Author: Rebecca Ritter

Considering Newtons's 2nd Law: \begin{equation*} {m} \vec{a} = \vec{F} \end{equation*} We get for the Mass: \begin{equation*} {M} = \rho \delta{x}\delta{y}\delta{z} \end{equation*} And the material derivative stands for the acceleration: \begin{equation*} \frac{D\vec{u}}{Dt}\end{equation*} Therefore the momentum equation looks like: \begin{equation*} \rho \delta{x}\delta{y}\delta{z}\frac{D\vec{u}}{Dt} = \vec{F} \end{equation*}

$ \vec{F} $ = net force on an elementary fluid parcel of infinitesimal dimensions in the three coordinate directions (the parcel is moving with velocity $ \vec{u} $).

We destinguish three different forces, which are part of the momentum equation:
1. The gravitational force $ \vec{F}_G $
2. The Pressure (compressive stress) $ \vec{F}_P $
3. The frictional force $ \vec{F}_R $ (in oceanic flows, frictional effects are negligible except close to boundaries)

\vec{F}_G = {M}\vec{g} = \rho \delta{x}\delta{y}\delta{z} \vec{g} $|

\vec{F}_P = -\nabla p \delta{x}\delta{y}\delta{z} $|

\frac{D\vec{u}}{DT} = \frac{1}{\rho\delta{x}\delta{y}\delta{z}} \Bigl( \vec{F}_G + \vec{F}_P + \vec{F}_R \Bigr) $ \\ \\ $ \frac{D\vec{u}}{DT} = \vec{g} - \frac{1}{\rho}\nabla {p} + \frac{1}{\rho} \nabla\tau $|

• H. Wernli, S. Pfahl (2013), Script: Introduction to Environmental Fluid Dynamics.

Write out the equations in $x,y,z$ coordinates. Try to explain the equation in words.