Dear all,

For many it would be obvious, but after reading some papers I couldn’t figure out why the equilibrium distribution has that name.

Thanks.

Dear all,

For many it would be obvious, but after reading some papers I couldn’t figure out why the equilibrium distribution has that name.

Thanks.

Raul,

I cannot give you a conclusive answer, except it is the discrete analogue of the Maxwell distribution which is the equilibrium of a gas of certain density and temperature.

I would like to add another question to yours: Why does the equ. distribution the exact form for the D2Q9 and other descriptors? And how do I calculate it for other configurations? Can’t I define any other equ. distribution that agrees with a given u and rho, and the LBM will also converge to this one?

Peer

Hello peer,

I do not completely understand your questions. Please clarify!

Timm

My question is, why

feq,i= rho wi (1 + 3/c^2*u.ei + 9/2c^4 (u.ei)^2 - 3/2c^2 u.u)

has exactly this form?

There are 9 unknowns (feq,i i=0…8) and 3 equations (rho, ux, uy) in the 2D case. This leaves 6 degrees of freedom. Which additional constraints lead to the above equation?

I am sorry, I didn’t want to distract from Rauls question. But maybe the answer to my question explains his, too.

Hope my question is clearer now.

Peer

Hello,

in this case you can see it as a taylor expansion of the maxwell distribution for small mach numbers. It is therefore seen as the local equilibrium probability distribution function to have a particle with velocity i, at point x, at time t.

In this case the hypothesis is made that in the discrete case the equilibrium distribution function is the same as in the continuous case.

Do you know a paper that shows this derivation from the Taylor expansion in a way that is easily understandable? In principle I have no problem to work with the Taylor expansion of any multidimensional function, but I have not been able to perform this derivation.

Peer