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I'am now studying Langevin model and Fokker-Planck equation with the lecture notes by Borghini Topics in Nonequilibrium Physics (NB: PDF).

On page 92, he talks about the Markovian property of the velocity of a Brownian particle following Langevin equation.

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We have $$v(t+\Delta t)=v(t)+\frac{dv}{dt}\Delta t+O({\Delta t}^2),$$ by plugging in the Langevin equation $$\frac{dv}{dt} = -M\gamma v + F_L(t),$$ the velocity drift from time $t$ to $t+\Delta t$ is $$v(t+\Delta t)-v(t)=\left(-\gamma v(t) + \frac{1}{M}F_L(t)\right)\Delta t+O({\Delta t}^2)\tag{1}$$

According to the notes mentioned above, if the time scale we are considering is much larger than the autocorrelation time, $\tau_c$, of the Langevin force $F_L(t)$, the Langevin force $F_L(t)$ has nothing to do with the past. Then the velocity drift $v(t+\Delta t)-v(t)$ is a Markov process.

But in the right-hand side of equation (1), there is also a $-\gamma v(t)\Delta t$ term. $v(t)$ should be correlated with the past as the autocorrelation time $\tau_{\gamma}$ of the velocity is much larger than $\tau_c$. ($\langle v(t)v(t^\prime)\rangle=\exp[-\gamma(t-t^\prime)]$)

This is further demonstrated in discussion of the position of the Brownian particle as a Markov process. (page 116)

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As for $x$, $x(t+\Delta t)=x(t)+v(t)\Delta t+O({\Delta t}^2)$. $x(t+\Delta t)-x(t)$ is in general not a Markov process because $v(t)$ depends on the past time of $t$. A coarse grain approximation should be made, that the time scale we are considering is much larger than $\tau_{\gamma}$, if we want to make $x(t+\Delta t)-x(t)$ a Markov process.


As a generalization of my question, why is $v(t+\Delta t)-v(t)$ a Markov process, considering there is $v(t)$ term in the right hand side of equation (1)? In the velocity case, the time scale is much larger than $\tau_c$, but not larger than $\tau_{\gamma}$.

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