Understanding the derivation of the Newtonian limit of GR I'm having difficulty understanding some of this derivation. The relevant information, as I understand it is:
We've let the lorentzian metric be a perturbation of the minkowski metric:
$$g_{\mu\nu} = \eta_{\mu\nu}+h_{\mu\nu},$$
where apparently if $h_{\mu\nu}$ is small enough, we can approximate the inverse as:
$$g^{\mu\nu} = \eta^{\mu\nu}-h^{\mu\nu}.$$
By some assumptions, the stated results are:
$$ \Delta \bar{h}_{00} = -2\kappa T_{00}$$
$$ \Delta \bar{h}_{0i}  = 0 $$
$$ \Delta \bar{h}_{ij}  = 0 $$
where $\bar{h}_{\mu \nu} = h_{\mu \nu} - \frac{1}{2}h_\beta^\beta\eta_{\mu\nu}$.
Which we can substitute into the Newtonian gravitation equation to yield:
$$ \bar{h}_{00}  = \frac{-\kappa c^2}{2\pi G} \Phi $$
$$ \bar{h}_{0i}  = 0 $$
$$ \bar{h}_{ij}  = 0 $$
We then 'invert' the perturbation to: $ h_{\mu \nu} = \bar{h}_{\mu \nu} - \frac{1}{2}\bar{h}_\beta^\beta\eta_{\mu\nu}$.
and obtain from this:
$$h_{00} = \frac{1}{2} \bar{h}_{00} =  \frac{-\kappa c^2}{2\pi G} \Phi $$
$$h_{0i} = 0$$
$$h_{ij} = \frac{1}{2} \delta_{ij} \bar{h}_{00} = -\delta_{ij}\frac{\kappa c^2}{2\pi G} \Phi$$
I can follow most of it alright, I'm just struggling to figure out the working behind the last step if anyone could explain this.
 A: I refer to some steps for derivation of the Newtonian limit of GR in the book General relativity written by S. Caroll, so some notations are quite different.
We want to interpret LHS of Einstein equation as gravitational potential, and RHS as mass density. Thus, approximations of $h_{\mu\nu}, T_{\mu\nu}$ in Newtonian limit are necessary.
Start with little perturbation of metric.
$$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} , \; |h_{\mu\nu}| \ll 1 $$
Calculate Christoffel symbol in the first order of $h_{\mu\nu}$. Only consider $00$ component to derive Newtonian limit. (We can find other components as linearzed gravity, too)
$$ \Gamma^{\mu}_{00} = \frac{1}{2} \eta^{\mu \lambda} ( 0 + 0 - \partial_{\lambda} h_{00} ) $$
Put it in geodesic equation.
$$ \frac{d^2 x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{00} \frac{dx^0}{d\tau} \frac{dx^0}{d\tau} = 0$$
$$ \frac{d^2 x^{\mu}}{d\tau^2} = \frac{1}{2} \eta^{\mu \lambda} \partial_{\lambda} h_{00}  (\frac{dt}{d\tau})^2 $$
$$\frac{d^2 x^{\mu}}{dt^2} = \frac{1}{2} \partial_{\mu} h_{00} $$
LHS is acceleration, so we can interpret $- \frac{1}{2} h_{00}$ as gravitational potential.
$$ h_{00} = - 2\Phi$$
Non-zero components of Riemann curvature tensor and Ricci tensor are only,
$$R^i \; _{0j0} = \partial_j \Gamma^i_{00} + 0 $$
$$R_{00} = R^i \; _{0i0} = \partial_i (-\frac{1}{2} \eta^{i \lambda} \partial_{\lambda} h_{00}  ) = -\frac{1}{2} \nabla^2 h_{00} $$
Particle moves slowly in Newtonian limit, so stress-energy tensor become,
$$T_{00} = \rho$$
$$T = g^{00}T_{00} = -\rho$$
Now, put them all in Einstein field equation ($c=1$)
$$ R_{\mu \nu} = 8\pi G (T_{\mu\nu} - \frac{1}{2} T g_{\mu\nu} )$$
$$ -\frac{1}{2} \nabla^2 h_{00} = 8 \pi G \frac{1}{2}\rho $$
$$ \nabla^2 \Phi = -4\pi G \rho  $$
which is Poisson equation in Newtonian-gravity.
A: 
I'm just struggling to figure out the working behind the last step

Note that $\overline{h}=h-\frac12h4=-h$ in $4$-dimensional spacetime, so $h_{\mu\nu}=\overline{h}_{\mu\nu}+\frac12h\eta_{\mu\nu}=\overline{h}_{\mu\nu}-\frac12\overline{h}\eta_{\mu\nu}$. What's more, $\overline{h}=\overline{h}_{00}=-\frac{\kappa c^2}{2\pi G}\Phi$. Now use $\eta_{00}=1,\,\eta_{0i}=0,\,\eta_{ij}=-\delta_{ij}$.
