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Rotating frame: a question of interpretation

  1. May 16, 2014 #1

    Philip Wood

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    I'm reading Lanczos: 'The variational Principles of Mechanics'. I need help resolving a paradox - which is probably trivial…

    Lanczos (page 100 Dover edition) introduces a system, S', rotating at angular velocity [itex]\vec \Omega[/itex] about an axis through a fixed point with respect to inertial system S. The radius vectors [itex]\vec R[/itex] and [itex]\vec R'[/itex] in the two systems are, he says, the same: [itex]\vec R = \vec R'[/itex]. [I don't have trouble with this: it's fundamental to the idea of a vector that the same vector can be expressed in terms of different basis vectors, in this case, [itex]\vec i, \vec j, \vec k[/itex] and [itex]\vec i', \vec j', \vec k'[/itex].]

    Nevertheless, he says, the velocities and accelerations measured in both systems differ from each other because rates of change observed in the two systems are different. If a certain vector [itex]\vec B[/itex] is constant in S' it rotates with the system and thus, if observed in S, undergoes in the time dt an infinitesimal change [itex]d \vec B = (\vec \Omega \times \vec B) dt[/itex]. [Again, easily seen - especially for [itex]\vec R[/itex] itself.]

    Hence [itex]\frac{d \vec B}{dt} = (\vec \Omega \times \vec B)[/itex] while at the same time, Lanczos says, [itex]\frac{d' \vec B}{dt} = 0 [/itex].

    Here, Lanczos has introduced the notation [itex]\frac {d'}{dt}[/itex] which refers to the operation of observing the rate of change of a quantity in the moving system S'.

    If you've read as far as this, well done and thank you. Now here's my problem. Regarding [itex]\vec R[/itex] and [itex]\vec R'[/itex] as functions of time, we can surely write:
    [tex]\vec R (t) = \vec R' (t).[/tex]
    But because [itex]\frac{d \vec B}{dt} = (\vec \Omega \times \vec B)[/itex]
    [tex]\vec R (t + dt) = \vec R (t) + (\vec \Omega \times \vec R (t)) dt[/tex]
    while because [itex]\frac{d' \vec B}{dt} = 0 [/itex]
    [tex]\vec R' (t + dt) = \vec R' (t) + 0.[/tex]
    It would therefore seem that at time (t + dt), it is no longer the case that [itex]\vec R = \vec R'[/itex]. Whereas [itex]\vec R'[/itex] has changed over dt, [itex]\vec R'[/itex] has stayed constant.
    Where is my reasoning wrong?
     
    Last edited: May 16, 2014
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  3. May 16, 2014 #2

    Andrew Mason

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    I am not clear on what [itex]\vec{R} \text{ and } \vec{R'}[/itex] are. Are they the vectors from the inertial point to the origins S and S'? In other words, S and S' have a common origin?

    AM
     
  4. May 16, 2014 #3

    UltrafastPED

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    I checked my copy of Lanczos; yes, they have a common origin. The discussion is for rotations with no translation.
     
  5. May 16, 2014 #4

    UltrafastPED

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    Lanczos uses the "superposition principle of infinitesimal processes", while you are not.

    Also note that there is a typo on eq. (45.9) of the fourth edition; there should be a "dot" over the Ω in the final term.
     
  6. May 16, 2014 #5

    Philip Wood

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    Thanks for replying, but with respect, UPED, I don't think either of your points solves my problem. L uses the superposition principle when moving on from a vector which is constant in S' to one which is not necessarily constant in S'. My concern is merely with a vector which is constant in S'. [Incidentally I don't much like L's appeal to a superposition principle; there are, imo, nicer treatments in Kibble or in Synge which differentiate [itex]x_i \vec e_i[/itex] as a product.] Equation 45.9 is also beyond the point where my problem lies.
     
  7. May 19, 2014 #6

    Philip Wood

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    No more ideas? Is this because my question is not understood, or is it because it's baffling you, too?

    [I apologise for the typo in the last-but-one sentence of post 1. This should read: Whereas [itex]\vec R[/itex] has changed over dt, [itex]\vec R'[/itex] has stayed constant.]
     
  8. May 20, 2014 #7

    Andrew Mason

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    I am still not clear on [itex]\vec R[/itex] and [itex]\vec R'[/itex]. I don't have a copy of Lanczos' book.


    Can you post the page in question?

    AM
     
  9. May 20, 2014 #8
    When you say R = R' does that mean they have the same numerical values or does that mean they represent the same physical vector? You can't have it both ways. You're trying to have it both ways and that's why you're getting in trouble.
     
  10. May 21, 2014 #9

    Philip Wood

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    AM and Dauto. Thank you for your interest.

