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.

  • 1
    $\begingroup$ Perhaps references in this answer could help. $\endgroup$
    – Roger V.
    Commented Nov 29, 2022 at 9:31

1 Answer 1


A couple of comments

  1. ``and the four-velocity of the fluid satisfies the relativistic hydrodynamic equation(s).'' This is not an assumption. The equations of fluid dynamics can be derived from kinetic theory by expanding about a stationary solution (solving the kinetic equation in the limit of slowly varying $f$).

  2. Note that the particle current only exists if the collision term conserves particle number. This is not the case, for example, if $f$ is the distribution function of gluons in a (quark) gluon plasma. If particle number is conserved, then your definition of $f$ is correct.

  3. Note that the definition of fluid velocity is not unique. There are infinitely many possible choices. Two of them are frequently adopted in the literature. Following Eckart, we can use the particle current to define $u_\mu$. This is what you do. Or, following Landau, we can use the energy current $T_{0\mu}$ to define the velocity. Of course, the two fluid dynamic descriptions are equivalent.

Postscript (How to define the Landau frame): Define the energy momentum tensor $$ T_{\mu\nu} = \sum_{p_\mu}\frac{1}{p_0} p_\mu p_\nu f(x^i,\vec{p}) $$ and require that $u^\mu T_{\mu\nu} = eu_\nu$, where $e$ is the equilibrium energy density. This is the statement that the fluid velocity is defined by the energy current $T_{0i}\sim e u_i$.

  • $\begingroup$ 2. What if the particle current is not conserved? What will be the expression then? $\endgroup$
    – SCh
    Commented Nov 29, 2022 at 1:29
  • $\begingroup$ 3. I forgot to mention that initially. I've edited my question accordingly. I am wanting for both the cases: Eckhart's one and also for Landau's one. $\endgroup$
    – SCh
    Commented Nov 29, 2022 at 1:31

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