- #1

aman02

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## Homework Statement

For the weak field metric

[tex]g_{00}=-1+\epsilon h_{00}+O(\epsilon^2)[/tex]

[tex]g_{\alpha\beta}=\delta_{\alpha\beta}+\epsilon h_{\alpha\beta}+O(\epsilon^2)[/tex]

Prove

[tex]R_{00}=-\frac{1}{2}\epsilon\frac{\partial^2h_{00}}{\partial x^\alpha\partial x^\beta}+O(\epsilon^2)[/tex]

## Homework Equations

The hint was to use:

[tex]R^q_{lmn}=\Gamma^q_{ln,m}-\Gamma^q_{lm,n}+\Gamma^p_{ln}\Gamma^q_{pm}-\Gamma^p_{lm}\Gamma^q_{pn}[/tex]

[tex]R_{ijkl;m}+R_{ijlm;k}+R_{ijmk;l}=0[/tex]

But I couldn't figure out how, so I included

[tex]\Gamma^i_{jk}=\frac{1}{2}g^{il}(g_{jl,k}+g_{lk,j}-g_{kj,l})[/tex]

[tex]R_{jl}=R^i_{jil}[/tex]

## The Attempt at a Solution

I tried to do it the long way without the Bianchi identity, and started with the basic definition of the Ricci tensor.

[tex]R_{00}=R^\alpha_{0 \alpha 0}=\Gamma^\alpha_{00,\alpha}-\Gamma^\alpha_{0 \alpha,0}+\Gamma^p_{00}\Gamma^\alpha_{p \alpha}-\Gamma^p_{0 \alpha}\Gamma^\alpha_{p0}[/tex]

Then using the definition of the Christoffel symbols,

[tex]\Gamma^\alpha_{00}=\frac{1}{2}g^{\alpha \alpha}(g_{0\alpha,0}+g_{\alpha 0,0}-g_{00,\alpha})[/tex]

[tex]g_{\alpha \alpha}\Gamma^\alpha_{00}=\frac{1}{2}(4)(-1+\epsilon h_{00}+O(\epsilon^2))_{, \alpha}[/tex]

[tex]g_{\alpha \alpha}\Gamma^\alpha_{00}=-2\epsilon (\frac{\partial h_{00}}{\partial x^\alpha})+O(\epsilon^2)[/tex]

Similarly, I found that

[tex]g_{\alpha \alpha}\Gamma^0_{0 \alpha}=2\epsilon (\frac{\partial h_{00}}{\partial x^\alpha})+O(\epsilon^2)[/tex]

It seems from here, the solution is fairly close, but I don't know what to do with the [tex]g_{\alpha \alpha}[/tex], or if I should even have them. If not, I don't know how to deal with the [tex]g^{\alpha \alpha}[/tex] in the original definition (I just lowered the indices and got the trace of the metric, 4 for our 4-dimensional space-time). Could using Bianchi help? I'm not quite sure how to proceed, any help is appreciated.

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