Source: Thomas Moore's A General Relativity Workbook

Equation 1: $R= g^{\mu\nu}R_{\mu\nu} = R^\nu{}_\nu$

Equation 2: $R^{\mu\nu}=g^{\mu\beta}g^{\nu\sigma}R_\beta\sigma$

Question: Does $R$ also equal $g_{\mu\nu}R^{\mu\nu}$?

The Ricci Scalar $R$ is simply the trace of the Ricci Tensor $R^{\mu\nu}$ with respect to the metric $g_{\mu\nu}=g^{\mu\nu}$, so I don't see why the Curvature Scalar $R$ cannot be written as $g_{\mu\nu}R^{\mu\nu}$.

However, I am struggling to use Equation 2 to turn $R=g_{\mu\nu}R^{\mu\nu}=g_{\mu\nu}g^{\mu\beta}g^{\nu\sigma}R_\beta\sigma$ into Equation 1.

I was working on transforming $R^{\mu\nu}-(1/2)g^{\mu\nu}R+\Lambda g^{\mu\nu}=\kappa T^{\mu\nu}$ into $-R+\Lambda g^{\mu\nu}=\kappa T$, and I ran into this question.


Using the equality $g_{\mu\nu}g^{\nu\sigma}=\delta_{\mu}^\sigma$ and the symmetry of the metric tensor, $g^{\sigma\beta}=g^{\beta\sigma}$, we have that \begin{align} g_{\mu\nu}R^{\mu\nu}&=g_{\mu\nu}g^{\mu\beta}g^{\nu\sigma}R_{\beta\sigma} \\ &=g^{\mu\beta}g_{\mu\nu}g^{\nu\sigma}R_{\beta\sigma} \\ &=g^{\mu\beta}\delta_{\mu}^\sigma R_{\beta\sigma} \\ &=g^{\sigma\beta} R_{\beta\sigma} \\ &=g^{\beta\sigma} R_{\beta\sigma}\\ &=R \end{align} Thus, $$R=g_{\mu\nu}R^{\mu\nu}$$

  • $\begingroup$ Brilliant, absolutely brilliant. I completely forgot about the ability to switch the order of the metric multiplication you did in the first step! Thank you!:D $\endgroup$ – Lopey Tall Jul 7 '19 at 18:45

Yes, $g_{\mu\nu} R^{\mu\nu} = g^{\mu\nu} R_{\mu\nu}$. You should check that $g^{\sigma\rho} = g_{\mu\nu} g^{\mu\sigma} g^{\nu\rho}$.


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