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In Schutz says When we have weak gravitaional fields then the line element ds is $$ ds^{2}=-(1+2\phi)dt^{2}+(1-2\phi)(dx^{2}+dy^{2}+dz^{2}) $$ so the metric is
$$ {g_{\alpha\beta}} =\eta_{\alpha\beta}+h_{\alpha\beta}= \left( \begin{array}{cccc} -(1+2\phi) & 0 & 0 & 0\\ 0 & (1-2\phi) & 0 & 0\\ 0 & 0 & (1-2\phi) & 0\\ 0 & 0 & 0 & (1-2\phi)\end{array} \right) $$ where $$ \phi=\frac{M}{r} $$ so h is

$$ {h_{\alpha\beta}} = \left( \begin{array}{cccc} -2\phi & 0 & 0 & 0\\ 0 & -2\phi & 0 & 0\\ 0 & 0 & -2\phi & 0\\ 0 & 0 & 0 & -2\phi\end{array} \right) $$ the element

$$ h_{00}= -2\phi $$

and the elements out of the diagonal are zero because the condition weak gravitational fields it implies

$$ T_{i,j}=0 $$

but i dont get it how in the book do

$$ h_{xx}=h_{yy}=h_{zz}=-2\phi $$

i believe they use de definition of trace reverse

$$ \bar h^{\alpha\beta}=h^{\alpha\beta}-\frac{1}{2}\eta^{\alpha\beta}h $$ and the trace definition

$$ h = h^{\alpha}_{\alpha} $$ but how they do? what im missing?

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  • $\begingroup$ The case you consider is the limiting case of the Schwarzschild solution which is axially symmetric, so the result follows. $\endgroup$ – Andrey Feldman Nov 28 '16 at 8:31
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When doing perturbation theory in GR it is custom to define a trace-reversed metric $\bar{h}$ and pick a certain gauge, for instance $\partial^\mu \bar{h}_{\mu \nu} = 0$, so that Einsteins equations reduce to $\partial^\rho \partial_\rho \bar{h}_{\mu \nu} = -16 \pi T_{\mu \nu}$.

You know need to make some assumptions regarding your matter sources -- for the newtonian limit you assume that your matter is non-relativistic (i.e. almost stationary), so that $T_{\mu \nu} << T_{0 0}$, and from the linearized field equations, the same holds for $\bar h_{\mu \nu}$ (i.e. $\bar h_{0 0} >> \bar h_{ij}$).

Hence for the trace we get $h= - \bar{h} = - \eta^{\mu \nu} \bar{h}_{\mu \nu} = \bar{h}_{00} + \mathrm{small}$.

But also we have $h_{i j} = \bar{h}_{ ij} - \frac{1}{2} \eta_{ij} \bar{h}$ (from inverting the definition of the trace-reversed metric), and hence $h_{ij} = - 2 \Phi \delta_{ij}$. (using that $h_{00} = - 2 \Phi$ is equivalent to $\bar{h}_{00} = 2 h_{00} = -4 \Phi$, from the definition of the trace-reversed metric.)

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