Usually in Brownian dynamics, we consider the Brownian particle size to be much-much larger than the size of the particles of the fluid on which the Brownian particle is immersed in. In this scenario the Langevin equation describes the motion of the Brownian particle. My doubt is, is it possible to apply Langevin equation onto a system were all the particles are of the same size (fluid particles as well as the Brownian particles) ?


1 Answer 1


The difference between the Langevin equation and Newton's equation is essentially the noise term representing the many collisions with the surrounding solvent. When one throws out the inertial term, one gets Brownian dynamics. Brownian dynamics is only valid when the length scales are much larger than the diameter of a typical solvent molecule. On shorter length scales (and corresponding time scales), the velocity autocorelation function is actually oscillatory (see fig 13). Using Brownian dynamics there is no autocorelation in the velocity. Even when one keeps the inertial term, the velocity autocorelation will be strictly positive, and not oscillatory. Of course you could probably achieve a reasonable velocity autocorelation function by throwing in a ton of particles, keeping the inertial term, and keeping the noise term relatively small. The better method would however be to use the Nose-Hoover thermostat (or some other thermostat for atomistic molecular dynamics).

  • $\begingroup$ The generalized Langevin equation does have memory, as in general anything Mori-Zwanzig projected will, regardless of the size of things. $\endgroup$
    – alarge
    Commented Oct 14, 2016 at 21:43
  • $\begingroup$ The Langevin equation does indeed have memory. Langevin dynamics has a velocity autocorrelation function which decays exponentially. The question specifies Brownian dynamics, where one throws away the inertial term (essentially the limit of the Langevin equation with the mass going to zero). $\endgroup$ Commented Oct 15, 2016 at 21:29
  • $\begingroup$ The framework is as follows \\ $\frac{dv(t)}{dt} = -m{\gamma}v(t) + {\eta}(t)$ $<{\eta}(t)> = 0$ and $<{\eta}(t){\eta}(t')> = {\Gamma}{\delta}(t-t')$ $<{X(t)}^2> = {\frac{2K_{B}T}{m{\gamma}^2}}[{\gamma}t-1+exp({-\gamma}t)]$ \\ $<v(t)v(t')> = \frac{K_{B}T}{m}exp(-{\gamma}|t-t'|) $ \\ $m{\gamma} = 6{\pi}a{\nu}$ Inorder to model Brownian motion, condition is that ${\gamma}t{\rightarrow}{\infty}$ so that the mean square displacement becomes proportional to time. i.e. $<{X(t)}^2> = 2Dt$ ,$D = \frac{K_{B}T}{m{\gamma}}$. \\ $\endgroup$
    – user1844
    Commented Oct 17, 2016 at 8:45
  • $\begingroup$ Sorry for the above comment, I'm completely new here. Usually we set the Brownian limit when the fluid is highly viscous, i.e. the $\nu$ is too large. The question is whether we can use this equation to model the Diffusion of particles which have the size comparable to that of the fluid molecules. Now the 'a' and 'm' are modified. $\endgroup$
    – user1844
    Commented Oct 17, 2016 at 9:02
  • $\begingroup$ I am also super new here. I would say the original answer pretty much still stands. On length scales comparable to the fluid particles' diameter (and corresponding time scales, e.g. when $\gamma$ is comparable to $t$), the motion will be unphysical. On longer length and time scales you get normal Brownian diffusion (the mean squared displacement is proportional to $t$) in both the physical system and the simulation. I am not a big fan of Brownian/Langevin dynamics since they both (generally) ignore hydrodynamic interactions, but that's beyond the scope of the question at hand. $\endgroup$ Commented Oct 17, 2016 at 16:29

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