2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.