Taking the zero entry of the following

[PhyAmateur]

Hello!

Given: g_{ρσ} = g_{μ\nu}\Lambda^{μ}_{ρ}\Lambda^{\nu}_{σ} equation (1)

It was then mentioned to take the 00 entry of equation (1):

So, it went like 1= \Lambda^{ρ}_{0}g_{ρσ}\Lambda^{σ}_{0} equation (2)

then equation (2) was set equal to (\Lambda^{0}_{0})^{2} - (\Lambda^{i}_{0})^{2} equation (3)


I didn't understand how did equation (3) show up, I thought it might be related to s^{2} = x^{0}^{0} - x^{i}x^{i} but then what does the vector x^{μ} have to do with \Lambda^{μ}_{μ}.

 

[PeterDonis]

Hi, PhyAmateur, and welcome to PF!

First, a general question: is there a reference you are getting these equations from? If so, giving the reference is helpful.

I didn't understand how did equation (3) show up, I thought it might be related to s^{2} = x^{0}^{0} - x^{i}x^{i}

It is. Substitute $g_{\rho \sigma} = \eta_{\rho \sigma}$ into $\Lambda^{\rho}_0 g_{\rho \sigma} \Lambda^{\sigma}_0$ (i.e., assume that the metric is the Minkowski metric). What do you get?

 

[PhyAmateur]

Thank you, PeterDonis for your welcoming and reply. Yes, this is taken from P. Ramond's book of Quantum Field Theory.

I am new to differential geometry that's why I am finding difficulty in trusting my intuition. So, answering your question, if we substitute it we will get (\Lambda^{0}_{0})^{2} since it will be the case of dummy indices as far as I know..

 

[PeterDonis]

if we substitute it we will get (\Lambda^{0}_{0})^{2}

That will be the 0-0 term, yes, since $\eta_{00} = 1$ (with the sign convention you're using). But $\eta_{11} = \eta_{22} = \eta_{33} = -1$, so the full summation will be $( \Lambda^0_0 )^2 - ( \Lambda^1_0 )^2 - ( \Lambda^2_0 )^2 - ( \Lambda^3_0 )^2$.

 

[PhyAmateur]

Thank you very much for taking the time to clear this!