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
