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I am confused about the concepts of micro-reversibility and entropy production and how they go together. In the micro-canonical ensemble (constant total energy), the entropy becomes maximal at equilibrium and to my understanding increases on the way towards equilibrium via irreversible processes. Let's say we consider now a box of classical molecules and run a molecular dynamics simulation in the micro-canonical ensemble, i.e. we are solving the Newton equations which are time-reversible (micro-reversibility). How can entropy increase so that we move the system towards equilibrium if the equations are reversible?

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This is possible because even though the number of microstates increasing entropy is the same as the number of microstates decreasing entropy, the latter are very improbable to observe and extremely hard to prepare in real world (we can prepare them in computer simulation by reverting velocities of molecules at appropriate point of time but this is not possible in reality).

Entropy is a function of some kind of "detailed macrostate" $D$ - a set of numbers we agree to use to distinguish apparently different macroscopic states of the system, including not only the equilibrium thermodynamic state but also many non-equilibrium states. For example, $D$ could be understood as a list of pairs $[(E_i,N_i)]$ where $E_i$ is energy in volume cell $i$, and $N_i$ is number of particles in volume cell $i$ (all cells having the same volume $\Delta V$).

Let's assume we observe in repeated experiments that the macrostate $D$ evolves into macrostate $D'$. We will show, in line with Jaynes' argument, that if the system obeys Hamiltonian mechanics, then due to the way entropy is defined and its relation to phase space weight distribution of ensembles, entropy of $D'$ can't be lower than entropy of $D$. The basic idea is that if macrostate $D$ reliably turns into macrostate $D'$ in laboratory, then Hamiltonian evolution of microstates from an ensemble most probable for $D$ must bring them to an ensemble compatible with macrostate $D'$, but not necessarily the most probable one for $D'$; and thus entropy of $D'$ is likely higher.

Let's assume by entropy of macrostate $D$ we mean the Gibbs entropy: $$ S(D) = \sum_k -w^*_{D,k} \ln w^*_{D,k}\tag{**} $$ where for each phase space cell $k$, $w^*_{D,k}$ is the ensemble weight of microstates that are in the phase space cell $k$. How do we find these weights from $D$? There are many different weight distributions $\{w_k\}$ that would be compatible with $D$, but due to laws of probability, the most probable one, and very sharply (other ones are negligibly probable) is that which maximizes the information entropy functional $$ I = \sum_k -w_k \ln w_k $$ and also satisfies the constraints of $D$.

So the most probable weight distribution of macrostate $D$ in the phase space is such that no other distribution compatible with $D$ can have greater information entropy. We denote it $w^*_{D,k}$. Similar statement is true for the most probable weight distribution of macrostate $D'$ and we denote it $w^*_{D',k}$.

Now imagine we let our ensemble of microstates, initially compatible with $D$, evolve in time according to the system Hamiltonian. Our ensemble weight distribution will evolve from $w^*_{D,k}$ to another distribution $w_{k}'$. When the ensemble reaches a state consistent with $D'$, the evolved weight distribution does not have to be equal to $w^*_{D',k}$; due to the Hamiltonian evolution, the phase density initially given by $w^*_{D,k}$ will be mixing into new regions of phase space in a complicated way, resulting often in a much different weight distribution than what the most probable distribution for $D'$ is (Hamiltonian evolution can't erase the initial condition from the weight distribution). Due to Liouville's theorem, value of the information entropy for this evolved weight distribution $w'_k$ is the same as value of information entropy for the initial distribution $w^*_{D,k}$. Which means it cannot be the maximum possible information entropy for the macrostate $D'$; the maximum is reached only by the most probable distribution for $D'$ which is smooth and different from $w'_k$. So we see that Hamiltonian evolution of weight distribution, preserving information entropy, requires that entropy of the macrostate $D'$ that reliably occurs after $D$, cannot be lower; it is either the same or higher.

This shows that entropy non-decrease when going from $D$ to $D'$ is really due to the assumption that the macrostate $D'$ is a reliable successor of the macrostate $D$. This makes the microstates in $D'$ evolved from $D$ very probable and the corresponding reverted microstates, although their amount is the same, very improbable. This also means that if entropy was observed to decrease in laboratory (macrostate $D$ evolving into macrostate $D''$ with lower entropy), it could not be a consistent phenomenon, only a rare fluctuation. If it somehow happened consistently, either the above argument use of the probability laws is incorrect, or the argument would imply the system does not obey Hamiltonian mechanics (ensemble evolution would not conserve information entropy).

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  • $\begingroup$ Quite complicated for me to understand, but it seems like the right explanation. What would that mean for us when we solve the equations in inverse time direction so from D' to D. In that case would the entropy of D be higher than D'? So by deciding on a time-direction we are kind of giving in some information that reduces entropy (some kind of non-equilibrium bias) and this entropy must be recovered (information lost) during the equilibration process? $\endgroup$
    – Guiste
    May 17 '21 at 17:35
  • $\begingroup$ Entropy is a function of macrostate, and for macrostates $D,D'$ as defined above ($D$ consistently evolves into macrostate $D'$), we have $S(D) <= S(D')$ regardless of what we do with equations for microstates and their solutions. "Solving in inverse time direction from $D'$ to $D$" is very hard to do if we only know the macrostate $D'$. Only very small part of initial conditions consistent with $D'$ will evolve into state consistent with $D$; these are entropy-decreasing trajectories. Most initial conditions will evolve into states consistent with even higher entropy than $D'$. $\endgroup$ May 17 '21 at 22:48
  • $\begingroup$ If we have microstate trajectory $\mu_{inc}(t)$ that brings $D$ to $D'$, we immediately know one microstate trajectory $\mu_{dec}(t)$ that brings $D'$ to $D$: we take the end microstate of $\mu_{inc}$ consistent with $D'$ and revert its velocities. The resulting trajectory will decrease entropy. However, without having specific $\mu_{inc}(t)$ first, it is difficult to find any such entropy-decreasing trajectory; most initial conditions consistent with $D'$ won't do such thing. $\endgroup$ May 17 '21 at 22:53

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