# 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. – Robin Ekman Apr 17 '15 at 20:00
• Look up the Chapman Enskog equations. – tpg2114 Apr 17 '15 at 21:02
• @RobinEkman, not surprising... everything is in Landau and Lifshitz. I especially enjoy their recipe for banana bread. – hft Apr 17 '15 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? – Name YYY Apr 20 '15 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. – honeste_vivere Oct 9 '15 at 11:36

## 1 Answer

I think you were right. The viscous term in the NS equations cannot be derived from the Boltzmann equations. If you derive the conservation laws from the Boltzmann equations using first order approximation, you will get an force term, which should include the pressure, viscous forces and external forces shown in the NS equations.

I think the approximation of the viscous term in the NS equations (viscous stress related to velocity gradient) were constructed from a continuum perspective, with the the tensor form satisfying certain symmetric properties of a stress tensor. See for example "An Introduction to Fluid Dynamics" by G. K. Batchelor for a nice discussion.

However, what I have seen is the derivation of the viscosity by assuming a velocity profile from the linearized Boltzmann equation. It is a question from the textbook "Statistical Physics of Particles" by Kardar, Ch. 3 questions 9.