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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.

Qmechanic
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Andrew Paul
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2 Answers2

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It’s a very fair question. The answer is that that’s not the Coriolis force, but it is related.

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.

So to be clear, what you derived is just “how you look at motion in an inertial reference frame, in polar coordinates, with polar unit vectors.” But it allows you to very quickly see what happens in a rotating reference frame.

In terms of physical meaning, if your motion stays approximately circular as you slowly change radius, then $v = r\dot \theta$ and $\dot v=\dot r\dot\theta+r\ddot\theta$, if we imagine you keep the same velocity $v$ then this would massage into an equation$$ \ddot\theta = -{r\dot\theta\over r},\\ \Delta\dot \theta=-{v~\Delta r\over r^2}= v~\Delta\left({1\over r}\right),$$ which you already may have known, because if I had just told you “$v, r_1,$ and $v$ is purely in the angular direction” you could have told me, “oh, then $\dot\theta_1 = v/r_1,$ that's just how we define our measure of angles in radians, $v$ is a distance (per time) along a circumference and radians (per time) are distance along circumference, divided by radius.” So this $\dot r\dot\theta$ term is literally just physically telling you that the same speed subtends smaller angles at higher radii, which only matters if both are changing simultaneously.

CR Drost
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In an inertial reference frame using polar coordinates, the radial and tangential unit vectors (magnitude 1) do not have a fixed direction and change position as the polar angle changes. This leads to a term in the tangential acceleration that is sometimes called the "Coriolis acceleration"; it appears because the unit vectors do not have a constant direction. (Cartesian unit vectors do not change direction.)

Just as the Coriolis force is associated with a non-inertial rotating reference frame, this "Coriolis acceleration" is associated with unit vectors that rotate with position even in an inertial reference frame.

John Darby
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