Deciphering and generalizing Demiański–Janis–Newman algorithm

Abstract

In the case of vanishing cosmological constant, Demiański has shown that the Janis–Newman algorithm can be generalized in order to include a NUT charge and another parameter c, in addition to the angular momentum. Moreover it was proved that only a NUT charge can be added for non-vanishing cosmological constant. However despite the fact that the form of the coordinate transformations was obtained, it was not explained how to perform the complexification on the metric function, and the procedure does not follow directly from the usual Janis–Newman rules. The goal of our paper is threefold: explain the hidden assumptions of Demiański’s analysis, generalize the computations to topological horizons (spherical and hyperbolic) and to charged solutions, and explain how to perform the complexification of the function. In particular we present a new solution which is an extension of the Demiański metric to hyperbolic horizons. These different results open the door to applications on (gauged) supergravity since they allow for a systematic application of the Demiański–Janis–Newman algorithm.

Appendices

Appendix 1: Original Demiański’s solution

In this appendix we recall the Demiański’s result, amended to fix some errors [29]. They follow from Sect. 4 with $\kappa = 1$ and $q = 0$. This gives also the opportunity to present prettier formulas.

The equations are

$$\begin{aligned} 2 F&= G^{{\prime }{\prime }}+ \cot \theta \; G^{\prime }, \end{aligned}$$

(84a)

$$\begin{aligned} n&= F + K, \end{aligned}$$

(84b)

$$\begin{aligned} 0&= \varLambda F^{\prime }\end{aligned}$$

(84c)

with

$$\begin{aligned} 2 K = F^{{\prime }{\prime }}+ \cot \theta \, F^{\prime }. \end{aligned}$$

(85)

The function $\tilde{f}$ is

$$\begin{aligned} \tilde{f} = 1 - \frac{2m r - q^2 + 2 F (F + K)}{r^2 + F^2} - \frac{\varLambda }{3}\, (r^2 + F^2) - \frac{4 \varLambda }{3}\, F^2 + \frac{8 \varLambda }{3}\, \frac{F^4}{r^2 + F^2}. \end{aligned}$$

(86)

The two solutions for F and G are

• $\varLambda \ne 0$

  $$\begin{aligned} F(\theta ) = n, \qquad G(\theta ) = - 2 n \ln \sin (\theta ). \end{aligned}$$

  (87)

• $\varLambda = 0$

  $$\begin{aligned} F&= n - a \cos \theta + c \left( 1 + \cos \theta \, \ln \tan \frac{\theta }{2} \right) , \end{aligned}$$

  (88a)
  $$\begin{aligned} G&= a \cos \theta - 2 n \ln \sin \theta - c \cos \theta \, \ln \tan \frac{\theta }{2}. \end{aligned}$$

  (88b)

Appendix 2: Relaxing assumptions

1.1 Appendix 2.1: Metric function F-dependence

In Sect. 4.2.1 we obtained the Eq. (46b)

$$\begin{aligned} \kappa \, F + K = \kappa \, n, \qquad 2 K = F^{{\prime }{\prime }}+ \frac{H^{\prime }}{H}\, F^{\prime }\end{aligned}$$

(89)

by asking that the function (46d)

$$\begin{aligned} \tilde{f} = \kappa - \frac{2m r - q^2 + 2 F (\kappa \, F + K)}{r^2 + F^2} - \frac{\varLambda }{3}\, (r^2 + F^2) - \frac{4 \varLambda }{3}\, F^2 + \frac{8 \varLambda }{3}\, \frac{F^4}{r^2 + F^2} \end{aligned}$$

(90)

depends on $\theta$ only through $F(\theta )$.

A more general assumption would be that $\kappa F + K$ is some function $\chi = \chi (F)$

$$\begin{aligned} \kappa \, F + K = \kappa \, \chi (F). \end{aligned}$$

(91)

First if $F^{\prime }= 0$ then $K = 0$ and the definition of K implies

$$\begin{aligned} \chi = F = n. \end{aligned}$$

(92)

The $(t\theta )$- and $(\theta \phi )$-components give the equation

$$\begin{aligned} 4 \varLambda \, F^2 F^{\prime }= F^{\prime }\, \partial _F \chi . \end{aligned}$$

(93)

If $\varLambda = 0$ we find that

$$\begin{aligned} \partial _F \chi = 0 \Longrightarrow \chi = n \end{aligned}$$

(94)

which reduces to the case studied in Sect. 4.2.1, while if $F^{\prime }= 0$ this equation does not provide anything.

On the other hand if $F^{\prime }\ne 0$ and $\varLambda \ne 0$ then the previous equation becomes

$$\begin{aligned} \partial _F \chi = 4 \varLambda F^2 \end{aligned}$$

(95)

which can be integrated to

$$\begin{aligned} \chi (F) = n + \frac{4}{3}\, \varLambda F^3 \end{aligned}$$

(96)

(notice that the limit $\varLambda \rightarrow 0$ is coherent). Plugging this function into Eq. (91) one obtains

$$\begin{aligned} \kappa \, F + K = \kappa \left( n + \frac{4}{3}\, \varLambda F^3 \right) \end{aligned}$$

(97)

(remember that $F^{\prime }\ne 0$). This differential equation is non-linear and we were not able to find an analytical solution. Despite that this provides a generalization of the algorithm with non-constant F in the presence of a cosmological constant this is not sufficient for obtaining Kerr-(a)dS: the form of $g_{\theta \theta }$ given in (26) is not required one.

Nonetheless by inserting the expression of $\chi$ in $\tilde{f}$ we see that the last term is killed

$$\begin{aligned} \tilde{f} = \kappa - \frac{2m r - q^2 + 2 \kappa \, n\, F}{r^2 + F^2} - \frac{\varLambda }{3}\, (r^2 + F^2) - \frac{4 \varLambda }{3}\, F^2. \end{aligned}$$

(98)

One can recognize the function given by Demiański [6]. Then this function is valid at the condition that Eq. (46b) is modified to (97), but in this case the solution is not the general (A)dS-Taub-NUT anymore.

1.2 Appendix 2.2: Gauge field integration constant

In Sect. 4.2.1 we obtained a second integration constant $\alpha$ in the expression of the gauge field

$$\begin{aligned} \tilde{f}_A = \frac{q\, r}{r^2 + F^2} + \alpha \, \frac{r^2 - F^2}{r^2 + F^2}. \end{aligned}$$

(99)

One of the Maxwell equation gave $\alpha = 0$ if $F^{\prime }\ne 0$, but otherwise no equation fixes its value. For this reason we focus on the case $F^{\prime }= 0$ or equivalently $\varLambda = 0$ through equation (46c).

In this case the function $\tilde{f}$ is modified to

$$\begin{aligned} \tilde{f} = \kappa - \frac{2m r - q^2 + 2 F (\kappa \, F + K) + 4 \alpha ^2 F^2}{r^2 + F^2} - \frac{\varLambda }{3}\, (r^2 + F^2) - \frac{4 \varLambda }{3}\, F^2 + \frac{8 \varLambda }{3}\, \frac{F^4}{r^2 + F^2}. \end{aligned}$$

(100)

Equation (46c) is modified but it is still solved by $F^{\prime }= 0$ and all other equations are left unchanged (in particular $\kappa F + K$ is still given by the function $\chi (F)$ (96)). For $\chi (F) = n$ the configuration with $\alpha \ne 0$ provides another solution when $\varLambda \ne 0$ but it is not clear how to get it from a complexification of the function.