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Mmmm
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Homework Statement
This question is a slightly customised version of Q18(a) P212 of Schutz
Prove that
[tex]G_{\alpha \beta} + \Lambda g_{\alpha \beta} = 8 \pi T_{\alpha \beta}[/tex]
in the Newtonian limit reduces to
[tex]\nabla^2 \phi = 4\pi \rho + \Lambda[/tex]
(I found this result in another text, using it the remaining parts of the question in the book work nicely)
Homework Equations
for weak gravity
[tex]g^{\alpha \beta} = \eta_{\alpha \beta} + h_{\alpha \beta}[/tex]
where
[tex]\eta_{\sigma \alpha} h^\sigma _\beta = h_{\alpha \beta}[/tex]
using lorentz gauge for stationary T
[tex]G_{\alpha \beta} = -\frac{1}{2}\nabla^2 \overline{h} _{\alpha \beta}[/tex]
where
[tex]\overline{h}_{\alpha \beta} = h_{\alpha \beta}-\frac{1}{2} \eta_{\alpha \beta}{h^\lambda} _\lambda[/tex]
Newtonian limits
[tex]\left T_{00}\right > \left T_{0i}\right > \left T_{ij}\right[/tex]
[tex]\left \overline{h}_{00}\right > \left \overline{h}_{0i}\right > \left \overline{h}_{ij}\right [/tex]
[tex]T_{00} \approx \rho[/tex]
[tex]\overline{h}_{00} \approx -4\phi [/tex]
[tex]{h}_{00} \approx -2\phi [/tex]
The Attempt at a Solution
[tex]G_{\alpha \beta} + \Lambda g_{\alpha \beta} = 8 \pi T_{\alpha \beta}[/tex]
using the above:
[tex]\Rightarrow -\frac{1}{2}\nabla^2 \overline{h}_{\alpha \beta} + \Lambda (\eta_{\alpha \beta} + h_{\alpha \beta}) = 8 \pi T_{\alpha \beta}[/tex]
non trivial eqn whaen [itex]\alpha = \beta =0[/itex]
[tex]\Rightarrow -\frac{1}{2}\nabla^2 \overline{h}_{00} + \Lambda (\eta_{00} + h_{00}) = 8 \pi T_{00}[/tex]
Newtonian limit
[tex]\Rightarrow -\frac{1}{2}\nabla^2 (-4\phi) + \Lambda (-1 + -2\phi) = 8 \pi \rho[/tex]
[tex]\Rightarrow \nabla^2 (\phi) = 4 \pi \rho + \frac{1}{2}\Lambda +\Lambda \phi[/tex]
it should be
[tex]\nabla^2 \phi = 4\pi \rho + \Lambda[/tex]
as you can see something has gone a bit wrong somewhere.
if that [itex] h_{00}[/itex] were -1 it would work...
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