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Consider a car driving around in a circle lying in the plane and suppose we were interested in determining its acceleration as measured by an observer stationary on the "ground" or whatever. Introducing cylindrical coordinates is a traditional route for this end.

Thus we introduce cylindrical coordinates in order to describe the riders planar motion. We will have a unit vector $\textbf{e}_r$ along the radius and another unit vector $\textbf{e}_\theta$ perpendicular to $\textbf{e}_r$. These unit vectors "follow" along with the car in the sense that they are always rotating.

So in this system, and as well as for the stationary system of the observer, the acceleration of the car is given by:

$$ \boldsymbol{a} = (\ddot{r} - r \dot{\theta}^2) \textbf{e}_r + (r \ddot{\theta} + 2\dot{r}\dot{\theta})\textbf{e}_\theta $$

The term $2 \dot{r} \dot{\theta} \textbf{e}_\theta $ is the Coriolis acceleration.

Now, I wonder why it does in fact appear in the expression. Because from my understanding, a rotating system is not inertial. How can then measurements of the car's acceleration made in that rotating, non-inertial system perfectly account for and coincide with the car's acceleration as observed in the stationary observer's system?

What perplexes is me is that we are using a rotating and non-inertial system, i.e. the cylindrical coordinate system, and we make calculations in it that happen to satisfactorily describe the acceleration as the stationary observer's non-rotating and inertial system would measure them. What!?

Now clearly since the Coriolis acceleration is in fact apparent in the expression for acceleration tells us that the rotating system is inertial, so the observations made in that rotating system is the same as the stationary one. How? I am obviously missing something here. What?

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  • $\begingroup$ If the angle theta is measured relative to, say, North, then isn't the cylindrical system also inertial? $\endgroup$ – DJohnM Jan 19 '15 at 23:12
  • $\begingroup$ I'm not sure I understand exactly what you mean. For all I know, the system is rotating and that stirs things up a bit. I guess it does turn out to be inertial, I just do not understand how. $\endgroup$ – larrydavid Jan 19 '15 at 23:14
  • $\begingroup$ Note: You have a sign error. The Coriolis acceleration is $-2\dot r \dot \theta \mathbf e_\theta$. This might be the source of your confusion. $\endgroup$ – David Hammen Jan 20 '15 at 0:46
  • $\begingroup$ Hm, really? Look at this wiki article: en.wikipedia.org/wiki/Polar_coordinate_system#Vector_calculus $\endgroup$ – larrydavid Jan 20 '15 at 0:55
  • $\begingroup$ Really. You're looking at things from the wrong perspective. Every term on the righthand side except the $\ddot r \mathbf e_r$ term needs to be moved to the lefthand side. Do that and you'll get $a - 2\dot r \dot \theta \mathbf e_\theta + \text{other terms} = \ddot r \mathbf e_r$. That's the source of your sign error in Coriolis acceleration. (Note well: Your expression for $a$ is correct. It's just what you called Coriolis acceleration that's wrong.) $\endgroup$ – David Hammen Jan 20 '15 at 1:47
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Let's be clear here.

You have a car rotating around a center in a circle at constant angular velocity and, of course, constant radius.

You use a Cartesian frame of reference with its origin at the center of the circle, the z-axis vertical, the x-axis East and the y-axis North. Call it Frame #1

In this frame of reference, all of Newton's laws hold true, and any accelerations in the x, y, or z directions at any particular moment can be explained by the x, y, and z components at that moment of all of the real forces that are plainly present and visible: gravity, the friction between the tires and the road, the reaction force of the (possibly) banked ground. You can write out all the equations of motion, and they will be true for the observed motion and forces.

Now you chose to use a cylindrical polar frame of reference. Call it Frame #2. In this frame of reference, the z axis vertical, and angles are measured counter-clockwise from East. In this frame of reference, the motion of the car is easier to describe: z = 0, r= Constant, and the angular velocity is constant.

This is not a rotating frame of reference.

There are transformations that allow you to replace any instance of x or its derivatives with a function of only r, $\theta$ and z and their derivatives. The same is true of y and z. So the Cartesian versions of Newton's Laws of Motion would each be converted into a purely Cylindrical Polar version of the same equations. Newton's Laws would still hold.

