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According to the Langevin model, we have, for the motion of Brownian particles, $$\frac{dv}{dt} = -M\gamma v + \zeta(t)$$ with $\zeta(t)$ the random force acting on the particle due to fluctuations.

Then I was told that the autocorrelation time $\tau_{c}$ of this fluctuating force is typically of the order of the time interval between two collisions of the fluid particles on the Brownian particle.

This statement puzzles me. I know these two terms $-M\gamma v$ and $\zeta(t)$ come from the collision (scattering) between the Brownian particles and the fluid particles But I wonder does the statement mean that the autocorrelation function of the fluctuating force $\langle\zeta(t)\zeta(t+\Delta t)\rangle$ have the mathematical relation that $$\langle\zeta(t)\zeta(t+\Delta t)\rangle \sim e^{-\gamma t} $$ and $\tau_{c} \sim 1/\gamma$?

If there is such a kind of relation, why? And if not, how to show

that the autocorrelation time $\tau_{c}$ of this fluctuating force is typically of the order of the time interval between two collisions of the fluid particles on the Brownian particle.

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It is possible to write down a generalized Langevin equation, in which the random force term has a memory: $$ \frac{dv(t)}{dt} = -\int_0^t dt' \Gamma(t-t') v(t') + \zeta(t) $$ where I have set $M=1$ for simplicity. In this case we have $$ \Gamma(t) = \frac{\langle \zeta(0)\zeta(t)\rangle}{\langle v^2\rangle} . $$ So, if you want to have an exponentially decaying correlation function for $\zeta(t)$, $$ \Gamma(t) = \frac{\gamma}{\tau_c} \exp(-t/\tau_c) $$ you must also have an integral of the frictional term involving a memory kernel based on that correlation function. In the limit that $\Gamma(t)=\gamma\delta(t)$ (which comes from the above formula by allowing $\tau_c\rightarrow 0$) you recover the Langevin equation $$ \frac{dv}{dt} = -\gamma v(t) + \zeta(t) . $$ You can find some discussion of this in various books on statistical mechanics, but also in this online account.

The physical arguments are briefly given in the answer to how to interprete that the random forces in Langevin Equation are assumed to be delta-correlated. For the classical case of Brownian motion, the Brownian particle is taken to be much heavier than the surrounding molecules, so this naturally leads to the notion that the interaction forces vary on a timescale which is very short compared with the motion of the Brownian particle. So one can try to justify a short correlation time.

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  • $\begingroup$ Thank a lot for your kind help! I have one further question. The question is that whether the above statement is for the equilibrium case or equilibrium case. I learned this in a nonequilibrium statistical mechanics. For the above discussion, I didnt see any precondition or assumption that the system is already at equilibrium state. $\endgroup$ – FaDA Dec 8 '18 at 8:46
  • $\begingroup$ The key condition is the fluctuation-dissipation theorem, connecting the statistical properties of the random force with the friction term This guarantees that the equilibrium (Boltzmann) distribution is stationary, for this equation of motion. However, there is no need to assume that the system is already at equilibrium. $\endgroup$ – user197851 Dec 8 '18 at 9:21

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