# Weinberg quick question on chapter 2 zero mass

1. Jul 23, 2011

### Heffernana

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

$(Wt)^{\mu}(Wt)_{\mu}=t^{\mu}t_{\mu}=-1$

Cheers

2. Jul 23, 2011

### Fredrik

Staff Emeritus
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 $W^T\eta W=\eta$, and this implies $$(Wt)^2=(Wt)^T\eta(Wt) =t^TW^T\eta Wt=t^T\eta t=t^2=-1.$$ You don't have to use the assumption about the components of t for anything other than the final "=-1".