Learning Notes

# Definition of Problem

Consider 1D Brownian motion of a system with infinite amount of particles in equilibrium state. Derive the profile of particle density $f(x, t)$ where x is the position and t is time.

# Annotation

k = Boltzmann’s constant

T = Temperature

η = Viscosity constant of fluid

a = Diameter of particles considered (Assumed spherical shape)

X = Random variable which denotes the a fluctuating force

v = First time derivative of x , namely velocity

# Derivation

Consider two force acting on the particles:

1) A varying force X.

2) A drag force caused by viscosity dependent on particle velocity v:

$\displaystyle \Phi(v) = -6\pi\eta a v$

Through statistical mechanics, it is well established that the mean K.E. in equilibrium is proportional to the temperature of the system [Eq.1].

$\displaystyle \left< \frac{1}{2} mv^2 \right> = \frac{1}{2} kT$

Then we can write Newton’s second law of motion as follow:

$\displaystyle m\frac{d^2 x}{dt^2} = -6\pi\eta a \frac{d x}{dt} + X$

Inspired by [Eq.1], multiply both sides by x and get [Eq.2]:

$\displaystyle m\frac{d^2 x}{dt^2} x = -6\pi\eta a \frac{d x}{dt}x + Xx$

consider the following equations:

$\displaystyle x \frac{d^2 x}{dt^2} =\frac{d}{dt}\left( x\frac{dx}{dt}\right) - v^2 = \frac{1}{2}\frac{d^2 x^2}{dt^2} - v^2$

$\displaystyle x\frac{dx}{dt} = \frac{1}{2}\frac{d x^2}{dt}$

and taking time average of both sides (reducing X term to zeros) of [Eq.2], we have:

$\displaystyle \frac{m}{2}\frac{d^2 \left}{dt^2} -mv^2 = -3\pi\eta a \frac{d \left}{dt} + \left$

$\displaystyle \frac{m}{2}\frac{d^2}{dt^2}\left +3\pi\eta a \frac{d }{dt}\left =kT$

This equation is called stochastic differential equation. The general solution of the equation is:

$\displaystyle \frac{d \left}{dt} = C\exp{(-6\pi \eta a t/m)} + kT/(3\pi\eta a)$

Observations by Langevin suggest the exponential term of the equation approaches zeros rapidly with a time constant of order 10^-8, so it is insignificant if we are considering time average. To recover Einstein’s result, integrate this one more time:

$\displaystyle \left - \left = \frac{kTt}{3\pi\eta a}$

# Reference

Gardiner, Crispin W. Stochastic methods. Springer-Verlag, Berlin–Heidelberg–New York–Tokyo, 1985.