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?

  • $\begingroup$ 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? $\endgroup$ Commented Oct 27, 2019 at 19:16
  • $\begingroup$ I follow all the definitions of Chandrasekhar for Christoffel symbols, etc. $\endgroup$ Commented 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. $\endgroup$ Commented 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. $\endgroup$ Commented 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. $\endgroup$ Commented Oct 27, 2019 at 22:28

1 Answer 1


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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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