Boltzmann Law in moving fluids? In my research, I am concerned with the analysis of systems which operate essential like this:
There is a tube, say of radius $r$. In this simplification in can be infinitely long. Air moves along it at a velocity $w$, and suspended in the air are particles (i.e. pollen or viral aeresols). Now, there is also a gravitational potential present, not necessarily parallel to the axis of the tube. My question is as follows, if a system in thermal equilibrium would follow the Boltzmann Law that
\begin{eqnarray}\rho(\vec{r}) \propto e^{-V(r)/kT}\end{eqnarray} then how does one approach the problem when in a moving fluid? Preferably, answers will not necessarily solve the problem but instead point me in the direction of the important principles in fluid mechanics and thermodynamics as i've really never studied fluids in detail before.
 A: The Boltzmann distribution in a moving fluid is
$$
n(x,p) = \exp\left(\frac{\mu(x)-V(x)}{k_BT(x)}
-\frac{(\vec{p}-m\vec{u}(x))^2}{2mk_BT(x)}\right)
$$
where $n(x,p)$ is the particle distribution, $p$ is the momentum of the particle, and $m$ is the particle mass. Also, $\mu(x)$ is the chemical potential of the fluid, $T(x)$ the temperature, and $\vec{u}$ is the fluid velocity. The expression $n(x,p)$ is a local equlibrium distribution provided $\mu(x),T(x)$ and $\vec{u}(x)$ satisfy the equation of fluid dynamics (the continuity, and Euler/Navier-Stokes equation). Depending on your setup, you may be able to assume that $\mu,T$ and $\vec{u}$ are approximately spatially constant.
Also note that the local density
$$
\rho(x)=mn(x)=\int d^3p\, n(x,p)
\sim (2mk_BT(x))^{3/2}
\exp\left(\frac{\mu(x)-V(x)}{k_BT(x)}\right)
$$
is not affected by the fluid velocity (because $m\vec{u}$ is just a shift in $\vec{p}$).
A: A moving fluid has non-zero mean velocity. Therefore, you'll have a kinetic component to the energy. So my guess would be that the density profile should instead obey
\begin{equation}
\rho(\vec{r}) \propto \exp\left(-{{m v^2(r)/ 2} + V(r) \over k T}\right),
\end{equation}
where $m$ is the mass of each air molecule and $v$ the fluid average speed.
If the fluid velocity is independent of the radius, which should be true for a non-viscous fluid, then $v(r) = v_0$ is a constant and we recover your expectation $\rho \propto \exp \left(-{V(r) \over k T}\right)$.
