# Issue deriving Fokker-Planck equation starting from Boltzmann's equation

I was trying to derive the Fokker-Planck equation starting from the Boltzmann's equation and I run into some issue while trying to do so.

Starting from Boltzmann and using the notation $$f \equiv f(x, v, t)$$ the normalized density:

$$\dfrac{\partial f}{\partial t} + v\dfrac{\partial f}{\partial x} + \dfrac{F}{m}\dfrac{\partial f}{\partial v} = \dfrac{f_{eq} - f}{\tau}\tag{1}$$

where I used the usual ansatz for the collision kernel.

Then I define $$P\equiv P(x, t) = \int f(x, v, t) dv$$ and itegrate the equation (1) to get an equation for $$P(x, t)$$ the density of probability at some point in space which, to me, is the unknown of the Fokker-Plank equation:

$$\dfrac{\partial P}{\partial t} + \dfrac{\partial \int v f dv}{\partial x} + \dfrac{F}{m}\int\dfrac{\partial f}{\partial v}dv = \dfrac{P_{eq} - P}{\tau}\tag{2}$$

The current of probability is defined as $$J = \int vf dv$$. Moreover, the last integral on the LHS is zero because $$f(x, v = \pm \infty, t) = 0$$. Putting these together, equation (2) becomes:

$$\dfrac{\partial P}{\partial t} + \dfrac{\partial J}{\partial x} = \dfrac{P_{eq} - P}{\tau}\neq 0\tag{3}$$

If the RHS was $$0$$, we would have the Fokker-Planck equation. But there, in the RHS, we have a non zero term taking into account the deviation to equilibrium which is strange to me. Why is that so?

The only issue I might see is the definition of $$J$$ as $$J = \int v f dv$$ which might not be the same as the $$J$$ appearing in the Fokker-Planck equation. But when we calculate the current directly by multiplying (1) by $$v$$ and integrating as done here (1) using the so called Drift-Diffusion Model. It seems to me that this current, is indeed very reminiscent of the $$J$$ in the Fokker-Planck equation: $$J_{FP} = \left[\mu (x,t)p(x,t)\right]-{\frac {\partial}{\partial x}}\left[D(x,t)p(x,t)\right]$$

(1) Drift-Diffusion Model: Introduction by Dragica Vasileska

• Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. Jun 11, 2023 at 16:23
• It is difficult to know what the "deviation from equilibrium" is if $f_{eq}$ is not defined in the first place.
– Javi
Jun 11, 2023 at 21:57
• @Javi can't we just assume a maxwell-boltzmann distribution? Jun 12, 2023 at 7:47
• Sep 20, 2023 at 17:32

the equilibrium distribution $$f^{eq}$$ is chosen such that it locally represents the same particle density as the non-equilibrium distribution f.
So effectively $$\displaystyle{\int f^{eq}(x, v,t)dv = \int f(x, v,t)dv}$$