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Weinberg quick question on chapter 2 zero mass

  1. Jul 23, 2011 #1
    In Weinberg QFT vol 1, page 69, when deriving the little group structure for the case of zero mass can anyone explain the following:

    The transformation W by definition leaves [itex] k^{\mu}=(0,0,1,1)[/itex] invariant, i.e. [itex]W^{\mu}_{\nu}k^{\nu} = k^{\mu} [/itex]. Why can you immediately deduce from this that for a time-like vector [itex] t^{\mu}=(0,0,0,1)[/itex] whose length (noting Weinberg's bizarre placing of the time component at the end of the 4-vector) is [itex]t^2=-1[/itex] is invariant under the previous W transformation, i.e. that:

    [itex](Wt)^{\mu}(Wt)_{\mu}=t^{\mu}t_{\mu}=-1 [/itex]

  2. jcsd
  3. Jul 23, 2011 #2


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    W is defined to be a member of a certain subgroup of the Lorentz group, so it's a Lorentz transformation. By definition, a Lorentz transformation W satisfies [itex]W^T\eta W=\eta[/itex], and this implies [tex](Wt)^2=(Wt)^T\eta(Wt) =t^TW^T\eta Wt=t^T\eta t=t^2=-1.[/tex] You don't have to use the assumption about the components of t for anything other than the final "=-1".
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