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Appendix B

In this appendix we derive the Smoluchowski equation  directly from the Fokker-Planck equation . This appendix will be rather technical.

In a potential field the Fokker-Planck equation reads

$\displaystyle { \frac{\partial {\mathcal{G}}}{\partial t} = -\sum_{i=1}^3 v_i \...
\Phi}{\partial r_i} \frac{\partial}{\partial v_i} {\mathcal{G}} }$
    $\displaystyle + \xi \sum_{i=1}^3 \frac{\partial}{\partial v_i} (v_i \mathcal{G}) + \xi
\frac{kT}{m} \sum_{i=1}^3 \frac{\partial^2}{\partial v_i^2} \mathcal{G}.$ (4.52)

On the Smoluchowski timescale we are interested in
$\displaystyle G(\vec{r},\vec{r}_{0};t)$ = $\displaystyle \int d^{3}v\bar{G}(\vec{r},\vec{v},\vec{r}
_{0};t)$ (4.53)
$\displaystyle \bar{G}(\vec{r},\vec{v},\vec{v}_0;t)$ = $\displaystyle \int d^3v_0
\left\{ \frac{m}{2\pi kT}\right\}^{\frac{3}{2}} \exp \{ -\frac{1}{2}\beta
mv_0^2\} \mathcal{G}(\vec{r},\vec{v};\vec{r}_0,\vec{v}_0;t)$ (4.54)

We first notice that $\bar{G}$ also satisfies Eq. (4.52), but with initial value condition

 \begin{displaymath}\lim_{t \rightarrow 0} \bar{G}(\vec{r},\vec{v},\vec{r}_0;t) =...
...i kT}\right\}^{\frac{3}{2}} \exp \{-\frac{1}{2
}\beta m v^2\}.
\end{displaymath} (4.55)

For large values of $\xi$ we expect that the velocities will continually be in thermal equilibrium, i.e. that the equilibration time of the velocities is small on the Brownian timescale. We therefore expect that

\begin{displaymath}\bar{G}(\vec{r},\vec{v},\vec{r}_0;t) \approx \left\{\frac{m}{...
...}\beta mv^2\} \int d^3v \bar{G}(
\end{displaymath} (4.56)

From Eq. (4.55) we see that this certainly holds true in the limit of t going to zero.

We now define the projection operator  P by

\vec{v})=\left\{ \frac{m}{2\pi kT}\right\} ^{\frac{3}{2}}\exp \{-\frac{1}{2}
\beta mv^{2}\}\int d^{3}vF(\vec{v})
\end{displaymath} (4.57)

where $F(\vec{v})$ may be any function of $\vec{v}$. We write the Fokker-Planck equation as
$\displaystyle \frac{\partial \bar{G}}{\partial t}$ = $\displaystyle \xi \mathcal{L}_{1}\bar{G}+\mathcal{L}
_{2}\bar{G}$ (4.58)
$\displaystyle \mathcal{L}_{1}$ = $\displaystyle \sum_{i=1}^{3}\frac{\partial }{\partial v_{i}}v_{i}+\frac{
kT}{m}\sum_{i=1}^{3}\frac{\partial ^{2}}{\partial v_{i}^{2}}$ (4.59)
$\displaystyle \mathcal{L}_{2}$ = $\displaystyle -\sum_{i=1}^{3}v_{i}\frac{\partial }{\partial r_{i}}
...rac{1}{m}\frac{\partial \Phi }{\partial r_{i}}\frac{
\partial }{\partial v_{i}}$ (4.60)

The projection can easily be shown to have the following properties
P2 = P (4.61)
$\displaystyle \mathcal{L}_{1}P$ = 0 (4.62)
$\displaystyle P\mathcal{L}_{1}$ = 0 (4.63)
$\displaystyle P\mathcal{L}_{2}P$ = 0 (4.64)

We shall now derive the equation of motion of $P\bar{G}$. The Smoluchowski equation, i.e. the equation of motion of $G(\vec{r},\vec{r}_{0};t)$ then follows at once from $P\bar{G}=\{m/2\pi kT\}^{3/2}\times \exp \{-\frac{1}{2}
\beta mv^{2}\}G$