    AM Please find extracts.

    Dauto: I've wondered the same thing. Yet R' = R seems to imply same value.
     

    Attached Files:

  11. May 21, 2014 #10

    Andrew Mason

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    The problem seems to be that we are not sure what [itex]\vec{R}[/itex] and [itex]\vec{R}'[/itex] are.

    Lanczos seems to be using 2 dimensional polar co-ordinates. So [itex]\vec{R}(r,θ)[/itex] differs from [itex]\vec{R}'(r,θ')[/itex] only by angle: i.e. the second co-ordinate - angle θ. If that is the case, his statement:

    [itex]\vec{R}= \vec{R}'[/itex] cannot be true, generally.

    Perhaps he meant to say [itex]|\vec{R}| = |\vec{R}'|[/itex] (i.e. the length (the first co-ordinate, r) is the same for both vectors).

    AM
     
  12. May 21, 2014 #11

    Philip Wood

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    AM Thank you. Yet he has R and R' in bold and refers to them (second para in extracts) as radius vectors And he does seem to be pretty scrupulous about nomenclature. Also by starting the third para in the extract with "Nevertheless" he seems to be pointing to the peculiarity of radius vector behaving differently from velocity and acceleration vectors: there wouldn't be anything worth remarking about if radius wasn't to be considered a vector.
     
  13. May 24, 2014 #12

    Andrew Mason

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    Laczos appears to be referring to [itex]\vec{R} \text{ and } \vec{R}'[/itex] at a particular time e.g. [itex]\vec{R}(0) = \vec{R}'(0)[/itex]. If [itex]\vec{R}'(t) = \vec{R}'(0)[/itex] for all t, then [itex]\vec{R}'[/itex] is just a fixed vector in S'. So, while [itex]\vec{R}(0) = \vec{R}'(0)[/itex], [itex]\vec{R}(t) \ne \vec{R}'(t)[/itex] (unless [itex]\dot{Ω}t = n2\pi[/itex]).

    AM
     
  14. May 24, 2014 #13

    Philip Wood

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    AM Yes, this interpretation is free of inconsistentency. I was loathe to accept it because to claim that [itex]\vec R = \vec R'[/itex] (only) at one particular time seemed such a weak claim. Thanks to your post, I'll try to stop worrying about the issue. I have a great gift for getting stuck when trying to learn; it has not faded with advancing years.

    Edit: Fickle to the last, I've now gone for a different interpretation. See post 15. Many thanks!
     
    Last edited: May 24, 2014
  15. May 24, 2014 #14

    D H

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    Surely you cannot.

    Imagine a merry-go-round with a hole in the middle, big enough so that you can stand on the ground in the middle of the merry-go-round and watch the horses move around you. The horses are moving. From this perspective, the position of some particular horse as a function of time is ##R(t) = R\cos(\omega t)\hat x + R\sin(\omega t)\hat y##.

    Alternatively, you can jump up on the merry-go-round, put a plank over that hole, and position yourself exactly as before, except now you are rotating with the merry. From this perspective, that horse is stationary: ##R'(t) = R\hat x'##. The horse's velocity is zero from this rotating perspective. It's obviously not zero from the perspective of the non-rotating observer.


    I don't have Lanczos, but the use of "infinitesimal rotations" at the bottom of that page you posted makes it looks like he's about to do some "physics math" hand waving in his derivations. Hand waving is standard fare in this regard, even in many graduate level texts. Even Goldstein waved his hands.
     
  16. May 24, 2014 #15

    Philip Wood

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    DH Thank you. I like your very clear scene-setting with the merry-go-round. I also like your use of unit vectors.

    I'm now going to turn your argument on its head and postulate that
    [tex]\textbf {R'} (t) = \textbf{R} (t)[/tex]
    in which case, with your notation,
    [tex]R \mathbf {\hat{x'}}=R\ \text {cos} \omega t \ \mathbf {\hat{x}}+R\ \text {sin} \omega t \ \mathbf {\hat{y}}[/tex]
    This is the familiar business of expressing the same vector on two different basis sets. It does, though, demand that we regard vectors fixed in the rotating system, such as [itex]\mathbf {\hat{x'}}[/itex], as functions of time.

    I've been labouring under the misapprehension that this is inconsistent with Lanczos's equation (in the extract in post 9):
    [tex]\frac {\text{d'} \mathbf{B}}{\text{d} t} = 0[/tex]
    He says that [itex]\frac {\text{d'} }{\text{d} t}[/itex] refers to the rate of change of a quantity in the moving system, S'. I now realise that he's not saying that [itex]\frac {\text{d} \mathbf{B'}}{\text{d} t} = 0[/itex]. What he means, I think, by [itex]\frac {\text{d'}\mathbf{B}}{\text{d} t}[/itex] is the rate of change of vector B' if we regard the rotating unit vectors [itex]\mathbf {\hat{x'}}[/itex] etc. as constant; in other words a sort of partial differentiation.

