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I have the proposed solution stated as:

$R_{\mu \nu} -\frac{1}{2} g_{\mu\nu}R=\kappa T_{\mu\nu}$ (4.43)

Caroll says:"note that contracting both sides of (4.43) yields (in four dimensions)"

$R = - \kappa T$, which I should get to.

But I get it differently. Those are my steps:

$g^{\mu\nu}(R_{\mu \nu} -\frac{1}{2} g_{\mu\nu}R)=\kappa g^{\mu\nu}T_{\mu\nu}$

$g^{\mu\nu}R_{\mu\nu} - \frac{1}{2}g^{\mu\nu}g_{\mu\nu}R=\kappa g^{\mu\nu}T_{\mu\nu} $

Making the tensors contractions:

$R - \frac{1}{2}R = \kappa T$

$R = 2 \kappa T$

Please, what I'm doing wrong?

Thanks in advance!

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    $\begingroup$ $g^{\mu\nu} g_{\mu\nu} = 4$ $\endgroup$
    – Prahar
    Commented Apr 9, 2015 at 0:38
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    $\begingroup$ To answer your question (which I think you deleted) - $g^{\mu\nu}$ is the inverse of the matrix $g$. In matrix notation, then $g^{\mu\nu} g_{\mu\nu} = \text{tr} \left( g^{-1} g \right) = \text{tr} \left( {\bf 1} \right) = 4$. $\endgroup$
    – Prahar
    Commented Apr 9, 2015 at 0:45
  • $\begingroup$ But the trace of the metric tensor would be 4, only if it's Minkowsky metric, isn't it? In here it's a general metric... $\endgroup$
    – Edison Cesar
    Commented Apr 9, 2015 at 0:46
  • $\begingroup$ For any general metric, $g^{\mu\nu}$ is defined to be the inverse matrix of $g_{\mu\nu}$. Then $g^{\mu\nu} g_{\mu\nu}$ is always computing $\text{tr}(g^{-1}g)$ which is the dimension of the space-time, namely 4. $\endgroup$
    – Prahar
    Commented Apr 9, 2015 at 0:48
  • $\begingroup$ Ok, got it. Thanks a lot for the help and patience! $\endgroup$
    – Edison Cesar
    Commented Apr 9, 2015 at 0:51

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Note that Carroll says "in four dimensions". Recall that raising an index on the metric tensor gives the Kronecker delta: $g^{\rho\mu}g_{\mu\sigma}=\delta^\rho{}_\sigma$. The delta has $n$ entries of one on the diagonal in $n$ dimensions. So in spacetime we have a tensor with 4 ones along the diagonal. Thus, the trace, $g^{\rho\mu}g_{\mu\rho}=\delta^\rho{}_\rho$ is equal to 4. This gives $$g^{\mu\nu}R_{\mu\nu}-\frac{1}{2}g^{\mu\nu}g_{\mu\nu}R=R-\frac{1}{2}\cdot 4R=-R$$ whence $R=-\kappa T$, as was to be shown.

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  • $\begingroup$ But the trace of the metric tensor would be 4, only if it's Minkowsky metric, isn't it? In here it's a general metric... $\endgroup$
    – Edison Cesar
    Commented Apr 9, 2015 at 0:45
  • $\begingroup$ @EdisonCesar See Eq. (2.39) in Carroll. When we say "trace", we mean "raise one index so you have a (1,1) tensor and then take the trace". $\endgroup$
    – Ryan Unger
    Commented Apr 9, 2015 at 0:46
  • $\begingroup$ Ok, got it. Thanks a lot for the help and patience! $\endgroup$
    – Edison Cesar
    Commented Apr 9, 2015 at 0:49

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