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.

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    $\begingroup$ It seems to be done in Landau and Lifshitz 10, Chapter 1. $\endgroup$ Commented Apr 17, 2015 at 20:00
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    $\begingroup$ Look up the Chapman Enskog equations. $\endgroup$
    – tpg2114
    Commented Apr 17, 2015 at 21:02
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    $\begingroup$ @RobinEkman, not surprising... everything is in Landau and Lifshitz. I especially enjoy their recipe for banana bread. $\endgroup$
    – hft
    Commented Apr 17, 2015 at 21:30
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    $\begingroup$ @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? $\endgroup$
    – Name YYY
    Commented Apr 20, 2015 at 6:49
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    $\begingroup$ @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. $\endgroup$ Commented Oct 9, 2015 at 11:36

2 Answers 2


To address specifically a more recent question that was closed as a duplicate of this one:

  • Boltzmann equation is indeed based on the assumption of molecular chaos, which means that the velocities of particles are uncorrelated before the collision, but does not exclude collisions and energy transfer, leading to the dissipation (in particular, to viscocity).
  • From the point of view of the BBGKY hierarchy of equations, Boltzmann equation and molecular chaos mean trancating this hierarchy and assuming a specific form of the collision integral in the first order equations. However, one occasionally uses term Boltzmann equation for the first equation in this hierarchy without explicitly specifying the form of the collision term, which, in principle, allowes deriving the results that are beyond the molecular chaos assumption - this seems to be the view taken in the Wikipedia article on the Champan-Enskog theory.
  • Under the assumption of local thermodynamic equilibrium the collision integral is zero, and such collisionless Boltzmann equation results in hydrodynamic equations without viscous terms (i.e., Euler equations).
  • Relaxing the assumption of local thermodynamic equilubrium, i.e., accounting for the local relaxation to the local equilibrium distribution, allows for deriving the Navier-Stokes equations. See, e.g., these notes for a brief introduction and this review for a more rigorous treatment.

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.


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