# Why does the Coriolis force appear when deriving the forces on a particle in polar coordinates?

I considered a particle in polar coordinates, $$(r,\theta)$$, with mass $$m$$. The standard basis vectors in polar coordinates are: $$\mathbf{\hat{r}}=\cos{\theta}\mathbf{\hat{x}}+\sin{\theta}\mathbf{\hat{y}}$$ And: $$\boldsymbol{\hat{\theta}}=\frac{\partial\mathbf{\hat{r}}}{\partial\theta}=-\sin{\theta}\mathbf{\hat{x}}+\cos{\theta}\mathbf{\hat{y}}$$ Differentiating the vector $$\mathbf{r}$$ to the particle twice, we find that: $$\mathbf{\ddot{r}}=(\ddot{r}-r\dot{\theta}^2)\mathbf{\hat{r}}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\boldsymbol{\hat{\theta}}$$ From which it follows that the radial component of force on this particle is $$F_r=m(\ddot{r}-r\dot{\theta}^2)$$ and the tangential component is $$F_\theta=m(2\dot{r}\dot{\theta}+r\ddot{\theta})$$.

I was able to understand three out of four of the terms in this pair of equations by considering the particle undergoing radial and circular motion (in which case $$\dot{\theta}=0$$ and $$\dot{r}=0$$, respectively).

Incidentally, however, the $$2m\dot{r}\dot{\theta}$$ term is the Coriolis force. But isn't this force fictitious and only observable in a non-inertial reference frame? Was I working in a non-inertial reference frame during this derivation? Does what I'm asking even make sense?

I think I primarily need some clarification of how inertial/non-inertial reference frames come into play in this derivation.

• Assume the radius of circle is $r$ which is constant. Then the tangential component of the acceleration is $a_{t}=r\frac{d\omega}{dt}$ where $\omega=\frac{d\theta}{dt}$ is the angular velocity - and the radial component of the acceleration is $a_{r}=\omega^{2}r$. $a_{r}$ is also known as the centripetal acceleration - it's the fictitious force. You know you're in an non-inertial frame when the origin of your coordinate system is the axis of rotation of the particle you're tracking. A inertial frame is one that moves at constant velocity and doesn't accelerate. – Cinaed Simson Jun 17 at 5:12
• And just to be clear, the magnitude of $\hat r=1$, which is constant, hence $\dot r=\ddot r=0$, which yields $F_r=-mr\dot{\theta}^2$ and $F_\theta=mr\ddot{\theta}$ - which are the correct answers. There is no Coriolis force in 2 dimensions - and there's no vector cross product in 2 dimensions either. $F_r=-mr\dot{\theta}^2$ is the centripetal acceleration - a fictitious force. – Cinaed Simson Jun 17 at 7:35
• @CinaedSimson This should be an answer, not a comment. – J. Murray Jun 17 at 13:45
• @CinaedSimson, what I'm considering is not necessarily circular motion though. The position vector is $\mathbf{r}=r\mathbf{\hat{r}}$. The standard basis vector's magnitude never changes, but $\dot{r}$ may be nonzero because the motion of the particle is arbitrary and so can be radial. – Andrew Paul Jun 17 at 15:45

Suppose you were to now examine the same particle in coordinates rotating about the origin with angular velocity $$\omega$$, this would perform a shift $$\dot\theta\mapsto \dot\theta -\omega$$ while leaving $$r, \dot r, \ddot\theta$$ invariant. As a consequence we would find that $$\mathbf {\ddot r}' = \mathbf{\ddot r} + 2 \omega \left(r \dot\theta ~\hat r-\dot r ~\hat \theta\right) - r \omega^2~\hat r.$$
The first of these terms is the actual Coriolis force $$-2m~\vec\omega\times\vec v$$. The final term is the similarly fictitious centrifugal force.
• This makes sense, thanks! So then, if we imagine $\dot{\theta}=0$ (in the inertial frame), and let a rotating frame have angular velocity $\omega$, then the shift becomes $0\mapsto -\omega$ and hence the term $2m\dot{r}\dot{\theta}$ is just $2m\boldsymbol{\omega}\times\mathbf{v}$, the Coriolis force observed from the rotating frame, as I now understand it. – Andrew Paul Jun 17 at 3:41