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I Is polar coordinate system non inertial?

  1. Mar 30, 2016 #1
    Studying the acceleration expressed in polar coordinates I came up with this doubt: is this frame to be considered inertial or non inertial?

    [itex] (\ddot r - r\dot{\varphi}^2)\hat{\mathbf r} + (2\dot r \dot\varphi+r\ddot{\varphi}) \hat{\boldsymbol{\varphi}} [/itex] (1)

    I do not understand what is the correct explanation for the presence of terms that seem the ones found in non inertial frames, like [itex]2\dot r \dot\varphi[/itex], mathematically equal to the Coriolis term present in the expression of acceleration for non inertial frames of reference, which I write down here. Does this mean that the polar coordinate system is non inertial?

    [itex] \vec{a}= \vec{a}'+2\vec{\omega}\times\vec{v}'+\dot{\vec{\omega}}\times\vec{r}'+\vec{\omega}\times(\vec{\omega}\times\vec{r}')[/itex] (2)

    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 [itex]x[/itex] axis (that is the axis oriented as the unit vector [itex]\hat{\mathbf r}[/itex] ), which is rotating as far as the particle move away from the radial direction. Is this the correct way to see it?

    I found on Wikpedia (https://en.wikipedia.org/wiki/Polar_coordinate_system#Centrifugal_and_Coriolis_terms) this answer to my question.
    I highlighted the things that confuses me the most. In particular here it is claimed that such coordinate system can be "used in inertial frame of reference", which is obviously against my idea of polar coordinates as non inertial frame. Now my question is: if the polar coordinate system is inertial, then how to interpret the terms that appear in the expression for acceleration (1)?

    Here it says 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 frames? In order to derive the expression for acceleration in non inertial frames (2) a differentiation (which takes in account the variation of unit vectors) is done, nothing more than that.

    Did I misunderstand Wikipedia or am I wrong to consider polar coordinates a non inertial frame of reference? If this is the case, how can I interpret in the proper way those terms in (1)?
     
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  3. Mar 30, 2016 #2

    PeroK

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    I think you have two fundamental misunderstandings here.

    1) Inertial and non-inertial refers to a physical property of a reference frame. The test for this is a physical real-world test. A reference frame is inertial if Newton's first law applies.

    2) Polar coordinates are a mathematical system of assigning a coordinate to each point in a plane. This does not imply any rotation. Moreover, if you use polar coordinates in a physical situation that may or may not be within an inertial reference frame. The same for Cartesian coordinates. There is no mathematical test for "inertial". Cartesian coordinates are neither inertial nor non-inertial in themselves; nor is any mathematical coordinate system.

    Regarding the centrifugal and coriolis components. Think mathematically and kinematically. If a body is accelerating in a polar coordinate system, then you can simply draw a vector in the direction of the acceleration. However, to describe this acceleration in terms of the unit vectors:

    ##\hat{\mathbf r}## and ##\hat{\boldsymbol{\varphi}}##

    Leads to the the formula you have given.

    Now, there is a relation between this formula and the situation in a rotating reference frame. Note that "rotating" here implies a physical property of the system. Mathematically, you can have a Cartesian (or Polar) coordinate system on the surface of the Earth: neither is inertial or non-inertial, per se. But, if the Earth is physically rotating, then any coordinate system at rest with respect to the surface of the Earth must be non-inertial. And, if the Earth was not rotating, then any such coordinate system would be inertial.

    In a rotating reference frame, something at rest or moving with constant velocity (in that frame) must be accelerating (physically). So, that thing (apparently at rest or moving with constant velocity) experiences the forces needed to accelerate it. These are mathematically related to the terms in the polar acceleration formula you are looking at.

    In short, you need to separate: the reference frame (a physical thing whose properties depend on the physics of the situation) and a coordinate system, which simply provides coordinates for each point in your reference frame, hence is neither inertial nor non-inertial, per se.
     
  4. Mar 30, 2016 #3
    Thanks a lot for this answer, it helped a lot to see better the difference between reference frames and coordinates systems, which I did not get before!

