# Dependance of diffusion coefficient on size?

What's the dependance of the diffusion coefficient on size? More explicitly, suppose I have a particles of characteristic length $l$, dissolved in a liquid. How does $D$, the diffusion coefficient of the particles, scale with $l$? I would like to see the answer (and derivation) for 3-dimensions and 2-dimensions (I suspect it changes with dimension).

The diffusion constant of small spheres in a three dimensional fluid was found by Einstein $$D = \frac{T}{6\pi\eta a}$$ where $\eta$ is the viscosity of the fluid, and $a$ is the radius of the particles. This is based on the Stokes result for the drag on a sphere, which does not immediately generalize to 2-d (sometimes called Stokes paradox). The Stokes paradox was resolved by Oseen. Using his result, I get $$D = \frac{T}{4\pi\eta}\log\left(\frac{4\eta}{a\rho}\sqrt{\frac{2M}{T}}\right)$$ where $\rho$ is the density of the fluid, and $M$ is the mass of the disks.

Derivation: This is based on two ingedients. The first is the Langevin equation $$\frac{d}{dt}\vec{p}=-\eta_D\vec{p}+\vec{\xi}$$ where $\eta_D$ is the drag force and $\xi$ is a stochastic force. Averaging this equation over the noise, and using $$M\big\langle \big(\frac{d\vec{x}}{dt}\big)^2\big\rangle = dT$$ where $M$ is the mass of the particles, and $d$ is the number of dimensions, I get $$\langle \vec{x}^2\rangle = \frac{2dT}{m\eta_D}\, t.$$ Comparing to the mean square deviation from the diffusion equation $\partial_0n-D\nabla^2 n=0$ I get $$D =\frac{T}{m\eta_D}$$ independent of $d$. This is the Einstein relation.

The second ingredient is to use the drag force predicted by Stokes drag. This is the drag on a sphere/disk in a viscous fluid in the limit of small Reynolds number. In 3d this is the well known result $$\eta_D = \frac{6\pi\eta a}{M}$$ and in 2d it is the more subtle result (due to Oseen) $$\eta_D = \frac{4\pi\eta}{M}\frac{1}{\log(4/Re)}$$ where $Re=ua\rho/\eta$. Here $u$ is the relative velocity of the disk and the fluid, and $\rho$ is the density of the fluid.

Comments: The result in 3d is simple and intuitive. $D\sim T/(\eta a)$, so diffusion is faster if the temperature is higher, the viscosity of the solvent is smaller, and the particles are smaller. The 2d result is unusual, because even though we find the same dependence on $T/\eta$, the diffusion constant is only weakly (logarithmically) dependent on the particle size $a$.

We could have guessed this on purely dimensional grounds. There is no formula for the drag force that scales as $\eta a$. Indeed, if instead of diffusion of macroscopic particles (experiencing drag) I study microscopic particles experiencing quantum mechanical scattering with scattering length $a_s$ I find a power law $D\sim \frac{T}{n}\frac{1}{\sqrt{mT}a_S^2}$ in 3d, and log dependence $D\sim \frac{T}{n}\log(\frac{1}{mTa_s^2})$ in 2d.

What is also peculiar is that the Langevin equation in 2d has logarithmic drag $p\log(p)$. I think that this is indeed ultimately related to a breakdown of the diffusion approximation in 2d. This is well known, see http://www.sciencedirect.com/science/article/pii/0378437178900419. This does not mean that the $\log(a)$ dependence is wrong, it means that there are long time tails in the correlation function.

• Just as an aside, there is also a slip correction factor for highly dilute systems like aerosols when the size of the particle is less than a micron or so. – Greg Petersen Feb 11 '16 at 21:07
• hmmm, also, the diffusion coefficient decreases with $a$ in three dimensions, but increases with $a$ in three dimensions... that's weird. Are you sure? – becko Feb 14 '16 at 20:07
• Decreases with $a$ in both cases, doesn't it? Note that the argument of the log has to be >1 (this is the small Re approximation). This condition also applies to the 3d result (nothing is obviously wrong for very large $a$, but the result is not reliable). The $M$ dependence was upside down -- fixed now. – Thomas Feb 14 '16 at 21:55
• No, in your expressions $\eta_D$ increases with $a$ in 3D, but decreases with $a$ in 2D. – becko Feb 14 '16 at 23:21
• Sorry, my mistake. The formula for D was right, but the formula for the drag should read $\log\to1/\log$. – Thomas Feb 15 '16 at 0:10