# Size of a Brownian particle

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) ?

• 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. $<{X(t)}^2> = 2Dt$ ,$D = \frac{K_{B}T}{m{\gamma}}$. \\ Commented Oct 17, 2016 at 8:45
• 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. Commented Oct 17, 2016 at 9:02
• 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. Commented Oct 17, 2016 at 16:29