Computational Physics

14.2 Derivation of the Navier–Stokes equations 453

will now be calculated using(14.15):

noneqαβ =

∫

mvαvβnnoneqd^3 v

=−τ

[∫

mvαvβ

∂neq

∂t

d^3 v+

∫

mvαvβv·∇rneqd^3 v

]

. (14.16)

Before we proceed to work out(14.16)further, we note that the tensornoneqαβ has
an important property: its trace vanishes. This can be seen by writing out this trace:

∑

α

noneqαα =

∫

v^2 nnoneq(r,v,t)d^3 v. (14.17)

Realizing that this expression represents the change in the average kinetic energy
due to the collisions, we immediately see that it vanishes as the (instantaneous)
collisions leave the total energy invariant:

Trnoneq=0. (14.18)

For the calculation of the nonequilibrium stress tensor,Eq. (14.16), we use
the following equations, which can easily be seen to hold for the equilibrium
distribution:
∫
mneq(r,v)d^3 v=ρ(r); (14.19a)
∫
m(vα−uα)(vβ−uβ)neq(r,v)d^3 v=ρ
kBT
m

δαβ=Pδαβ; (14.19b)

u ̇α=−

∑

β

uβ∂βuα−

1

ρ

(∂αP); (14.19c)

where in the last equation it is understood that the velocities are those evaluated for
the equilibrium distribution: this equation is the Euler equation, (14.12).
We first work out the first term in the square brackets on the right hand side in
(14.16), which can also be written as∂teqαβ(we use∂tto denote a derivative with
respect tot). After some manipulation, usingEqs. (14.10),(14.11a)and(14.11b)
(or(14.19c)), this can be written as

∂teqαβ=∂t(Pδαβ+ρuαuβ)

=P ̇δαβ−

∑

γ

[∂γ(ρuγ)uαuβ+ρuαuγ(∂γuβ)+ρuβuγ(∂γuα)]

−uβ∂αP−uα∂βP. (14.20)