# Help with Newtonian Gravity as Limit case of General Relativity

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?

• The case you consider is the limiting case of the Schwarzschild solution which is axially symmetric, so the result follows. Nov 28 '16 at 8:31

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.)