Why are spatial components equal to h_00 in weak field metric?

[needved]

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 don't 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 I am missing?

Thanks

 

[Quantum_gravity]

We are just pulling each diagonal term out of h. h₁ = h_x. Repeated subscript is an element along the diagonal.