# Could Navier-Stokes equation be derived directly from Boltzmann equation?

I know how to derive Navier-Stokes equations from Boltzmann equation in case where bulk and viscosity coefficients are set to zero. I need only multiply it on momentum and to integrate it over velocities.

But when I've tried to derive NS equations with viscosity and bulk coefficients, I've failed. Most textbooks contains following words: "for taking into the account interchange of particles between fluid layers we need to modify momentum flux density tensor". So they state that NS equations with viscosity cannot be derived from Boltzmann equation, can they?

The target equation is $$\partial_{t}\left( \frac{\rho v^{2}}{2} + \rho \epsilon \right) = -\partial_{x_{i}}\left(\rho v_{i}\left(\frac{v^{2}}{2} + w\right) - \sigma_{ij}v_{j} - \kappa \partial_{x_{i}}T \right),$$ where $$\sigma_{ij} = \eta \left( \partial_{x_{[i}}v_{j]} - \frac{2}{3}\delta_{ij}\partial_{x_{i}}v_{i}\right) + \varepsilon \delta_{ij}\partial_{x_{i}}v_{i},$$ $w = \mu - Ts$ corresponds to heat function, $\epsilon$ refers to internal energy.

Edit. It seems that I've got this equation. After multiplying Boltzmann equation on $\frac{m(\mathbf v - \mathbf u)^{2}}{2}$ and integrating it over $v$ I've got transport equation which contains objects $$\Pi_{ij} = \rho\langle (v - u)_{i}(v - u)_{j} \rangle, \quad q_{i} = \rho \langle (\mathbf v - \mathbf u)^{2}(v - u)_{i}\rangle$$ To calculate it I need to know an expression for distribution function. For simplicity I've used tau approximation; in the end I've got expression $f = f_{0} + g$. An expressions for $\Pi_{ij}, q_{i}$ then are represented by $$\Pi_{ij} = \delta_{ij}P - \mu \left(\partial_{[i}u_{j]} - \frac{2}{3}\delta_{ij}\partial_{i}u_{i}\right) - \epsilon \delta_{ij}\partial_{i}u_{i},$$ $$q_{i} = -\kappa \partial_{i} T,$$ so I've got the wanted result.

• It seems to be done in Landau and Lifshitz 10, Chapter 1. Apr 17, 2015 at 20:00
• Look up the Chapman Enskog equations. Apr 17, 2015 at 21:02
• @RobinEkman, not surprising... everything is in Landau and Lifshitz. I especially enjoy their recipe for banana bread.
– hft
Apr 17, 2015 at 21:30
• @RobinEkman : But I don't see the derivation there. There is only derivation of Boltzmann equation with tension tensor. Should it be multiplied on $\frac{mv^2}{2}$ and integrated over $v$for getting hydrodynamics equation with viscosity? Apr 20, 2015 at 6:49
• @NameYYY - All the fluid equations are effectively moments of the Boltzmann equation. The Navier-Stokes equations are just the combined effects of the zeroth to the second or third moment equations, depending on the problem. So I guess I am a little confused. Viscosity is just another way of saying off-diagonal terms in a pressure tensor or that there is j-momentum transported through the i-th plane. Oct 9, 2015 at 11:36