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Trace reverse perturbation

  1. Nov 22, 2012 #1
    From the tensor, ##\bar{h}^{ij}=h^{ij}-1/2\eta^{ij}h##
    Where, h=##h^i_i##,
    Prove that ##\bar{h}=-h##,
    Where, ##\bar{h}=\bar{h}^i_i##
     
    Last edited: Nov 23, 2012
  2. jcsd
  3. Nov 22, 2012 #2

    TSny

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    Hello, and welcome to PF!

    Part of the rules of this forum is that you must show some work towards a solution before help is given. You can click on the "Rules" tab at the top of the page for more information on posting questions.

    Your Latex formating will work if you use ## instead of $.
     
  4. Nov 23, 2012 #3
    just contract both sides with ηij.
     
  5. Nov 23, 2012 #4
    I tried it but I guess I made a mistake since I got h(bar) on the left side and on the right side I got h-1/2h. Please help.
     
  6. Nov 23, 2012 #5
    No,you get on the right side h-(1/2)(4)h=h-2h=-h,can you verify it?
     
  7. Nov 23, 2012 #6
    I had to find the absolute value of the ##n_{ij}## matrix. Is this is where I went wrong. Thank you very much
     
    Last edited: Nov 23, 2012
  8. Nov 23, 2012 #7
    Not necessarily,after contraction you get
    ηijηij00η0011η11
    22η2233η33,
    all others are zero,right.now this is =(1*1)+(-1*-1)+(-1*-1)+(-1*-1)=4,it does not depend on your signature.you can take as well as(-,+,+,+).
     
  9. Nov 23, 2012 #8
    Thanks a lot andrien.
     
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