# Diffusion 2D on a surface : diffusion coefficient and surface friction

We have a particle that is diffusing actively (meaning that the source of energy is a motor; the diffusion is like a Brownian motion, the only difference is that the diffusion coefficient is much higher) on a 2D flat surface in a fluid. The diffusion coefficient is $$D$$ so that $$\langle x^2\rangle=4Dt$$.

The reason the particle is not diffusing in the volume is because the particle is too heavy: $$\rho_{particle}>\rho_{fluid}$$.

My question is about the relation between the relation between the friction coefficient, the diffusion coefficient and the size of the particle. Since the friction coefficient is proportional to the gravity, $$F_{friction}=\mu_s F$$, the bigger the particle, the more friction there will be, and the lesser will be the diffusion coefficient.

Is there a way to formalize that with a relation $$D(D(r_0),\mu_s,r,\rho)$$ with $$D(r_0)$$ the diffusion coefficient for a particle of size $$r_0$$, $$\mu_s$$ the coefficient connecting gravity and $$\rho$$ the density of the object.

I'm going to leave all of these reservations aside (as well as the fact that these are active particles) and restate the problem in its most elementary form. Diffusive behaviour comes from solving the Langevin equation, which includes a drag force (proportional to velocity, $$-\zeta v$$, with a friction coefficient $$\zeta$$), and a random force or white noise $$R(t)$$ (whose statistical properties are related to $$\zeta$$). From this, one can deduce that the mean-squared displacement is proportional to time $$t$$, hence defining a diffusion coefficient $$D=k_BT/\zeta$$. One can also derive an equivalent Fokker-Planck equation. Optionally a systematic external force $$F_{\text{ext}}$$ can be added.
You want to add to this a "dry friction" force (sometimes called Coulomb friction), so the full Langevin equation will become (in one dimension for simplicity) $$m\frac{dv}{dt} = -\zeta v - F_{\text{friction}} \sigma(v) +F_{\text{ext}} +R(t) .$$ In the extra term, $$\sigma(v)$$ is the sign of the velocity, and $$F_{\text{friction}}$$ represents the magnitude which, in your case, would be given by the formula involving $$\mu_s$$ and a buoyancy-corrected particle weight. You can set $$F_{\text{ext}}=0$$, but it is sometimes helpful to discuss the mobility of the particle by allowing it to take a constant non-zero value.