# Diffusion properties of hard spheres in Monte Carlo simulation

In standard Monte Carlo simulations, say for hard sphere systems, how should one compute the mean-squared displacement of the spheres in order to extract dynamical properties such as the diffusion coefficient of the spheres? First problem that comes to mind is, how do we map our Monte Carlo steps to a real physical time, such that the resulting MSD and the extracted diffusion coefficient turn out to be physically meaningful and correct?

I'm just trying to learn how this can work out in standard Monte Carlo, though I've heard that to study such properties, Brownian dynamics or Kinetic Monte Carlo may be more suited, but that's another discussion.

• P.S. probably what you'd want for dynamics of hard spheres is something like an event-driven molecular dynamics simulation: introcs.cs.princeton.edu/java/assignments/collisions.html (This was just the first thing that showed up, to illustrate it - really you should see the literature, some of which is cited here: arxiv.org/abs/physics/0405089 or Frenkel and Smit's Understanding Molecular Simulation). – AJK Nov 20 '17 at 3:00

To my knowledge, conventional Monte Carlo simulations cannot tell you anything about the time-dependent properties of a system. The progression from state to state in, say, Metropolis Monte Carlo has nothing to do with the system's evolution in time.

Let's contrast with equilibrium molecular dynamics simulation. In MD, you actually evolve particles in time according to their microscopic equations of motion. In the course of evolving one time step, you have to ask dynamical questions like, "What is each particles' current velocity?" and "What is the net force acting on each particle?" In contrast, with each Monte Carlo step, you're singling out one particle, teleporting it and asking, "Did this move bring the system closer to or further from equilibrium?" Note my word choice here: "teleporting" in Monte Carlo versus "evolving" in MD.

These are two routes to the same goal: getting accurate estimates for equilibrium ensemble-averaged quantities like the pressure, pair distribution function, or free energy. In Monte Carlo, the averaging happens over several static configurations, each assumed statistically independent from one another. In MD, the averaging happens over a long time interval. If the system under study is ergodic, these two methods should give the same answers.

However, for time-dependent quantities, you have to care about how the system gets from point A to point B. In MD, the reason for making each move is unambiguous: we're just obeying Newton's second law (within some numerical accuracy). In Monte Carlo, you choose small moves based on convenience. No physical law prevents you from trying to totally reshuffle the particles every step; it's just really inefficient because there's a high probability of having to reject that move later on. Experience (and statistical mechanics) shows small moves are better.

In each case, we end up with the situation that sequential steps are correlated with one another. In the MD case, those correlations are meaningful, and we can use them to compute things like diffusion coefficients and viscosities and so on. In the Monte Carlo case, the correlations are artificial; sequential steps are correlated because we designed them to be so.

To see this, let's say you try to extract the mean-square displacement from a Monte Carlo sequence. The answer you get will depend on what you set to be the maximum trial displacement!

• This is pretty much right, especially for hard spheres. However, as a caveat: sometimes for systems with a simple-to-define energy, but less clear dynamics, a Monte Carlo procedure is used to create a dynamic simulation. In this case, time is measured in "Monte Carlo steps." The problems you mention still apply (the dynamics depend on the details of how you propose moves), but sometimes this is the best you can do. Simple example, e.g. is looking at the kinetics of an Ising model by measuring time in the number of attempts made to flip spins. – AJK Nov 20 '17 at 2:41