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Taking the zero entry of the following

  1. Sep 14, 2014 #1
    Hello!!

    Given: g[itex]_{ρσ}[/itex] = g[itex]_{μ\nu}[/itex][itex]\Lambda[/itex][itex]^{μ}[/itex][itex]_{ρ}[/itex][itex]\Lambda[/itex][itex]^{\nu}[/itex][itex]_{σ}[/itex] equation (1)

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

    So, it went like 1= [itex]\Lambda[/itex][itex]^{ρ}[/itex][itex]_{0}[/itex]g[itex]_{ρσ}[/itex][itex]\Lambda[/itex][itex]^{σ}[/itex][itex]_{0}[/itex] equation (2)

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


    I didn't understand how did equation (3) show up, I thought it might be related to s[itex]^{2}[/itex] = x[itex]^{0}[/itex][itex]^{0}[/itex] - x[itex]^{i}[/itex]x[itex]^{i}[/itex] but then what does the vector x[itex]^{μ}[/itex] have to do with [itex]\Lambda[/itex][itex]^{μ}[/itex][itex]_{μ}[/itex].
     
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  3. Sep 14, 2014 #2

    PeterDonis

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    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?
     
  4. Sep 14, 2014 #3
    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 ([itex]\Lambda[/itex][itex]^{0}_{0}[/itex])[itex]^{2}[/itex] since it will be the case of dummy indices as far as I know..
     
  5. Sep 14, 2014 #4

    PeterDonis

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    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##.
     
  6. Sep 14, 2014 #5
    Thank you very much for taking the time to clear this!
     
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