Learning Notes

Stochastic – Brownian Motion Langevin’s Derivation

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<x^2\right>}{dt^2} -mv^2 = -3\pi\eta a \frac{d \left<x^2\right>}{dt} + \left<Xx\right>

\displaystyle \frac{m}{2}\frac{d^2}{dt^2}\left<x^2\right> +3\pi\eta a \frac{d }{dt}\left<x^2\right> =kT

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

\displaystyle \frac{d \left<x^2\right>}{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<x^2\right> - \left<x^2_0\right> = \frac{kTt}{3\pi\eta a}

Reference

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

See Also

Stochastic – Brownian Motion Einstein Derivation
Stochastic – Python Example of a Random Walk Implementation

 

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