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I Worked example on a covariant vector transformation

  1. Mar 20, 2017 #1
    Hello.
    I would like to check my understanding of how you transform the covariant coordinates of a vector between two bases.
    I worked a simple example in the attached word document.
    Let me know what you think.
     

    Attached Files:

  2. jcsd
  3. Mar 20, 2017 #2

    fresh_42

    Staff: Mentor

    You should know, that to download a word document from unknown origin is a big hurdle not many of us want to risk. Definitely not me, as I additionally consider it as bad behavior to force me to take action rather than trying to concentrate on help. But this is my personal attitude towards users who don't take the effort of posting their questions adequately while simultaneously demanding efforts from others. I think you should know this and consider to type in the entire question instead. We have a LaTex library in place that helps a lot to even type in complicated formulas.
     
  4. Mar 20, 2017 #3

    jedishrfu

    Staff: Mentor

    Here's a PDF version:
     

    Attached Files:

  5. Mar 20, 2017 #4

    jedishrfu

    Staff: Mentor

    Here's the content of the PDF (limited to 10 image files per post so had to use latex codes for some stuff) via copy and paste (equations are images)

    ----------------------// example.docx as pdf //----------------------------------------------

    The formula for the covariant vector transformation from the ##R^2## coordinate system to the ##B## coordinate system is:

    upload_2017-3-20_21-10-33.png

    For our example, the vector upload_2017-3-20_21-10-33.png , where upload_2017-3-20_21-10-33.png and upload_2017-3-20_21-10-33.png are the basis for upload_2017-3-20_21-10-33.png .

    Our new coordinate system upload_2017-3-20_21-10-33.png ; spanned by

    upload_2017-3-20_21-10-33.png

    upload_2017-3-20_21-10-33.png




    To find ##V_1## in the ##B## basis in terms of ##V_i## in ##R^2##:

    upload_2017-3-20_21-10-33.png

    upload_2017-3-20_21-10-33.png

    upload_2017-3-20_21-10-33.png
     
  6. Mar 21, 2017 #5
    First, it is not a good idea to use superscripts for components of a covector. Simple way to restore these formulas is as follows. Vector's coordinates form the column: ##x=(x^1,\ldots, x^m)^T## and covector's coordinates form the row ##\xi=(\xi_1,\ldots, \xi_m)##. The pairing ##\xi x=\xi_ix^i## is a scalar.

    If vector's coordinates transform with the help of a matrix ##C## by the rule ##x'=Cx## and the covector's coordinates transform as follows ##\xi'=\xi D##
    then formula ##\xi'x'=\xi x## implies ##D=C^{-1}##
     
    Last edited: Mar 21, 2017
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