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What's the common term for PMF in statistical physics?

Is it plausible to use the chemists' PMF even in non-equilibrium systems that don't follow the canonical ensemble rules or Maxwell-Boltzmann distribution?

Is writing the following valid?

$$ \mathcal{P}(x) = \frac{\int d\mathbf{q} \, \delta[x-x(\mathbf{q})] \exp(-E(\mathbf{q})/k_BT)}{\int d\mathbf{q} \, \exp(-E(\mathbf{q})/k_BT)} \equiv \frac{\mathcal{Q}(x)}{Q} $$

Where $ \mathcal{Q}(x)$ is the partition function.

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Independently of being part of the physicists or chemists community, PMF has not a unique definition. Some forms of PMF have been introduced many decades ago in different contexts and I would advice to cross check what is the exact definition used in a given context.

Whatever is its definition, the potential of mean force embodies a statistical average which depends on the kind of macroscopic state (equilibrium or out of equilibrium) and even for equilibrium states, for finite size systems it depends on the chosen ensemble. Therefore, use of a PMF obtained in an equilibrium ensemble for non-equilibrium calculations is not a fully consistent procedure, although it could be seen as an approximation.

Finally, the expression for the probability density $\mathcal{P}(x)$ is perfectly valid. Its intuitive meaning is to sum the probability of all the states such that the function $x({\bf q})$ is equal to $x$.

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  • $\begingroup$ What about calling it the $ -KT log$ of some histogram over an order parameter as a generalized free energy? $\endgroup$ – 0x90 Apr 13 at 15:52
  • $\begingroup$ @0x90 Yes, it looks perfectly acceptable, provided you may find a suitable order parameter. $\endgroup$ – GiorgioP Apr 13 at 16:03
  • $\begingroup$ mathematically, it seems like the probability space isn’t well defined in a non-equilibrium system since we don’t sample the entire phase space. That’s why I find it a bit hand-wavy. $\endgroup$ – 0x90 Apr 13 at 16:57
  • $\begingroup$ @0x90 Well, that is a problem, but how important is strictly depending on the kind of non-equiiibrium. Just to mention three well separated cases, there is the slightly perturbed equilibrium (treatable within linear response theory), steady state non-equilibrium fluxes, completely chaotic systems. It is unlikely that the same approximation could be equally good in all the cases, although the problem is less related to the possibility of a well defined probability space. $\endgroup$ – GiorgioP Apr 13 at 19:25
  • $\begingroup$ Can't we use $e^ { -\Delta F / k T} = \overline{ e^{ -W/kT } }$ some how to justify this: $Q(x=x')∝exp⁡^{(-A(x = x' )/(K_B T))}$ where $A$ is "free energy"? $\endgroup$ – 0x90 May 15 at 22:00

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