Ricci tensor of schwarzschild metric

  • Thread starter PhyAmateur
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  • #1
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In schwarzschild metric:

$$ds^2 = e^{v}dt^2 - e^{u}dr^2 - r^2(d\theta^2 +sin^2\theta d\phi^2)$$
where v and u are functions of r only
when we calculate the Ricci tensor $R_{\mu\nu}$ the non vanishing ones will only be $$R_{tt}$$,$$R_{rr}$$, $$R_{\theta\theta}$$,$$R_{\phi\phi}$$
But when u and v now depend on r and t, we get an extra term of Ricci tensor which is the
$$R_{tr}$$ I thought that if our matrix is diagonal we should not get a non diagonal Ricci tensor and all the Ricci tensors must be diagonal. Am I mistaken?
 

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  • #2
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You are mistaken that all Ricci tensors must be diagonal. The Ricci tensor is symmetric, but it does not have to be diagonal.
 
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  • #3
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Aha, is there a way to find out which are the nonvanishing? Or I will have to try it for every single pair?
 
  • #4
ChrisVer
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No, a diagonal metric tensor doesn't have to lead to diagonal ricci tensor. That is because it may be true that [itex]g_{\mu \nu}[/itex] is diagonal, in practice only [itex]g_{\mu \mu}[/itex] is non-vanishing, however the Ricci Tensor also has indices that come from derivatives of [itex]g_{\mu \nu}[/itex], and since the last can depend on any [itex]x^\rho[/itex] then its derivatives don't have to vanish in general. In other word you can't write [itex] R_{\mu \nu} = (A) g_{\mu \nu} [/itex] with [itex]A[/itex] an index free operator with derivatives etc...

As for whether you can determine the non-vanishing, I think you have to look at the Ricci tensor's form before determining it. In most books I've seen they always start and derive everything up to Ricci scalar...
 
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  • #5
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because it's a vacuum space-time

I'm not sure the OP was only talking about the vacuum Schwarzschild metric; based on the line element he wrote down, I think he intended the more general usage of "Schwarzschild-type metric", meaning a metric for a spherically symmetric spacetime, not necessarily vacuum, that uses Schwarzschild coordinates (i.e., spherical coordinates with ##r## defined as circumference divided by ##2 \pi## ). This usage is not terribly common, AFAICT, but it is found, for example, at some points in MTW.
 
  • #6
pervect
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Ooops, that's the interior Schwarzschild metric...
 
  • #7
pervect
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I'm not sure the OP was only talking about the vacuum Schwarzschild metric;
Yeah, my bad
 

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