4. Brownian particles
LAST TIME
Started to discuss Fokker-Planck equations on the form



J=Probabiliy current
A(x)P(x,t)=drift term or transport term
$\frac{1}{2}\frac{\partial}{\partial x}B(x)P(x,t)=$ diffusion term Linear $\Rightarrow A(x)=A_0+xA_1,\; B(x)=const.$
Ornstein-Uhlenbeck process $\Rightarrow A_1<0$
$A(x), \; B(x)$ obtained from jump moments


Today: Examples

MESOSCOPIC PARTICLE IN STATIONARY FLUID
$\delta t=$ "infinitesimal" time
$\delta t>>\tau _c=$ mean time for collisions
$\delta t>>\tau_v=$ velocity relaxation time
(see later)
$\delta t<< t=$ time of interest
f=external force on particle
$\mu =$ mobility $\langle v\rangle /f$
v= velocity jump moments:

$$a_1=\frac{\langle \delta x\rangle}{\delta t}=f \mu$$

$$a_2=\frac{\langle (\delta x)^2 \rangle }{\delta t}$$

FOKKER-PLANCK EQUATION



BROWNIAN MOTION $f=0,\;a_1=0$

$$\frac{\partial P(x,t)}{\partial t}=\frac{\langle (\delta x)^2 \rangle }{\delta t}\frac{\partial^2P}{\partial x^2}$$

This is just the diffusion equation, hence require

$$D=\frac{\langle (\delta x)^2 \rangle }{2\delta t}$$

Fokker-Planck equation does not admit a stationary (steady) state but can find



Gaussian processes like the boxed expression are also called Wiener processes

PARTICLE IN FIELD OF GRAVITY
We now have $a_1=Mg \mu$. Presence of field changes a₂ little
Why is this assumption reasonable?
The resulting Fokker-Planck equation is

$$\frac{\partial P(x,t)}{\partial t}=Mg\mu \frac{\partial P}{\partial x} +D\frac{\partial^2P}{\partial x^2}$$

No stationary solution for $-\infty<x<+\infty$.
We can solve time dependent equation:
Go to moving frame

$$y =x+Mg\mu t$$

$$P(x,t)=\Pi(x+Mg\mu t,t)$$

$$\frac{\partial P}{\partial t}=\frac{\partial\Pi}{\partial t}+Mg\mu\frac{\partial\Pi}{\partial x}$$

Left with diffusion equation for $\Pi$!



REFLECTING BOUNDARY
Steady state exists if $0<x<\infty$!
Vessel with diffusing particle has a bottom.
Require that probability current vanishes in steady state

$$J=0=Mg\mu P_s(x)+D\frac{d P_s}{d x}$$

with solution

$$P_s(x)=const. \exp-[\frac{Mg\mu x}{D}]$$

At equilibrium must have

$$P_e(x)=\frac{Mg}{kT} \exp-[\frac{Mgx}{kT}]$$

Get Einstein relation



For time dependent problem apply zero current condition only at x=0!

RAYLEIGH PARTICLE
Consider fine time scale:
$\delta t>>$ time between collisions
$\delta t<<$ time for relaxation of velocity
velocity v independent variable
Macroscopic law $M\dot{v}=F-\frac{1}{\mu}v$
E.g. for spherical Stokes particle

$$\mu=\frac{1}{6\pi\eta r}$$

r=radius, $\eta=$ viscosity
Jump moments

$$a_1=\frac{\delta v}{\delta t}=-\frac{1}{M\mu}v$$

Assume

$$a_2=\frac{\langle (\delta v)^2\rangle}{\delta t}\approx const$$

No external force!

Get Fokker Planck-equation

$$\frac{\partial P(v,t)}{\partial t}=\frac{1}{M\mu}\frac{\partial vP}{\partial v}+\frac{a_2}{2}\frac{\partial^2 P}{\partial v^2}$$

Stationary state (J(v)=0!)

$$\frac{1}{M\mu} vP+\frac{a_2}{2}\frac{\partial P}{\partial v}$$

with solution

$$P_s(v)\propto \exp-\left[\frac{v^2}{a_2M\mu}\right]$$

Maxwell-Boltzmann distribution:




Comparing expressions we find

$$\frac{a_2}{2}=\frac{kT}{M^2\mu}=\frac{(kT)^2}{M^2D}$$

RAYLEIGH EQUATION

$$\frac{\partial P(v,t)}{\partial t}=\frac{1}{M\mu}\left[ \frac... ...partial v}+\frac{kT}{M}\frac{\partial^2 P}{\partial v^2}\right]$$

Equation is linear and describes an Ornstein-Uhlenbeck process. Solution:

$$P(v,t\vert v_o,0)=\frac{\exp-\left[\frac{ M(v-v_0e^{- t/\ta... ...tau_v})}\right]} {\sqrt{\frac{2\pi kT}{M}(1-e^{-2 t/\tau_v})}}$$

$\tau_v=M\mu=$ velocity relaxation time
We have if $v=v_0\;for\;t=0$

$$\langle v(t)\rangle=v_0e^{-t/\tau_v};\; \langle v(t)^2\rangle -\langle v\rangle^2= \frac{kT}{M}(1-e^{-t/\tau_v})$$

Can also calculate equilibrium velocity-velocity correlation

$$\langle v(t)v(0)\rangle=\int vdv\int v_0dv_0P(v,t\vert v_0,0)P_e(v_0)$$

We find



GENERALIZE TO ARBITRARY FORCE



Overdamped motion in force field.
Macroscopic equation of motion Aristotelian:

$$v=\dot {x}=\mu f$$

not Newtonian.

$$M\frac{dv}{dt}=f-\frac{v}{\mu}$$

Aristotle ok if $f\approx const$ if $\delta t\approx\tau_v$

$$\frac{d f}{dx}v\tau_v<<f$$

Substituting $\mu f=v,\;\tau_v=M\mu$ we find for the boxed expression to be valid

$$\frac{df}{dx}<<\frac{1}{M\mu^2}$$

If condition

$$\frac{df}{dx}<<\frac{1}{M\mu^2}$$

not satisfied both v and x independent variables!
Jump moments

$$\frac{\langle \delta x\rangle}{\delta t}=v;\;\;\frac{\langle \delta v\rangle}{\delta t}= \frac{f}{M}-\frac{v}{\mu}$$

$$\frac{\langle (\delta x)^2\rangle}{\delta t}=v^2\delta t\appr... ... v\rangle}{\delta t}=v[\frac{\mu f -v} {M\mu}]\delta t\approx 0$$

$$\frac{\langle (\delta v)^2\rangle}{\delta t}=\frac{kT}{M^2\mu}$$

Fokker-Planck equation:
KRAMERS EQUATION




SUMMARY

• We have considered examples of equations of type
  
  $$\frac{\partial P(x,t)}{\partial t}=-\frac{\partial}{\partial ... ...P(x,t)]+\frac{1}{2}\frac{\partial^2}{\partial x^2}[a_2 P(x,t)]$$
  
  

• If no force $\Rightarrow a_1=0$. Wiener process. Diffusion
• Linear force $\Rightarrow$ Ornstein-Uhlenbeck process
• Reflecting boundaries $\Rightarrow$ J=0
• Einstein relation for equilibrium systems $\Rightarrow D=\mu kT$
• Rayleigh equation described decay of velocity correlations
• Had Aristotelian or Newtonian picture depending on relative magnitude of df/dx and $1/M\mu^2$.
• Kramers equation applies when velocity correlations decay slowly.

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