path: root/source/know/concept/euler-equations/index.md

---
title: "Euler equations"
sort_title: "Euler equations"
date: 2021-03-31
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---

The **Euler equations** are a system of partial differential equations
that govern the movement of **ideal fluids**,
i.e. fluids without [viscosity](/know/concept/viscosity/).



## Incompressible fluids

In a fluid moving according to the velocity field $$\va{v}(\va{r}, t)$$,
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:

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

Where $$m$$ and $$\dd{V}$$ are the particle's mass and volume,
and $$\rho$$ is the fluid density, which we assume
to be constant in space and time in this case.
Now, 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 $$\partial V$$.
Using the divergence theorem, we then find:

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

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:

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

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

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

Since $$V$$ is arbitrary,
the integrand must vanish by itself,
leading to the **continuity relation**:

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

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 $$\rho$$:

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



## Compressible fluids

If the fluid is compressible,
the condition $$\nabla \cdot \va{v} = 0$$ no longer holds,
so to update the equations we demand that mass is conserved:
the mass evolution of a volume $$V$$
is equal to the mass flow through its boundary $$\partial V$$.
Applying the divergence theorem again:

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

Since $$V$$ is arbitrary, the integrand must be zero.
The new **continuity equation** is therefore:

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

When the fluid gets compressed in a certain location, thermodynamics
states that the pressure, temperature and/or entropy must increase there.
For simplicity, let us assume an *isothermal* and *isentropic* fluid,
such that only $$p$$ is affected by compression, and the
[fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/)
reduces to $$\dd{E} = - p \dd{V}$$.

Then the pressure is given by a thermodynamic equation of state $$p(\rho, T)$$,
which depends on the system being studied
(e.g. the ideal gas law $$p = \rho R T$$).
However, the quantity in control of the dynamics
is not $$p$$, but the internal energy $$E$$.
Dividing the fundamental thermodynamic relation by $$m \: \mathrm{D}t$$,
where $$m$$ is the mass of $$\dd{V}$$:

$$\begin{aligned}
    \frac{\mathrm{D} e}{\mathrm{D} t}
    = - p \frac{\mathrm{D} v}{\mathrm{D} t}
\end{aligned}$$

With $$e$$ and $$v$$ the specific (i.e. per unit mass)
internal energy and volume.
Using that $$\rho = 1 / v$$,
and substituting the above continuity relation:

$$\begin{aligned}
    \frac{\mathrm{D} e}{\mathrm{D} t}
    = - p \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{1}{\rho} \Big)
    = \frac{p}{\rho^2} \frac{\mathrm{D} \rho}{\mathrm{D} t}
    = - \frac{p}{\rho} \nabla \cdot \va{v}
\end{aligned}$$

It makes sense to see a factor $$-\nabla \cdot \va{v}$$ here:
an incoming flow increases $$e$$.
This gives us the time-evolution of $$e$$ due to compression,
but its initial value is another equation of state $$e(\rho, T)$$.

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

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

What happens if the fluid is actually incompressible,
so $$\nabla \cdot \va{v} = 0$$ holds again? Clearly:

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

So $$e$$ is constant, which is in fact equivalent to saying that $$\nabla \cdot \va{v} = 0$$.
The equation for $$\rho$$ enforces conservation of mass
for inhomogeneous fluids, i.e. fluids that are "lumpy",
but where the size of the lumps is conserved by incompressibility.



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