# Cauchy momentum equation

The Cauchy momentum equation is a vector partial differential equation put forth by Cauchy that describes the non-relativistic momentum transport in any continuum:[1]

$\rho \frac{D \mathbf{v}}{D t} = \nabla \cdot \boldsymbol{\sigma} + \mathbf{f}$

or, with the material derivative expanded out,

$\rho \left[\frac{\partial \mathbf{v}}{\partial t} + (\mathbf{v} \cdot \nabla) \mathbf{v}\right] = \nabla \cdot \boldsymbol{\sigma} + \mathbf{f}$

where $\rho$ is the density of the continuum, $\boldsymbol{\sigma}$ is the stress tensor, and $\mathbf{f}$ contains all of the body forces per unit volume (often simply density times gravity). $\mathbf{v}$ is the velocity vector field, which depends on time and space.The expression $\mathbf{v} \cdot \nabla \mathbf{v}$ denotes the vector field with components $v^j \partial_j v^i$ and $\nabla \cdot \sigma$ stands for the vector field with components $\partial_j \sigma_j^i$.

The stress tensor is sometimes split into pressure and the deviatoric stress tensor:

$\boldsymbol{\sigma} = -p\mathbb{I} + \mathbb{T}$

where $\scriptstyle \mathbb{I}$ is the $\scriptstyle 3 \times 3$ identity matrix and $\scriptstyle \mathbb{T}$ the deviatoric stress tensor. The divergence of the stress tensor can be written as

$\nabla \cdot \boldsymbol{\sigma} = -\nabla p + \nabla \cdot\mathbb{T}.$

All non-relativistic momentum conservation equations, such as the Navier–Stokes equation, can be derived by beginning with the Cauchy momentum equation and specifying the stress tensor through a constitutive relation.

## Derivation

Applying Newton's second law ($i^{th}$ component) to a control volume in the continuum being modeled gives:

$m a_i = F_i\,$
$\rho \int_{\Omega} \frac{d u_i}{d t} \, dV = \int_{\Omega} \nabla_j\sigma_i^j \, dV + \int_{\Omega} f_i \, dV$
$\int_{\Omega} (\rho \frac{d u_i}{d t} - \nabla_j\sigma_i^j - f_i )\, dV = 0$
$\rho \dot{u_i} - \nabla_j\sigma_i^j - f_i = 0$

where $\Omega$ represents the control volume. Since this equation must hold for any control volume, it must be true that the integrand is zero, from this the Cauchy momentum equation follows. The main step (not done above) in deriving this equation is establishing that the derivative of the stress tensor is one of the forces that constitutes $F_i$.[2]

## Expression in coordinates

### Cartesian coordinates

\begin{align} x:\;\; \rho \left(\frac{\partial u_x}{\partial t} + u_x \frac{\partial u_x}{\partial x} + u_y \frac{\partial u_x}{\partial y} + u_z \frac{\partial u_x}{\partial z}\right) &= -\frac{\partial P}{\partial x} + \frac{\partial \tau_{xx}}{\partial x} + \frac{\partial \tau_{xy}}{\partial y} + \frac{\partial \tau_{xz}}{\partial z} + \rho g_x \\ y:\;\; \rho \left(\frac{\partial u_y}{\partial t} + u_x \frac{\partial u_y}{\partial x} + u_y \frac{\partial u_y}{\partial y}+ u_z \frac{\partial u_y}{\partial z}\right) &= -\frac{\partial P}{\partial y} + \frac{\partial \tau_{yx}}{\partial x} + \frac{\partial \tau_{yy}}{\partial y} + \frac{\partial \tau_{yz}}{\partial z} + \rho g_y \\ z:\;\; \rho \left(\frac{\partial u_z}{\partial t} + u_x \frac{\partial u_z}{\partial x} + u_y \frac{\partial u_z}{\partial y}+ u_z \frac{\partial u_z}{\partial z}\right) &= -\frac{\partial P}{\partial z} + \frac{\partial \tau_{zx}}{\partial x} + \frac{\partial \tau_{zy}}{\partial y} + \frac{\partial \tau_{zz}}{\partial z} + \rho g_z. \end{align}

### Cylindrical coordinates

$r:\;\;\rho \left(\frac{\partial u_r}{\partial t} + u_r \frac{\partial u_r}{\partial r} + \frac{u_{\phi}}{r} \frac{\partial u_r}{\partial \phi} + u_z \frac{\partial u_r}{\partial z} - \frac{u_{\phi}^2}{r}\right) = -\frac{\partial P}{\partial r} + \frac{1}{r}\frac{\partial {(r{\tau_{rr})}}}{\partial r} + \frac{1}{r}\frac{\partial {\tau_{\phi r}}}{\partial \phi} + \frac{\partial {\tau_{z r}}}{\partial z} - \frac {\tau_{\phi \phi}}{r} + \rho g_r$
$\phi:\;\;\rho \left(\frac{\partial u_{\phi}}{\partial t} + u_r \frac{\partial u_{\phi}}{\partial r} + \frac{u_{\phi}}{r} \frac{\partial u_{\phi}}{\partial \phi} + u_z \frac{\partial u_{\phi}}{\partial z} + \frac{u_r u_{\phi}}{r}\right) = -\frac{1}{r}\frac{\partial P}{\partial \phi} +\frac{1}{r}\frac{\partial {\tau_{\phi \phi}}}{\partial \phi} + \frac{1}{r^2}\frac{\partial {(r^2{\tau_{r \phi})}}}{\partial r} + \frac{\partial {\tau_{z \phi}}}{\partial z} + \rho g_{\phi}$
$z:\;\;\rho \left(\frac{\partial u_z}{\partial t} + u_r \frac{\partial u_z}{\partial r} + \frac{u_{\phi}}{r} \frac{\partial u_z}{\partial \phi} + u_z \frac{\partial u_z}{\partial z}\right) = -\frac{\partial P}{\partial z} + \frac{\partial {\tau_{z z}}}{\partial z} + \frac{1}{r}\frac{\partial {\tau_{\phi z}}}{\partial \phi} + \frac{1}{r}\frac{\partial {(r{\tau_{rz})}}}{\partial r} + \rho g_z.$

By expressing the shear stress in terms of viscosity and fluid velocity, and assuming constant density and viscosity, the Cauchy momentum equation will lead to the Navier–Stokes equations. By assuming inviscid flow, the Navier–Stokes equations can further simplify to the Euler equations.