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Noncoordinate basis vector fields

  1. Oct 27, 2006 #1
    I'm self-studying Schutz Geometric methods of mathematical physics, having problems with ex. 2.1. page 44.


    show this is non-coordinate basis, i.e. show commutator non-zero.

    I try to apply his formula 2.7, assuming

    V1=cos(theta), V2=sin(theta)
    W1=-sin(theta), W2=cos(theta)
    x(r)=r cos(theta)
    y(r)=r sin(theta)

    I believe the component of x should be (sin(theta))/r, however I get (sin(theta) - r sin(theta))/r.

    would appreciate any help
  2. jcsd
  3. Dec 1, 2006 #2

    Chris Hillman

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    Science Advisor

    Vector fields

    Hi, desic,

    Vector fields, flows (in the sense of manifold theory), and first order linear homogeneous partial differential operators, are all tighly inter-related concepts, and it helps to able to freely intertranslate between these various representations.

    E.g. in [tex]\mathbold{R}^2[/tex] "translation along x" and "rotation around the origin" can be written
    [itex] \vec{T} = \partial_x[/itex]
    [itex] \vec{R} = \partial_\theta = -y \, \partial_x + x \, \partial_y [/itex]
    Now you can just compute [tex] \vec{T} \vec{R} f - \vec{R} \vec{T} f[itex] for an undetermined function f, from which you can read off the commutator.

    See Olver, Equivalence, Invariants, and Symmetry for more about the virtues of this style of thinking/computing.

    Chris Hillman
    Last edited: Dec 1, 2006
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