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  Quadratic order perturbation terms in the expansion of Ricci tensor

+ 1 like - 0 dislike
3611 views

I want to expand Einstein-Hilbert action for the metric

$$ g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} $$

up to quadratic order in $h_{\mu \nu}$. For this purpose I need to calculate the Ricci tensor at some stage. There is no problem in linear terms of the metric perturbation but there happens to be a problem when I intend to calculate the quadratic terms using

$$ R_{\mu \nu}^{(2)} = \partial_{\alpha} \Gamma^{\alpha} {} _{\mu \nu}^{(2)} - \partial_{\nu} \Gamma^{\alpha} {} _{\mu \alpha}^{(2)} + \Gamma^{\alpha} {} _{\beta \alpha}^{(1)} \Gamma^{\beta} {} _{\mu \nu}^{(1)} - \Gamma^{\alpha} {} _{\beta \nu}^{(1)} \Gamma^{\beta} {} _{\mu \alpha}^{(1)} $$

When I use

$$ \Gamma^{\alpha} {} _{\mu \nu}^{(1)} = \frac{1}{2} \left( \partial_{\mu} h^{\alpha} {} _{\nu} + \partial_{\nu} h^{\alpha} {} _{\mu} - \partial^{\alpha} h_{\mu \nu} \right) $$ $$ \Gamma^{\alpha} {} _{\mu \nu}^{(2)} = -\frac{1}{2}h^{\alpha \beta} \left( \partial_{\mu} h_{\beta \nu} + \partial_{\nu} h_{\mu \beta} - \partial_{\beta} h_{\mu \nu}\right) $$

in the above expression for $R_{\mu \nu}^{(2)}$ I suppose to find

$$ R_{\mu \nu}^{(2)} = \frac{1}{2}\Bigg[ \frac{1}{2}\partial_{\mu}h_{\alpha \beta}\partial_{\nu}h^{\alpha \beta} + \partial_{\beta}h_{\nu \alpha}\left( \partial^{\beta}h^{\alpha}{}_{\mu} - \partial^{\alpha}h^{\beta}{}_{\mu} \right) + h_{\alpha \beta} \left( \partial_{\mu}\partial_{\nu}h^{\alpha \beta} + \partial^{\alpha}\partial^{\beta}h_{\mu \nu} - \partial^{\beta}\partial_{\nu}h^{\alpha}{}_{\mu} - \partial^{\beta}\partial_{\mu}h^{\alpha}{}_{\nu} \right) - \left( \partial_{\alpha}h^{\alpha \beta} - \frac{1}{2}\partial^{\beta}h\right)\left( \partial_{\mu} h_{\nu \beta} + \partial_{\nu} h_{\mu \beta} - \partial_{\beta} h_{\mu \nu}\right) \Bigg] $$

but I do not. I tried it over and over again but somehow I cannot get the correct result. It seems I am missing something but I don't know what. I will be glad if someone can help.


This post imported from StackExchange Physics at 2015-02-13 11:37 (UTC), posted by SE-user sahin

asked Feb 12, 2015 in Theoretical Physics by sahin (5 points) [ revision history ]
retagged Feb 13, 2015
Most voted comments show all comments
There isn't much of a learning curve, it's really easy.

This post imported from StackExchange Physics at 2015-02-13 11:37 (UTC), posted by SE-user JamalS
I hope so. I will take it serious indeed, thanks again.

This post imported from StackExchange Physics at 2015-02-13 11:37 (UTC), posted by SE-user sahin
Let us continue this discussion in chat.

This post imported from StackExchange Physics at 2015-02-13 11:37 (UTC), posted by SE-user sahin
I didn't do the calculation. However, there seems to be a mistake in your $\Gamma^{(2)}$ definition. The first two terms inside the bracket have an index $\alpha$ which should be changed to $\beta$. Maybe this helps (I will not edit it, so you can have a look at it).

This post imported from StackExchange Physics at 2015-02-13 11:37 (UTC), posted by SE-user Clever
It was just a typo, I corrected it now. I also checked my calculations again and the indices are correctly placed there. So, the problem is not about the indices. But, thanks for the warning of course.

This post imported from StackExchange Physics at 2015-02-13 11:37 (UTC), posted by SE-user sahin
Most recent comments show all comments
I don't use Mathematica. Besides I need to do this derivation by hand to add it into the Appendix part of my thesis. Thanks for the advice anyway.

This post imported from StackExchange Physics at 2015-02-13 11:37 (UTC), posted by SE-user sahin
I understand. It's a good tool to check calculations, though.

This post imported from StackExchange Physics at 2015-02-13 11:37 (UTC), posted by SE-user JamalS

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