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Form of Lorentz Transformation Using West-Coast Metric

  1. Aug 14, 2012 #1
    This is a fairly trivial question I think. I'm only asking it here because after some googling I was unable to find its answer. I was at one point led to believe that the form of the Lorentz-transformation matrix is dependent on the convention used for the Minkowski metric. Specifically it was my understanding that

    [γ, βγ, 0, 0]
    [βγ, γ, 0, 0]
    [0, 0, 1, 0]
    [0, 0, 0, 1]

    was the transformation matrix when the West-coast metric, diag(1, -1, -1, -1), is used. This is the inverse of the more commonly encountered

    [γ, -βγ, 0, 0]
    [-βγ, γ, 0, 0]
    [0, 0, 1, 0]
    [0, 0, 0, 1]

    which I know is correct at least for the East-coast metric, diag(-1, 1, 1, 1). I was working a problem recently and got a result using the former of these transformation matrices that was clearly incorrect. Does the Lorentz transformation's form actually depend on the convention for the metric, or did I concoct this entire distinction? Perhaps I misunderstood one of my professors.
     
  2. jcsd
  3. Aug 14, 2012 #2

    PAllen

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    Gold Member

    The form of the Lorentz transform does not depend on whether you use +t metric or -t metric. The two transform matrices you give are simply inverses of each other corresponding to boosts in opposite directions.
     
  4. Aug 14, 2012 #3
    As a group the lorentz transformations, [itex]\Lambda^{\mu}_{\nu}[/itex] are defined by:

    \begin{equation}
    \Lambda^{\mu}_{\nu} \eta_{\mu \rho} \Lambda^{\rho}_{\sigma} = \eta_{\nu \sigma}
    \end{equation}

    Thus, the metric should not affect the transform because you can multiply both sides of the equation to switch metric convention.
     
  5. Aug 14, 2012 #4
    Haha, I feel silly now--thanks, guys.
     
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