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Ricci tensor of schwarzschild metric

  1. Dec 16, 2014 #1
    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?
     
  2. jcsd
  3. Dec 16, 2014 #2

    PeterDonis

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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.
     
  4. Dec 16, 2014 #3
    Aha, is there a way to find out which are the nonvanishing? Or I will have to try it for every single pair?
     
  5. Dec 16, 2014 #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...
     
  6. Dec 16, 2014 #5

    PeterDonis

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    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.
     
  7. Dec 16, 2014 #6

    pervect

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    Ooops, that's the interior Schwarzschild metric...
     
  8. Dec 16, 2014 #7

    pervect

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    Yeah, my bad
     
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