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I am wondering about exact vacuum solutions to the field equations of GR. The field equations of Newtonian gravity are given by

$$ \nabla^2 \phi = -4\pi G \rho, $$

and the associated field equations of GR are given by

$$ G_{\mu\nu} = \kappa T_{\mu\nu}. $$

For vacuum solutions we have $G_{\mu\nu} \equiv 0.$

Now my question is regarding whether or not we can draw any conclusions from the following situation:

Say I have a potential $\phi(\rho,\theta,z)$ that satisfies the Laplace equation in cylindrical coordinates given by

$$ \nabla^2 \phi(\rho, \theta, z) = 0.$$

Then say I look for vacuum solutions of the field equations where the metric components are explicit functions of the potential given in the Laplace equation namely,

$$ g_{\mu\nu} = g_{\mu\nu} (\phi), $$

so that the equations generated from the Einstein are ODE's and not PDE's.

Can we draw any conclusions from the specific set up mentioned above? i.e. is there any simplifications we can place on the system?

OR

Is there any relation between a potential satisfying the Laplace equation and the Einstein tensor when the metric components are functions of the potential?

OR

Is there any process that’s been developed for such a set up to help in solving the Einstein field equations?

NOTE:

I have just arbitrarily chose cylindrical coordinates.

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  • $\begingroup$ The EFE (Einstein field equations) in GR (general relativity) reduce to the field equations of Newtonian gravity when the field is weak, but you can not infer the opposite. $\endgroup$ – Michele Grosso Apr 3 '18 at 16:21
  • $\begingroup$ when the field is weak and quasistatic $\endgroup$ – Slereah Apr 4 '18 at 7:32
  • $\begingroup$ where the metric components are explicit functions of the potential what does that mean? Do you mean like $$ds^2=e^{2\phi}dt^2-e^{-2\phi}\gamma_{ab}dx^a dx^b?$$ If so, then yes, EFE do produce Laplace equation on $\phi$. $\endgroup$ – A.V.S. Apr 4 '18 at 16:34
  • $\begingroup$ @A.V.S. Almost the metric will take the form $ds^2 = e^{2 a(\phi)} dt^2 - e^{2b(\phi)} \gamma_{ab} dx^a dx^b $. Any ideas there? $\endgroup$ – Rumplestillskin Apr 4 '18 at 21:33

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