Why aren't other Einstein equation components derived in the Newtonian limit?

[Dale]

Expressing that differential equation as a Poisson equation in the one unknown function does not neglect any information in the EFE; that's all the information there is in the case you are talking about.

That seems like a pretty clean approach. If the OP is not interested, I am. Do you have a reference?

I see the reason for doing it the Carroll way, since that is not limited to spherical symmetry, but pedagogically the approach you outline seems clean.

 

[PeterDonis]

If the OP is not interested, I am. Do you have a reference?

The fact that a general static spherically symmetric spacetime has only two unknown functions of $r$ in the metric is demonstrated in MTW, and also (IIRC) Wald. It's probably covered in other texts as well.

The fact that, if you add vacuum to the specifications, you're down to one function of $r$, is just Birkhoff's theorem. (The only wrinkle here is that you don't actually need the "static" assumption if you have spherical symmetry and vacuum; Birkhoff's theorem derives the "static" part--more precisely, the fact that there is a fourth KVF--for you.)

Those items, plus Carroll's general discussion of how to derive the Poisson equation, which is a differential equation for one unknown function ($h_{00}$), are sufficient to demonstrate what I said. If there is only one unknown function in the metric, and you have a differential equation that gives a solution for it, that equation must contain all the information that the EFE contains, since the metric is the solution of the EFE.

 

[PeterDonis]

This is a too famous result of the standard analysis of GR in the weak field limit

Ah, ok, this clarifies what you were referring to. It's more or less equivalent to what I was thinking (measuring orbital parameters of Keplerian orbits at large $r$ is equivalent to measuring the quantity $\mu$ in the metric coefficients you give).

This is wrong because if such a source of gravity doesn't exist there, then we would just have $R_{\mu \nu \lambda \rho}=0$ and there wouldn't be gravity in such a case.

You are incorrect. If you were correct, the Schwarzschild solution, and indeed any vacuum solution that is not Minkowski spacetime, would not exist. Since they do exist, your statement is disproved.

The right answer is simply that in such a spacetime we have a M source of gravity whose information is inside the metric

Now you're moving the goalposts; you're defining "a source of gravity" to mean that the metric is not Minkowski, i.e., that there is curvature present. But, as I showed above, the metric can be curved in a vacuum solution, and "vacuum" means $T_{\mu \nu} = 0$ everywhere, i.e., no "source of gravity" present.

I think you didn't understand what I said

I understand enough of what you are saying to see that you have made a number of incorrect statements about standard GR. I would advise taking some time to learn why those statements are incorrect, so that you have a proper understanding of standard GR, before trying to find some alternative theory that you think "works better".

 

[Markus Hanke]

I think it is easier to think about the "M" as a parameter for a family of metrics. Physically speaking, M should be a global property of the entire space-time, not "just" an isolated object in it. That wouldn't even make sense, since there are plenty of vacuum solutions which are everywhere source-free. I think the OP was questioning how a space-time that is globally source-free can still be non-trivial; my amateur answer to that would be that the energy-momentum tensor via the Einstein equations does not in fact determine all aspects of local geometry, but only a certain combination of components of the Riemann tensor, being the Einstein tensor ( which, I believe, is the contracted double dual of Riemann ). All other information needed to uniquely fix the local geometry comes from boundary conditions imposed when solving the system of differential equations - and that's where the M comes from, not the ( vanishing ) energy-momentum tensor.

 

[PeterDonis]

I think it is easier to think about the "M" as a parameter for a family of metrics. Physically speaking, M should be a global property of the entire space-time, not "just" an isolated object in it.

This is correct; $M$ appears in the metric everywhere in the spacetime, so it is a global property, not a local one; and there is an infinite family of metrics parameterized by their value of $M$ (flat Minkowski spacetime is the limiting case of $M = 0$).

I think the OP was questioning how a space-time that is globally source-free can still be non-trivial

This can happen because the Einstein Field equation is nonlinear.

the energy-momentum tensor via the Einstein equations does not in fact determine all aspects of local geometry, but only a certain combination of components of the Riemann tensor, being the Einstein tensor ( which, I believe, is the contracted double dual of Riemann ).

This is correct if we put in the word "locally"; the EFE equates the Einstein tensor and the stress-energy tensor at the same event. The rest of the Riemann tensor (the Weyl tensor) is determined by how spacetime curvature "propagates" (this isn't really the right word, but it's the best I can do) from one event to another.

All other information needed to uniquely fix the local geometry comes from boundary conditions imposed when solving the system of differential equations - and that's where the M comes from

The problem with looking at it this way, if we are looking at the idealized case where the spacetime is vacuum everywhere, is: at which boundary do we impose the condition that gives the value of $M$? As I noted above, since flat Minkowski spacetime is a member of this family of spacetimes (the one with $M = 0$), whatever boundary condition you are referring to can't distinguish between flat Minkowski spacetime and curved Schwarzschild spacetime.

There is a way of defining the "mass" of an asymptotically flat spacetime by looking at the limiting behavior of the metric coefficients as $r \rightarrow \infty$, called the ADM mass. (Actually, there are two ways of doing this, depending on whether you choose spacelike infinity, as the ADM mass does, or future null infinity, as the Bondi mass does.) Wikipedia has a very brief discussion:

https://en.wikipedia.org/wiki/Mass_...ndi_masses_in_asymptotically_flat_space-times

But this isn't really a "boundary condition".

 

[Markus Hanke]

But this isn't really a "boundary condition".

Ok, point taken, and I think I understand what you are attempting to say. These are the finer subtle points that I, as an amateur, are often missing. Thank you !