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Space with a diagonal metric

  1. Oct 19, 2012 #1
    Hey guys! I am considering a space with a diagonal metric, which is maximally symmetric.

    It can be proven that in that case of a diagonal metric the following equations for the Christoffel symbols hold:
    [tex] \Gamma^{\gamma}_{\alpha \beta} = 0 [/tex]
    [tex] \Gamma^{\beta}_{\alpha \alpha} = -(1/g_{\beta\beta})\partial_{\beta}g_{\alpha\alpha} [/tex]
    [tex] \Gamma^{\beta}_{\alpha \beta} = \partial_{\alpha}\ln(\sqrt{|g_{\beta\beta}|}) [/tex]
    [tex] \Gamma^{\alpha}_{\alpha \alpha} = \partial_{\alpha}\ln(\sqrt{|g_{\alpha\alpha}|}) [/tex]

    Furthermore: for a maximally symmetric space we have for the Riemann tensor:
    [tex] R_{\rho\sigma\mu\nu} = R/12(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu})[/tex], where R is the Ricci scalar.

    Given these equations, I come accross a contradiction. From the above equation for the Riemann tensor we easily see that if it has three different indices, it must be zero (IF the metric is diagonal). However, if I plug in the Christoffelsymbols into the definition of the Riemanntensor expressed in Christoffel symbols and their derivatives, I do not find that this is zero in general. Does anyone know what is going wrong here?

    EDIT: For a maximally symmetric space, three indices cannot be unequal, but for a diagonal space this might not be the case. How is this all related?
    Last edited: Oct 19, 2012
  2. jcsd
  3. Oct 19, 2012 #2


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    If I put the following diagonal line element into GrTensor, for coordinates (t,x,y,z),


    I get the following (just a partial set).

    [tex]\Gamma^t{}_{tt} = \frac{1}{2} \frac{\partial_t \, p}{p} [/tex]
    [tex]\Gamma^x{}_{tt} = -\frac{1}{2} \frac{\partial_x \, p}{q} [/tex]
    [tex]\Gamma^t{}_{xt} = \frac{1}{2} \frac{\partial_x \, p}{p} [/tex]

    They look similar, except for a missing factor of 1/2. The ln's complicate things a bit - are they really that useful?

    I was rather surprised to see an apparently non_zero term pop out for R_txty myself. I'm not sure if it's really nonzero, or just didn't simplify:

    [tex]R_{txty} = 1/4\,{\frac {-2\, \left( {\frac {\partial ^{2}}{\partial x\partial y}}
    p \left( t,x,y,z \right) \right) p \left( t,x,y,z \right) q \left( t,
    x,y,z \right) r \left( t,x,y,z \right) + \left( {\frac {\partial }{
    \partial y}}p \left( t,x,y,z \right) \right) \left( {\frac {
    \partial }{\partial x}}p \left( t,x,y,z \right) \right) q \left( t,x,
    y,z \right) r \left( t,x,y,z \right) + \left( {\frac {\partial }{
    \partial x}}p \left( t,x,y,z \right) \right) \left( {\frac {
    \partial }{\partial y}}q \left( t,x,y,z \right) \right) p \left( t,x,
    y,z \right) r \left( t,x,y,z \right) + \left( {\frac {\partial }{
    \partial y}}p \left( t,x,y,z \right) \right) \left( {\frac {
    \partial }{\partial x}}r \left( t,x,y,z \right) \right) p \left( t,x,
    y,z \right) q \left( t,x,y,z \right) }{p \left( t,x,y,z \right) q
    \left( t,x,y,z \right) r \left( t,x,y,z \right) }}
  4. Oct 19, 2012 #3


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    When you say "maximally symmetric space" and "diagonal space," you're talking about two different kinds of things. The symmetry is intrinsic, but the diagonal form of the metric is coordinate-dependent.

    I think the only maximally symmetric spaces in 3+1 dimensions are Minkowski space, de Sitter space, and anti de Sitter space. Have you tried, for example, checking your calculations in the special case of de Sitter space?

    What if the different components of the metric are unequal?

    Are you thinking that the different components of the metric have to be equal because it's maximally symmetric? That's not true. The symmetry doesn't have to be manifest when you write the metric in certain coordinates.
  5. Oct 19, 2012 #4


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    Given the line element, I don't see any requirement that there be ANY Killing vectors, much less the maximum possible number (which would be a requirement to be maximally symmetric space, if I understand correctly).
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