# Earth Cubed

## The Cross Product in Non Orthogonal Coordinate Systems

The form of the cross product I’ve shown in my post Coriolis Forces is:

$\boldsymbol{\Omega \times v} = \begin{pmatrix} \Omega_y v_z - \Omega_z v_y \\ \Omega_z v_x - \Omega_x v_z \\ \Omega_x v_y - \Omega_y v_x \end{pmatrix}$

The components of this cross product can be written as follows:

$\left[ \boldsymbol{\Omega \times v} \right]_x= \left[ \begin{matrix}\Omega_x&\Omega_y&\Omega_z \end{matrix} \right]\left[ \begin{matrix}0&0&0 \\ 0&0&1 \\ 0&-1&0\end{matrix} \right]\left[ \begin{matrix}v_x\\v_y\\v_z \end{matrix} \right]$

$\left[ \boldsymbol{\Omega \times v} \right]_y= \left[ \begin{matrix}\Omega_x&\Omega_y&\Omega_z \end{matrix} \right]\left[ \begin{matrix}0&0&-1 \\ 0&0&0 \\ 1&0&0\end{matrix} \right]\left[ \begin{matrix}v_x\\v_y\\v_z \end{matrix} \right]$

$\left[ \boldsymbol{\Omega \times v} \right]_z= \left[ \begin{matrix}\Omega_x&\Omega_y&\Omega_z \end{matrix} \right]\left[ \begin{matrix}0&1&0 \\ -1&0&0 \\ 0&0&0\end{matrix} \right]\left[ \begin{matrix}v_x\\v_y\\v_z \end{matrix} \right]$

We will abbreviate these relationships as follows:

$\left[ \boldsymbol{\Omega \times v} \right]_x=\Omega^T[\times]_xv$

$\left[ \boldsymbol{\Omega \times v} \right]_y=\Omega^T[\times]_yv$

$\left[ \boldsymbol{\Omega \times v} \right]_z=\Omega^T[\times]_zv$

Now define the coordinate transform:

$\boldsymbol{r'}=T \boldsymbol{r}$
where
$\boldsymbol{r}=\left[\begin{matrix} x \\ y \\ z \end{matrix} \right]$

Then the cross product components can be written as follows:

$\left[ \boldsymbol{\Omega \times v} \right]_x=(\Omega^TT^T)((T^{-1})^T[\times]_xT^{-1})(Tv)$

$\left[ \boldsymbol{\Omega \times v} \right]_y=(\Omega^TT^T)((T^{-1})^T[\times]_yT^{-1})(Tv)$

$\left[ \boldsymbol{\Omega \times v} \right]_z=(\Omega^TT^T)((T^{-1})^T[\times]_zT^{-1})(Tv)$

Now Right multiplying the matrix $\left[ \boldsymbol{\Omega \times v} \right]$ by the transform $T$ gives:

$\left[ \boldsymbol{\Omega \times v} \right]_{x'}=\sum_{k \in \{x,y,z\}}T_{1,k}(\Omega^TT^T)((T^{-1})^T[\times]_kT^{-1})(Tv)$

$\left[ \boldsymbol{\Omega \times v} \right]_{y'}=\sum_{k \in \{x,y,z\}}T_{2,k}(\Omega^TT^T)((T^{-1})^T[\times]_kT^{-1})(Tv)$

$\left[ \boldsymbol{\Omega \times v} \right]_{z'}=\sum_{k \in \{x,y,z\}}T_{3,k}(\Omega^TT^T)((T^{-1})^T[\times]_kT^{-1})(Tv)$

Which can be written in this form:
$\left[ \boldsymbol{\Omega \times v} \right]_{x'}=\Omega'^T[\times]_{x'}v'$

$\left[ \boldsymbol{\Omega \times v} \right]_{y'}=\Omega'^T[\times]_{y'}{v'}$

$\left[ \boldsymbol{\Omega \times v} \right]_{z'}=\Omega^T[\times]_{z'}{v'}$

Where:

$\left[[\times]_{x'}\right]_{i,j}=\sum_{k \in \{x,y,z\}}T_{1,k}[((T^{-1})^T[\times]_kT^{-1})]_{i,j}$

$\left[[\times]_{y'}\right]_{i,j}=\sum_{k \in \{x,y,z\}}T_{2,k}[((T^{-1})^T[\times]_kT^{-1})]_{i,j}$

$\left[[\times]_{z'}\right]_{i,j}=\sum_{k \in \{x,y,z\}}T_{3,k}[((T^{-1})^T[\times]_kT^{-1})]_{i,j}$

$\Omega'=T\Omega$
$v'=Tv$

September 7, 2009 Posted by | Math | 2 Comments