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Diffusion equation with an external force 
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#1
May2613, 11:16 PM

P: 11

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: [itex] \int \chi (z) dz = 1 [/itex] [itex] \int z\chi (z) dz = 0 [/itex] [itex] \int z^2\chi (z) dz = a^2 [/itex] For a particle to go from x' at t to x at t+Δt the step, l(t), must be xx'. This happens with probability χ(xx') times the probability density ρ(x', t) that it started at x'. [itex] \rho (x, t+\Delta t) = \int \rho (x', t)\chi (xx') dx' [/itex] [itex] \rho (x, t+\Delta t) = \int \rho (xz, t)\chi (z) dz [/itex] Assume ρ is broad and do a Taylor series. [itex] \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 [/itex] Using the moments, we get: [itex] \rho (x, t+\Delta t) ≈ \rho (x, t)+1/2\frac{\partial^2 \rho}{\partial x^2}a^2 [/itex] Now assume ρ is slow and so changes little during this time step: [itex] \frac{\partial \rho}{\partial t}\Delta t ≈ \rho (x, t+\Delta t)  \rho (x, t) [/itex] So we get: [itex] \frac{\partial \rho}{\partial t} = D\frac{\partial^2 \rho}{\partial x^2} [/itex] where [itex] D = \frac{a^2}{2\Delta t} [/itex]  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 [itex] \frac{\partial \rho}{\partial t} = γF\frac{\partial \rho}{\partial x}+D\frac{\partial^2 \rho}{\partial x^2} [/itex] But I am having trouble getting that result. It seems that the second derivative term is more like [itex] D+F^2γ^2Δt/2. [/itex] 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! 


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