# Weak-field Einstein's equation approximation

For Einstein's equation $$G_{\mu \nu} + \Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$ with $$G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} R \, g_{\mu \nu}$$ where $$R_{\mu \nu}$$ is the Ricci tensor, $$R$$ is the scalar curvature, $$g$$ is the metric tensor and $$T$$ is the stress-energy tensor, the weak-field approximation is proposed; $$g$$ is a locally flat metric, so $$g_{\mu \nu} \approx \eta_{\mu \nu} + h_{\mu \nu}$$ with $$h << 1$$, $$\Lambda$$ is ignored, then the following equation can be eventually derived $$\nabla^2 \phi = 4 \pi G \rho$$ where $$\bar{h}^{00} = - 2 \phi \,$$ (here $$\bar{h}^{\mu \nu} = h^{\mu \nu} - \frac{1}{2} \eta^{\mu \nu} h^\alpha_{\;\; \alpha}$$ is the trace inverse) and $$T^{00} = \rho$$ and so it seems we get a regular Poisson equation for a Newtonian potential following the mass distribution.

However, the form $$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$ is only a local approximation, in other words, for a general $$g$$ $$g = g_{\mu \nu} \, \mathrm{d} x^\mu \otimes \mathrm{d} x^\nu$$ we need to introduce special coordinates - some kind of normal coordinates, that will fix this form around some specific point $$P$$ on the manifold. For example, the Riemann normal coordinates $$\xi$$ given to the third order approximation give the mapping $$\xi \to x$$ explicitly as $$x^\mu = x_0^\mu + \xi^\mu - \frac{1}{2} \Gamma^\mu_{\;\; \alpha \beta} (P) \xi^\alpha \xi^\beta - \frac{1}{3!} \left( \Gamma^\mu_{\;\; \alpha \beta, \gamma} (P) - 2 \Gamma^\mu_{\;\; \alpha \lambda} (P) \Gamma^\lambda_{\;\; \beta \gamma} (P) \right) \xi^\alpha \xi^\beta \xi^\gamma + \cdots$$ where $$\Gamma$$ are Christoffel symbols and their derivatives expressed at point $$P$$ and $$x_0^\mu$$ is the coordinate expression in the unprimed coordinates of the point $$P$$.

Only in these coordinates, we find, that $$g = g_{\mu \nu} \, \mathrm{d} x^\mu \otimes \mathrm{d} x^\nu = \left( \eta_{\mu \nu} - \frac{1}{3} R_{\mu \alpha \nu \beta} (P) \, \xi^\alpha \xi^\beta + \cdots \right) \mathrm{d} \xi^\mu \otimes \mathrm{d} \xi^\nu$$ where $$R$$ is the Riemann tensor in the unprimed coordinates at point $$P$$.

Now we can identify $$h_{\mu \nu}$$ as $$- \frac{1}{3} R_{\mu \alpha \nu \beta} (P) \, \xi^\alpha \xi^\beta$$ but this is small only in the close vicinity of the point $$P$$ (i.e. small $$\xi$$) and how quickly this blows up depends on size of $$R_{\mu \alpha \nu \beta}$$ at $$P$$. I then wonder what the Poisson equation for $$\phi$$ really means; if we mean $$\left( \frac{\partial^2}{\partial \xi_1^2} + \frac{\partial^2}{\partial \xi_2^2} + \frac{\partial^2}{\partial \xi_3^2} \right) \phi (\xi) = 4 \pi G \, \rho (\xi)$$ (here we neglect the small part of $$g$$ when raising and lowering indices, since we are only interested in the lowest order)

then this is only valid when $$\xi_1, \xi_2, \xi_3$$ are around 0, corresponding to some region around the point $$P$$, otherwise the term $$h_{\mu \nu}$$ is eventually bound to grow (even if somehow $$R$$ is zero at $$P$$, there are higher order terms beyond $$h_{\mu \nu}$$ that will overtake the growth).

Does that mean we actually have several sets of $$\xi$$-like variables, for small chunks of space and we sow the solutions to the Poisson equation together, like so?

$$\hskip1in$$

In that case we cannot even impose the standard boundary condition $$\phi (|\xi| \to \infty) \to 0$$, because no specific $$\xi$$ can grow like that, because at some point we lose the approximation $$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$!

Or am I overthinking this and is it just some symbolic manipulation to show how Einstein's equation is related to Newtonian gravity, without actually claiming that the equation $$\nabla^2 \phi = 4 \pi \rho$$ extends to an arbitrarily large region? I would understand why you might not be able to come "too close" to the source generating $$\phi$$, because then $$\phi$$ is no longer a "weak field", but this construction of normal variables shows that you cannot even drag $$\xi$$ to infinity, where there are no sources, without losing the approximation that $$h$$ is small.

While what I presented in my question is correct in general, whenever the Newtonian limit of Einstein's equation is discussed, the metric $$g$$ is assumed to be of the form $$\eta + h$$ where $$h$$ is assumed to be small in the whole space, by definition, in some default provided coordinates. Indeed, the weak field limit, when reflected by the metric, presents itself as $$g = -(1+2 \phi) \mathrm{d} t \otimes \mathrm{d} t + (1 - 2 \phi) \left( \mathrm{d} x \otimes \mathrm{d} x + \mathrm{d} y \otimes \mathrm{d} y + \mathrm{d} z \otimes \mathrm{d} z \right)$$ where $$\phi \ll 1$$ by definition, in the whole space (or most of it, outside some finite region), and so the equation $$\nabla^2 \phi = 4 \pi \rho$$ can be solved in the whole space (or most of it, outside some finite region with some boundary condition at the boundary).
If $$h$$ being small does not apply, we can still force $$g$$ to be locally $$\eta + h$$ where $$h$$ is small, however, this would only be true in a small region around some point we pick, in special (normal) coordinates. With the metric above, it would be unwise to do so, because while $$h$$ is globally small, normal coordinates would spoil this property$${}^1$$, so a better approach is to keep the default coordinates in which $$h \ll \eta$$ in the whole spacetime.
$${}^1$$With a caveat: the normal coordinates wouldn't actually spoil it if we summed all terms beyond $$\eta$$, but the quadratic term itself, equal to $$- \frac{1}{3} R_{\mu \alpha \nu \beta} \xi^\alpha \xi^\beta$$ eventually blows up. This is similar to how for $$f(x) = 1 + 0.01 \sin x$$, term $$0.01 \sin x$$ is small compared to $$1$$, if we take the second lowest-order approximation $$f (x) \approx 1 + 0.01 x$$, this eventually blows up and $$0.01 x$$ will no longer be much smaller than 1. This shows that the choice of coordinates might, as usual, simplify or complicate things, depending on the circumstances.