# A question on relaxation time approximation

I was learned that the form of collision term in relaxation time approximation is set to be:

$$\left (\frac{\partial f}{\partial t}\right)_c=-\frac{f-f^0}{\tau}$$

in with $$f^0$$ is local equilibrium distribution function.

Many places use the simplest case to explain relaxation time approximation, that is spatially homogeneous distribution，so the Boltzmann equation becomes:

$$\frac{\partial f}{\partial t}=-\frac{f-f^0}{\tau}$$

It is very clear in this situation that $$f^0$$ is just equilibrium distribution function which is already known，and the steady solution of this equation is exactly $$f^0$$.

But it seems more complete in more general condition:

$$\frac{\partial f}{\partial t}+\mathbf v\cdot\frac{\partial f}{\partial \mathbf r}=-\frac{f-f^0}{\tau}$$

My question is:

What is $$f^0$$ (i.e. the local equilibrium distribution function) in this condition? I guess: it is the local equilibrium distribution function corresponding to the initial distribution, and will change with time and approach the limit of equilibrium distribution. If so, it seems not simpler to use relaxation time approximation when spatial distribution is not homogeneous initially.

As can be seen from the name, it is a "local equilibrium" distribution function. In a non-relativistic context, I mean: $$f^0(t, \boldsymbol{x})= A(t,\boldsymbol{x})\,e^{-\frac{m}{2 K_B T(t, \boldsymbol{x})}\big(\boldsymbol{v}-\boldsymbol{u}(t, \boldsymbol{x})\big)^2}$$ where $$A(t, \boldsymbol{x})$$ is the normalization factor depending on local density and local temperature. Note that in "global equilibrium", $$T(t,\boldsymbol{x})=T=\text{constant}$$ and $$\boldsymbol{u}(t, \boldsymbol{x})=\boldsymbol{u}=\text{constant}$$; obviously, $$A$$ will Also be a constant.
The surprising point is that this function doesn't solve the Boltzmann equation you wrote: $$\partial_t f^0(t,\boldsymbol{x})+\boldsymbol{v}.\boldsymbol{\nabla}f^0(t,\boldsymbol{x}) \ne 0$$ However, it turns out that the left hand side depends only on $$\boldsymbol{\nabla}\boldsymbol{u}$$ and $$\boldsymbol{\nabla}T$$.
To proceed, now we need to include the "long wavelength limit assumption". This is the limit of $$\ell_{mfp}\ll \lambda$$, where $$\lambda$$ is the scale over which the local quantities $$\boldsymbol{u}$$ and $$T$$ vary appreciably. Therefore $$\frac{\ell_{mfp}\boldsymbol{\nabla}T}{T}$$ and $$\frac{\ell_{mfp}\boldsymbol{\nabla}u}{u}$$ are small quantities. This suggests to expand $$f$$ in a derivative expansion: $$f=f^0+f^1+\cdots$$ where $$f^n\sim \mathcal{O}\big((\ell_{mfp}\boldsymbol{\nabla})^n\big)$$. The local distribution $$f^{0}$$ is the leading order solution in derivatives, because: $$\partial_t f^0(t,\boldsymbol{x})+\boldsymbol{v}.\boldsymbol{\nabla}f^0(t,\boldsymbol{x}) = \mathcal{O}(\boldsymbol{\nabla})$$ To find the sub-leading solution, taking $$f=f^0+f^1$$, we arrive at $$\partial_t f^0(t,\boldsymbol{x})+\boldsymbol{v}.\boldsymbol{\nabla}f^0(t,\boldsymbol{x}) = -\frac{f^1}{\tau}$$ which simply gives $$f^1$$. Similarly we can go through the derivative expansion to find higher order solutions $$f^n$$ with $$n>1$$.
In summary, $$f^0$$ describes a state close to equilibrium. The further away from equilibrium, the shorter the wavelength, and the greater the number of terms in $$f=f^0+f^1+\cdots$$ to describe the fluid state.