A third possibility is to use CylPol co-ordinates in which the angles are measured counter-clockwise from the position of the car! Now you are really in trouble. You have all the real forces acting on the car, they don't balance out, and the car is sitting absolutely positively still. No motion, no acceleration. Now you need some fictitious forces...

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  • $\begingroup$ So the key idea is that every vector $(r,\theta)$ given in the cylindrical can always be converted to a vector $(x,y)$ in the Cartesian system i.e. the non-rotating frame of reference. But is this not always true, whether the rotating frame of reference be a cylindrical one or whatnot? Does this mean that no rotating frame of reference is a rotating frame of reference simply because there always are transformations back to the Cartesian system? By the way, fantastic outline in your whole answer. I think I am on the verge of finally understanding this, just need to figure a few things out $\endgroup$ – larrydavid Jan 20 '15 at 0:11
  • $\begingroup$ How is it that the cylindrical polar frame of reference is not a rotating frame of reference? Sure, all coordinates in the polar frame can be transformed into the Cartesian frame but why does that mean it is not a rotating frame? Take any rotating frame and you can transform it's coordinates into Cartesian coordinates, so when exactly is the frame actually a rotating reference? $\endgroup$ – larrydavid Jan 20 '15 at 19:15
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So in this system, and as well as for the stationary system of the observer, the acceleration of the car is given by:

$$ \boldsymbol{a} = (\ddot{r} - r \dot{\theta}^2) \textbf{e}_r + (r \ddot{\theta} + 2\dot{r}\dot{\theta})\textbf{e}_\theta $$

The term $2 \dot{r} \dot{\theta} \textbf{e}_\theta $ is the Coriolis acceleration.

The car's acceleration in the rotating system and stationary system are not the same. The terms involving $\dot r$ and $\ddot r$ vanish if the car is going in a circle and the rotating observer is at the center of the circle. The acceleration in this case is identically zero in the rotating system but is nonzero in the stationary system.

The stationary observer only needs to know the frictional forces at the wheels of the car, the aerodynamic drag on the body of the car, and the acceleration from the car's engine to explain the car's motion. The stationary observer sees no Coriolis effect. That's a fictitious effect needed only by the rotating observer, and only if the rotating observer wants to use Newton's second law to explain the car's motion. This is one of several fictitious forces that arise in non-inertial frames.

What perplexes is me is that we are using a rotating and non-inertial system, i.e. the cylindrical coordinate system, and we make calculations in it that happen to satisfactorily describe the acceleration as the stationary observer's non-rotating and inertial system would measure them. What!?

There's no magic here. Those fictitious forces were specifically defined in a manner that allows non-inertial observers to describe motion via Newton's second law.


Update: Using Newton's Second Law in a Non-Inertial Frame

Suppose an inertial observer (I'm ignoring the rotation of the Earth) is in an observation stand, situated right above the circular track. The inertial observer knows about the individual forces acting on the car (wheel friction from turning the steering wheel, aerodynamic drag, force from engine torque (which also acts through the wheels)), sums these forces vectorially, and uses $\mathbf F = m\mathbf a$ to find the acceleration of the car.

Suppose another observer at the center of the track rotates so that the car appears to be stationary. Although there is a net horizontal force on the car, the car's acceleration from the rotating observer's perspective is zero (the car is stationary). Even worse, the rotating observer sees the inertial observer as circling around the track, opposite the rotating observer's rotation. Obviously a naive application of $\mathbf F = m\mathbf a$ doesn't work for the rotating observer. Newton's second law can be made to work by adding some fictional forces.

