This is from Carrol's book, page 13 -
This Sort of notation is new to me, and i'm having trouble understanding the claim on the bootom part (second sentence from the end).
Does $(\rho,\sigma)= (0,0) \Rightarrow \rho=\sigma=0$?
And how can we show that $|\Lambda_{0}^{0'}|\geq 1$?
Does it refer to the RHS or the LHS of 1.29?
Also, can anyone give an example to a time reversal? Why don't we allow those?
For $\eta=diag(-1,1,1,1)$ take the zeroth component of $\eta_{\rho \sigma}=\Lambda^{\mu}_{\,\rho} \Lambda^{\nu}_{\,\sigma} \eta_{\mu \nu}$ setting $\rho=\sigma=0$:
$$-1=\eta_{0 0}=\Lambda^{\mu}_{\,0} \Lambda^{\nu}_{\,0} \eta_{\mu \nu}=\Lambda^{0}_{\,0} \Lambda^{0}_{\,0} \eta_{0 0} + \sum_{i=1}^3 \Lambda^{i}_{\,0} \Lambda^{i}_{\,0} \eta_{ii} = -\Lambda^{0}_{\,0} \Lambda^{0}_{\,0} + \sum_{i=1}^3 \Lambda^{i}_{\,0} \Lambda^{i}_{\,0}.$$ Combining the very left and the very right one obtains $$(\Lambda^{0}_{\,0})^2 =\sum_{i=1}^3 ( \Lambda^{i}_{\,0})^2 +1.$$
Hence $|\Lambda^{0}_{\,0}|\geq 1$.
which means $\left(\Lambda^{0}{}_{0}\right)^{2}=-1$?
– proton Sep 01 '16 at 05:58