1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple indice question -- metric, traces

  1. Dec 18, 2014 #1
    Mod note: OP warned about not using the homework template.
    Let ##g_{ac}## be a 3-d metric.
    So the trace of a metric is equal to its dimension so I get ##g_{ac}g^{ac}=3##

    But I'm a tad confused with the expression : ##g^{ac}g_{ad}##=##delta^{c}_{d}##
    I thought it would be ##3delta^{c}_{d}###

    Because ##delta^{c}_{d}## implies that ##g^{ac}g_{ad}## is only non-zero when c=d, in which case don't you just get: ##g_{ac}g^{ac}=3##

    Cheers in advance.
     
    Last edited by a moderator: Dec 18, 2014
  2. jcsd
  3. Dec 18, 2014 #2
    The summation rule applies, in the last expression c is being summed over, but in delta-expression it is not. If you contract the indices in it, you get trace of the identity matrix, which is 3 in three dimensions:

    ##3=\delta^c_c=g^{ac}g_{ac}\neq\delta^c_d=g^{ac}g_{ad}##

    The first is a scalar, the second is a (1,1) tensor. Btw, you need a blackslash \ in front of delta you want to write ##\delta##
     
  4. Dec 18, 2014 #3
    Another way of thinking about this is that you are raising an index:
    [tex]g^{ac}g_{ad}=g^{ca}g_{ad}=g^c_d=\delta^c_d[/tex]
    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simple indice question -- metric, traces
  1. Trace question (Replies: 2)

  2. Simple Simple question (Replies: 1)

  3. Simple Indices (Replies: 8)

  4. Trace prove question (Replies: 5)

Loading...