PHY.8:5.3 - Derive transport from persistent microscopic motion

Particles move on an unbounded line with speed v>0. Each reverses direction at independent Poisson events of rate alpha>0. Initially each particle is at the origin, with either direction equally likely. This is a physical stochastic model for the motion; its validity must come from the selected mechanism and regime.

Let p_+(x,t) and p_-(x,t) describe position probabilities with positive and negative velocity. The point preparation leaves probability atoms at the unreversed fronts x=+vt and x=-vt; interpret the following density equations in the distributional sense. The transition and transport laws give

partial_t p_+ = -v partial_x p_+ - alpha p_+ + alpha p_-,

partial_t p_- = v partial_x p_- + alpha p_+ - alpha p_-.

Define total density n=p_++p_- and probability current j=v(p_+-p_-). Adding and subtracting give

partial_t n = -partial_x j,

partial_t j = -v² partial_x n - 2 alpha j.

The current retains the directional persistence. Eliminating it gives

partial_tt n + 2 alpha partial_t n = v² partial_xx n.

For times long compared with 1/(2 alpha) and spatial variation slow enough for the current to relax, use j approximately -D partial_x n, with D=v²/(2 alpha). The resulting diffusion equation is an approximation with physical grounds, not a consequence of fitting a bell-shaped histogram.

The mean position stays zero. For the stated initial preparation, the mean-square displacement is

E[x(t)²] = (v²/alpha) [t - (1-exp(-2 alpha t))/(2 alpha)].

With v=2 cm/s and alpha=1/s, it is about 0.03746 cm² at t=0.1 s. A diffusion calculation would give 0.4 cm², more than ten times as much. At t=20 s, the values are about 78 and 80 cm²: diffusion overestimates this observable by about 2.56% relative to the microscopic model. A 3% tolerance admits the later approximation but not the earlier one.

Changing the requested time therefore changes the needed description. Retain the density-current equations for the early response; use the diffusion reduction when its error is adequate for the requested observable. Boundary arrival probabilities would need their own comparison, since this mean-square agreement does not validate every feature of the distribution.

A simulation that flips velocity with the specified rate implements physical time. A different Markov chain designed to sample a position distribution has no such interpretation merely because its histogram agrees.