Coriolis Field - Mathematical Expression

Mathematical Expression

Being is the angular velocity vector of the rotating frame, the speed of a test particle used to meassure the field, and using the expression of the acceleration in a rotating reference frame, it is known that the acceleration of the particle in the rotating frame is:

$\mathbf{a}_{\mathrm{r}} = \mathbf{a}_{\mathrm{i}} - 2 \boldsymbol\omega \times \mathbf{v} - \boldsymbol\omega \times (\boldsymbol\omega \times \mathbf{r}) - \frac{d\boldsymbol\omega}{dt} \times \mathbf{r}$

the Coriolis force is assumed to be the fictitious force that compensates the second term:

$\mathbf{F}_{\mathrm{Coriolis}} = -2m ( \boldsymbol\omega \times \mathbf{v}) = -2 ( \boldsymbol\omega \times \mathbf{p})$

Where denotes the linear momentum. It can be seen that for any object, the coriolis force over it is proportional to its momentum vector. As vectorial product can be expressed in a tensorial way using the Hodge dual of :

$\mathbf{F}_{\mathrm{Coriolis}} = -2(\mathbf{\omega} \times \mathbf{p}) = -2(\mathbf{\omega} \times) \mathbf{p} = \begin{bmatrix}\,0&\!-2\omega_3&\,\,2\omega_2\\ \,\,2\omega_3&0&\!-2\omega_1\\-2\omega_2&\,\,2\omega_1&\,0\end{bmatrix}\begin{bmatrix}p_1\\p_2\\p_3\end{bmatrix}$

This matrix can be seen as a constant tensor field, defined in the whole space, that will yield coriolis forces when multiplied by momentum vectors.

Read more about this topic: Coriolis Field

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