The Teukolsky-Starobinsky identites relate solutions of the two Newman-Penrose scalars $\Psi_0$ and $\Psi_4$ around Kerr black holes. For example (here I am quoting Theorem 1 in Sec 81 of Chandrasekhar's book on black holes):

\begin{equation} \Delta^2D_0^+D_0^+D_0^+D_0^+\Delta^2R_{+2}\propto R_{-2}, \end{equation}

where $R_+$ and $R_-$ are related to $\Psi_0$ and $\Psi_4$, respectively, and the terms $\Delta$ and $D_0^+$ are derivative operators that depend on the Kerr geometry. For more context, the above relation is assuming we write, e.g.

\begin{equation} \Psi_0(t,r,\theta,\phi)=e^{i\omega t+im\phi}S(\theta)R_+(r). \end{equation}

My question is: are there versions of the Teukolsky-Starobinsky identities written out in terms of the Newman-Penrose scalars and derivative operators? What I would like is an expression of the form

\begin{equation} \Delta \delta etc \times \Psi_4 = \Psi_0, \end{equation}

instead of expressions in terms of the $R$ functions. Either an answer here or a link to references would both be helpful!


An operator in the form that you request, does not exist in general. This stems from the fact that the first equation you quote is not an equality, but a proportionality. In his book, Chandrasekhar goes on to determine this constant. This constant depends on $m$, $\omega$, and the eigenvalue of the radial Teukolsky equation for $R_{\pm 2}$.

Consequently, the operator that maps $\Psi_0$ into $\Psi_4$ cannot depend on the NP scalars and derivatives alone.

What is true, is that regardless of the constant of proportionality, the operator will map solution of the vacuum Teukolsky equation to solutions of the vacuum Teukolsky equation with opposite spin. If this is the property you are after, then we can write a general operator that does this. This is simply (see e.g. Lousto and Whiting, 2002m arxiv:gr-qc/0203061)

$$\Delta^2 \hat{\Delta}\hat{\Delta}\hat{\Delta}\hat{\Delta} \Delta^2 \bar{\phi}_{+2} = \phi_{-2},$$

where $\phi_{\pm2}$ are solutions of the spin $\pm2$ (vacuum) Teukolsky equations $\hat{\Delta}=n^\mu\partial_\mu$, and $\Delta= r(r-2M)+a^2$ (note that this is not a differential operator as incorrectly stated in the question), and the over bar denotes complex conjugation. It is however important to remember that while this operator will map vacuum solution of the spin +2 Teukolsky equation to a valid vacuum solution of the spin -2 Teukolsky equation, the spin -2 solution you get when applying it to $\Psi_0$ will not be the corresponding $\Psi_4$, but a completely different solution.


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