# Earth Cubed

## Coriolis Forces

A derivation  for coriolis forces can be found be found on the wikipedia page for fictitious forces.

In general for an accelerating reference frame in rectangular coordinates the factious forces are given by:

$\mathbf{F}_{\mbox{fictitious}} =-m\ \mathbf{a}_{AB} -2m\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} - m\ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ .$

Where:

he first term is the Coriolis force, the second term is the centrifugal force, and the third term is the Euler force. When the rate of rotation doesn’t change, as is typically the case for a planet, the Euler force is zero.

Looking specifically at the Coriolis force:

$- 2 m \boldsymbol \omega \times \mathbf{v}$

which gives in (east-west,north-sourth, height) coordinates:

$\boldsymbol{ \Omega} = \omega \begin{pmatrix} 0 \\ \cos \varphi \\ \sin \varphi \end{pmatrix}\ ,$ $\boldsymbol{ v} = \begin{pmatrix} v_e \\ v_n \\ v_u \end{pmatrix}\ ,$

$\boldsymbol{ a}_C =-2\boldsymbol{\Omega \times v}= 2\,\omega\, \begin{pmatrix} v_n \sin \varphi-v_u \cos \varphi \\ -v_e \sin \varphi \\ v_e \cos\varphi\end{pmatrix}\ .$

Where $\phi$ is the latitudinal coordinate (equator=zero latitude).

In general the cross product for a coordinate with orthonormal direction vectors is given by:

$\boldsymbol{\Omega \times v} = \begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\ \Omega_x & \Omega_y & \Omega_z \\ v_x & v_y & v_z \end{vmatrix}\ = \begin{pmatrix} \Omega_y v_z - \Omega_z v_y \\ \Omega_z v_x - \Omega_x v_z \\ \Omega_x v_y - \Omega_y v_x \end{pmatrix}\ ,$
http://en.wikipedia.org/wiki/Coriolis_effect#Formula

since the basis direction vectors are orthogonal in hopkins and simmons coordinates write write:

$\boldsymbol{\Omega \times v} = \begin{vmatrix} \boldsymbol{e_{\sigma}}&\boldsymbol{e_{\mu}}&\boldsymbol{e_{\lambda}} \\ \Omega_{\sigma} & \Omega_{\mu} & \Omega_{\lambda} \\ W & V & U \end{vmatrix}\ = \begin{pmatrix} \Omega_{\mu} U - \Omega_{\lambda} V \\ \Omega_{\lambda} W - \Omega_{\sigma} U \\ \Omega_{\sigma} V - \Omega_{\mu} W \end{pmatrix}\ ,$

(note with regards to weather the system is right handed we can choose the direction of the logitudanal cordinate $\lambda$ to make it right handed.)

Just to recall from the post (Hoskins and Simmons (1974) Coordinate System):

$\mu = sin( \theta )$ where theta is the latitude.
$\sigma = pressure/P_*$ Where $P_*$ is the surface pressure and $\sigma$ is the vertical coordinate.
$\lambda$ is the longitude.

U is the longitudinal component of the velocity
V is the latitudinal component of the velocity
W is the vertical component of the veolicty (not used in Hopkins and Simmons 1974)

Now the angular velocity of the earth in Hopkins and Simmons is given by:

$\boldsymbol{ \Omega} = \omega \begin{pmatrix} \cos \varphi \\ \sin \varphi \\ 0 \end{pmatrix}\ = \omega \begin{pmatrix} \mp \sqrt{1-\mu^2} \\ \mu \\ 0 \end{pmatrix}\,$ $\boldsymbol{ v} = \begin{pmatrix} W \\ V \\ U \end{pmatrix}\ ,$

Where the sign of $\pm$ is positive for the northern hemisphere and negative for the southern hemisphere.

Therefore:

$\boldsymbol{\Omega \times v} = \begin{pmatrix} \mu U \\ \pm \sqrt{1-\mu^2} U \\ \mp \sqrt{1-\mu^2} V - \mu W \end{pmatrix} \$

The result obtained is essentialy the same result that one would get if, they took the(east west,north south, altitude) coordinate system and replaced $\phi$ with $asin(\mu)$ .

The only differences are the order and sign of the components.  These are the only differences because both coordinate contain the same unit vectors.  In my example of a Hopkins and Simon’s like coordinate system I used a different order for the components then was used in my example for the (east-west, north south altitude) coordinate system. This will effect the sign in the cross product.

I wrote the z component of the angular velocity as $\mp \sqrt{1-\mu^2}$ to emphasize that the positive direction for the z component in Simpons coordinate system  is downward. However, the actual angular rotation of the earth in simons coordinate system still have a postive $\mu$ component depending on the which direction is defined as positive for the longitudinal  coordinate.

The order which we specify the coordinates determines the right handedness of the coordinate system.  Therefore, righthandedness is not inherently a geometric property because it depends on the order of the coordinates. For instance, in standard Cartesian coordinates $e_z \times e_y = -e_x$

In our case the first coordinat,e $\sigma$ , was specified in the downward direction, our second coordinate,  $\mu$, points south, now using the right hand  rule means that gives the positive direction for the third coordinate $\lambda$ in the east direction.

It is for these reasons that differences can arrise, and therefore it is very important when doing cross products to clearly express the postive direction of the coordinate unit vectors and the order of the coordinates.

August 29, 2009 - Posted by | GCM (General Circulation Models)