# Is polar coordinates frame non-inertial?

Consider the acceleration expressed in polar coordinates.

$\left( \ddot r - r\dot\varphi^2 \right) \hat{\mathbf r} + \left( r\ddot\varphi + 2\dot r \dot\varphi \right) \hat{\boldsymbol{\varphi}} \$

I do not understand what is the correct explanation for the presence of these terms. I have the idea that polar coordinates are just a particular case of a non-inertial rotating frame. The "special" thing about it is that the point is constantly on the $x$ axis (that is the axis oriented unit vector $\hat{\mathbf r}$ ), which is constantly rotating. Is this the correct way to see it?

I found on Wikpedia this answer to my question.

The term $r\dot\varphi^2$ is sometimes referred to as the centrifugal term, and the term $2\dot r \dot\varphi$ as the Coriolis term. Although these equations bear some resemblance in form to the centrifugal and Coriolis effects found in rotating reference frames, nonetheless these are not the same things. In particular, the angular rate appearing in the polar coordinate expressions is that of the particle under observation, $\dot{\varphi}$, while that in classical Newtonian mechanics is the angular rate $Ω$ of a rotating frame of reference. The physical centrifugal and Coriolis forces appear only in non-inertial frames of reference. In contrast, these terms, that appear when acceleration is expressed in polar coordinates, are a mathematical consequence of differentiation; these terms appear wherever polar coordinates are used. In particular, these terms appear even when polar coordinates are used in inertial frames of reference, where the physical centrifugal and Coriolis forces never appear.

I highlighted the things that confuses me the most. In particular here it is claimed that these terms are not to be interpreted as caused by fictitious forces, but that they just come from differentiation. That is true, but isn't it the same for the (real ?) non inertial frame? In order to derive the expression for acceleration in non inertial frames a differentiation (which takes in account the variation of unit vectors) is done.

Moreover it says that polar coordinates "are used in inertial frame of reference", which is obviusly against my idea of polar coordinates as I said.

Did I misunderstand Wikipedia or am I wrong to consider polar coordinates a non inertial frame of reference?

• I guess this can be argued, since in "inertial frame" is a very physical thing, while what I am about to say is mathematical, but to me a frame is inertial if and only if the basis vectors are parellel transported along themselves, or alternatively, if all christoffel symbol/connection form components vanish. This is clearly not the case for polar coordinates, so the coordinate frame of polar coordinates is a noninertial frame. – Bence Racskó Mar 27 '16 at 20:59

The equation you wrote does not assume that the polar coordinate system is rotating. The derivation of the equation you have written starts with expressing a position vector drawn from the origin to the moving particle in the following form:$$\vec{r}=r\hat{r}\tag{1}$$where the radial unit vector $\hat{r}$ is a function of $\theta$, and where r and $\theta$ are functions of time t. This is always the equation for an arbitrary position drawn from the origin to a point in the x-y plane (since it is always pointing in the direction of the unit vector in the radial direction).

If we take the derivative of Eqn. 1 for the position vector with respect to time, we get the velocity vector: $$\vec{v}=\frac{d\vec{r}}{dt}=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{d\theta}\frac{d\theta}{dt}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}\tag{2}$$ If we take the derivative of this equation for the velocity with respect to time, we obtain your equation for the acceleration.

The first thing to do is convince yourself that for a particle in uniform motion, neither it's radius nor it's polar angle can be expected to be constant (in general, radial motion will have constant $\theta$ but that is a special case) or to have a constant first derivative.

• Draw a $x$-$y$ origin on a piece of paper. This is just a place to measure $r$ from and a orientation to use for measuring $\theta$ (counterclockwise from the $+x$ axis as usual).
• For each successive mark on the line (indexed $i$) use a protractor and ruler to find $r_i$ and $\theta_i$. These are the polar coordinates of the object at each successive tick of some clock.
• Plot $r_i$-versus-$i$ and $\theta_i$-versus-$i$ and see that they are not straight lines even though the motion is both straight and uniform. This should be enough to convince you that unlike Cartesian coordinates polar coordinates do not represent uniform motion with constant-derivative equations of motion.