In the derivation of the (Klein)-Kramers equation it is possible to start from the differential equations: $$\frac{d\vec v(t)}{dt}=\vec F(\vec r)-\zeta(\vec r) \vec v(t)+\sqrt{2\zeta(\vec r)k_B T(\vec r)} u(t)$$ $$\frac{d \vec r}{dt}=\vec v$$ where $\vec u$ is a Gaussian white-noise of unit strength. The derivation follows that using the the Kramers-Moyal expansion for the Fokker-Planck equation with multiplicative noise. Except as stated in (van Kampen, 1992; pg215) we have: $$\langle \Delta v\rangle_{\vec X,\vec V} =\left\{ \frac{\vec F(\vec X)}{M}-\gamma \vec V\right\} \Delta t$$ For the analogous expression (for $\langle \Delta r\rangle$) when just looking at the high-friction Langevin equation you would get a term $\propto \Theta(0)$ in this expression. This is due to the expansion of $\zeta(\vec r)T(\vec r)$ under the square root. This $\Theta(0)$ does not appear here - why?

i.e. why does the Ito vs Stratonovich dilemma not occur for the Kramers equation?


1 Answer 1


Short Answer

The term due to the Taylor Expansion actually vanishes here rather then appearing as a $\Delta t$ term as it does in the high friction limit.

Long Answer

The term that we are concerned with is: $$\left< \int^{t+\Delta t}_tdt_1 \sqrt{2\zeta(\vec r(t_1))k_B T(\vec r(t_1))} u(t_1)\right>$$ $$\sim \left< \int^{t+\Delta t}_tdt_1\Delta r(t_1) u(t_1) \right>$$ $$\sim \left< \int^{t+\Delta t}_tdt_1\int^{t_1}_t dt_2 q(t_2) u(t_1) \right>$$ $$\sim \left< \int^{t+\Delta t}_tdt_1\int^{t_1}_t dt_2 \int^{t_2} dt_3 u(t_3)u(t_1)\right>$$ $$\sim \int^{t+\Delta t}_tdt_1\int^{t_1}_t dt_2 \int^{t_2} dt_3 \delta(t_3-t_1)$$ $$\sim \int^{t+\Delta t}_tdt_1\int^{t_1}_t dt_2 \Theta(t_2-t_1)=0$$

using $\sim$ as not keeping track of constants.


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