7.6: Momentum Equation

Newton’s Second Law relates the change of the momentum of a fluid mass due to an applied force. The change is: \[\frac{D (m \mathbf{v})}{Dt} = \mathbf{F} \nonumber \]

where \(\mathbf{F}\) is force, \(m\) is mass, and \(\mathbf{v}\) is velocity. I have emphasized the need to use the total derivative because we are calculating the force on a particle. We can assume that the mass is constant, and \((\PageIndex{1})\) can be written: \[\frac{D \mathbf{v}}{Dt} = \frac{\mathbf{F}}{m} = \mathbf{f}_{m} \nonumber \]

where \(\mathbf{f}_{m}\) is force per unit mass.

Four forces are important: pressure gradients, Coriolis force, gravity, and friction. Without deriving the form of these forces (the derivations are given in the next section), we can write \((\PageIndex{2})\) in the following form. \[\frac{D \mathbf{v}}{Dt} = - \frac{1}{\rho} \nabla p - 2 \mathbf{\Omega} \times \mathbf{v} + \mathbf{g} + \mathbf{F_{r}} \nonumber \]

Acceleration equals the negative pressure gradient minus the Coriolis force plus gravity plus other forces. Here \(\mathbf{g}\) is acceleration of gravity, \(\mathbf{F_{r}}\) is friction, and the magnitude \(\Omega\) of \(\mathbf{\Omega}\) is the Rotation Rate of earth, \(2 \pi\) radians per sidereal day or \[\boxed{\Omega = 7.292 \times 10^{-5} \ \text{radians/s}} \nonumber \]

Momentum Equation in Cartesian coordinates

Expanding the derivative in \((\PageIndex{3})\) and writing the components in a Cartesian coordinate system gives the Momentum Equation: \[\begin{align} \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z} &= - \frac{1}{\rho} \frac{\partial p}{\partial x} + 2 \Omega \ v \sin \varphi + F_{x} \\ \frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} + w \frac{\partial v}{\partial z} &= - \frac{1}{\rho} \frac{\partial p}{\partial y} - 2 \Omega \ u \sin \varphi + F_{y} \\ \frac{\partial w}{\partial t} + u \frac{\partial w}{\partial x} + v \frac{\partial w}{\partial y} + w \frac{\partial w}{\partial z} &= - \frac{1}{\rho} \frac{\partial p}{\partial z} + 2 \Omega \ u \cos \varphi - g + F_{z} \end{align} \nonumber \]

where \(F_{i}\) are the components of any frictional force per unit mass, and \(\varphi\) is latitude. In addition, we have assumed that \(w << v\), so the \(2 \Omega \ w \cos \varphi\) term has been dropped from equation \((\PageIndex{5})\).

The momentum equation appears under various names. Leonhard Euler (1707–1783) first wrote out the general form for fluid flow with external forces, and the equation is sometimes called the Euler equation or the acceleration equation. Louis Marie Henri Navier (1785–1836) added the frictional terms, and so the equation is sometimes called the Navier-Stokes equation.

The term \(2 \Omega \ u \cos \varphi\) in \((\PageIndex{7})\) is small compared with \(g\), and it can be ignored in ocean dynamics. It cannot be ignored, however, for gravity surveys made with gravimeters on moving ships.

Figure \(\PageIndex{1}\): Sketch of flow used for deriving the pressure term in the momentum equation.

Derivation of Pressure Term

Consider the forces acting on the sides of a small cube of fluid (figure \(\PageIndex{1}\)). The net force \(\delta F_{x}\) in the \(x\) direction is \[\begin{align*} \delta F_{x} &= p \ \delta y \ \delta z - (p + \delta p) \delta y \ \delta z \\ \delta F_{s} &= -\delta p \ \delta y \ \delta z \end{align*} \]

But \[\delta p = \frac{\partial p}{\partial x} \delta x \nonumber \]

and therefore \[\begin{align*} \delta F_{x} &= -\frac{\partial p}{\partial x} \delta x \ \delta y \ \delta z \\ \delta F_{x} &= -\frac{\partial p}{\partial x} \delta V \end{align*} \]

Dividing by the mass of the fluid \(\delta m\) in the box, the acceleration of the fluid in the \(x\) direction is: \[a_{x} = \frac{\delta F_{x}}{\delta m} = -\frac{\partial p}{\partial x} \frac{\delta V}{\delta m} \nonumber \] \[\boxed{a_{x} = -\frac{1}{\rho} \frac{\partial p}{\partial x}} \nonumber \]

The pressure forces and the acceleration due to the pressure forces in the \(y\) and \(z\) directions are derived in the same way.

The Coriolis Term in the Momentum Equation

The Coriolis term exists because we describe currents in a reference frame fixed on earth. The derivation of the Coriolis terms is not easy. Henry Stommel, the noted oceanographer at the Woods Hole Oceanographic Institution, even wrote a book on the subject with Dennis Moore (Stommel & Moore, 1989).

Usually, we just state that the force per unit mass, the acceleration of a parcel of fluid in a rotating system, can be written: \[\mathbf{a}_{fixed} = \left(\frac{D \mathbf{v}}{Dt}\right)_{fixed} = \left(\frac{D \mathbf{v}}{Dt}\right)_{rotating} + (2 \mathbf{\Omega} \times \mathbf{v}) + \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{R}) \nonumber \]

where \(\mathbf{R}\) is the vector distance from the center of Earth, \(\mathbf{\Omega}\) is the angular velocity vector of earth, and \(\mathbf{v}\) is the velocity of the fluid parcel in coordinates fixed to Earth. The term \(2 \mathbf{\Omega} \times \mathbf{v}\) is the Coriolis force, and the term \(\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{R})\) is the centrifugal acceleration. The latter term is included in gravity (figure \(\PageIndex{2}\)).

Figure \(\PageIndex{2}\): Downward acceleration \(\mathbf{g}\) of a body at rest on Earth’s surface is the sum of gravitational acceleration between the body and Earth’s mass, \(\mathbf{g}_{f}\), and the centrifugal acceleration due to Earth’s rotation, \(\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{R})\). The surface of an ocean at rest must be perpendicular to \(\mathbf{g}\), and such a surface is close to an ellipsoid of rotation. Earth’s ellipticity is greatly exaggerated here.

The Gravity Term in the Momentum Equation

The gravitational attraction of two masses \(M_{1}\) and \(m\) is: \[\mathbf{F}_{g} = \frac{G M_{1} m}{R^{2}} \nonumber \]

where \(R\) is the distance between the masses, and \(G\) is the gravitational constant. The vector force \(\mathbf{F}_{g}\) is along the line connecting the two masses.

The force per unit mass due to gravity is: \[\frac{\mathbf{F}_{g}}{m} = \mathbf{g}_{f} = \frac{G M_{E}}{R^{2}} \nonumber \]

where \(M_{E}\) is the mass of Earth. Adding the centrifugal acceleration to \((\PageIndex{10)\) gives gravity \(g\) (figure \(\PageIndex{2}\): \[\mathbf{g} = \mathbf{g}_{f} - \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{R}) \nonumber \]

Note that gravity does not point toward Earth’s center of mass. The centrifugal acceleration causes a plumb bob to point at a small angle to the line directed to Earth’s center of mass. As a result, Earth’s surface including the ocean’s surface is not spherical, but it is a oblate ellipsoid. A rotating fluid planet has an equatorial bulge.