# Entropy in the Klimontovich Picture of Kinetic Theory

In the Klimontovich picture of kinetic theory, a classical $$N$$-body system is fully specified by the phase-space distribution $$\rho(\vec X,t) = \sum_{i=1}^N\delta[\vec X-\vec X_i(t)] ,$$ where $$\vec X_i=(\vec q_i,\vec p_i)$$ is the phase-space coordinate of particle $$i$$, and $$\delta$$ is the Dirac delta distribution. This distribution can be accurately approximated by certain computer simulation methods, e.g., molecular dynamics. Suppose I wanted to compute the entropy of a particular configuration obtained by such a simulation. My natural inclincation is to compute $$S = -k_B\left<\ln\rho\right>$$ with the angle brackets denoting $$\left = \frac{\int A(\vec X) \rho\, d\vec X}{\int \rho \,d\vec X} = \frac{1}{N}\sum_{i=1}^N A(\vec X_i) ,$$ i.e., just the sample mean.

For most observables, such a computation makes perfectly good sense and is simple to do given the output of an $$N$$-body simulation. But naively going through the same motions with $$A=\ln\rho$$ as the observable leads to strange attempts to take the logarithm of Dirac deltas. On paper, I have trouble making sense of this, much less the matter of computing it from simulation data.

My experience has taught me these kinds of problems are a sign that I have missed something at the big-picture scale. Is the Klimontovich distribution incompatible with the usual notion of entropy in statistical mechanics? If so, what is the disconnect? Essentially, does it make sense to compute the entropy of a phase-space configuration in isolation?

• What if the particles have finite size, does this take care of the problem? – Maxim Umansky Dec 17 '18 at 19:58
• I don't believe so. The deltas represent that particles have sure (non-random) coordinates and momenta, not that they are point particles. That is, even a system of hard spheres may be characterized by the motion of each particles' center. – Endulum Dec 18 '18 at 16:11