand canonical)? Can it be connected to the partition function in some way? Free energies typically correspond to some conditions of equilibrium (e.g. for Helmholtz these are constant T and P), are there similar conditions for the PMF?

Explanations or pointers to resources would both be great.

## 2 Answers

I'll start by assuming a canonical ensemble, but the ideas apply equally well to other ensembles. Suppose you can measure the probability distribution function of a degree of freedom, i.e. a coordinate $$x$$. Call this $$\mathcal{P}(x)$$. Formally it is defined as $$\mathcal{P}(x) = \left\langle \delta[x-x(\mathbf{q})] \right\rangle$$ Here, $$x$$ is a particular value of the coordinate; $$x(\mathbf{q})$$ is the coordinate expressed as a function of all the instantaneous particle coordinates $$\mathbf{q}$$ in the system. The angle brackets represent an ensemble average (which involves integrating over all the $$\mathbf{q}$$ coordinates), including a weight function such as a Boltzmann factor.

The thing being averaged is a Dirac delta function: just think of this as a histogram of the frequency of occurrence of values of $$x$$ that occur in the statistical ensemble, resolved into very narrow bins.

If we write down this average in (say) the canonical ensemble it looks like this: $$\mathcal{P}(x) = \frac{\int d\mathbf{q} \, \delta[x-x(\mathbf{q})] \exp(-E(\mathbf{q})/k_BT)}{\int d\mathbf{q} \, \exp(-E(\mathbf{q})/k_BT)} \equiv \frac{\mathcal{Q}(x)}{Q}$$ Here, $$Q$$ is the partition function; $$\mathcal{Q}(x)$$ can also be thought of as a partition function, for a system which is constrained to lie on a surface defined by the equation $$x(\mathbf{q})=x$$.

This is the link with free energy. From the general formula $$F=-k_BT\ln Q$$, and a similar equation for $$\mathcal{Q}(x)$$, we can define a quantity which is often called the "Landau free energy": $$\mathcal{F}(x) = F-k_BT\ln\mathcal{P}(x)$$ Since $$F$$ is just a constant here, and the variation of $$\mathcal{F}(x)$$ with $$x$$ is of most interest, the $$F$$ term is sometimes omitted from this equation.

Apart from the restriction to a given value of $$x$$, the usual conditions of equilibrium apply. The coordinate $$x$$ might be the total magnetisation of an Ising system, for instance, or a chemical reaction coordinate. For a simple liquid, the pair distribution function can be interpreted as $$\mathcal{P}(x)$$, and the log of this is sometimes called the potential of mean force. Often one can map out a free energy barrier as a function of $$x$$ between two stable, or metastable, states of the system. This type of function is often of interest; for instance, the classical theory of nucleation involves a free energy calculated as a function of the radius of the growing nucleus.

Some additional manipulations can be used to show that the quantity $$-d\mathcal{F}(x)/dx$$ is equal to the ensemble average of the mechanical force $$\langle -d E/dx\rangle_x$$, where the subscript indicates that the ensemble average is evaluated at the chosen value of $$x$$. This gives rise to the name "potential of mean force".

Hope this will do as a general introduction. The potential of mean force is covered in some statistical mechanics books. It is mentioned in Introduction to Modern Statistical Mechanics by D Chandler, and in Statistical Mechanics: Theory and Molecular Simulation by M Tuckerman.

LonelyProf gave a nice general description of the potential of mean force, but to get a basic feel for what it "is", I think it's helpful to look at the simple case of a system of $$N$$ point particles in thermal equilibrium with temperature $$T$$. That is, we'll just look at the canonical ensemble right now. The probability distribution of the system is $$f(\vec r_1,\vec p_1,\dots,\vec r_N,\vec p_N) \propto e^{-H/kT} = e^{-p_1^2/2mkT}\dots e^{-p_N^2/2mkT} e^{-U/kT} ,$$ where $$U = U(\vec r_1,\dots,\vec r_N)$$ is the total ($$N$$-body) potential energy. Let's now think about the force on any given particle, say particle 1, which is just $$-\nabla_1U$$. We can compute the mean value of this force by averaging over all the other particles: \begin{align} \overline{-\nabla_1U} &= \frac{ \int d\vec r_2 d\vec p_2\dots d\vec r_N d\vec p_N (-\nabla_1 U) f }{ \int d\vec r_2 d\vec p_2\dots d\vec r_N d\vec p_N f } \\ &= \frac{ \int d\vec r_2\dots d\vec r_N (-\nabla_1 U) e^{-U/kT} }{ \int d\vec r_2 \dots d\vec r_N e^{-U/kT} } \\ &= \frac{ -kT \nabla_1 \int d\vec r_2\dots d\vec r_N e^{-U/kT} }{ \int d\vec r_2 \dots d\vec r_N e^{-U/kT} } \end{align} Now, integrals appearing in the numerator and denominator are just proportional to probability that a particle lies at position $$\vec r_1$$. If we call that distribution, $$g_1(\vec r_1)$$, we can write $$\overline{-\nabla_1U} = -kT\nabla_1 \ln g_1(\vec r_1)$$ So that the potential corresponding to this mean force is just $$\phi_1(\vec r) = -kT\ln g_1(\vec r)$$. Nothing fancy here.

However, nothing stops us from holding other particles' positions fixed as well. Let's not integrate over particle 2's position. Then we'll find $$\overline{-\nabla_1U} = -kT\nabla_1\ln g_2(\vec r_1,\vec r_2)$$ where $$g_2(\vec r_1,\vec r_2)$$ is the pair distribution function, the probability of finding one particle at position $$\vec r_1$$ and another at position $$\vec r_2$$. Then we can readily imagine defining a two-body potential of mean force $$\phi_2(\vec r,\vec r') = -kT\ln g_2(\vec r,\vec r')$$. This one is a bit more subtle to interpret compared to the one-body potential. We are still considering the force on one particle only, but there is second spectator particle present as well. This means that after averaging over the other $$N-2$$ particles, $$\phi_2(\vec r,\vec r')$$ "looks like" an effective interaction potential between a particle at $$\vec r$$ and a particle at $$\vec r'$$. There's of course no reason to stop at two particles. It is probably clear how one can define a three-body potential of mean force, which will have the expected relationship to the three-body distribution function. And so on for higher orders.

For a brief textbook reference along these same lines, I recommend Terrell L. Hill's "An Introduction to Statistical Thermodynamics," wherein the potential of mean force is introduced on pp. 313 and generalized to the grand canonical ensemble on pp 314.