I have seen Fick's first law of diffusion derived for a homogeneous material many times, however I am struggling to find a satisfactory proof for inhomogeneous, particularly for particle diffusion. Why does it take the form:

$$D {\partial \phi \over \partial x}$$

Where $D$ is the diffusion coefficient, And the same form:

$$D(x) {\partial \phi \over \partial x}$$

When $D$ is a function of $x$? For example wiki gives this (http://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion) but no explanation as to why? What assumptions are made that lead to this? And under what circumstances is the Fokker–Planck diffusion law a better model?

  • $\begingroup$ Why would you expect the two to be different? All that's changed between the two is the spatial dependence on the diffusion coefficient. $\endgroup$
    – Kyle Kanos
    Commented Sep 28, 2015 at 21:27

2 Answers 2


First of all, this is a reasonable question. If I derive the diffusion equation only for D=const, then I can obviously not distinguish between $$ \vec\jmath =-D\vec\nabla \phi \quad {\rm and} \quad \vec\jmath =-\vec\nabla (D\phi) , $$ which are cleary different.

1) The diffusion constant (like other transport coefficients, for example the shear viscosity $\eta$ and the thermal conductivity $\kappa$) is a function of the thermodynamic variables $T$ and $P$ (or $T$ and $\mu$). This means that $D$ acquires a dependence on $x$ via $P$ and $T$, that is $D=D(T(x),P(x))$. This arises from the standard coarse graining employed in deriving macroscopic equations.

2) This is seen explicitly if we compute $D$ in kinetic theory. The diffusion constant is obtained by linearizing the Boltzmann equation around the local equilibrium distribution (this is called the Chapman-Enskog procedure) $$ f(x,p)=\exp(-(E_p-\mu(x))/T(x)) $$ so that we obtain $D=D(\mu(x),T(x))$.

3) This means that the two versions of Fick's law given above differ by gradients of $T$ and $P$. However, the most general form of Fick's law does contain such terms explicitly. For example, Landau and Lifshitz define $$ \vec\jmath =-D\left[\vec\nabla \phi +k_T\vec\nabla\log T +k_P\vec\nabla\log P\right], $$ where $k_T$ and $k_P$ are the "thermal diffusion" and "baro-diffusion" coefficient.

4) This implies that I should measure (or compute) $D$ for the most general driving term, containing gradients of $\phi,T,P$. This $D$ will still be a fucntion of $T(x)$ and $P(x)$, and so are $k_T$ and $k_P$. The diffusion equation is $$ \partial_t\phi + \vec\nabla\cdot\vec\jmath=0 $$ and gradients of $\vec\jmath$ will act on any spatial variation of $D$.


From a mesoscopic point of view the net number of particles going across a section with area $\delta y \delta z $ in a time $\tau$ and located about $x$ is given by

\begin{equation} \Delta N(x,\tau) = \phi(x)\delta y \delta z v_0 \tau - \phi(x+\delta x)\delta y \delta z v_0 \tau \end{equation}By expanding $\phi(x+\delta x) \approx \phi(x)+\partial_x \phi(x)\delta x$, we end up with the current density $j \equiv \Delta N/(\delta y \delta z\tau)$ giving:

\begin{equation} j(x,t) = -\delta x \: v_0 \frac{\partial \phi}{\partial x} \end{equation}Now, $v_0$ can be interpreted (up to some unimportant prefactor) as the ratio of the mean free path of the tracer particles $\lambda$ divided by the collision time $\tau$. We can end up rewriting the whole equation for the current as being:

\begin{equation} j(x,t) = -D \frac{\partial \phi}{\partial x} = -\frac{\lambda^2}{\tau} \frac{\partial \phi}{\partial x} \end{equation}

Now, in an inhomogeneous medium, the mean free path of tracer particles and the typical time between collisions will be both dependent on position $\lambda \rightarrow \lambda(x)$ and $\tau \rightarrow \tau(x)$, such that the diffusion coefficient of the tracer particles becomes effectively position dependent:

\begin{equation} j(x,t) = -D(x) \frac{\partial \phi}{\partial x} \end{equation}

Now, in crowded environments where there can be strong local inhomogeneities (not because of an external medium but because the tracer particles themselves have a stationary density profile), it is often the case that $D(x) = D(\phi(x))$, which often simplifies the derivation of general results on these problems.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.