# Diffusion equation with an external force

1. May 26, 2013

### tehrv

I thought I would do a little physics for fun, since it has been over 20 years since I last did any. I picked up Sethna's Stat Mech book. He gives a derivation of the diffusion equation that goes as follows:

A particle makes random steps, ie x(t+Δt)=x(t) + l(t). The steps l(t) are given according to a distritubion χ whose moments are:

$\int \chi (z) dz = 1$

$\int z\chi (z) dz = 0$

$\int z^2\chi (z) dz = a^2$

For a particle to go from x' at t to x at t+Δt the step, l(t), must be x-x'. This happens with probability χ(x-x') times the probability density ρ(x', t) that it started at x'.

$\rho (x, t+\Delta t) = \int \rho (x', t)\chi (x-x') dx'$

$\rho (x, t+\Delta t) = \int \rho (x-z, t)\chi (z) dz$

Assume ρ is broad and do a Taylor series.

$\rho (x, t+\Delta t) ≈ \int \left[\rho (x, t)-z\frac{\partial \rho}{\partial x}+z^2/2\frac{\partial^2 \rho}{\partial x^2} \right]\chi (z) dz$

Using the moments, we get:

$\rho (x, t+\Delta t) ≈ \rho (x, t)+1/2\frac{\partial^2 \rho}{\partial x^2}a^2$

Now assume ρ is slow and so changes little during this time step:

$\frac{\partial \rho}{\partial t}\Delta t ≈ \rho (x, t+\Delta t) - \rho (x, t)$

So we get:

$\frac{\partial \rho}{\partial t} = D\frac{\partial^2 \rho}{\partial x^2}$

where

$D = \frac{a^2}{2\Delta t}$

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So far so good. But he says if we add an external force F whereby x(t+Δt)=x(t) + FγΔt + l(t), then we get

$\frac{\partial \rho}{\partial t} = -γF\frac{\partial \rho}{\partial x}+D\frac{\partial^2 \rho}{\partial x^2}$

But I am having trouble getting that result. It seems that the second derivative term is more like
$D+F^2γ^2Δt/2.$

Seeing as this is my first Taylor series expansion in over 20 years, I was wondering if one of you could help me out and tell me if I have made a mistake. It shouldn't take you very long.

Thanks!