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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

Syrocco
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    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. – Qmechanic Jun 11 '23 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 '23 at 21:57
  • @Javi can't we just assume a maxwell-boltzmann distribution? – Syrocco Jun 12 '23 at 07:47
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    Related: https://physics.stackexchange.com/q/601296/226902, https://physics.stackexchange.com/q/434723/226902, https://physics.stackexchange.com/q/712416/226902, https://physics.stackexchange.com/q/258089/226902 – Quillo Sep 20 '23 at 17:32

1 Answers1

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I found an explanation online here. It's never stated explicitly, but the ansatz (called the BGK approximation) is constructed to verify the conservation of mass (my question), velocity, temperature and higher moments. So that the hydrodynamics equations we can derive from it are consistent and looks like the Navier-Stokes one. To do this it is required that:

the equilibrium distribution $f^{eq}$ is chosen such that it locally represents the same particle density as the non-equilibrium distribution f.

So effectively $\displaystyle{\int f^{eq}(x, v,t)dv = \int f(x, v,t)dv}$

Syrocco
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