    Here you mentioned a relation between the terms appearing in (1) and the situation in a (physically) rotating frame. I'm still not sure about what this relation really is. I mean the terms in (1) comes out in case of rotational motion (where some forces, apparently absent, are needed) and if I consider some examples I can explain to myself the reason of the presence of them, including the Coriolis term. However, if I may ask, would you be so kind as to highlight this general relation between the terms in (1) and (2)?
     
  5. Mar 30, 2016 #4

    PeroK

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    Take the simpler example of the centripetal force for rotational motion:

    a) Imagine you are observing an object rotate at a distance ##R## from the centre. There must be a force acting on that object to keep it in a circular path. That force is:

    [itex] - mR\dot{\varphi}^2 \hat{\mathbf r} [/itex]

    b) Now imagine you are in a rotating reference frame and you observe an object at rest. It must have the same real force acting on it:

    [itex] - mR\dot{\varphi}^2 \hat{\mathbf r} [/itex]

    Holding it in place.

    c) If that object has no real forces acting on it, then it will move inertially with constant velocity. But, in the rotating reference frame it will have fictitious centrifugal and coriolis forces acting on it:

    [itex] +mr\dot{\varphi}^2 \hat{\mathbf r} - m2\dot r \dot\varphi \hat{\boldsymbol{\varphi}}[/itex]
     
    Last edited: Mar 30, 2016
  6. Apr 3, 2016 #5

    Thanks so much for the reply, sorry if I answer now, I thought about this for a while, but I'm still concerned about it. In the end I do not get if there is a substantial difference, for example, between Coriolis terms in (1) and in (2). I thought that, after reading your examples, the Coriolis term in (1) can be seen as a particular case of the term in (2). Nevertheless, while reading Kleppner Kolenkow, page 34 I found this explanation which made me doubt of this.

    Both on Wikipedia page reported above and in the book a distinction is made between the Coriolis terms in (1) and (2), but I still don't see clearly the difference. Could you please highlight this difference?
     
  7. Apr 3, 2016 #6

    PeroK

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    What don't you understand about K & K's explanation? I'm not sure how I can improve on that. The terms are mathematically the same, but they occur in two very different scenarios.
     
  8. Apr 3, 2016 #7
    What is not clear is the difference between the scenarios. I understood that polar coordinates are a just a mathematical way to describe the position on plane, hence they are not inertial or not inertial per se. But take a motion on a spiral: if we describe it in polar coordinates the point moves only radially and [itex]\theta[/itex] changes over time so, supposing to know the acceleration [itex]\vec{a}[/itex] if we solve (1) for [itex]\ddot{r}[/itex] we find that [itex]\ddot{r}[/itex] is equal to [itex]\vec{a}[/itex] plus a centrifugal and a "coriolis" term (suppose that angular velocity does not change). Now if we start again (same spiral motion) but we choose a cartesian coordinate system, but rotating at the same rate (in such way that the point moves only on the [itex]x[/itex] axis), the expression we find for [itex]\ddot{x}[/itex] is exactly the same as to [itex]\ddot{r}[/itex].
    That's not a general thing of course, but, when it comes to motion like spirals and the coriolis and centrifugal term do appear, can (1) be considered as a special case of (2)? If this is not true I think I really miss this (physical) difference between, for istance, the Coriolis term in (1) and the one in (2).
     
  9. Apr 3, 2016 #8

    PeroK

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    Okay. You talk about motion in a spiral. That's scenario 1. You can use the general formula for acceleration in polar coordinates to study that motion. That formula includes a "Coriolis" term.

    Now, you are sitting at rest on the earth which is rotating. That is scenario 2. But, if you start to move, you feel a Coriolis force.

    In particular if you move directly upwards, the expression for this force is similar to the Coriolis term in the first scenario. With ##\dot{r}## your upward speed and ##\dot{\phi}## the angular velocity of the earth.

    Note: Not quite directly upwards unless you are at the equator.

    In general, the Coriolis force applies to 3D motion so has a more general form, which is equation (2) in your first post.
     
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