Derivation of Euler’s equations of motion for a perfect fluid from Newton’s second law

Having read a number of highly technical derivations of Euler’s equations of motion for a perfect fluid I feel that the mathematical meanderings tend to obscure the underlying physics. In this note I want to explore the derivation from a more physically intuitive point of view. The dynamics of a fluid element of mass $m$ are governed by Newton’s second law which says that the vector sum of the forces acting on the fluid element is equal to the mass of the element times its acceleration. Thus,

$\vec{F} = m \vec{a}$

The net force $\vec{F}$ can be decomposed into two distinct types of forces, so-called body forces $\vec{W}$ that act on the entire fluid element (e.g., the fluid element’s weight due to gravity) and stresses $\vec{S}$ such as pressures and shears that act upon the surfaces enclosing the fluid element. For the purposes of deriving the differential form of Euler’s equation we will focus on the net force per unit volume acting on the fluid element, $\vec{f}$, which we will decompose into a weight per unit volume $\vec{w}$ and a net stress force per unit volume $\vec{s}$. The weight per unit volume is simply obtained as

$\vec{w} = \rho \vec{g}$

where

$\rho = \frac{m}{V}$

is the mass density of the fluid (i.e., mass per unit volume) and $\vec{g}$ is the acceleration due to gravity. In index notation, the equation for the $i$-th component of the weight per unit volume is

$w^i = \rho g^i$

The net stress force per unit volume, $\vec{s}$, is a little more complicated to derive since it involves the rank-2 stress tensor $\tau^{ij}$. This tensor contains nine components and is usually represented as a $3 \times 3$ symmetric matrix. In Cartesian coordinates the components along the main diagonal, namely $\tau^{xx}$, $\tau^{yy}$ and $\tau^{zz}$, represent normal stresses, i.e., forces per unit area acting orthogonally to the planes whose normal vector is identified by the first superscript, as indicated in the diagram below. (Note that a stress is a force per unit area, so to convert a stress tensor component $\tau^{ij}$ into a force it would be necessary to multiply it by the area over which it acts).

In the diagram each normal stress is shown as a tension, i.e., a normal stress pointing away from the surface. When a normal stress points towards the surface it acts upon, it is called a pressure.

The off-diagonal components of the stress tensor represent shear stresses, i.e., forces per unit area that point along the sides of the fluid element, parallel to these sides rather than normal to them. These shear stresses are shown in the following diagram.

Shear stresses only arise when there is some kind of friction in the fluid. A perfect fluid is friction-free so there are no shear stresses. Euler’s equation only applies to perfect fluids so for the derivation of the equation we can ignore the off-diagonal components of the stress tensor.

The normal stresses along the main diagonal are usually written as

$\tau^{xx} = - p^x$

$\tau^{yy} = - p^y$

$\tau^{zz} = - p^z$

where $p$ stands for pressure and the negative sign reflects the fact that a pressure points in the opposite direction to a tension.

In a perfect fluid the pressure is isotropic, i.e., the same in all directions, so we have

$p^x = p^y = p^z = p$

Therefore the stress tensor of a perfect fluid with isotropic pressure reduces to

$\tau^{ij} = -p \delta^{ij}$

where $\delta^{ij}$ is the Kronecker delta (and may be thought of here as the metric tensor of Cartesian 3-space).

Now suppose we consider the net stress (force per unit area) in the y-direction of an infinitesimal volume element.

The stress on the right-hand face can be approximated using a Taylor series expansion as being equal to the stress on the left plus a differential adjustment based on its gradient and the length $dy$. If we take the stress on the right to be pointing in the positive direction and the one on the left as pointing in the negative (opposite) direction, the net stress in the y-direction is given by

$\tau^{yy} + \frac{\partial \tau^{yy}}{\partial y} dy - \tau^{yy} = \frac{\partial \tau^{yy}}{\partial y} dy$

Similarly, the net stresses in the $x$ and $z$-directions are

$\frac{\partial \tau^{xx}}{\partial x} dx$

and

$\frac{\partial \tau^{zz}}{\partial z} dz$

To convert these net stresses to net forces we multiply each one by the area on which it acts. Thus, the net forces on the fluid element (in vector form) are

$\big(\frac{\partial \tau^{xx}}{\partial x} dxdydz\big) \vec{i}$

$\big(\frac{\partial \tau^{yy}}{\partial y} dxdydz\big) \vec{j}$

$\big(\frac{\partial \tau^{zz}}{\partial z} dxdydz\big) \vec{k}$

The total net force on the fluid element is then

$\big(\frac{\partial \tau^{xx}}{\partial x} \ \vec{i} + \frac{\partial \tau^{yy}}{\partial y} \ \vec{j} + \frac{\partial \tau^{zz}}{\partial z} \ \vec{k}\big) dxdydz$

Switching from tensions to pressures using $\tau^{ij} = -p \delta^{ij}$ and dividing through by the volume $dxdydz$ we finally get the net stress force per unit volume to be

$\vec{s} = -\big(\frac{\partial p}{\partial x} \ \vec{i} + \frac{\partial p}{\partial y} \ \vec{j} + \frac{\partial p}{\partial z} \ \vec{k}\big)$

In index notation, the equation for the $i$-th component of this net pressure per unit volume is written as

$s^i = -\frac{\partial p}{\partial x^i}$

We have now completed the analysis of the net force on the left-hand side of Newton’s second law.

On the right-hand side of Newton’s second law we have mass times acceleration, where acceleration is the change in velocity with time. To obtain an expression for this we observe that the velocity of a fluid element may change for two different reasons. First, the velocity field may vary over time at each point in space. Second, the velocity may vary from point to point in space (at any given time). Thus, we consider the velocity field to be a function of the time coordinate as well as the three spatial coordinates, so

$\vec{v} = \vec{v}(t, x, y, z) = v^x(t, x, y, z) \ \vec{i} + v^y(t, x, y, z) \ \vec{j} + v^z(t, x, y, z) \ \vec{k}$

Considering the $i$-th component of this velocity field, the total differential is

$dv^i = \frac{\partial v^i}{\partial t} \ dt + \frac{\partial v^i}{\partial x} \ dx + \frac{\partial v^i}{\partial y} \ dy + \frac{\partial v^i}{\partial z} \ dz$

so the total derivative with respect to time is

$\frac{dv^i}{dt} = \frac{\partial v^i}{\partial t} \ dt + v^x \frac{\partial v^i}{\partial x} + v^y \frac{\partial v^i}{\partial y} + v^z \frac{\partial v^i}{\partial z}$

where I have used

$v^x = \frac{dx}{dt}$

$v^y = \frac{dy}{dt}$

$v^z = \frac{dz}{dt}$

We can write this more compactly using the Einstein summation convention as

$\frac{dv^i}{dt} = \frac{\partial v^i}{\partial t} + v^j \frac{\partial v^i}{\partial x^j}$

This is then the $i$-th component of the acceleration vector on the right-hand side of Newton’s second law. In component form, therefore, we can write mass times acceleration per unit volume for the fluid element as

$\rho \big(\frac{\partial v^i}{\partial t} + v^j \frac{\partial v^i}{\partial x^j}\big)$

This completes the analysis of the mass times acceleration term on the right-hand side of Newton’s second law.

In per-unit-volume form, Newton’s second law for a fluid element is

$\vec{w} + \vec{s} = \rho \vec{a}$

and writing this in the component forms derived above we get the standard form of Euler’s equations of motion for a perfect fluid:

$\rho g^i - \frac{\partial p}{\partial x^i} = \rho \big(\frac{\partial v^i}{\partial t} + v^j \frac{\partial v^i}{\partial x^j}\big)$