Let' make the track oval instead of circular. Now the rotating observer does see some acceleration. The car gets closer to and further from the observer as the car goes around the track. The acceleration as observed by the rotating observer is $\mathbf a_\text{rotating} = \ddot r \mathbf e_r$. The acceleration of the car as observed by the inertial observer, transformed to the rotating observers reference frame, is $\mathbf a_\text{inertial} = (\ddot r - r\dot\theta^2)\mathbf e_r + (r\ddot\theta + 2\dot r \dot\theta)\mathbf e_\theta$. (Note that this is the same expression in the opening question). The very first term, $\ddot r \mathbf e_r$, is the acceleration observed by the rotating observer. Using Newton's second law, this can be rewritten as $$\frac {\mathbf F_\text{ext}} m = \mathbf a_\text{rotating} - r\dot\theta^2\mathbf e_r + (r\ddot\theta + 2\dot r \dot\theta)\mathbf e_\theta$$ or $$\mathbf F_\text{ext} + mr\dot\theta^2\mathbf e_r - mr\ddot\theta\mathbf e_\theta - 2m\dot r \dot\theta)\mathbf e_\theta = m\mathbf a_\text{rotating}$$ Denote $$\begin{aligned} &\mathbf F_\text{centrifugal} && \equiv mr\dot\theta^2\mathbf e_r \\ &\mathbf F_\text{coriolis} && \equiv -2m\dot r \dot\theta)\mathbf e_\theta \\ &\mathbf F_\text{euler} && \equiv -mr\ddot\theta\mathbf e_\theta \\ &\mathbf F_\text{tot} && \equiv \mathbf F_\text{ext} + \mathbf F_\text{centrifugal} + \mathbf F_\text{coriolis} + \mathbf F_\text{euler} \end{aligned}$$

With this, the expression that relates force and acceleration observed by the rotating observer simplifies to $$\mathbf F_\text{tot} = m\mathbf a_\text{rotating}$$ Newton's second law!

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  • $\begingroup$ But it is written in terms of $\textbf{e}_r$ and $\textbf{e}_\theta$ which are the unit vectors for the rotating cylindrical system. Why then is e.g. the Coriolios term present? Should it not vanish? Is that fictious force observed by the rotating or non-rotating system? For one of them it has to vanish no? $\endgroup$ – larrydavid Jan 19 '15 at 23:55
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Having pondered this for a while, I think the simplest answer must be that the formula $$\boldsymbol{a} = (\ddot{r} - r \dot{\theta}^2) \textbf{e}_r + (r \ddot{\theta} + 2\dot{r}\dot{\theta})\textbf{e}_\theta$$ can only be produced when we differentiate with respect to the inertial observer. It is only due to our differentiation of the terms in the inertial system that we produce the Coriolis terms and so forth.

So basically, it is in fact an inertial result since it was differentiated in the inertial system, despite the fact that the position vector was written in the rotating polar coordinate system.

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In the Euclidean space a point is often described with Cartesian coordinates.This coordinates have unit vectors not changing with time and position: $\bf{e}_{x}$, $\bf{e}_{y}$ and $\bf{e}_{z}$. There are alternative descriptions to describe this point more conveniently depending on the problem.

Cylindrical coordinates: Wikipedia link (Cylindrical)

With unit vectors $\bf{e}_{\theta}$, $\bf{e}_{r}$ and $\bf{e}_{z}$

Spherical coordinates: Wikipedia link (Spherical)

With unit vectors $\bf{e}_{\theta}$, $\bf{e}_{r}$ and $\bf{e}_{z}$

We can note that this unit vectors do change with time and with position respect to the origin.

If we describe the position of a point in space with Cartesian or with Cylindrical:

$$P=x {\bf{e}_{x}}+y {\bf{e}_{y}} + z {\bf{e}_{z}}= r {\bf{e}_{\theta}}+\theta {\bf{e}_{r}}+z \bf{e}_{z}$$

You can look at this link for the derivation of $\dot{P}$ and $\ddot{P}$, where:

$$\dot{P}=\dot{x} {\bf{e}_{x}}+\dot{y} {\bf{e}_{y}} + \dot{z} {\bf{e}_{z}}= \dot{r} {\bf{e}_{\theta}}+r\dot{\theta} {\bf{e}_{r}}+\dot{z} \bf{e}_{z}$$

$$\ddot{P}=\ddot{x} {\bf{e}_{x}}+\ddot{y} {\bf{e}_{y}} + \ddot{z} {\bf{e}_{z}}= (\ddot{r} - r\dot{\theta}^2) {\bf{e}_{\theta}}+(2\dot{r}\dot{\theta} +r\ddot{\theta}){\bf{e}_{r}}+\ddot{z} \bf{e}_{z}$$

As you can see I'm only describing the kinematics of a point with respect to a static origin. That means that I'm in an inertial reference frame measuring the position, velocity and acceleration only with a new kind of ruler.

