Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taking the zero entry of the following

  1. Sep 14, 2014 #1

    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].
  2. jcsd
  3. Sep 14, 2014 #2


    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?
  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


    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##.
  6. Sep 14, 2014 #5
    Thank you very much for taking the time to clear this!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook