Prove Weak Field Metric Homework Statement

[aman02]

Homework Statement

For the weak field metric
$$g_{00}=-1+\epsilon h_{00}+O(\epsilon^2)$$
$$g_{\alpha\beta}=\delta_{\alpha\beta}+\epsilon h_{\alpha\beta}+O(\epsilon^2)$$
Prove
$$R_{00}=-\frac{1}{2}\epsilon\frac{\partial^2h_{00}}{\partial x^\alpha\partial x^\beta}+O(\epsilon^2)$$

Homework Equations

The hint was to use:
$$R^q_{lmn}=\Gamma^q_{ln,m}-\Gamma^q_{lm,n}+\Gamma^p_{ln}\Gamma^q_{pm}-\Gamma^p_{lm}\Gamma^q_{pn}$$
$$R_{ijkl;m}+R_{ijlm;k}+R_{ijmk;l}=0$$
But I couldn't figure out how, so I included
$$\Gamma^i_{jk}=\frac{1}{2}g^{il}(g_{jl,k}+g_{lk,j}-g_{kj,l})$$
$$R_{jl}=R^i_{jil}$$

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.
$$R_{00}=R^\alpha_{0 \alpha 0}=\Gamma^\alpha_{00,\alpha}-\Gamma^\alpha_{0 \alpha,0}+\Gamma^p_{00}\Gamma^\alpha_{p \alpha}-\Gamma^p_{0 \alpha}\Gamma^\alpha_{p0}$$
Then using the definition of the Christoffel symbols,
$$\Gamma^\alpha_{00}=\frac{1}{2}g^{\alpha \alpha}(g_{0\alpha,0}+g_{\alpha 0,0}-g_{00,\alpha})$$
$$g_{\alpha \alpha}\Gamma^\alpha_{00}=\frac{1}{2}(4)(-1+\epsilon h_{00}+O(\epsilon^2))_{, \alpha}$$
$$g_{\alpha \alpha}\Gamma^\alpha_{00}=-2\epsilon (\frac{\partial h_{00}}{\partial x^\alpha})+O(\epsilon^2)$$
Similarly, I found that
$$g_{\alpha \alpha}\Gamma^0_{0 \alpha}=2\epsilon (\frac{\partial h_{00}}{\partial x^\alpha})+O(\epsilon^2)$$
It seems from here, the solution is fairly close, but I don't know what to do with the $$g_{\alpha \alpha}$$, or if I should even have them. If not, I don't know how to deal with the $$g^{\alpha \alpha}$$ 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.

 

[kdv]

Homework Statement

For the weak field metric
$$g_{00}=-1+\epsilon h_{00}+O(\epsilon^2)$$
$$g_{\alpha\beta}=\delta_{\alpha\beta}+\epsilon h_{\alpha\beta}+O(\epsilon^2)$$

that's a bit confusing. You mean that alpha and beta are only 1,2,3 in the second equation?

Prove
$$R_{00}=-\frac{1}{2}\epsilon\frac{\partial^2h_{00}}{\partial x^\alpha\partial x^\beta}+O(\epsilon^2)$$

Homework Equations

The hint was to use:
$$R^q_{lmn}=\Gamma^q_{ln,m}-\Gamma^q_{lm,n}+\Gamma^p_{ln}\Gamma^q_{pm}-\Gamma^p_{lm}\Gamma^q_{pn}$$
$$R_{ijkl;m}+R_{ijlm;k}+R_{ijmk;l}=0$$
But I couldn't figure out how, so I included
$$\Gamma^i_{jk}=\frac{1}{2}g^{il}(g_{jl,k}+g_{lk,j}-g_{kj,l})$$
$$R_{jl}=R^i_{jil}$$

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.
$$R_{00}=R^\alpha_{0 \alpha 0}=\Gamma^\alpha_{00,\alpha}-\Gamma^\alpha_{0 \alpha,0}+\Gamma^p_{00}\Gamma^\alpha_{p \alpha}-\Gamma^p_{0 \alpha}\Gamma^\alpha_{p0}$$
Then using the definition of the Christoffel symbols,
$$\Gamma^\alpha_{00}=\frac{1}{2}g^{\alpha \alpha}(g_{0\alpha,0}+g_{\alpha 0,0}-g_{00,\alpha})$$

This does not make sense since there are three alphas on the right. be careful with the indices

$$g_{\alpha \alpha}\Gamma^\alpha_{00}=\frac{1}{2}(4)(-1+\epsilon h_{00}+O(\epsilon^2))_\alpha$$

This does not make sense. What does $1_\alpha$ mean? Or what is $(h_{00})_\alpha$??


I am not sure about the fastest way to proceed but if you do it the long way, why not find $g^{\mu \nu }$ to first order? You should find easily that it is $\eta^{\mu \nu} - \epsilon h^{\mu \nu}$

 

[aman02]

that's a bit confusing. You mean that alpha and beta are only 1,2,3 in the second equation?

Sorry, yes, that is what I meant.

This does not make sense since there are three alphas on the right. be careful with the indices

I may have skipped a step - I noticed the only non-zero elements in $$g_{ab}$$ and $$g^{ab}$$ are those on the diagonal, so I let $$l=\alpha$$. That put a few more alphas in on the right.

This does not make sense. What does $1_\alpha$ mean? Or what is $(h_{00})_\alpha$??

Sorry, another typo...that should be $$(-1+\epsilon h_{00}+O(\epsilon^2))_{ , \alpha}$$, as in partial derivative of all that with respect to the alpha coordinate. Then $1_\alpha$ becomes 0, and $(h_{00})_\alpha$ is the partial with respect to alpha.

I am not sure about the fastest way to proceed but if you do it the long way, why not find $g^{\mu \nu }$ to first order? You should find easily that it is $\eta^{\mu \nu} - \epsilon h^{\mu \nu}$

You're correct, in fact that fact is a different question on the assignment that I did solve, but I wasn't sure what the $$h^{\mu \nu}$$ would do, because the final solution didn't have anything in it of that form.

 

[kdv]

Sorry, yes, that is what I meant.

Then it is easier to use the notation $$g_{\alpha \beta} = \eta_{\alpha \beta} + \epsilon h_{\alpha \beta}$$ for all the components or if you give only the spatial part, use latin indices (i,j,k)

I may have skipped a step - I noticed the only non-zero elements in $$g_{ab}$$ and $$g^{ab}$$ are those on the diagonal, so I let $$l=\alpha$$. That put a few more alphas in on the right.

why do you assume that g has only diagonal elements?
In any case, even if it was true, you should never write something with three identical indices!

Sorry, another typo...that should be $$(-1+\epsilon h_{00}+O(\epsilon^2))_{ , \alpha}$$, as in partial derivative of all that with respect to the alpha coordinate. Then $1_\alpha$ becomes 0, and $(h_{00})_\alpha$ is the partial with respect to alpha.


You're correct, in fact that fact is a different question on the assignment that I did solve, but I wasn't sure what the $$h^{\mu \nu}$$ would do, because the final solution didn't have anything in it of that form.

Well, if you rewrite your Christoffel symbols with the correct notation for the indices it will be easier to help with the remaining steps.

 

[aman02]

Alright, thanks for the help kdv - I've gone through with correct indices and reduced the problem to showing spatial derivatives dominate time derivatives...why would this be true?