I'm not in a rotating reference frame, if you are interested in that kind of non-inertial reference frame go to this link.

In your example with the car, if the car is rotating respect a point. I can describe his motion with cylindrical coordinates, but as I said earlier I'm only using other kind of ruler. I can also use Cartesian coordinates, but things will be much more complex.

Saying this, as I'm in an stationary inertial reference frame newton laws hold with no corrections. This Coriolis acceleration is a True as you can see from the derivation, if you want to know why is sometimes called a fictitious acceleration read this link. But that is when I'm in a non-inertial reference frame, in particular rotating.

This might be a misunderstanding if you think this is a fictitious acceleration in all reference frames, is like saying that I measure all the times in Cartesian coordinates a "Fictitious" acceleration when I'm in an accelerated Train. That is because I call it "Train acceleration" and has the same mathematical notation as the true real acceleration measured from an inertial system.

Fictitious train acceleration: $\vec{a}_{\text{tr}}=\ddot{\vec{a}}$

Inertial acceleration: $\vec{a}=\ddot{\vec{a}}$

They are the same because the letter at the end is the same and I call it Train acceleration and it's fictitious, so it's allays present in Cartesian coordinates. FAKE

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Let's be clear on the terminology here, because taken literally, your question is a bit like asking whether angular momentum is green or non-green.

Polar coordinates exist in space. Curves of constant $r$ are circular and curves of constant $\theta$ are radial, in the usual way. Taking their tangent vectors and normalizing them, we get $\hat{\mathbf{e}}_\theta$ and $\hat{\mathbf{e}}_r$ as usual. They don't "follow" the car. They're not even single vectors, because they vary according to location in the Euclidean plane. Therefore, they're vector fields over the entire space.

Therefore, polar coordinates constitute neither an inertial frame nor a non-inertial frame because they're not a frame.

A frame would consist of vector fields over spacetime, but because time is really simple in Newtonian mechanics, we can just think of this as mapping from time $t$ to a pair of smooth vector fields over space. If we want to use polar coordinates here, then: $$t\mapsto (\hat{\mathbf{e}}_r(t),\hat{\mathbf{e}}_\theta(t))\text{.}$$ Think of the frame as the assignment from time to vector fields over space. This assignment can be inertial or non-inertial.

Now, I wonder why [the Coriolis term $2 \dot{r} \dot{\theta}\hat{\mathbf{e}}_\theta$] does in fact appear in the expression. Because from my understanding, a rotating system is not inertial. How can then measurements of the car's acceleration made in that rotating, non-inertial system perfectly account for and coincide with the car's acceleration as observed in the stationary observer's system?

So, you're essentially asking why the equation $\ddot{\mathbf{x}} = (\ddot{r} - r\dot{\theta}^2) \hat{\mathbf{e}}_r + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\mathbf{e}}_\theta$ is correct either way. The reason is simple: it's correct for polar coordinates in general.

In a co-rotating frame with polar coordinates around the center of the track, the car has $r\dot\theta = 2\dot{r}\dot{\theta} = 0$ because $\dot\theta = 0$ for the car's trajectory. Hence, we have to add centrifugal and Coriolis forces to our Newton's second law in order to "fix" this.

Now clearly since the Coriolis acceleration is in fact apparent in the expression for acceleration tells us that the rotating system is inertial, ...

It does not, because the expression is completely frame-agnostic. It doesn't concern frames at all.

... so the observations made in that rotating system is the same as the stationary one. How? I am obviously missing something here. What?

In the above, I have intentionally avoided equating the coordinate acceleration $\ddot{\mathbf{x}}$ with the acceleration $\mathbf{a}$, because I wanted reserve the latter for the absolute acceleration. If we assign those interpretations for these symbols, then (1) in an inertial frame, $\mathbf{a} = \ddot{\mathbf{x}}$, while (2) in the co-rotating frame, $\mathbf{a} \neq \ddot{\mathbf{x}}$ because you have to add the centrifugal and Coriolis terms of the car (frame) to Newton's second law.

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