This is a question about solving Einstein's Field Equations for some simple cases, but I'm not sure how to explain this without analogy. So bear with me for a brief discussion.(adsbygoogle = window.adsbygoogle || []).push({});

Consider the electrostatic equation:

[tex] \nabla^2 V = \rho / \epsilon_0[/tex] which show explicitly how the charges source the scalar field.

Or we can solve for the vacuum solutions easier, which are of the form

[tex] \nabla^2 V = 0[/tex].

Now, if for some reason we knew how the fields should fall of towards infinity, one can use this "boundary" condition plus our vacuum solution to get a full solution without even dealing with the source term. But I wouldn't find this very satisfactory without knowing how to solve for the equations with the source term explicitly at least for some simple cases.

Turning now to GR, we can start with a simple form of the Hilbert action depending on the matter terms + the Ricci curvature scalar times some constant. Vary with respect to the metric and we get Einstein's field equations. To fix the constant, we can take the Newtonian limit and relate it to G.Now that we've fixed that constant, there are no "free" parameters left in the field equations.We shouldn't ever need to touch Newton's law again to solve these equations.

However, in all derivations I've seen of the metric outside a static stationary spherical mass, to fix some integration constants, theyagainuse the Newtonian limit. More-over, it is difficult to even define what is meant by mass in GR, and one of the more useful definitions (Komar mass) again refers to the vacuum region. So most of the simple solutions never refer to the actual source term in GR! They solve / define those integration constants some other way.

I'm really having trouble with this. It almost seems like the source term doesn't matter. I mean, if there was a theory of gravity which only gave vacuum solutions of the form [itex]R_{\mu\nu} = 0[/itex] just like GR (without a cosmological constant), could we even experimentally distinguish it from GR? I really feel the answer should be yes, but since the means of fixing the integration constants above never even refer to the source term ... it seems to point to no. So what is going on here? Since we can't even say (in general) what the mass content in a specific spacetime region is in GR, does this make GR's equivalent of the Poisson equation no more useful than its equivalent of the Laplace equation? This seems to me like if in EM we couldn't tell the difference between [itex] \nabla^2 V = \rho / \epsilon_0[/itex] or [itex] \nabla^2 V = a(\rho / \epsilon_0)^3[/itex].

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EDIT:

Alright, said more succinctly in one example. How do I know this isn't a solution to Einstein's Field Equations? (units such that c=1)

[tex]{d \tau}^{2} =

\left(1 - \frac{G M}{r} \right) dt^2 - \frac{dr^2}{\displaystyle{1-\frac{G M}{r}}} - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)[/tex]

It satisfies the vacuum equations. But it claims an event horizon doesn't form till r=GM instead of r=2GM.

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# Vacuum solutions, and Explicit source terms

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