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This is from Carrol's book, page 13 - enter image description here

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).

  1. Does $(\rho,\sigma)= (0,0) \Rightarrow \rho=\sigma=0$?

  2. And how can we show that $|\Lambda_{0}^{0'}|\geq 1$?

  3. 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?

Qmechanic
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proton
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1 Answers1

2

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 $|\Lambda^{0}_{\,0}|\geq 1$.

Statics
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