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3D Vectors

  1. Nov 23, 2008 #1
    1. Variables

    Given a generalized basis in three dimensions: [tex]e_{1},e_{2},e_{3}[/tex] and the standard Kronecker delta [tex]\delta_{ij}[/tex], and using Einstein summation.
    With the vector [tex]\textbf{x},\textbf{y},\textbf{z}[/tex] i'm trying to simplify this problem:

    2. Problem
    [tex]\delta_{il} . \delta_{jm} . x_{j}[/tex]

    3. My attempt
    [tex]\delta_{il}.\delta_{jm} . \textbf{x} . e_{j}
    = (e_{i}. e_{l}) . (e_{j} . e_{m}) . \textbf{x} . e_{j}
    = (e_{j}. e_{j}) . (e_{l} . e_{m} . e_{j}) . \textbf{x}
    = 1 . (e_{l} . e_{m} . e_{j}) . \textbf{x} [/tex]

    Surely this leads to [tex]\delta_{il} . \delta_{jm} . x_{j} = 0[/tex] as [tex]e_{l} , e_{m} , e_{j}[/tex] are all orthagonal ?

    Ultimately I'm trying to prove that
    [tex](\delta_{il} . \delta_{jm} - \delta_{jl} . \delta_{im}).x_{j}.y_{l}.z_{m}
    = y_{i}.x_{j}.z_{j} - z_{i}.x_{j}.y_{j}[/tex]
  2. jcsd
  3. Nov 24, 2008 #2


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    Homework Helper

    Welcome to PF!

    Hi timscully! Welcome to PF! :smile:

    (have a delta: δ :wink:)

    Forget about the basis vectors!

    All δij does is replace i by j (or vice versa) in anything else.

    So δijxj = xi, δijxi = xj.

    So just plug 'n' play! :biggrin:
  4. Nov 24, 2008 #3
    Thanks for the welcome.

    It looks like a great forum.

    So, is [tex]\delta_{ij} . x_{m}[/tex] zero, because m is neither i nor j ?
  5. Nov 25, 2008 #4


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    not a sum


    δijxj is a sum over all values of j, so it only depends on i: δijxj = xi.

    But δijxm is not a sum; there is no "contraction"; it still depends on i j and m: δijxm = xm if i = j and = 0 if i≠j. :smile:
  6. Nov 25, 2008 #5
    Got it. Much appreciated.
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