    This interpretation rings true for me. Many thanks to those who've came to my aid.
     
    Last edited: May 24, 2014
  17. May 24, 2014 #16

    D H

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    Before I start delving into your last post, Philip, it's important to realize that time derivatives of vector quantities are frame dependent. It's certainly true for translation. Why wouldn't this be the case for rotation?

    To illustrate that it's true for translation, suppose you and your best buddy decided to take the long weekend off with a trip to Vegas. The only problem: Who's car? You like your vintage Dodge Charger with its 426 cu. in. engine, he likes his vintage AMC Matador with its smaller but zestier 401 cu. in. engine. He comes up with the perfect solution: "Let's race!"

    You're neck and neck as your cross the California-Nevada state line. From your perspective, his velocity as you cross into Nevada is nearly zero. The same is true regarding your velocity from your buddy's perspective. Unfortunately for both of you, you just zoomed past a Nevada highway patrol officer sitting on a lawn chair in the freeway median strip. That officer (along with the radar gun he was holding just before he called you in) had a slightly different perspective on your velocities than did the two of you.


    The derivatives of those unit vectors is frame-dependent. From the perspective of the non-rotating observer, the ##\hat x, \hat y, \hat z## unit vectors are stationary (time derivatives are zero) while the ##\hat x', \hat y', \hat z'## unit vectors are rotating (time derivatives are non-zero). It's the other way around for the rotating observer.

    From the above, it's obvious that ##\hat x' = \cos(\omega t)\hat x + \sin(\omega t) \hat y##. Let's finish off the relationship between those unit vectors. The two observers share the same z axis: ##\hat z' = \hat z##. This means that to complete a right-handed system, we must have ##\hat y' = \hat z' \times \hat x' = -\sin(\omega t) \hat x + \cos(\omega t)\hat y##.

    These relationships form the transformation matrix that transforms a vector as represented in the rotating frame to the representation of the same vector in the non-rotating frame:

    [tex]
    \begin{bmatrix} x \\ y \\z \end{bmatrix} =
    \begin{bmatrix}
    \cos(\omega t) & -\sin(\omega t) & 0 \\
    \sin(\omega t) & \phantom{-} \cos(\omega t) & 0 \\
    0 & 0 & 1
    \end{bmatrix}
    \;
    \begin{bmatrix} x' \\ y' \\z' \end{bmatrix}
    [/tex]
    Denoting that transformation matrix as ##T_{R\to I}##, the above becomes ##\vec R(t) = T_{R\to I}(t) \, \vec R'(t)##. Note this applies to all vectors, not just position vectors. Any vector quantity ##\vec q## can be transformed from its rotating frame representation to its non-rotating frame representation.

    Taking the time derivative of above results in ##\dot{\vec R}(t) = T_{R\to I}(t) \, \dot{\vec R}'(t) + \dot T_{R\to I}(t) \, \vec R'(t)##. What's the time derivative of this transformation matrix? From the above, that time derivative is
    [tex]
    \dot T_{R\to I}(t) =
    \begin{bmatrix}
    -\omega\sin(\omega t) & -\omega\cos(\omega t) & 0 \\
    \phantom{-}\omega\cos(\omega t) & -\omega\sin(\omega t) & 0 \\
    0 & 0 & 0
    \end{bmatrix}
    [/tex]
    Pre-multiplying by the transpose of the transformation matrix yields
    [tex]
    T_{R\to I}(t)^T \, \dot T_{R\to I}(t) =
    \begin{bmatrix} 0 & -\omega & 0 \\ \omega & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
    [/tex]
    Note that the right hand side is the skew symmetric cross product matrix corresponding to the angular velocity vector ##\vec \omega = \omega \hat z'##.

    My hand wave is the unproven claim (unproven here, that is) that ##T_{R\to I}(t)^T \, \dot T_{R\to I}(t) = \text{Sk}(\vec \omega)## where ##\vec \omega## is the angular velocity vector as represented in the rotating frame is always true for rotations in three dimensional space. This is a consequence that the set of all 3x3 proper real transformation matrices is the Lie group SO(3). The Lie algebra for this group is the set of 3x3 skew symmetric matrices. Proving this claim either requires a class in differential geometry and Lie groups, or five pages of a math.

