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Mechanics, question about generalized coordinates

  1. Feb 19, 2015 #1
    I can start explaining the problem but a more quicker way would be to open this link:

    http://onlinelibrary.wiley.com/doi/10.1002/9783527627486.app2/pdf

    and check the paragraph resulting in expression (B.5).

    Note that I don't really care about the kinetic energy they talk about in this link, just the expression B.5 and the paragraph that precedes it.

    So it seems that as long as the constraints between the particles do not depend on time explicitly (scleronomic constraints), cartesian coordinates only depend on the generalized coordinates explicitly.

    However, then I have a question about what choices are allowed for the generalized coordinates. What if I pick another coordinate frame moving with a constant velocity relative to mine. And then I call those coordinates the generalized coordinates. Then obviously, no constraints are present but still the transformation between my cartesian coordinates and the generalized coordinates are explicitly time dependent.

    However in class we clearly stated that if no time dependent constraints are presents all the position vectors in our frame depend only on the generalized coordinates explicitly.

    Where am I thinking wrong?
     
  2. jcsd
  3. Feb 19, 2015 #2
    What do you mean by "picking another frame relative to mine"? What is "your" coordinate frame? If you have a collection of free particles, there is no intrinsic coordinate frame until you've picked one.

    For whatever reference frame you choose, the Cartesian coordinates of a point in that reference frame will not be time-dependent.
     
  4. Feb 19, 2015 #3
    Alright so let me pick one frame F. From this frame all the particles are described by my cartesian coordinates.

    Assuming I have no constraints, I can pick a set of numbers that are equal to the amount of cartesian coordinates which contain information about my system as well. More specifically, I have to be able to transform from the carthesian frames to these numbers and vice versa. These ''numbers'' are what I understand generalized coordinates.

    Now I can always pick these numbers to be the positions for the particles that would be measured from a hypothetical frame moving away from me at some velocity. (This frame doesn't exist I just imagine it to move way from me and imagine what one would measure from that frame.) If I know the amount of time passed, and all those numbers , I can uniquely transform back and forth from these numbers to my carthesian coords and vice versa.

    So this ''weird'' choice seems to behave like I'd like generalized coordinates to behave. Why isn't this a valid choice for the generalized coordinates?
     
  5. Feb 19, 2015 #4
    If I understand correctly, our entire conversation is about the following claim made in the PDF, which I have slightly paraphrased:

    If the constraints do not explicitly depend on time, then the coordinate transformation which describes the generalized coordinates in terms of the Cartesian coordinates will not explicitly depend on time either.

    Let's see if this is true for your proposed counter-example.

    Your example is a system of free particles, such that there are no constraints. Clearly, then, the constraints do not depend on time, since there are none.

    For your generalized coordinates {q1, …, qN}, you have picked the Cartesian coordinates of some reference frame F' = {x'1, …, x'N}, which moves with a constant velocity v with respect to some other reference frame F = {x1, …, xN}.

    You point out that the coordinate transformation between your generalized coordinates and the Cartesian coordinates of F is time-dependent. This is true but irrelevant to the claim in the PDF. What meaning does frame F have? So far you have done nothing to even specify the meaning of F.

    But the way I read it, the statement in bold above refers to the coordinate transformation between your generalized coordinates and the Cartesian coordinates of the frame F'. Which is simply
    { q1 = x'1
    { ⋮
    { qN = x'N,​
    which is time-independent.

    You wrote that the frame F' is a "hypothetical frame moving away from me at some velocity. (This frame doesn't exist I just imagine it to move way from me and imagine what one would measure from that frame)". I disagree. If you have defined your generalized coordinates in the frame F', then that is the frame which is relevant to you. It is frame F which is hypothetical and imaginary. You haven't done anything to give any physical relevance to frame F, and it bears no consideration in your analysis of the relationship between your generalized coordinates and Cartesian coordinates.
     
  6. Feb 19, 2015 #5
    Thanks for taking the time to elaborate. I will certainly re-read it a few times to fully grasp the clue of your explanation.

    I just realize I over complicate the point of my question with my example. Do you mind thinking about one more ''counter -example''?

    Consider a 1D motion of a point particle on the x-axis. The cartesian coordinate is this ##x##.

    Now I define my generalized coordinate as ##q=x+t## just for the fun.

    Since I've never seen any formal definition of what generalized coordinates have to obey in class, I don't see why I'm not allowed to make this choice for q.
     
  7. Feb 19, 2015 #6
    No, I don't mind at all, but this is actually not a different counter-example from your old one, just a more specific case. Here, you have again chosen a frame F = {x}, and then defined frame F' = {x'}, where x' = xvt. You've just specified that v=–1.

    So once again, the relationship between the generalized coordinate you have chosen and the Cartesian coordinate of the frame F' is time-independent. The relationship between your generalized coordinate and the Cartesian coordinate of the frame F is not, but that doesn't matter, frame F would never have even entered the conversation in the first place if you hadn't artificially introduced it.

    You are absolutely allowed to make this choice for q. But the generalized coordinate that you have chosen by making that choice is simply the Cartesian coordinate of F', and F' is the frame in which the relationship between the generalized coordinate and Cartesian coordinate is time-independent.
     
  8. Feb 19, 2015 #7
    So there is no way I can construct a time dependent conversion between generalized and cartesian coordinates for a system of free particles without constraints? The only way to do so is having time dependent constraint functions on the generalized variables?
     
  9. Feb 19, 2015 #8
    Not without me being able to find a different frame in which your conversion is no longer time-dependent.
     
  10. Feb 19, 2015 #9
    It might help to discuss an examples of how generalized coordinates actually get used in the wild.

    Consider a system of 1 free particle in 3 dimensions, where I choose spherical coordinates (r, θ, φ) for my generalized coordinates. There are no constraints, so the constraints are trivially time-independent. The coordinate transformation between the generalized coordinates I've chosen and the Cartesian coordinates is given by
    [tex]
    \begin{cases}
    x(r,\theta,\phi) = r\sin\theta\cos\phi \\
    y(r,\theta,\phi) = r\sin\theta\sin\phi \\
    z(r,\theta,\phi) = r\cos\theta,
    \end{cases}
    [/tex]
    which you'll notice is time-independent. So the statement is confirmed: Time-independent constraints ⇒ time-independent coordinate transformation.

    Of course, if I ask for the coordinate transformation between these generalized coordinates and the Cartesian coordinates of some other arbitrary reference frame, that will depend on time. But that would be a strange thing to ask for, and it doesn't change the fact that the relationship between my generalized coordinates and the Cartesian coordinates of the relevant reference frame is time-independent.

    The example still works if I introduce a constraint that the particle is confined to the surface of a cone, rather than free. The constraint can be written as θ = some constant α, which is time-independent. My coordinate transformation is still the same as before, so again: Time-independent constraints ⇒ time-independent coordinate transformation.
     
  11. Feb 19, 2015 #10
    Yeah I've used them in the wild for a bit already, and indeed they always were time independent for scleronomic constraints but I never really thought about what the theory behind it was . Thanks a lot for taking the time it's a lot clearer now!
     
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