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?

$\hskip1in$ enter image description here

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.


1 Answer 1


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.


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