How to get the four-velocity of a fluid in terms of its microscopic distribution function $f(x^{i},\vec{p})$?
For the sake of initial simplicity, the fluid can be thought of to be single component.
The distribution function $f(x^{i},\vec{p})$ follows the Boltzmann's equation: $$p^{\mu}\dfrac{\partial f(x^{i},\vec{p})}{\partial x^{\mu}} - \Gamma^{\mu}_{\nu \lambda}p^{\nu}p^{\lambda} \dfrac{\partial f(x^{i},\vec{p})}{\partial p^{\mu}} = C[f]$$
and the four-velocity of the fluid satisfies the relativistic hydrodynamic equation(s).
My attempt: Particle current, $N^{\mu} = \displaystyle\sum_{p^{\mu}}{\dfrac{1}{p^{0}}p^{\mu} f(x^{i},\vec{p})}$
Also $N^{\mu} = \rho u^{\mu}$
So, $u^{\mu} = \dfrac{1}{\rho} \displaystyle\sum_{p^{\mu}}{\dfrac{1}{p^{0}}p^{\mu} f(x^{i},\vec{p})}$.
Is it conceptually and physically okay? What if the particle number in the system isn't conserved?
I looked here: arxiv.org/abs/hep-ph/0609056, but unable to find any such direct relationship between $u^{\mu}$ and $f(x^{i},\vec{p})$, which I'm looking for.
EDIT: Just a small modification in my question. I found that it is not possible to derive the expression of $f(x^{i},\vec{p})$ (and hence the Boltzmann's equation) from $u^{\mu}$, as $u^{\mu}$ contains much smaller information about the system than $f(x^{i},\vec{p})$. So I am wanting $f(x^{i},\vec{p})$ to $u^{\mu}$.
Here, while calculating $u^{\mu}$ from $f(x^{i},\vec{p})$, calculate it in both cases: Eckhart's consideration and also for the case of consideration by Landau.