The Kerr spacetime is of Petrov type D (see here for the Petrov classification of spacetimes). In the Newman-Penrose formalism, from the Goldberg-Sachs theorem we can conclude that there is a choice of null tetrad such that the following Newman-Penrose scalars are zero

\begin{equation} \kappa=\sigma=\nu=\lambda=0, \end{equation}

in addition to the Weyl scalars $\Psi_0=\Psi_1=\Psi_3=\Psi_4=0$ from the spacetime being type D.

For the Kerr spacetime, one can choose the Kinnersly tetrad such that the Newman-Penrose scalar $\epsilon=0$ in addition to the above quantities. Does this indicate that the Kerr spacetime is in some sense "more symmetric" than a "typical" type D spacetime? Or in general can one set one more Newman-Penrose scalar to zero in a type D spacetime beyond the scalars that can be zero due to the Goldberg-Sachs theorem?

EDIT: in addition to the answer given below, reading through Kinnersley's original paper I realize that in fact he sets $\epsilon=0$ for a general type D metric.

  • $\begingroup$ Well, essentially all Petrov D space-times are in the Plebanski-Demianski class, and Kerr is the only asymptotically flat and vacuum one... $\endgroup$
    – Void
    Commented Nov 15, 2019 at 0:10
  • $\begingroup$ @Void why would being asymptotically (A)dS make it any less symmetric? $\endgroup$
    – TimRias
    Commented Nov 15, 2019 at 8:10
  • $\begingroup$ @mmeent It is more the "vacuum" I am talking about, of course once you allow for electro+$\Lambda$-vacuum (with appropriate regular asymptotics) then the class gets enlarged to Kerr-Newman-(A)dS. The asymptotic flatness (regular asymptotics) refers more to eliminating NUT and acceleration. $\endgroup$
    – Void
    Commented Nov 15, 2019 at 12:41

1 Answer 1


The Kerr spacetime is more symmetric then other Petrov type D spacetimes, but not for the reason stated.

Choosing a null tetrad that satisfies $\Psi_0=\Psi_1=\Psi_3=\Psi_4=0$ in a Petrov type D spacetime requires only 4 of the 6 degrees of freedom in choosing a null tetrad. One remaining degree of freedom is the ability to use a boost to rotate the two principle null vectors ($l^\mu$ and $n^\mu$ in usual notation) into each other. The other is the freedom to rotate the other two legs of the tetrad ($m^\mu$ and $\bar{m}^\mu$) leaving $l^\mu$ and $n^\mu$ invariant. These freedoms can always be used to set $\epsilon=0$, for any Petrov type D spacetime. So this particular property does not set Kerr apart.

Kerr does however have an additional symmetry not shared by all Petrov type D spacetimes. Kerr possesses a (non-trivial) Killing-Yano tensor. This hidden symmetry is responsible for many special properties of the Kerr solution such as intergrability of the geodesic equations and separability of various wave equations. Kerr is not the only Petrov type D spacetime that has such a hidden symmetry, but there are Petrov type D spacetimes that don't (e.g. the C-metric).

  • $\begingroup$ Well, the whole Kerr-NUT-(A)dS class (the Carter class) has a Killing-Yano tensor, it is only the acceleration that ruins it, and even there you have a conformal Killing-Yano tensor. $\endgroup$
    – Void
    Commented Nov 15, 2019 at 12:37

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