$\displaystyle \mathcal{V}$ = $\displaystyle P \bar{G}$ (4.65)
$\displaystyle \mathcal{W}$ = $\displaystyle (1-P) \bar{G}$ (4.66)

we get from Eq. (4.58)
$\displaystyle \frac{\partial \mathcal{V}}{\partial t}$ = $\displaystyle P \mathcal{L}_2 \bar{G} = P
\mathcal{L}_2 (1-P) \bar{G} = P \mathcal{L}_2 \mathcal{W}$ (4.67)
$\displaystyle \frac{\partial \mathcal{W}}{\partial t}$ = $\displaystyle \xi (1-P) \mathcal{L}_1 \bar{G}
+ (1-P) \mathcal{L}_2 \bar{G}$  
  = $\displaystyle \{ \xi \mathcal{L}_1 + (1-P)\mathcal{L}_2 \}\mathcal{W} + \mathcal{L}
_2 \mathcal{V} .$ (4.68)

We now solve Eq. (4.68) using $\mathcal{W}(0)=0$, and introduce the result in Eq. (4.67):

 \begin{displaymath}\frac{\partial \mathcal{V}}{\partial t} = \int_0^t d\tau P \m...
...P)\mathcal{L}_2 )(t-\tau) \} \mathcal{L}_2
\end{displaymath} (4.69)

This result is exact, and therefore not very useful.

We shall now simplify Eq. (4.69) by means of some approximations based on physical grounds. First, since we want to describe the Smoluchowski timescale, we let $\xi$ be large, obtaining

\begin{displaymath}\frac{\partial \mathcal{V}}{\partial t} = \int_0^t d\tau P \m...
...\xi \mathcal{L}_1 (t-\tau) \} \mathcal{L}_2 \mathcal{V}(\tau).
\end{displaymath} (4.70)

Next we invoke the Markov property; the coordinates develop slowly, meaning that the system quickly loses its memory of the past, and $\partial \mathcal{
V} / \partial t$ can only depend on the value of $\mathcal{V}$ at time tand not on its values at earlier times

 \begin{displaymath}\frac{\partial \mathcal{V}}{\partial t} = \int_0^t d\tau P \m...
... (\xi \mathcal{L}_1 (t-\tau) \} \mathcal{L}_2 \mathcal{V }(t).
\end{displaymath} (4.71)

It is a simple but rather tedious task to prove that

\begin{displaymath}\mathcal{L}_1\mathcal{L}_2 \mathcal{V} = -\mathcal{L}_2 \mathcal{V} .
\end{displaymath} (4.72)

Using this in Eq. (4.71) yields
$\displaystyle \frac{\partial \mathcal{V}}{\partial t}$ = $\displaystyle \int_0^t d\tau P \mathcal{L}
_2 \exp \{ -\xi (t-\tau) \} \mathcal{L}_2 \mathcal{V}(t)$  
  = $\displaystyle P \mathcal{L}_2 \frac{1}{\xi} (1-e^{-\xi t}) \mathcal{L}_2 \mathcal{V}
  = $\displaystyle P \mathcal{L}_2 \frac{1}{\xi} \mathcal{L}_2 \mathcal{V}(t)$ (4.73)

where in the last step we have used the overdamped condition $\xi t \gg 1$. Introducing the definition of P and $\mathcal{L}_2$, and working through all derivations and integrals one obtains
$\displaystyle { \frac{\partial {\mathcal{V}}}{\partial t} = \left\{ \frac{m}{2\pi kT}
\right\}^{3/2} \exp \{ -\frac{1}{2}\beta mv^2 \} }$
$\displaystyle \times \sum_i \frac{\partial}{\partial r_i} \{ D \frac{\partial}{\partial r_i
} G + \frac{1}{m\xi} \frac{\partial \phi}{\partial r_i} G \}$     (4.74)

which is recognized as the Smoluchowski equation after using $\mathcal{V} =
P \bar{G} = \{ m/2\pi kT \}^{3/2} \exp \{ -\frac{1}{2}\beta mv^2 \}G$.

next up previous contents index
Next: Appendix C Up: Stochastic processes Previous: Appendix A
W.J. Briels