The CFT states that for a finite, classical system, coupled to a set of baths, each characterized by a constant intensive variable , for which the dynamics is stochastic, Markovian and microscopically reversible, and for which the entropy production is odd under time-reversal, the following identity holds:

$\frac{P_F(\omega)}{P_R(-\omega)} = e^{+\omega} $.

Here $\omega$ is the entropy production of the system over some finite time interval. $P_F(\omega)$ gives the probability distribution of this entropy production (because the system is stochastic $\omega$ will be different for each run). $P_R(-\omega)$ gives the distribution for the time-reversed process. What this typically means is that the control parameter $\lambda_F(\tau)$ for the driving force is run in reversed order, $\lambda_R(\tau) = \lambda_F(t-\tau)$, where $t$ is the length of the time interval.

This theorem is derived from the assumption of microscopic reversibility, and this is stated in the form

$\frac{\mathcal{P}[x(+t)|\lambda(+t)]}{\mathcal{P}[\bar{x}(-t)|\bar{\lambda}(-t)]} = \exp{(-\beta Q[x(+t)|\lambda(+t)])}$,

where $\mathcal{P}[x(+t)|\lambda(+t)]$ is the likelihood of seeing the microscopic trajectory $x(+t)$ (corresponding to e.g. positions and momenta of particles) if we apply the control parameter $\lambda(t)$, and the bar indicates the time-reversed processes. $Q$ is the heat produced in the process.

Now consider a process where the control parameter is not varied over time, i.e. $\lambda(\tau) = \lambda$. There seems to be no reason why the theorem should no longer hold. But then there should also be no heat transfer, $Q=0$ and no work done on the system. That would imply that the forward and reverse processes are equally likely. However, consider for instance the mixing of a gas, where initially two different types of particles are separated by a barrier at $t=0$. The barrier is released and the gas molecules are free to move in the whole container. The entropy of the system increases, and the process is clearly irreversible. Yet the Crooks Fluctuation Theorem would still predict equal probabilities of finding the forward and reverse processes.

Is there anyone else that finds this strange?

Crooks' original paper: https://arxiv.org/abs/cond-mat/9901352


2 Answers 2


At thermodynamic equilibrium the probability of observing any trajectory is the same as observing the time reversed trajectory. You are just as likely to see an unmixed gas mix (very uncommon initial state), as you are to observe a mixed gas unmix (very uncommon final state).

Also you have misstated microscopic reversibility. Once you write things out in terms of heat (instead of work), you need to condition on initial states. And the heat flow along a particular trajectory need not be zero for an unperturbed system. The probability of reverse and converse trajectories, given the initial states, are not the same. It is the probability of reverse and converse trajectories given that the initial states are sampled from equilibrium that are the equal.

The problem with the barrier example is that the system does not start in equilibrium. Each side is locally in equilibrium, but the global equilibrium would have the gases already mixed.

See http://threeplusone.com/GECthesis.pdf page 42.


I think that your scenario is invalid, since forward and backward process should both start at equilibrium. Quoting from the paper:

As a concrete example of a system for which the above theorem is valid, consider a classical gas confined in a cylinder by a movable piston. The walls of this cylin- der are diathermal so that the gas is in thermal contact with the surroundings, which therefore act as the con- stant temperature heat bath. The gas is initially in equi- librium with a fixed piston. The piston is then moved inwards at a uniform rate, compressing the gas to some new, smaller volume. In the corresponding time-reversed process, the gas starts in equilibrium at the final volume of the forward process, and is then expanded back to the original volume at the same rate that it was compressed by the forward process.

  • $\begingroup$ Yes, but this is the case in my scenario too. There is nothing driving the system out of equilibrium. Let's forget about the barrier. Suppose the gas is in equilibrium, then the state where the two types of particles are completely separated is allowed. Suppose that this is the initial state. Crooks' theorem predicts that any two microtrajectories are equally likely. Hence the probability of seeing the particles mix is the same as the reverse of unmixing. There is no work, heat exchange, entropy change or whatsoever. It's just that the initial state is unlikely in equilibrium. $\endgroup$ Aug 15, 2016 at 12:39
  • 1
    $\begingroup$ But when the two types are separated, you are not in equilibrium. Since in equilibrium the density of the gas is constant (or more generally $\rho(x) \sim e^{-\beta H(x)}$) $\endgroup$
    – user29918
    Aug 15, 2016 at 12:42
  • $\begingroup$ The equilibrium density is an averaged quantity. The most likely state of the system is indeed one where the particles are homogeneously distributed. However, it should be clear that from this state, without external input of work or heat the system can reach the separated state spontaneously. Hence even in equilibrium this state is allowed. Only in the long time average of the system the particles are homogeneously distributed. $\endgroup$ Aug 15, 2016 at 13:38
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    $\begingroup$ Every micro-state is possible in the canonical ensemble with the probability $P(x,p) \propto e^{-\beta H}$. You cannot "set" the micro-state. To test this theorem you should take many identical systems, with initial canonical distribution, then move your piston (or not). The backward process should also start at equlibrium (many systems, distributed in canonical equilibrium with the piston in the final position). Only then you can test probability distribution. $\endgroup$
    – user29918
    Aug 15, 2016 at 14:07
  • $\begingroup$ Let's assume the energy stays constant over the course of the evolution, so $P(x,p) $ is constant. It might not be experimentally feasible to control the microstate of the system, but suppose we could (in a simulation, or simply by observing for a very long time and selecting the one we want). Then Crooks' theorem would still predict that in a situation where you don't move the piston and everything stays in equilibrium, any trajectory is equally likely to its reverse. $\endgroup$ Aug 15, 2016 at 14:20

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