0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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, $$R=g_{\mu\nu}R^{\mu\nu}$$

$\endgroup$
  • $\begingroup$ 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 at 18:45
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.