    Another way to write this relation between the transformation matrix, its time derivative, and the angular velocity vector is ##\dot T_{R\to I}(t) = T_{R\to I}(t) \text{Sk}(\vec \omega(t))##. With this, the time derivative becomes ##\dot{\vec R}(t) = T_{R\to I}(t) \, \dot{\vec R}'(t) + T_{R\to I}(t)\,\text{Sk}(\vec \omega(t)) \, \vec R'(t)##, or
    [tex]\dot{\vec R}(t) = T_{R\to I}(t) \left(\dot{\vec R}'(t) + \vec \omega(t) \times \vec R'(t)\right)[/tex]
     
  18. May 24, 2014 #17

    Andrew Mason

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    Laczos must be using polar coordinates.

    Using polar coordinates (where [itex]\hat{k}[/itex] is the unit vector in the direction perpendicular to the plane of rotation ), the orthogonal basis vectors in S are [itex]\hat{r} \text{ and } \hat{Ω}[/itex] where [itex]\hat{Ω} = \hat{k} \times \hat{r}[/itex]. In S' the basis vectors are: [itex]\hat{r}' \text{ and } \hat{Ω}'[/itex] where [itex]\hat{Ω}' = \hat{k} \times \hat{r}'[/itex]. However, since [itex]\hat{r} = \vec{R}/|\vec{R}| = \vec{R}'/|\vec{R}'| = \hat{r}'[/itex], [itex] \hat{Ω}' = \hat{k} \times \hat{r}' = \hat{k} \times \hat{r} = \hat{Ω}[/itex], so the unit basis vectors for S and S' are identical.

    The only difference between [itex]\vec{R}[/itex] and [itex]\vec{R}'[/itex] is in the values for Ω and Ω'.

    [itex]\vec{R} = r\hat{r}+ Ω\hat{Ω}[/itex] and

    [itex]\vec{R}' = r'\hat{r}'+ Ω'\hat{Ω}' = r\hat{r}+ Ω'\hat{Ω}[/itex]

    Subtracting:

    [itex]\vec{R}' - \vec{R} = Ω'\hat{Ω}' - Ω\hat{Ω} = (Ω'- Ω)\hat{Ω}[/itex]

    Differentiating a rotating vector R in S with respect to time results in:

    [itex]\dot{\vec{R}} = \dot{r}\hat{r} + r\dot{Ω}\hat{Ω} = r\dot{Ω}\hat{Ω} [/itex]

    but in co-rotating frame S', [itex]\dot{Ω} = 0[/itex] so the derivative with respect to time is:

    [itex]\dot{\vec{R}'} = \dot{r}'\hat{r} + r'\dot{Ω}'\hat{Ω} = 0\hat{r} + r0\hat{Ω} = 0[/itex]

    AM
     
    Last edited: May 24, 2014
  19. May 24, 2014 #18

    Philip Wood

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    DH Thank you for this interesting piece. The case that concerned me was of a vector fixed to the rotating frame, so I think you'll agree that [itex]\left(\begin{array}{cc}x'\\y'\\z'\ \end{array}\right)[/itex] is constant, though the unit vectors [itex]\hat{\mathbf{x'}}[/itex] etc., are functions of time, when expressed in terms of [itex]\hat{\mathbf{x}}[/itex] etc.. This, I believe, is what Lanczos means by [itex]\frac{\text{d'} \mathbf{B}}{\text{d}t} = 0[/itex].
     
  20. May 24, 2014 #19

    D H

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    No, he's not. Almost assuredly. He's using two reference frames, one of which is rotating with respect to another. It doesn't really matter what kind of coordinates you use. This an issue of the time derivative of some vector quantity as observed in one reference frame versus the time derivative as observed in another reference frame.

    What he's doing is developing what some call the dynamical transport theorem. You don't need coordinate systems at all to express this theorem:
    [tex]\left(\frac{d\vec q}{dt}\right)_A = \left(\frac{d\vec q}{dt}\right)_B + \vec \omega \times \vec q[/tex]
    The subscripts on the derivatives denote the time derivative as observed by two observers, call them observer A and observer B. Observer B is rotating with respect to observer A with an angular velocity of ##\vec \omega##.
     
  21. May 24, 2014 #20

    Andrew Mason

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    But by using polar coordinates you can use the same basis vectors for both the rotating and non-rotating reference frames. It is just much easier to do the math.

    Without the pages prior to page 100 I am just guessing at what the statement [itex]\vec{R} = \vec{R}'[/itex] means. It just occurred to me that he may mean [itex]\vec{R} = R\hat{r} = R\hat{r}' = \vec{R}'[/itex]. The [itex]r\hat{r}[/itex] component then is always the same using polar coordinates. That may be what he is referring to.

    AM
     
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