A Solving Linearized EFE for Newtonian Potential Under Lorentz Gauge

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Under the Lorentz Gauge, the linearized Einstein Field Equations (EFE) are expressed as a wave equation involving the perturbation tensor. The discussion centers on understanding how the d'Alembert operator (□) simplifies to the Laplacian (∇²) in this context. Participants seek clarification on specific equations from a referenced text, particularly regarding the dominance of the 00 component of the stress-energy tensor and its implications for other components. The conversation highlights the assumption that the 00 component significantly outweighs the spatial components, leading to simplifications in calculations. Ultimately, the participants resolve their queries regarding the relationships between the various components of the perturbation tensor.
Arman777
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Under the Lorentz Gauge the Einstein Field Equations are given as

$$G^{\alpha \beta} = -\frac{1}{2}\square \bar{h}^{\alpha \beta}$$

Then the linearized EFE becomes,

$$\square \bar{h}^{\mu\nu} = -16 \pi T^{\mu\nu}$$

For the later parts, I ll share pictures from the book

1621685421634.png
1621685459266.png
I have couple of questions

1) I did not understand how the ##\square## becomes ##\nabla^2## for this case.

2) I did not understand equation 8.47 at all.

3) I also did not understand 8.49. Since he only defined $$\nabla^2\bar(h)^{0 0} = -16 \pi \rho$$.

Is there also terms like ##\nabla^2\bar(h)^{xx} = -16 \pi \rho##, ##\nabla^2\bar(h)^{yy} = -16 \pi \rho## and ##\nabla^2\bar(h)^{zz} = -16 \pi \rho## ?

This might be helpful for you guys
1621685866975.png


Please help. Thanks

For instance, by using 8.31 and 8.46 I can write,

$$h^{0 0} = -4\phi - \frac{1}{2} (-1) \bar{h}$$ but what is ##\bar{h}## here ?

If we only know (8.45), how can we calculate ##h^{xx}## ?
 

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1) The argument is that ##\partial_t## is of order ##v \partial_x##. Since ##v\ll 1##, the ##\partial_t## terms are negligible.

2) This is just computing the trace of ##h##. First step is the definition of the trace. Second step is using that ##\bar h## is the trace-reversed perturbation (take the trace of the definition of ##\bar h_{ab}##). Last step is using that the 00 component is assumed to completely dominate ##\bar h##.

3) No. The entire argumentation is based on the 00 component of the stress energy tensor dominating and all other terms therefore being negligible.
 
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Orodruin said:
1) The argument is that ##\partial_t## is of order ##v \partial_x##. Since ##v\ll 1##, the ##\partial_t## terms are negligible.

2) This is just computing the trace of ##h##. First step is the definition of the trace. Second step is using that ##\bar h## is the trace-reversed perturbation (take the trace of the definition of ##\bar h_{ab}##). Last step is using that the 00 component is assumed to completely dominate ##\bar h##.

3) No. The entire argumentation is based on the 00 component of the stress energy tensor dominating and all other terms therefore being negligible.
Thanks for your answer.

1) I understand this one

2-3) So you mean ##\bar{h}^{00} \gg \bar{h}^{11}, \bar{h}^{22} , \bar{h}^{33}## ?

If that's the case, then how can we calculate ##h^{xx}, h^{yy}## and ##h^{zz}## ?
 
Okay, solved it. Nvm
 
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