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+---
+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.
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