Inhomogeneous (poincare) lorentz transormation

  1. Dec 29, 2011 #1
    I'm reading a physics book and in the section on relativity they are using the Einstein summation convention, with 4vectors and matrices.

    They say that the transformations take the form:
    where it is required that [itex]\Lambda^{\mu}_{\nu}[/itex] satisfy the following relation:
    (note: I found the same thing on wikipedia, so you can see it in context if you like. it appears a tiny bit down from the section that the link takes you to.)

    My problem is that this seems impossible to satisfy by my current understanding, but I know I must be wrong, I just cannot see how.

    So we are summing over [itex]\mu[/itex] and [itex]\nu[/itex] in the above relation right? and we do this for all [itex]\alpha[/itex] and [itex]\beta[/itex] in order to satisfy all the components of the matrices.
    My problem is what happens when we get to the following situation?:
    [tex]\mu=0, \nu=1, \alpha=0, \beta=0[/tex]
    But, [itex]\eta_{01}=0[/itex], and [itex]\eta_{00}=-1[/itex]. So there is no possible values of the [itex]\Lambda[/itex]'s that will satisfy this because we now have 0=-1, which is a contradiction.

    Where did I go wrong with my thinking? Thanks.
  2. jcsd
  3. Dec 29, 2011 #2

    George Jones

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    In an inertial coordinate system,
    [tex]\eta_{\alpha\beta} = \eta_{\mu\nu}\Lambda^{\mu}_{\alpha}\Lambda^{\nu}_{ \beta} = -\Lambda^{0}_{\alpha}\Lambda^{0}_{ \beta} + \Lambda^{1}_{\alpha}\Lambda^{1}_{ \beta} + \Lambda^{2}_{\alpha}\Lambda^{2}_{ \beta} + \Lambda^{3}_{\alpha}\Lambda^{3}_{ \beta}[/tex]
  4. Dec 29, 2011 #3


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    You're not summing, you've just assumed 4 values for the 4 variables. Remember you have to sum over mu and nu.
  5. Dec 29, 2011 #4
    Right! I knew it would have to have been something stupidly simple >.<

    thanks guys.
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