# Kullback-Leibler divergence as a measure of irreversibilty?

I watched this recent KITP webinar on Nonequilibrium thermodynamics for active matter yesterday. I saw that KLD(Kullback-Leibler divergence) is used as a measure to quantify irreversibility in the active matter systems like biological ones. KLD is defined in our case as below:

$$D(P_f||Pr) = \int dp dq P_f(q,p;t) \ln \frac{P_f(q,p;t)}{P_r(q,-p;t)}$$

where $$q$$ stands for positions, $$p$$ stands for momenta, $$P_f$$ is the pdf of the forward trajectory, $$P_r$$ is the pdf of reverse trajectory and $$t$$ stands for the snapshot where the ensemble of trajectories considered to compute distributions. I am yet to digest these concepts properly. I tried imagining in the {coordinate, momenta} space to get a qualitative intuition.

My queries are below:

• For a reversible process that follows detailed balance, my intuition says that KLD measure is zero. Is this correct?

• Is the reverse trajectory a reflected one of forward trajectory? If that is the case why would both forward and reverse PDFs differ? I am certainly missing something.

• Is this KLD a time dependent quantity, or should the time dependence disappear after the integration over the momenta and coordinates? The first option seems more likely. – Roger Vadim May 9 '20 at 17:41
• @Vadim : I have to think about these. Not sure. – dexterdev May 10 '20 at 3:38

The reverse trajectory is a trajectory obtained by reversing the direction of time, which here means simply reversing the directions of all the momenta (more precisely, reversing the directions of the initial momenta and following the trajectory).

I do not have a complete answer to the first question, but two points are to consider:

• KL divergence is also known as relative entropy (in fact, its first term is just the negative entropy): $$-\int dpdq P_f(p,q,t)\log P_f(p,q,t)$$ Moreover, KL divergence is non-negative and takes zero value only when the two distributions are identical.
• The trajectory in question is the phase trajectory for the whole system, and under usual thermodynamic assumptions the probability density of this trajectory is constant (Liouville theorem): $$P_f = P_r = const$$. Which means that in tehrmodynamic equilibrium KL divergence is zero.
• Thanks for the insights. Okay I understand that if 2 distributions are identical KLD=0. When the distributions are constant you mean the NVE ensemble, right? What about thinking in terms of NVT ensemble? Even then the forward and reverse trajectories will be identical , right? Until now I was talking about equilibrium cases. Now think about non-equilibrium active systems, where the system is subject to some external flux. In this case certainly irreversibility is an outcome. But how can I think in terms of KLD? Even here reversed trajectory will have same distribution right? – dexterdev May 9 '20 at 16:38
• In a canonical ensemble I would think of it as Brownian motion - i.e. the probability distribution is the solution of the Fokker-Planck equation. Then, in equilibrium both distributions are stationary, while in non-equilibrium they depend on the initial conditions (i.e. the direction of the momenta) - hence my question about the time dependence. – Roger Vadim May 10 '20 at 6:10
• I think KLD is time-dependent. When distributions are evolving in time (due to the change in Potential function due to the external field or whatever reasons in non-equilibrium case), KLD will change with time I guess. – dexterdev May 12 '20 at 4:21

There are many different formulations of the detailed flutuation theorem. The one you are discussing, I believe, was introduced in

1. "Is the reverse trajectory a reflected one of forward trajectory? If that is the case why would both forward and reverse PDFs differ? I am certainly missing something." Under the forward process, the system is started from a Boltzmann distribution corresponding to the initial parameters $$\lambda(0)$$, and allowed to evolve in a Hamiltonian fashion under the influence of the driving dynamics $$\lambda(t)$$ over $$t\in[0,1]$$. Under the backward process, the system is started from a Boltzmann distribution corresponding to the final parameters $$\lambda(1)$$, and evolves under the time-reversed protocol $$\lambda(1-t)$$. Note that, in general, Boltzmann distribution corresponding to $$\lambda(1)$$ will not be equal to the final distribution of the forward process.
2. "Is the reverse trajectory a reflected one of forward trajectory? If that is the case why would both forward and reverse PDFs differ?" They are only the same if the system remains in equilibrium, so that the final distribution under the forward protocol, $$p_f(q,p,1)$$, is the same as the initial distribution under the backward protocol (the Boltzmann distribution for $$\lambda(1)$$).