Ricci tensor equals a constant times the metric

  • Thread starter Pacopag
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Main Question or Discussion Point

Hello;
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?
 

Answers and Replies

197
4
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
 
kdv
336
1
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
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})
[/tex]

By contracting, you can show that

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

[tex]
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]
 
197
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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