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

+ 8 like - 0 dislike
377 views

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.


This post imported from StackExchange Physics at 2016-02-10 14:08 (UTC), posted by SE-user Name YYY

asked Apr 17, 2015 in Theoretical Physics by NAME_XXX (1,010 points) [ revision history ]
edited Feb 10, 2016 by Dilaton
Most voted comments show all comments
@RobinEkman : do you mean "Kinetics"?

This post imported from StackExchange Physics at 2016-02-10 14:08 (UTC), posted by SE-user Name YYY
Yes, that is correct.

This post imported from StackExchange Physics at 2016-02-10 14:08 (UTC), posted by SE-user Robin Ekman
Look up the Chapman Enskog equations.

This post imported from StackExchange Physics at 2016-02-10 14:08 (UTC), posted by SE-user tpg2114
@RobinEkman, not surprising... everything is in Landau and Lifshitz. I especially enjoy their recipe for banana bread.

This post imported from StackExchange Physics at 2016-02-10 14:08 (UTC), posted by SE-user hft

To the Chapman-Enskog method I would also like to add my personal favorite: Grad's method of moments: Grad, H. (1949) "On the kinetic theory of rarefied gases". Israel-Müller-Stewart theory is just a relativistic extension of Grad's ideas.

Most recent comments show all comments
@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.

This post imported from StackExchange Physics at 2016-02-10 14:08 (UTC), posted by SE-user honeste_vivere
Since this question has been answered it showld be updated.

This post imported from StackExchange Physics at 2016-02-10 14:08 (UTC), posted by SE-user falematte

1 Answer

+ 1 like - 0 dislike

The book by Müller and Ruggeri  Rational extended thermodynamics, Springer 2013, contains a derivation of a relativistic version of the Navier-Stokes equations from the Boltzmann equation. From the relativistic equations one gets the Navier-Stokes equations by taking the limit $c\to\infty$.

answered Aug 3 by Arnold Neumaier (12,355 points) [ no revision ]

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