# Schwarzschild metric

Why, if the Schwarzschild metric is a vacuum solution ($T_{\mu\nu}=0$) , do textbooks state that $T=\rho c^{2}$ when approximating Poisson's Equation from the Einstein Field Equations?

Thank you.

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Because the approximation for Poisson's equation is weak field. In order to get a self-sourcing gravitational field, like the Schwartschild solution, you need a region with a strong field. You can always imagine that the source for the Schwarschild solution is right on the event horizon of the black hole it describes.

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@ Ron Maimon. Thanks, but what's the difference between a weak field and a self-sourcing field. It just seems odd that you can describe a weak field using two methods that are based on opposite assumptions. Please don't worry about making your explanation too simple. –  Peter4075 Sep 30 '11 at 13:50

There is a coordinate slicing known as the Kerr-Schild coordinate system where one can look at the Hamiltonian constraint $16\pi \rho = {}^{3}R - K^{ab}K_{ab} + K^{2}$, and find that the left hand side has the same singularity that you would find in $\nabla \cdot E = \rho$ when you put in the $E$ for a point charge.

So, you can interpret the Schwarzschild solution as having a delta function matter distribution. (though there are problems with this interpretation, too, and it's not something one should take too seriously)

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This is full of formulas and nonsense. There is no sense in which the part near the central singularity can be interpreted as a delta-function point source for matter, not in any formalism. –  Ron Maimon Sep 29 '11 at 3:59
@Ron: Do you want me to walk you through the integral? There is a $\nabla^{2}\left(\frac{M}{r}\right)$ term in ${}^{3}R$ that shows up clear as day. –  Jerry Schirmer Sep 29 '11 at 10:57
Yes, I want you to walk me through it, because you are wrong. Spacelike surfaces which define the three-manifold in the Schwarschild metric don't include the singularity, and $^3R$ is not divergent there. If you slice the Schwarschild metric at constant "t", t is spacelike on the interior, and there is no way for the putative source at r=0 to propagate the gravity out. If you deform the Schwarschild with a tiny charge, to resolve the singularity, the divergence at the center changes. THere is no sense in which your calculation is right, not even approximately. –  Ron Maimon Sep 29 '11 at 14:21
@Ron: First, I did make a mistake in my first statement. The slicing is done using Kerr coordinates, not Schwarzschild coordinates, so the spacelike/timelike issue is resolved that way, as Kerr coordiantes are timelike all the way to $r=0$. If you do this, you will find that ${}^{3}\gamma_{ab}={\rm diag}(1+\frac{2M}{r},r^{2},r^{2}\sin^{2}\theta)$. And, in this system ${}^{3}\gamma_{ab}$ totally does contain a $r=0$ divergence that formally looks identical to the $r=0$ divergence from a point charge. I don't have the time to TeX up the full thing right now, but that's the outline. –  Jerry Schirmer Sep 29 '11 at 18:31
I want to get rid of my downvote, because this answer is actually pretty reasonable, given that you choose this slicing, but you should say that in the answer. I wonder, what is the best way to understand this in a coordinate free way? The charged black hole "Kerr coordinates" would be instructive too. –  Ron Maimon Sep 29 '11 at 21:34