Langevin and Einstein formulas of Brownian motion

[mikeclinton]

Homework Statement

Demonstrate that Eq. (1.1) will convert to the Einstein relation Eq. (1.2) in the limit of t→∞ when we assume ξ=6πaμ.

Conversely, show that Eq. (1.1) will yield <x²> ~ t² in the limit of t→0. Confirm the consistency of the principle of equipartition of energy.

Homework Equations

Eq. (1.1) is <x²> = (2kT/ξ)[t-(m/ξ)(1-e^-tξ/m)]
Eq (1.2) is (MSD)² = 2Dτ

x is the displacement of a Brownian particle moving in a viscous liquid
m is the mass of the particle
ξ is the friction coefficient. We assume it is governed by Stoke's law which states that the frictional force decelerating a spherical particle of a radius and mass ms is ξ=6πaμ, where μ the viscosity of the surrounding liquid.

MSD (or MΔ², I don't know the HTML code for the overbar) is the mean square displacement of the particle
τ is the time interval of observation in which the particle will move as much as Δ
D is the diffusion coefficient
k is Boltzmann's constant
T is the absolute temperature I don't know where to start with this...any help will be much appreciated.

 

[Asim Riaz]

The first part involves taking the limit of Eq. (1.1) as t→∞, while the second part involves taking the limit of Eq. (1.1) as t→0.For the first part, we can use L'Hospital's rule to take the limit of Eq. (1.1) as t→∞:<x2> = lim t→∞ [ (2kT/ξ)[t - (m/ξ)(1-e-tξ/m)] ] = lim t→∞ [ (2kT/ξ)(t) - (2kT/ξ)(m/ξ)(1-e-tξ/m) ] = lim t→∞ [ (2kT/ξ)(t) ] = 2kT/ξ Using the assumed value of ξ = 6πaμ, we can substitute it in to get: <x2> = 2kT/(6πaμ) which is the same as the Einstein relation, Eq. (1.2). For the second part, we can take the limit of Eq. (1.1) as t→0:<x2> = lim t→0 [ (2kT/ξ)[t - (m/ξ)(1-e-tξ/m)] ] = lim t→0 [ (2kT/ξ)(t) - (2kT/ξ)(m/ξ)(1-e-tξ/m) ] = lim t→0 [ (2kT/ξ)(t) ] = 0 Since <x2> = 0 when t=0, then <x2> ~ t2 in the limit of t→0, which is consistent with the principle of equipartition of energy.