# Equilibrium distribution function

Hi,

For my Master’s Thesis I am working with the Lattice Boltzmann Method. When reading through literature I notice the equilibrium distribution function used for the LBGK model is
f[sub]eq,i[/sub] = rhot[sub]i[/sub] * (1 + c[sub]i[/sub]u/cs[sup]2[/sup] + (c[sub]i[/sub]u)[sup]2[/sup]/2cs[sup]4[/sup] - u[sup]2[/sup]/2cs[sup]2[/sup])
which is basically a truncated Taylor series expansion up to second order of the Maxwellian distribution function. Is there a specific reason why this truncated expansion is used instead of the full expression of the Maxwellian:
f[sub]M[/sub] = rho * (1/2
piRT)[sup]D/2[/sup] * exp(-(c-u)[sup]2[/sup] / R*T)

And, if it is indeed necessary to use a truncated Taylor series expansion, when would you use a higher order approximation?

I hope you can help. Thanks in advance.

Hello,

The truncated equilibrium is used because you discretize the velocity space. If you use the same equilibrium as the exponential one, the conservation laws is fulfilled. Actually truncation with the conservation laws fulfilled and the equilibrium functions forms come naturally from Hermite polynomial expansion. You can refer to works of Shan (2006) and Orestis Malaspinas thesis (located on this website). Higher order approximations are also coming from Hermite polynomials, you can use them to obtain better stability, accuracy and so on. However, they are not required to restore the Navier-Stokes equation.

Hopefully it will help,
Alex

Hello Alex,

So, if I understand correctly, due to the orthogonality of the Hermite polynomials the higher order terms that have been truncated have no influence on the conservation equations and, since the stress tensor is a second order moment of the distribution function, the truncated series has to retain terms of at least order 2 to correctly give the Navier-Stokes equation.

Would this, then, mean that I have to expand the distribution function up to third order if I were to simulate a thermal flow, so if I were to include the energy equation as well?

Please correct me if I’m wrong.

Elrohir

Hello,

you are absolutely right. There are two ways to simulate thermal flow:

1. You do what you suggest and take the third order moments for the energy flux as well.
2. The third order terms are neglected, and you take an additional set of populations for the temperature field instead.
Both approaches have advantages and disadvantages. I think that you will find some information here in the discussions.

Best,
Timm

Hi Elrohir,
Its not just sufficient, but actually necessary that the equilibrium distribution function be truncated somewhere. Because the goal is to calculate the lower order moments of the equilibrium distribution function exactly, and not the equilibrium distribution function itself. With the full exponential expression f[sub]M[/sub] (your post), no finite order Gauss-Hermite quadrature can calculate any lower order moment (even the 0th moment, which is the density) accurately.

Best
Anirban

Just one question about Anirban’s post above: the equilibrium distribution function in the LBGK equation is the local Maxwellian that gives the same density, mean velocity and temperature as the actual distribution function, right? So if the zeroth, first and second moment of the actual distribution function can be calculated by solving the LBGK equation numerically, isn’t the local Maxwellian uniquely defined then?

## jans Wrote:

With the full exponential expression fM (your post), no
calculate any lower order moment (even the 0th
moment, which is the density) accurately.

Actually, don’t the higher order terms cancel automatically due to the orthogonality of the Hermite polynomials?

Also

## Elrohir Wrote:

the equilibrium distribution function in the LBGK
equation is the local Maxwellian that gives the
same density, mean velocity and temperature as the
actual distribution function, right? So if the
zeroth, first and second moment of the actual
distribution function can be calculated by solving
the LBGK equation numerically, isn’t the local
Maxwellian uniquely defined then?

Dear Elrohir,

## Probably I haven’t understood you right. Could you please explain what do you mean by: Elrohir Wrote:

the equilibrium distribution function in the LBGK
equation is the local Maxwellian that gives the
same density, mean velocity and temperature as the
actual distribution function, right? So if the
zeroth, first and second moment of the actual
distribution function can be calculated by solving
the LBGK equation numerically, isn’t the local
Maxwellian uniquely defined then?

How do you want to calculate moments by solving LBGK equation if there is equilibrium function where which contains them? As soon as you move from the continuous space to discretized velocity space - you have to conserve moments and that can be done through Gauss quadrature rule and as soon as you identified it will give you certain view of equilibrium function. However, it’s a bit strict. If you start right away from certain shape of discretized equilibrium functions you have better flexibility to variate the speed of sound, weights, etc.

Hopefully it will help,
Alex

Hello Alex,

I have only some basic knowledge of LBM, as at my university we have just begun exploring the ‘realm’ of the LBM.

[quote=jans]Its not just sufficient, but actually necessary that the equilibrium distribution function be truncated somewhere. Because the goal is to calculate the lower order moments of the equilibrium distribution function exactly, and not the equilibrium distribution function itself. With the full exponential expression fM (your post), no finite order Gauss-Hermite quadrature can calculate any lower order moment (even the 0th moment, which is the density) accurately.
[/quote]
In the discrete case, the BGK equation is
f[sub]a[/sub](x + c[sub]a[/sub], t+1) = (1- omega) f[sub]a[/sub](x, t) + omega f[sup]eq[/sup] (x, t)
I can appreciate more or less that all terms in the continuous BGK equation
df/dt + \xi_i df/dx_i = - omega * (f - f[sup]M[/sup])
have to be projected on the same basis of Hermite polynomials and that therefore the local Maxwellian f[sup]M[/sup] will have to be approximated by the truncated series as well, thus giving f[sup]eq[/sup].
But, when the distribution function f is solved for numerically at time t, we can calculate the hydrodynamic moments, i.e. density rho, mean velocity u and internal energy e and from the latter also the temperature T.

So my question was rather, if these hydrodynamic moments are known at time t, isn’t the local Maxwellian is uniquely defined then at that time t? All required parameters seem to be available, because the local Maxwellian is calculated from
f[sub]M[/sub] = rho * (1/2piRT)[sup]D/2[/sup] * exp(-(c-u)[sup]2[/sup] / RT)

This can then be used to update the distribution function for time t+1, calculate moments, calculate the new local Maxwellian etc.

I hope this clarifies my question and that you can help.

Elrohir

Hi Elrohir,

Now it’s becoming more clear. Basically yes, from known conserved moments you can reconstruct the equilibrium distribution function. However, the substitution of the equilibrium function in a Maxwellian form into the lattice Boltzmann equation won’t help because the collision operator in this case won’t conserve density and momentum. The maxwellian conserves density and momentum in the continuum space as LB equilibrium function conserves it in the discretized space.

Hopefully it will help,
Alex

Thanks Alex.

I think I understand now. I was under the impression that the Maxwellian could also be written as an infinite Hermite series on the lattice, with the higher order terms dropping out when moments would be taken. But the Maxwellian distribution is actually ‘lumped’ into the finite number of velocity directions of the lattice (as opposed to the infinite number of directions in the continuous case), thus giving the discrete equilibrium function.

Many thanks!

Elrohir