Does solving Einstein's field equation depend on Newtonian equations?

Question

It seems that Einstein's field equation in a vacuum depends on Newton, to actually be solved.

$$R_{ij}=0$$

where we assume in the weak gravity limit, it reduces to Poisson's equation.

Now, I suspect that the addition of the Stress-Energy Tensor fixes that, because it adds a source to the field, but my knowledge is limited here. Is that the case? I never thought about using the Stress-Energy Tensor because I'm working outside a massive body, so in a vacuum. But, is the addition of it (Stress-Energy Tensor) completing the equation, so it doesn't depend on any Newtonian limit?

Edit: The question was somewhat unclear. What I meant is:

Are EFE enough in order to derive the metric? Because the derivation of schwarzschild metric depends on it reducing to Newtonian limit, and this might be due to the equation not having an energy tensor. So if the energy tensor is added, EFE can derive schwarschild metric without any Newtonian limit taken into account.

Answer 1

When Einstein derived his equations the Newtonian limit (and it may be taken also in the present of sources) was an important check for the theory. It is also useful to understand the limits of the Newtonian theory, and sometimes helps to work out the intuition about the relativistic case.

However to solve Einstein equations the Newtonian limit is not required and sometimes is not useful at all.

First, there are exact solutions of the Einstein equations: various black holes and the sufficiently symmetric stars, the cosmological solutions and some other situations. Second, you may use the perturbation theory near the exact solutions. The simplest such application is linearized gravity that describes gravitational waves on a flat Minkowski background. Note that contrary to your original post, this weak gravity limit is not Newtonian. The gravitational waves do not exist in the Newtonian theory.

Third, you may use the Newtonian limit and start with the Poisson equations. Then you may perturb this solution adding relativistic effects. This is,obviously, a method depending on the Newtonian theory. It is known as a post-Newtonian expansion and is different from a weak gravity limit.

You may also solve the Einstein equations numerically. This is e.g. is used to study the black hole collisions where different perturbation theories fail. This is also a situation where the Newtonian theory is useless.

Answer 2

I recently heard, that there is a derivation of the field equations not using the Newtonian limit. The one I know of uses it though. We get $g_{00}=1-\frac{2\Phi}{c^2}$ out of the Newtonian limit of the geodesic equations. We also get $T_{00}=\rho c^2$ out of the energy-stress tensor, which yields: \begin{equation} \Delta\Phi =4\pi G\rho \Rightarrow \Delta g_{00} =\frac{8\pi G}{c^4}T_{00} =\kappa T_{00}. \end{equation}

A Lorentz invariant generalization would be $\square g_{\alpha\beta}=\kappa T_{\alpha\beta}$, but applying the covariance principle yields $\square g_{\mu\nu}=g^{\kappa\lambda}\nabla_{\kappa}\nabla_{\lambda}g_{\mu\nu}=T_{\mu\nu}$ which can't work out since the left side (due to $\nabla_\lambda g_{\mu\nu}=0$) always vanishes.

We forgot about the fundamental difference to Newtonian gravity. Since gravity carries energy, it is its own source and therefore has to be considered as well with its own energy-stress tensor: \begin{equation} \square g_{\mu\nu} =\kappa\left(T_{\mu\nu}+t_{\mu\nu}^\text{grav}\right) \Leftrightarrow G_{\mu\nu} =\kappa T_{\mu\nu}. \end{equation} To determine this tensor $G_{\mu\nu}$, called the Einstein tensor, we need a few conditions (symmetry and being divergence free, are directly inherited from the energy-stress tensor), like being linear in the second quadratic, in first derivatives of the metric tensor. This only determines the tensor as $R_{\mu\nu}-\frac{R}{2}g_{\mu\nu}$ up to a scalar, which can be obtained with the last condition of the Newtonian limit dictating $G_{00}\approx\Delta g_{00}$ and giving us the field equations as: \begin{equation} R_{\mu\nu}-\frac{R}{2}g_{\mu\nu} =\kappa T_{\mu\nu}. \end{equation}

Answer 3

If, in some limit, GR didn't reduce down to $\kappa \nabla ^{2}\phi = \rho$ for some constant kappa, (where $\phi$ is a gravitational potential such that ${\vec g} = {\vec \nabla}\phi$), then it wouldn't be a very good generalization of Newtonian mechanics, so of course this happens.