---
title: "Navier-Stokes equations"
date: 2021-04-12
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---

While the [Euler equations](/know/concept/euler-equations/) govern *ideal* "dry" fluids,
the **Navier-Stokes equations** govern *nonideal* "wet" fluids,
i.e. fluids with nonzero [viscosity](/know/concept/viscosity/).


## Incompressible fluid

First of all, we can reuse the incompressibility condition for ideal fluids, without modifications:

$$\begin{aligned}
    \boxed{
        \nabla \cdot \va{v} = 0
    }
\end{aligned}$$

Furthermore, from the derivation of the Euler equations,
we know that Newton's second law can be written as follows,
for an infinitesimal particle of the fluid:

$$\begin{aligned}
    \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
    = \va{f^*}
\end{aligned}$$

$\mathrm{D}/\mathrm{D}t$ is the [material derivative](/know/concept/material-derivative/),
$\rho$ is the density, and $\va{f^*}$ is the effective force density,
expressed in terms of an external body force $\va{f}$ (e.g. gravity)
and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$:

$$\begin{aligned}
    \va{f^*}
    = \va{f} + \nabla \cdot \hat{\sigma}^\top
\end{aligned}$$

From the definition of viscosity,
the stress tensor's elements are like so for a Newtonian fluid:

$$\begin{aligned}
    \sigma_{ij}
    = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i)
\end{aligned}$$

Where $\eta$ is the dynamic viscosity.
Inserting this, we calculate $\nabla \cdot \hat{\sigma}^\top$ in index notation:

$$\begin{aligned}
    \big( \nabla \cdot \hat{\sigma}^\top \big)_i
    = \sum_{j} \nabla_j \sigma_{ij}
    &= \sum_{j} \Big( \!-\! \delta_{ij} \nabla_j p + \eta (\nabla_i \nabla_j v_j + \nabla_j^2 v_i) \Big)
    \\
    &= - \nabla_i p + \eta \nabla_i \sum_{j} \nabla_j v_j + \eta \sum_{j} \nabla_j^2 v_i
\end{aligned}$$

Thanks to incompressibility $\nabla \cdot \va{v} = 0$,
the middle term vanishes, leaving us with:

$$\begin{aligned}
    \va{f^*}
    = \va{f} - \nabla p + \eta \nabla^2 \va{v}
\end{aligned}$$

We assume that the only body force is gravity $\va{f} = \rho \va{g}$.
Newton's second law then becomes:

$$\begin{aligned}
    \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
    = \rho \va{g} - \nabla p + \eta \nabla^2 \va{v}
\end{aligned}$$

Dividing by $\rho$, and replacing $\eta$
with the kinematic viscosity $\nu = \eta/\rho$,
yields the main equation:

$$\begin{aligned}
    \boxed{
        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
        = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
    }
\end{aligned}$$

Finally, we can optionally allow incompressible fluids
with an inhomogeneous "lumpy" density $\rho$,
by demanding conservation of mass,
just like for the Euler equations:

$$\begin{aligned}
    \boxed{
        \frac{\mathrm{D} \rho}{\mathrm{D} t}
        = 0
    }
\end{aligned}$$

Putting it all together, the Navier-Stokes equations for an incompressible fluid are given by:

$$\begin{aligned}
    \boxed{
        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
        = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
        \qquad
        \nabla \cdot \va{v} = 0
        \qquad
        \frac{\mathrm{D} \rho}{\mathrm{D} t}
        = 0
    }
\end{aligned}$$

Due to the definition of viscosity $\nu$ as the molecular "stickiness",
we have boundary conditions for the velocity field $\va{v}$:
at any interface, $\va{v}$ must be continuous.
Likewise, Newton's third law demands that the normal component
of stress $\hat{\sigma} \cdot \vu{n}$ is continuous there.



## References
1.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.