---
title: "Euler equations"
firstLetter: "E"
publishDate: 2021-03-31
categories:
- Physics
- Fluid mechanics
- Fluid dynamics

date: 2021-03-31T19:04:17+02:00
draft: false
markup: pandoc
---

# Euler equations

The **Euler equations** are a system of partial differential equations
that govern the movement of **ideal fluids**,
i.e. fluids without viscosity.
There exist several forms, depending on
the surrounding assumptions about the fluid.


## Incompressible fluid

In a fluid moving according to the velocity vield $\va{v}(\va{r}, t)$,
the acceleration felt by a particle is given by
the **material acceleration field** $\va{w}(\va{r}, t)$,
which is the [material derivative](/know/concept/material-derivative/) of $\va{v}$:

$$\begin{aligned}
    \va{w}
    \equiv \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
    = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
\end{aligned}$$

This infinitesimal particle obeys Newton's second law,
which can be written as follows:

$$\begin{aligned}
    \va{w} \dd{m}
    = \va{w} \rho \dd{V}
    = \va{f^*} \dd{V}
\end{aligned}$$

Where $\dd{m}$ and $\dd{V}$ are the particle's mass volume,
and $\rho$ is the fluid density, which we assume, in this case, to be constant in space and time.
Then the **effective force density** $\va{f^*}$ represents the net force-per-particle.
By dividing the law by $\dd{V}$, we find:

$$\begin{aligned}
    \rho \va{w}
    = \va{f^*}
\end{aligned}$$

Next, we want to find another expression for $\va{f^*}$.
We know that the overall force $\va{F}$ on an arbitrary volume $V$ of the fluid
is the sum of the gravity body force $\va{F}_g$,
and the pressure contact force $\va{F}_p$ on the enclosing surface $S$.
Using the divergence theorem, we then find:

$$\begin{aligned}
    \va{F}
    = \va{F}_g + \va{F}_p
    = \int_V \rho \va{g} \dd{V} - \oint_S p \dd{\va{S}}
    = \int_V (\rho \va{g} - \nabla p) \dd{V}
    = \int_V \va{f^*} \dd{V}
\end{aligned}$$

Where $p(\va{r}, t)$ is the pressure field,
and $\va{g}(\va{r}, t)$ is the gravitational acceleration field.
Combining this with Newton's law, we find the following equation for the force density:

$$\begin{aligned}
    \va{f^*}
    = \rho \va{w}
    = \rho \va{g} - \nabla p
\end{aligned}$$

Dividing this by $\rho$,
we get the first of the system of Euler equations:

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

The last ingredient is **incompressibility**:
the same volume must simultaneously
be flowing in and out of an arbitrary enclosure $S$.
Then, by the divergence theorem:

$$\begin{aligned}
    0
    = \oint_S \va{v} \cdot \dd{\va{S}}
    = \int_V \nabla \cdot \va{v} \dd{V}
\end{aligned}$$

Since $S$ and $V$ are arbitrary,
the integrand must vanish by itself everywhere:

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

Combining this with the equation for $\va{w}$,
we get a system of two coupled differential equations:
these are the Euler equations for an incompressible fluid
with spatially uniform density $\rho$:

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

The above form is straightforward to generalize to incompressible fluids
with non-uniform spatial densities $\rho(\va{r}, t)$.
In other words, these fluids are "lumpy" (variable density),
but the size of their lumps does not change (incompressibility).

To update the equations, we demand conservation of mass:
the mass evolution of a volume $V$
is equal to the mass flow through its boundary $S$.
Applying the divergence theorem again:

$$\begin{aligned}
    0
    = \dv{t} \int_V \rho \dd{V} + \oint_S \rho \va{v} \cdot \dd{\va{S}}
    = \int_V \dv{\rho}{t} + \nabla \cdot (\rho \va{v}) \dd{V}
\end{aligned}$$

Since $V$ is arbitrary, the integrand must be zero.
This leads to the following **continuity equation**,
to which we apply a vector identity:

$$\begin{aligned}
    0
    = \dv{\rho}{t} + \nabla \cdot (\rho \va{v})
    = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho + \rho (\nabla \cdot \va{v})
\end{aligned}$$

Thanks to incompressibility, the last term disappears,
leaving us with a material derivative:

$$\begin{aligned}
    \boxed{
        0
        = \frac{\mathrm{D} \rho}{\mathrm{D} t}
        = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho
    }
\end{aligned}$$

Putting everything together, Euler's system of equations
now takes the following form:

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

Usually, however, when discussing incompressible fluids,
$\rho$ is assumed to be spatially uniform,
in which case the latter equation is trivially satisfied.



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