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For the derivation of Kramer's equation we use the multivariable Fokker-Planck equation: $$\frac{\partial P}{\partial t} = \frac{\partial (P A_{1})}{\partial x} + \frac{\partial (P A_{2})}{\partial v} +\frac{1}{2} \frac{\partial^{2} (P B_{11})}{\partial x^2} + \frac{\partial^{2} (P B_{12})}{\partial v \partial x} + \frac{1}{2} \frac{\partial^{2} (P B_{22})}{\partial v^2}$$ where, $$ A_{1} = \frac{<\Delta{x}>}{\Delta{t}} , A_{1} = \frac{<\Delta{v}>}{\Delta{t}} ,$$$$ B_{11} = \frac{<(\Delta{x})^2>}{\Delta{t}} ,B_{12} = \frac{<\Delta{x}\Delta{v}>}{\Delta{t}} ,B_{22} = \frac{<(\Delta{v})^2>}{\Delta{t}} $$ Final form was derived using the following conditions: $$ (\Delta{x})^2 \rightarrow 0$$ $$ (\Delta{x} \Delta{v}) \rightarrow 0$$ I couldn't find the reason behind these conditions. Can someone please explain it to me.

Qmechanic
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dexter
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  • You're missing the context this derivation was introduced for. Ask the guy who wants you to learn this derivation. He should give you reason for learning this and motivate the assumptions used. – Ján Lalinský Mar 03 '15 at 20:39
  • @JánLalinský:Actually i have to learn this equation to understand the dynamics of my biomolecule.I'm reading Van Kampen and Gardiner but it's hard for me to understand some concepts.Can you suggest some elementary books for a biologist. – dexter Mar 04 '15 at 06:00
  • link :this lecture notes explains the derivation of $A_{1}$ n other moments.I got the answer for my question. – dexter Mar 04 '15 at 18:58
  • dexter: "link :this lecture notes explains [...]" -- Note, btw., that the linked lecture notes are nowhere referring to "Kramer's equation", but instead for instance to the "Kramers-Moyal expansion", "Klein-Kramers equation" and "Kramers' theory"; all named for H. A. Kramers. – user12262 Apr 03 '15 at 15:53
  • @user12262: yes, you are rite.But my doubt was in the derivation of Fokker-Planck equation from Langevin.From there we'll start the "Kramer's" derivation. – dexter Apr 25 '15 at 17:29

1 Answers1

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$B_{12}$ and $B_{11}$ are proportional to $\Delta(t)^m$, so when we take small $\Delta(t)$ limit and they vanish.

But in case of $<(\Delta v)^2>$, the term proportional to $\Delta t$ so $B_{22}$ survives.

because $<dW\cdot dw> = \int\int dt'dt <\xi(t)\xi(t')> = (costant)\cdot\int\int dt'dt \delta(t-t') = (constant)\int dt = (constant)\cdot\Delta t$

( $W(t)$ is also called Wigner process. )

Here, i use Langevin equation $m \dot{v} = f(x) -\gamma v + \xi (t) $ where $\xi $ is uncorrelated Gaussian noise $<\xi(t)> = 0$, $<\xi(t)\xi(t')>=(constant)\delta(t-t')$. And this constant can be determined by Einstein relation.

Anyway from below relations you can derive Kramer's equation
$d x = v dt $

$d v = f(x)dt-\gamma v dt + \xi dt = f(x)dt-\gamma v dt + dW(t) $

Another way is that if the stochastic process satisfies below conditions then you can prove higher terms are gone away 1) $lim_{\delta t\rightarrow 0} \frac{1}{\delta t} \int _{x-z<\eeta}dx(x-z)p(x,t+\delta t|z,t) = A(z,t) + \bigo{\eeta} $ 2) $lim_{\delta t\rightarrow 0} \frac{1}{\delta t} \int _{x-z<\eeta}dx(x-z)p(x,t+\delta t|z,t) = B(z,t) + \bigo{\eeta}$

Detail derivation is in Handbook of stochastic method chater 3.4

sdfsdf
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  • It seems to me that OP says that $(\Delta v)^2$ survives, so this answer might be better if you addressed that aspect. – Kyle Kanos Jun 28 '17 at 10:05