# Rigorous derivation of Fick's first law

I am looking for a rigorous derivation of Fick's law, i.e. that the current density $\mathbf{j}$ satisifies

$\mathbf{j} = - D \nabla u$

where $u$ is e.g. some concentration and $D$ the diffusion constant. I know how it could be done in one dimension, as outlined in this wikipedia article https://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion#Fick.27s_first_law.

However, when transferring the proof to the three dimensional case, there is of course the issue of the shape of the test volume. Is there a proof which overcomes this?

• Well, that's the issue, I don't want to take a little cube. Say, in the derivation of the continuity equation, you could also consider a little cube, but you could also do a proper proof using Gauss' theorem. I'm looking for something similar. Jan 18 '16 at 22:50

2) The most basic argument is still one of the best. If the concentration is inhomogeneous we expect to see a current, and $\vec\nabla u$ is the only vector available. Then $$\vec\jmath = - D\vec\nabla u + O(\nabla^2)\, ,$$ which is Fick's law. The modern view of this is called effective (field) theory: Simply write down the most general expression for the current as a power series in gradients of the thermodynamic variables.
3) The standard kinetic theory derivation goes roughly like this: Consider a chemical potential $\mu$ conjugate to $u$. Then the distribution function is $f_0\sim \exp(-\mu(x)/T)$, where we have allowed for gradients in the concentration. If the inhomogeneity is weak, we can seek solutions of the Boltzmann equation of the form $f_0+\delta f$. Insert into the Boltzmnann equation and expand. We get $$\frac{1}{T}\vec{v}\cdot\vec\nabla\mu \,f_0 = -\frac{\delta f}{\tau}$$ where I write the linearized collision operator in terms of its smallest eigenvalue, the inverse collision time $1/\tau$. ($v$ is the quasi-particle velocity). I can directly solve for $\delta f =-\frac{\tau}{T}\vec{v}\cdot\vec\nabla\mu\, f_0$. Now compute the current generated by $\delta f$: $$\vec\jmath = \int d^3p \,\vec{v}\delta f = - \frac{\tau}{3mT}\left(\int d^3p \,v^2 f_0\right) \vec\nabla\mu$$ Use thermodynamic identities to relate $\nabla\mu$ to $\nabla u$, and compute the integral in terms of the total pressure or energy of the gas. Presto, we get Fick's law, plus an explicit result for $D$.