A Solving Linearized EFE for Newtonian Potential Under Lorentz Gauge

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

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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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