Is the Ricci tensor zero in black holes because of the Kasner solution (http://en.wikipedia.org/wiki/Kasner_metric)? I'm asking this to help resolve an uncertainty of mine as to which of two cosmological models (Aguirre & Gratton's "steady state eternal inflation" and Poplawski's "cosmology with torsion") is more plausible.
Question:
Is the Ricci tensor zero in black holes because of the Kasner solution?
Wyatt: 3 days ago
Answer:
Joshua: 3 days ago
No, $R_{\mu \nu} = 0$ is just Einstein's equations in vacuum.
Can be seen as follows:
$$ R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = 8 \pi G T_{\mu \nu} = 0 $$
multiply (and contract) both parts by $g^{\mu \nu}$:
$$ R - \frac{1}{2} R \cdot 4 = R - 2 R = - R = 0 $$ $$ R_{\mu \nu} - \frac{1}{2} \cdot 0 \cdot g_{\mu \nu} = R_{\mu \nu} = 0; $$ $$R_{\mu \nu} = 0.$$
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