# How would we explain meteorological phenomena in an inertial reference frame?

The Coriolis effect is used to explain the formation of typhoons, hurricanes, and cyclones and other phenomena where we take the Earth as a rotating reference frame. How could we explain this phenomena in an inertial reference frame?

Disclaimer: borrowed from wikipedia with minor modifications.

The acceleration is the second time derivative of position, $$\mathbf{a}_{\mathrm{i}} \ \stackrel{\mathrm{ }}{=}\ \left( \frac{d^{2}\mathbf{r}}{dt^{2}}\right)_{\mathrm{i}} = \left( \frac{d\mathbf{v}}{dt} \right)_{\mathrm{i}} = \left[ \left( \frac{d}{dt} \right)_{\mathrm{r}} + \boldsymbol\Omega \times \right] \left[ \left( \frac{d\mathbf{r}}{dt} \right)_{\mathrm{r}} + \boldsymbol\Omega \times \mathbf{r} \right] \ ,$$ where subscript $i$ means the inertial frame of reference. Carrying out the differentiations and re-arranging some terms yields the acceleration in the rotating reference frame $$\mathbf{a}_{\mathrm{r}} = \mathbf{a}_{\mathrm{i}} - 2 \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}} - \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r}) - \frac{d\boldsymbol\Omega}{dt} \times \mathbf{r}$$ where $$\mathbf{a}_{\mathrm{r}} \ \stackrel{\mathrm{}}{=}\ \left( \frac{d^{2}\mathbf{r}}{dt^{2}} \right)_{\mathrm{r}}$$ is the apparent acceleration in the rotating reference frame, the term $$-\boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r})$$ represents centrifugal acceleration, and the term $$-2 \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}}$$ is the coriolis acceleration.

When the expression for acceleration is multiplied by the mass of the particle, the three extra terms on the right-hand side result in fictitious forces in the rotating reference frame, that is, apparent forces that result from being in a non-inertial reference frame, rather than from any physical interaction between bodies. Using Newton's second law of motion $\mathbf{F}=m\mathbf{a}$, we obtain:

the Coriolis force

$$\mathbf{F}_{\mathrm{Coriolis}} = -2m \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}}$$

the centrifugal force

$$\mathbf{F}_{\mathrm{centrifugal}} = -m\boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r})$$ and the Euler force

$$\mathbf{F}_{\mathrm{Euler}} = -m\frac{d\boldsymbol\Omega}{dt} \times \mathbf{r}$$ where $m$ is the mass of the object being acted upon by these fictitious forces.

• I should've remembered that equation, I just read it yesterday. – Oscar Flores Nov 10 '14 at 23:27
• Take it easy, I usually forget much more basic things :) – Wolphram jonny Nov 10 '14 at 23:32