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Components of metric perturbation in TT gauge

  • #1

Homework Statement



I'm just working through a text book and there's a line in which I'm clearly missing something. What I want to do is show that from:
[tex] \bar{h^{TT}_{\mu \nu}} = A^{TT}_{\mu \nu} cos(\omega (t-z)) [\tex]
to
[tex] h^{TT}_{\mu \nu} = B^{TT}_{\mu \nu} cos(\omega (t-z)) [\tex]


Homework Equations



[tex]\bar{h}_{\mu \nu} = h_{\mu \nu} - \frac{1}{2} \eta_{\mu \nu} h[\tex]
[tex]h = h^{\alpha}_{\alpha} = \eta^{\alpha \beta} h_{\alpha \beta}[\tex]


The Attempt at a Solution



There's just one line where I have an issue, i think.

I've used the above equations to say:

[tex] \bar{h} = \eta^{\mu \nu}\bar{h}_{\mu \nu}[\tex]
then substituted the equation for [tex]\bar{h}_{\mu \nu}[\tex] to get:

[tex] \bar{h} = h - \frac{1}{2} \eta^{\mu \nu} \eta_{\mu \nu} h[\tex]

which is fine, but for it to work this last line must equal [tex]-h[\tex], I think and I'm not sure i see where that comes from.

thanks.

ftj
 

Answers and Replies

  • #2
104
0
Your eta is the Minkowski metric, right?

eta_{alpha beta} eta^{mu nu} = delta_{alpha}^{mu} delta_{beta}^{nu}

Take it away.
 

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