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Group of translations by a fixed distance

  1. Feb 17, 2007 #1

    Mentz114

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    Gold Member

    I think I need this group ( if it exists) to help solve a physics problem.
    Not being a mathematician, what I'd really like is a matrix representation
    and a rule for getting the covariant derivative in the event of a broken symmetry.

    I'd be much obliged if any one can give me any information.

    M.
     
  2. jcsd
  3. Feb 19, 2007 #2

    HallsofIvy

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    To represent translation by a matrix multiplication, you will have to use a "projective geometry" representation: each point (x,y,z) is represented by (x,y,z,1) with the understanding that (ax, ay, az, a) (a not 0) represents the same point. Then a translation by <u, v, w> can be written as
    [tex]\left(\begin{array}{ccccccc}1 && 0 && 0 && u \\ 0 && 1 && 0 && v\\ 0 && 0 && 1 && w \\ 0 && 0 && 0&& 1\end{array}\right)\left(\begin{array}{c}x \\ y \\ z \\ 1\end{array}\right)= \left(\begin{array}{c}x+ u \\ y+ v \\ z+ w \\ 1\end{array}\right)[/tex]
     
  4. Feb 19, 2007 #3

    Mentz114

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    HallsofIvy, thank you very much. This looks promising. Seems to form a group under multiplication, with an identity if u=v=w=0.

    [tex] \left(\begin{array}{ccccccc}1 && 0 && 0 && u \\ 0 && 1 && 0 && v\\ 0 && 0 && 1 && w \\ 0 && 0 && 0&& 1\end{array}\right)\left(\begin{array}{ccccccc}1 && 0 && 0 && u \\ 0 && 1 && 0 && v\\ 0 && 0 && 1 && w \\ 0 && 0 && 0&& 1\end{array}\right) = \left(\begin{array}{ccccccc}1 && 0 && 0 && 2u \\ 0 && 1 && 0 && 2v\\ 0 && 0 && 1 && 2w \\ 0 && 0 && 0&& 1\end{array}\right)

    [/tex]
     
    Last edited: Feb 19, 2007
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