- #1

christodouloum

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[tex]\delta(h)=2 h h_{\alpha\beta}\delta(h^{\alpha\beta})[/tex]

can someone give a hint on how this is computed?

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- Thread starter christodouloum
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- #1

christodouloum

- 35

- 0

[tex]\delta(h)=2 h h_{\alpha\beta}\delta(h^{\alpha\beta})[/tex]

can someone give a hint on how this is computed?

- #2

- 13,238

- 994

[tex]\delta(h)=2 h h_{\alpha\beta}\delta(h^{\alpha\beta})[/tex]

can someone give a hint on how this is computed?

That's inaccurate

[tex] \delta(h^2)=2h \delta h = 2 h^2 h^{\alpha\beta}\delta h_{\alpha\beta}[/tex]

The indexed part in the last term of the muliple equality comes simply by considering the way one computes the inverse of a square matrix. The argument can be found in most GR books when discussing the HE action and, of course, in maths books when discussing integration on arbitrary manifolds.

- #3

christodouloum

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- #4

Avodyne

Science Advisor

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Start from det = exp Tr log.

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