# Taking the zero entry of the following

1. Sep 14, 2014

### 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}$.

2. Sep 14, 2014

### Staff: Mentor

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.

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?

3. Sep 14, 2014

### 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..

4. Sep 14, 2014

### Staff: Mentor

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 $( \Lambda^0_0 )^2 - ( \Lambda^1_0 )^2 - ( \Lambda^2_0 )^2 - ( \Lambda^3_0 )^2$.

5. Sep 14, 2014

### PhyAmateur

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