Computational Physics

462 The lattice Boltzmann method for fluid dynamics

Now we consider the moments of these equations. The zeroth moment of the
nonequilibrium and equilibrium distribution gives us the density:
∑

i

mni=

∑

i

mneqi =ρ (14.46)

and the first moment the mass flux:
∑

i

mceiαni=

∑

i

mceiαneqi =jα (14.47)

We furthermore define the second and third moments:
∑

i

mc^2 eiαeiβni≡αβ (14.48)

and
∑

i

mc^3 eiαeiβeiγni≡ (^) αβ γ (14.49)
We now take the zeroth moment ofEq. (14.45). Realising that this moment is
identical forneqi andn(i^2 ), we obtain
∂tρ+∂αjα=

(

τ−^12

)

t(∂t^2 ρ+ 2 ∂t∂αjα+∂α∂β(αβ^0 ))+O(t^2 ). (14.50)

For the first moment, we obtain

∂tjα+∂β(αβ^0 )=

(

τ−^12

)

t(∂t^2 jα+ 2 ∂t∂β(αβ^0 )+∂β∂γ (^) αβ γ(^0 ))+O(t^2 ).
(14.51)
Now we note that theO(t)term on the right hand side of the Eq. (14.50) can
be written as
∂^2 tρ+ 2 ∂t∂αjα+∂α∂β(αβ^0 )=∂t(∂tρ+∂αjα)+∂α(∂tjα+∂β(αβ^0 )). (14.52)
From Eqs. (14.50) and (14.51), the two terms on the right hand side of this equation
can easily be seen to be of order t, so we see that
∂tρ+∂αjα= 0 +O(t^2 ). (14.53)
We have now recovered the continuity equation to order t^2.
For the other moment equation, (14.51), we can argue along the same lines that
the first half of theO(t)term is close to zero, and we are left with
∂tjα+∂β(αβ^0 )=

(

τ−^12

)

t(∂t∂βαβ(^0 )+∂β∂γ (^) αβ γ(^0 ))+O(t^2 ). (14.54)