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I am currently working through Chapter 1 on Chandrasekhar's book Mathematical Theory of Black holes (where he lays out the Einstein equations in terms of the Newman-Penrose formalism). I am presently trying to relate components of the Riemann tensor to the Newman Penrose scalars using

\begin{equation} R_{abcd}=W_{abcd}+\frac{1}{2}\left(\eta_{ac}R_{bd}-\eta_{bc}R_{ad}-\eta_{ad}R_{bc}+\eta_{bd}R_{ac}\right)-\frac{1}{6}\left(\eta_{ac}\eta_{bd}-\eta_{ad}\eta_{bc}\right)R , \end{equation} where $R_{abcd}$ is the Riemann tensor, $W_{abcd}$ the Weyl tensor, $R_{ab}$ the Ricci tensor, and $R$ the Ricci scalar.

I use $\eta_{ab}$ above as the above equation is written in terms of the Newman-Penrose formalism, so that $\eta_{ab}$ is in fact the "tetrad metric" and equals \begin{equation} \eta_{ab} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \end{equation} (but the Riemann tensor, etc are not zero). The indices $a$, etc run over $1,2,3,4$. In his book (Eq. (293) in Chapter 1), Chandrasekhar (using the above two formulas) derives formulas such as \begin{equation} R_{1314}=\frac{1}{2}R_{11} . \end{equation} (using arguments I won't get into here, he shows $C_{1314}=0$). The problem I have is: I get the opposite sign for all the Ricci tensor/scalar terms in my equations. For example I instead get \begin{equation} R_{1314}=-\frac{1}{2}R_{11} , \end{equation} as $\eta_{34}=-1$. Does anyone know what I could be doing wrong (and then give a correct derivation of the above equation), or if there is a series of typos in Chandrasekhar's book about the signs of the Ricci tensor/scalar terms?

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  • $\begingroup$ What definition are you using for the Riemann tensor in terms of the Christoffel symbols and its derivatives? Is it the same as one that Chandrasekhar uses? $\endgroup$ Oct 27, 2019 at 19:16
  • $\begingroup$ I follow all the definitions of Chandrasekhar for Christoffel symbols, etc. $\endgroup$ Oct 27, 2019 at 19:32
  • $\begingroup$ Yes, on second thoughts I don't think the definition of the Riemann tensor is the issue. I consulted Chandrasekhar's book but I would need to look into it more to give a proper answer. $\endgroup$ Oct 27, 2019 at 20:04
  • $\begingroup$ I checked the calculations for $R_{2324}$ and $R_{3132}$ and both my results obtained by substituting the results of equation 291 and the components of $\eta_{ab}$ on page 41 using equation 282 into equation 287 differ from the results in equation 293 by a multiplicative factor of -1. I have not attempted to check the other results in equation 293. $\endgroup$ Oct 27, 2019 at 22:20
  • $\begingroup$ I just checked $R_{1212}$ and I got $R_{1212} = C_{1212} - R_{12} + R/6$ instead of the result $R_{1212} = C_{1212} + R_{12} - R/6$ that is stated in Chandrasekhar's book. $\endgroup$ Oct 27, 2019 at 22:28

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After some digging around, I think the problem may lie in the definition of the Ricci tensor. In the original Newman-Penrose article, they cite Eisenhart's book on Riemannian geometry. There he defines

\begin{equation} R_{ab}\equiv R^c{}_{abc}. \end{equation}

So that the Weyl tensor is defined by \begin{equation} R_{abcd}=W_{abcd}-\frac{1}{2}\left(\eta_{ac}R_{bd}-\eta_{bc}R_{ad}-\eta_{ad}R_{bc}+\eta_{bd}R_{ac}\right)+\frac{1}{6}\left(\eta_{ac}\eta_{bd}-\eta_{ad}\eta_{bc}\right)R \end{equation} which has the opposite sign for the Ricci tensor/scalar terms than what is presented in Chandrasekhar's book. Now while in my edition of Chandrasekhar's book he uses $R_{ab}=R^c{}_{acb}$, in an earlier edition (1983) that I dug out of the library it appears he may have originally used the other definition. Somehow maybe not everything was updated with that change.

As a bit of an aside, note then that if we use the definition

\begin{equation} R_{ab}\equiv R^c{}_{acb}, \end{equation}

then to match the Eqs (4.2) in the Newman-Penrose article (and in Chandrasekhar's book) we need to define the Ricci "scalars" with the opposite sign than they do (note I am still working in $+---$ signature as they do); e.g.

\begin{equation} \Phi_{00}=+\frac{1}{2}R_{11} \end{equation}

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