Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Quadratic order perturbation terms in the expansion of Ricci tensor

+ 1 like - 0 dislike
3569 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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...