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Ricci tensor equals a constant times the metric

  1. Mar 16, 2008 #1
    What does it mean physically if I have?
    [tex]R_{a b} = Ag_{a b}[/tex]

    I think it means that my manifold is an n-sphere (i.e. if A is positive),
    or it is AdSn (i.e. if A is negative). Is this correct?
  2. jcsd
  3. Mar 17, 2008 #2
    I think I got it. What I wrote above is correct for some particular values of A;
    [tex]n(n-1) \over r^2[/tex] where r is the radius of the n-sphere.
    Put a minus sign in front of this for AdSn
  4. Mar 17, 2008 #3


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    Are you sure?

    I thought that the equations were the following.

    For maximally symmetric spaces, we have (n= number os spacetime dimensions)

    [tex]R_{\rho \sigma \mu \nu} = C (g_{\rho \mu} g_{\sigma \nu} - g_{\rho \nu} g_{\sigma \mu})

    By contracting, you can show that

    [tex] C = \frac{R}{n(n-1)} [/tex]
    whre R is the Ricci scalar.

    R_{\rho \sigma \mu \nu} = \frac{R}{n(n-1)} (g_{\rho \mu} g_{\sigma \nu} - g_{\rho \nu} g_{\sigma \mu}) [/tex]

    Thsi agrees with eq 3.191 of Carroll

    Now, this means

    [tex] R_{\sigma \nu} = \frac{R}{n} g_{\sigma \nu} [/tex]
  5. Mar 18, 2008 #4
    To be honest, I don't really know what I'm talking about. But everything you wrote seems correct. But now when I find R for an n-sphere of radius r I find that
    [tex]R={n(n-1)\over r^2}[/tex]. So we get
    [tex]R_{a b}={n-1\over r^2}g_{a b}[/tex]
    Oh wait! That's not what I had above. I had an extra factor of n on the right-hand side.

    Thank you kdv
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