The Navier-Stokes equation is a set of differential equations for a space and time dependent velocity field. We relate the momentum averaged form of Boltzmann's transport equation to it.
We start from the momentum-averaged Boltzmann equation,
\[\frac\partial{\partial t}
\rho\overline v_j
+\sum_i
\frac\partial{\partial x_i}
\left(
\rho\overline v_i\overline v_j
+\rho\overline{\delta v_i\delta v_j}
\right)
-\rho a_j
=
\left.
\frac\partial{\partial t}
\rho\overline v_j
\right|_{\text{Streu}}.\]
Let’s split the fluctuation term containing $\overline{\delta v_i\delta
v_j}$ into a contribution from bulk forces (diagonal terms) and shear forces:
-
pressure
$\displaystyle P(x,t):=\frac13\sum_i\rho\overline{\delta v_i^2}$,
-
viscous tensor
$\displaystyle \Pi_{ij}(x,t):=P\delta_{ij}-\rho\overline{\delta v_i\delta v_j}$.
We will work from now on with the averaged quantities only and drop the
horizontal bar we have used to distinguish them from the original
phase-space variables used in Boltzmann’s formalism. Let’s continue rewriting
\[\begin{eqnarray*}
\sum_i
\frac\partial{\partial x_i}
\left(
\rho v_iv_j
\right)
&=&
\sum_i\left(
\frac{\partial\rho v_i}{\partial x_i}v_j
+\rho v_i\frac{\partial v_j}{\partial x_i}
\right)
\\
&=&
\left[
\left(\nabla\cdot\rho v\right) v
+\rho\left(v\cdot\nabla\right) v
\right]_j.
\end{eqnarray*}\]
With the continuity equation,
\[\partial_t\rho+\nabla\cdot\rho v=0,\]
we obtain for the first two terms
\[\begin{eqnarray*}
\frac\partial{\partial t}\rho v_j
+\sum_i\frac\partial{\partial x_i}\left(\rho v_iv_j\right)
&=&
\frac{\partial\rho}{\partial t}v_j
+\rho\frac{\partial v_j}{\partial t}
+\sum_i\left(\frac\partial{\partial x_i}\rho v_i\right)v_j
+\sum_i\rho v_i\frac{\partial v_j}{\partial x_i}
\\
&=&
\left(\partial_t\rho\right)v_j+\rho\partial_tv_j
+\left(\nabla\cdot\rho v\right)v_j
+\rho\left(v\cdot\nabla\right)v_j
\\
&=&
\left[\rho\partial_tv+\rho\left(v\cdot\nabla\right)v\right]_j
\end{eqnarray*}\]
such that we eventually get
\[\fbox{$\displaystyle
\rho\frac{\mathrm d}{\mathrm dt}v\equiv
\rho
\left[
\partial_t
+ \left(v\cdot\nabla\right)
\right] v
=
\nabla\cdot\Pi-\nabla P+\rho a
+\left.\partial_t(\rho v_j)\right|_\text{Streu}.
$}\]
Inserting the linear approximation to the viscosity tensor $\Pi$ described below yields
\[\rho
\left[
\partial_t
+ \left(v\cdot\nabla\right)
\right] v
=
\eta\Delta v-\nabla P+\rho a
+\left(\eta+\eta'\right)\nabla(\nabla\cdot v)
+\left.\partial_t(\rho v_j)\right|_\text{Streu},\]
which is a perhaps more familiar form of the Navier-Stokes equation with an additional collision term describing nonconserving processes. We note that this form of the equation, in particular the form of the viscous tensor, is correct only for velocity-independent external forces.
