For the continuous master equation in real space and time, we have for the distribution $f(x,t)$: $$\frac{\partial f(x,t)}{\partial t}=\int_{-\infty}^{\infty}[f(x',t)W(x',x)-f(x,t)W(x,x')]\mathrm{d}x'$$ In the statistical mechanics book by P.K. Pathria, the right hand side is Taylor expanded to obtain (keeping up to second order) $$\frac{\partial f(x,t)}{\partial t}=-\frac{\partial}{\partial x}[f(x,t)\int_{-\infty}^{\infty}\xi W(x;\xi)\mathrm{d}\xi]+\frac{1}{2}\frac{{\partial}^2}{\partial x^2}[f(x,t)\int_{-\infty}^{\infty}{\xi}^2 W(x;\xi)\mathrm{d}\xi],$$ where $\xi=x'-x$, and $W(x;\xi)=W(x,x')$, which is the transition probability density from $x$ to $x'$. This is pretty much confusing to me and I have the following questions:

  1. How is the expansion carried out? It seems that we should expand around $x'$, after which a change of integration variable is done. However, it doesn't seem to lead to the above expression.

  2. The difference $\xi$ between $x$ and $x'$ is not necessarily small and in fact as the integration variable it goes all the way to $\infty$. Then in this case, how is the keeping up to second order and neglect all the higher orders justified?

Any help is much appreciated.

  • $\begingroup$ $\uparrow$ Which page? $\endgroup$ – Qmechanic Aug 5 '15 at 13:03
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    $\begingroup$ Have a look at page 27 of physics.ucsd.edu/students/courses/fall2013/physics210b/LECTURES/… $\endgroup$ – jac Aug 5 '15 at 13:11
  • $\begingroup$ @Qmechanic the page number might be different for different editions, but it's in the chapter about fluctuations $\endgroup$ – M. Zeng Aug 5 '15 at 14:17
  • $\begingroup$ @jac Many thanks, this is exactly what I was looking for. $\endgroup$ – M. Zeng Aug 5 '15 at 14:24
  • $\begingroup$ @jac I tried to reproduce the steps and it turned out that mine differs from the standard form by a negative sign, which in my case comes from the change of variable from $x'=x-y$ to $y$. Did I miss anything? $\endgroup$ – M. Zeng Aug 6 '15 at 5:32

For the first question, we start by rewriting $$ \frac{\partial f(x,t)}{\partial t}=\int_{-\infty}^{\infty}[f(x',t)W(x',x)-f(x,t)W(x,x')]\, dx' $$ as $$ \begin{aligned} \frac{\partial f(x,t)}{\partial t} &=\int_{-\infty}^{\infty}[f(x',t)W(x';x-x')-f(x,t)W(x;x'-x)] \, d (x' - x) \\ &= \int_{-\infty}^{\infty} f(x-\xi,t) W(x-\xi;\xi) \, d \xi -f(x, t) \int_{-\infty}^{\infty} W(x; \xi) \, d \xi, \qquad (1) \end{aligned} $$ where $\xi \equiv x - x'$. For the first term, the change of the minus sign in $d(-\xi) \to d\xi$ is compensated by swapping the lower and upper limits of the integral: $$ \int_{-\infty}^\infty g(\xi) d(-\xi) = -\int_{\infty}^{-\infty} g(\xi) d\xi = \int_{-\infty}^{\infty} g(\xi) d\xi $$

Next we can expand the first term using the Kramers-Moyal expansion $$ f(x-\xi,t) W(x-\xi;\xi) = f(x,t) W(x;\xi) -\xi \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right] +\frac{ \xi^2 } {2!} \frac{\partial^2 }{ \partial x^2} \left[ f(x,t) W(x;\xi) \right] + \dots, \qquad (2) $$ in which the first term of right-hand side will cancel the second term of (1). So $$ \begin{aligned} \frac{\partial f(x,t)}{\partial t} &= \int_{-\infty}^\infty \left( -\xi \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right] +\frac{ \xi^2 } {2!} \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right] \right) \, d\xi, \\ &= -\frac{\partial }{ \partial x} \left[f(x,t) \int_{-\infty}^\infty \xi W(x;\xi) \, d\xi\right] + \frac{ 1 } {2!} \frac{\partial^2 }{ \partial x^2}\left[ f(x,t) \int_{-\infty}^\infty \xi^2 W(x;\xi) \, d\xi \right]. \end{aligned} $$

For the second question, you are absolutely correct to say that we cannot always justify the truncation. Van Kampen has a entire chapter devoted to this (chapter X) among complaints in other places. The formally correct expansion, Kramers-Moyal expansion, yields $$ \begin{aligned} \frac{\partial f(x,t)}{\partial t} &= \sum_{k = 1}^\infty \frac{(-1)^k}{k!} \frac{\partial^k }{ \partial x^k} \left[f(x,t) \int_{-\infty}^\infty \xi^k W(x;\xi) \, d\xi\right]. \end{aligned} $$ The problem is that this expansion is equivalent to the master equation, and it is generally too difficult to solve.

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    $\begingroup$ I see that you've made all your answers community wiki. Is there some particular reason? Usually this is reserved for truly collaborative efforts among multiple people (take for example this post), and it deprives you of all the reputation you deserve. $\endgroup$ – user10851 Dec 28 '15 at 2:03

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