chapman enskog expansion

Lattice Boltzmann Method LBM: Theory
#1

i want to understand why the chapman enskog expansion is used in LB and how it works , how its helps in simulation

#2

Basically, the Chapman-Enskog expansion is useful to relate the LB variables (particle populations) to the macroscopic flow variables (pressure, velocity, and potentially others). You can use these results to interpret you simulation: as an example, the C-E expansion is used to understand how the relaxation time is related to the fluid viscosity, or it is used to conclude that the stress tensor Pi is related to the velocity gradients. Reversely, you may use the results of the expansion to infer macroscopic information onto the particle populations. This is for example the case when you construct an initial or a boundary condition. You can find explanations on applying the C-E expansion to the continuum Boltzmann equation in just about any textbook on off-equilibrium statistical mechanics. This is often easier to read than articles in which C-E is directly applied to the LB scheme, because things rapidly grow somewhat messy in the latter case.

Anyway, the C-E expansion, as applied to the numerical BGK scheme, is introduced in my thesis, where I have tried as well as I could to explain things in simple terms (see Chapter 2):

http://www.unige.ch/cyberdocuments/theses2007/LattJ/meta.html

#3

Hello,

does anybody know a book or a paper, where the Chapman-Enskog analysis is performed up to third order in epsilon WITH reasonable intermediate steps? I already know the results, but I’d like to check the calculations.

Timm

#4

Hi Timm,

I found quite understandable the following paper but only for Advection-Diffusion Equation -

B. Servan-Camas, F. T. C. Tsai, Lattice Boltzmann method for two relax-
ation times for advection-diffusion equation: Third order analysis and stability
analysis. Adw. Water. Res., 31:1113-1126, 2008.

Good luck,
Alex

#5

Hello Alex,

thank you! This is not exactly what I am looking for, but it helps me a lot.
Does anybody know the result for the fourth moment $\sum_i c_ia c_ib c_ic c_id f_i^eq$ of the equilibrium populations?

Timm

#6

Hi Timm,
it depends on which equilibrium you are looking at… Is it the standard second order expansion of the maxwellian distribution? Or is it something else? If you are looking at standard lattices, then there is an isotropy problem for the computations. What I mean by this statement is that there is no nice formula for the tensor \sum_i t_i c_ia c_ib c_ic c_id c_ie c_if (which will enter the computation) as you have for second and fourth order tensors sum_i t_i c_ia c_ib and sum_i t_i c_ia c_ib c_ic c_id.

If you are looking at the continuum case then the formula is (take a deep breath)

int dxi xi_a xi_b xi_c xi_id f^eq = rho u_a u_b u_c u_d
+ rho theta (u_a u_b delta_cd + u_a u_c delta_bd + u_a u_d delta_cb + u_c u_b delta_ad
+ u_d u_b delta_ca + u_c u_d delta_ab)
+ rho theta^2 (delta_ab delta_cd +delta_ac delta_bd +delta_ad delta_cb )

I hope i did no mistake…

I hope this helps a bit…

#7

Hi Orestis,

I see what you mean. I tried to calculate (starting from the standard equilibrium which is second order in u) the fourth moment and I have noticed that every reasonable ansatz leads to some conflict in the expressions. As you say, the problem enters the calculations when I want to evaluate the term (c_ia u_a)^2 in the equilibrium. Then I have the sixth moment and something strange is happening.
Now there is the question: Does it make sense to compute the fourth moment of the equilibrium? Is there physical information in this moment?
Your post has helped me a lot, thank you!

Timm

#8

When you are using the standard equilibrium there is no point in computing the fourth order tensor since you are way beyond the possibilities of the model, since you are not even correctly recovering the third order moment (that’s why you are only in the incompressible approximation).

To have a correct third order moment you need an extended lattice (D2Q17, D3Q39) (see the paper by X. Shan “Kinetic theory representation of hydrodynamics: a way beyond the navier stokes equation”, J. Fluid Mech. 550, 413-441, (2006)). At this order of approximation you will get a correct momentum equation (including the compressibility) and the correct “form” of energy equation but you’ll still have something missing, but the energy flux will be wrong.

But if you want a correct 4-th order moment then you’ll need to extend the lattice once again (D2Q52 I think , and D3Q121). At this point you recover also the correct energy equation, with a correct energy flux term.

At each step you will be able to add more physics in your problem. But you also have to pay a huge price for it…

#9

I see that the standard equilibrium is only applicable to the zeroth, first and second moment. That’s true.
So I will have a look at the reference you’ve suggested.
I really find the discussions in this forum very helpful and illuminating. Thanks again!

Timm