The continuity equation
In the derivation of the Navier-Stokes equation, we have used the conservation of particle mass in the form of a continuity equation. Intuitively this is clear, however we can derive this as well from Boltzmann's equation. We multiply both sides with $m$ and integrate over the particle velocity,
$$
\int{\rm d}^3vm\frac\partial{\partial t}f
+\sum_i\int{\rm d}^3vmv_i\frac\partial{\partial x_i}f
+\sum_i\int{\rm d}^3vma_i\frac\partial{\partial v_i}f
=0.
$$
The integral over the collision term vanishes because collisions do not create or destroy particles. We can also write
$$
\partial_t\int{\rm d}^3vmf
+\nabla_x\cdot\int{\rm d}^3vvmf
+\int{\rm d}^3v\nabla_v\cdot amf
=0.
$$
The last term on the left-hand side is
$$
\int{\rm d}^3v\nabla_v\cdot amf
=
\int_{\partial V}{\rm d}A n\cdot amf
=0
$$
(Gauß' theorem), provided $f\to0$ for $|v|\to\infty$. This leaves us with
$$
\partial_t\rho+\nabla_x\cdot(\rho\overline v)=0.
$$
Viscosity due to shear forces
The velocity field itself doesn't produce any forces. We can interpret the viscous tensor Π as due to little deformations acting onto little "parcels" of the particles in motion. The dominant contributions to these deformations -- internal friction from shear forces -- are due to different parts of our fluid moving with different velocities. As long as these velocity differences are not too large this tensor can be expressed approximately by
$$
\Pi_{ij}=\eta
\left(
\frac{\partial v_i}{\partial x_j}
+\frac{\partial v_j}{\partial x_i}
\right)
+\eta'\delta_{ij}\nabla\cdot v
$$
introducing the viscosity $\eta$ and the second viscosity $\eta'$. The latter describes internal forces due to compression. The exact form of the expression above has a reason: For a constant rotational motion with angular velocity $\Omega$, we have $v=\Omega\times x$ and $\Pi_{ij}=0$. We can always split a linear combination of derivatives $\partial v_i/\partial x_j$ into a sum of a symmetric and an antisymmetric part. Because $\Omega\times x$ is antisymmetric, the velocity derivatives which vanish in this case are exactly the symmetric ones introduced above.
For the spatial derivatives we then obtain
$$
\begin{eqnarray*}
\left(\nabla\cdot\Pi\right)_j
&=&
\sum_i\frac\partial{\partial x_i}\Pi_{ij}
\\
&=&
\eta\sum_i\left(
\frac{\partial^2v_i}{\partial x_i\partial x_j}
+\frac{\partial^2v_j}{\partial x_i^2}
\right)
+\eta'\frac\partial{\partial x_j}\left(\nabla\cdot v\right)
\\
&=&
\eta\left[
\frac{\partial}{\partial x_j}\left(\nabla\cdot v\right)+\Delta v_j
\right]
+\eta'\frac\partial{\partial x_j}\left(\nabla\cdot v\right)
\\
&=&
\eta\Delta v_j+\left(\eta+\eta'\right)\left[\nabla(\nabla\cdot v)\right]_j.
\end{eqnarray*}
$$
For an incompressible fluid, again, the divergence term vanishes.
Stationary flow and relaxation-time approximation
We still have the collision term in our version of the Navier-Stokes
equation. In many cases it appears appropriate to replace this
complicated function by the drift velocity field multiplied by a constant
being the inverse of a characteristic time scale for momentum relaxing
collisions,
$$
\left.\partial_t(\rho v_j)\right|_\text{Streu}\to-\frac1{\tau_\text{MR}}\rho v
$$
which is commonly known as the relaxation-time approximation. For a stationary flow, $v$ is time independent, and we eventually obtain
$$
\eta\Delta v-\nabla P+\rho a
-\frac1{\tau_{\text{MR}}}\rho v
=0
$$
for an incompressible medium.
Particles with charge $q$ in a static electric field $E=-\nabla\phi$ are subject to the Lorentz force, and we set $\rho a=\rho(q/m)E$ to solve the equation for the velocity field $v$ subject to properly chosen boundary conditions. Ignoring the pressure gradient, this is
$$
\fbox{$\displaystyle
\frac\eta\rho\Delta v+\frac qmE-\frac1{\tau_{\text{MR}}}v=0.
$}
$$