Taylor expansion in Lattice Boltzmann Method derivation

Question

Currently I'm trying to understand Lattice Boltzmann Method for solving CFD problems. In its derivation BGK approximation is used to get rid of complicated collision integral. But when they come to descritization of equilibrium function (Maxwellian): $$ f^{eq} = \cfrac{\rho}{(2\pi RT)^{D/2}}\exp\left[-\cfrac{(\bf e - u)^2}{2RT}\right] $$ they use following Taylor expansion: $$ f^{eq} \approx \cfrac{\rho}{(2\pi RT)^{D/2}}\exp\left[-\cfrac{\bf e^2}{2RT}\right] \left(1 + \cfrac{\bf e \cdot u}{RT} + \cfrac{(\bf e \cdot u)^2}{2(RT)^2} - \cfrac{\bf u^2}{2RT}\right) $$ which is valid for low Mach numbers and constant temperature. So my question is: why do we use this expansion? why original expression of $f^{eq}$ is not used? And where does condition $Ma \ll 1$ come from? Do we need the condition just for expansion or it is needed somewhere else?

Thanks in advance!

Answer 1

So, in the future I would recommend (as a polite request) providing definitions for each variable so that a reader is not left sifting through the internet trying to deconvolve the symbols.

According to wiki, which used the same notation (and failed to clearly define many of the variables), we have the following definitions:
$\mathbf{e}$ $\equiv$ microscopic velocity;
$\mathbf{u}$ $\equiv$ macroscopic velocity (i.e., Avg. or bulk flow velocity which is often derived from the first velocity moment of the distribution function);
D $\equiv$ number of degrees of freedom;
R $\equiv$ gas constant;
T $\equiv$ Avg. temperature (i.e., derived from second velocity moment of the distribution function); and
$\rho$ $\equiv$ number(?) density.

In order to do this expansion, you would need to assume something like: $$ \frac{u}{R \ T} \sim small $$ which is where the small Mach number argument would come from (since RT $\propto$ thermal speed and/or sound speed, depending on definitions used). [Side Note I am a little bothered by the idea of a dyadic product in an exponent. What does an exponential tensor physically imply?] The other reason to make this assumption would be to decrease the total number of computations necessary. If you set up a high Mach number warm, dense flow, you would need to resolve several scales increasing your computations (the theorists/simulationists on this site would be able to clarify this in more detail).

As for why one would use such an expansion, I imagine it has to do with computational efficiency or ease. It is easier to compute a simple numerical fraction than the exponential of the square of a difference. Typically, one does a Taylor approximation like this to simplify the equation. The simplified equation can often be interpreted in a physically intuitive sense, whereas the original form may be too obscure to understand physically.

For instance, think of the Navier-Stokes equations and how would you try to understand the full system. Without approximations (e.g., assume inviscid or incompressible flow) the full equation can be very difficult to deal with, even for computers (7 dimensions [3 spatial, 3 velocity/momentum, and 1 time] for each computation). That is a lot of computations for just one time step.