2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.