- 2

- 0

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]

Cheers