Brownian Motion and Two Particle Collisions Leading to the MB Distribution

The equation for Brownian motion is the Langevin equation which may be converted into a Fokker Planck equation: d/dt f = -p/m df/dx + d/dp { [ dV/dx + g(x,p)p/m] f} + d/dp D(x,p) d/dp f This includes a friction term (the g term) and a Gaussian noise term (with D). For the stationary state df/dt=0, for D and g constant and D/g = T, one may obtain the MB distribution as a solution. This distribution, however, implies conservation of kinetic energy in a two particle collision. At first glance, however, Brownian motion does not seem to be related to conservation of energy. Yet somehow the stationary Brownian motion picture should be the same as an equilibrium gas picture in which two particles collide elastically. The Maxwell-Boltzmann (MB) distribution may be derived in a variety of ways. A traditional way makes use of the Boltzmann transport equation which seems equivalent to two particle elastic collisions . In such a case, f(p)f(p1)=f(p2)f(p4). Taking the natural log of both sides yields a conservation equation which one may identify with kinetic energy conservation and thus obtain the MB distribution. In (1), however, p2 is set equal to p+dp and with conservation of energy: p2 + p22 approx = p2 + 2pdp + p42. This leads to the equation: (df/dp 1/p) = constant which is essentially of the form of the stationary Brownian motion equation. Thus in this note, we try to examine the connection between the stationary Brownian motion case and an equilibrium MB gas