A Solving Linearized EFE for Newtonian Potential Under Lorentz Gauge

Arman777
Insights Author
Gold Member
Messages
2,163
Reaction score
191
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}## ?
 

Attachments

  • 1621685438572.png
    1621685438572.png
    28 KB · Views: 146
Physics news on Phys.org
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.
 
  • Like
Likes vanhees71 and Arman777
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
 
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
Back
Top