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Covariant derivatives in Wolfram Math

  1. Dec 27, 2009 #1
    In the Wolfram Mathworld section on spherical coordinates there's given a list of nine covariant derivatives. The derivatives are given with respect to radius, azmuth, and zenith using the usual symbols r, theta and phi. The question is: what would be examples of the vectors whose derivatives are taken, sybolized by A(subscript)r, A(subscript)theta, and A(subscript)phi. These vectors appear explicitly on the right side of the equations. I would have expected the covariant derivatives to be of the position vector parameterized by r, theta, and phi, but not so. Anyone?
     
  2. jcsd
  3. Dec 28, 2009 #2
    You may compute covariant derivative for any covarinat tensor, and in this case for [itex]A_i[/itex]. As expressions are in spherical coordinate system then subscript i must agree with names of coordinates, so then [itex] i \in {r,\theta, \phi}[/itex]. You may treat it as usual as with [itex] {x,y,z} [/itex]. The proper use of Christoffel symbols, and covariant derivatives is exactly for this - for computing with this coefficients as close as in Cartesian system.
    So You ask for example of vector You may put into this formulas. Here You are ( please make some picture): simple Culomb force, notice angular parts are vanish so, it is easy to compute with it just exactly in spherical coordinates:

    [itex]
    A_r = CQ/r^2
    [/itex]
    [itex]
    A_{\phi} = 0
    [/itex]
    [itex]
    A_{\theta} = 0
    [/itex]

    Here is picture of something similar: index.jpeg
    Of course You may substitute anything You want for [itex]A_{r},A_{\theta},A_{\phi}[/itex], then You may obtain interesting vector fields. If You have Sage computing environment You may create some pictures with it: http://www.sagenb.org/home/pub/216/ [Broken]

    Any vector [itex](A_x,A_y,A_z)[/itex] may be described in spherical coordinate system by taking simple and well known transformation between coordinate systems, but when there is no spherical symmetry this may be a paint to compute with it.

    Best regards ;-)
    Kazek
     
    Last edited by a moderator: May 4, 2017
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