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Often in a course in GR one can recover Newton's law of gravity under certain assumptions. weak field, slow moving particles etc. Is there a general method to recover Newton's laws of gravity for an arbitrary spherically symmetric, static and asymptotically flat spacetime metric? For example with the following line element

$$ (ds)^2 = A(r) dt^2 + 2B(r) dr dt - C(r) dr^2 -r^2 d\Omega^2, $$ where the functions $A,B,C$ are all determined and the above line element is an exact solution to the field equations.

Or, can we only recover Newton's law of gravity when the components of the metric tensor are of the form $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ where the $\eta_{\mu\nu}$ are the components of Minkowski's flat space plus a perturbation given by $h_{\mu\nu}$?

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A hint :

\begin{eqnarray} -A(r) dt^2 + B(r) dr dt - C(r) dr^2 -r^2 d\Omega^2 &=& (dt^2 - dr^2 -r^2 d\Omega^2)\\ &+& (A(r)-1) dt^2 + B(r) dr dt - (C(r) - 1) dr^2 \end{eqnarray}

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  • $\begingroup$ I'm not really sure I follow your hint? $\endgroup$ – Rumplestillskin Jul 11 '17 at 12:48
  • $\begingroup$ You can always write any metric in the form $g = \eta + h$, the only difference will be at which point you consider $h$ to be a small perturbation, which is decided by criterions of precision, not the form of the metric. $\endgroup$ – Slereah Jul 11 '17 at 12:53
  • $\begingroup$ Okay, I guess I am confused about how I would recover Newton's law from your hint? $\endgroup$ – Rumplestillskin Jul 11 '17 at 12:58
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    $\begingroup$ @Rumplestillskin You're right that we can only recover Newton's law of gravity when the components of the metric tensor are of the form $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ and $h_{\mu \nu}$ is small. However, note that you aren't required to write $\eta_{\mu \nu}$, the Minkowski metric, in Cartesian coordinates. It is equally valid to write $\eta = -\mathrm{d}t^2 + \mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2$. Perhaps this clarifies Slereah's answer. $\endgroup$ – gj255 Jul 11 '17 at 13:31
  • $\begingroup$ @gj255 ah of course!! Thanks for clearing that up. I notice that in the standard literature we are only concerned with the $00$ component of the metric tensor. Does the off diagonal term in the above line element play any role now that it's not orthogonal? $\endgroup$ – Rumplestillskin Jul 11 '17 at 14:36

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