Rotating Cosmological Models, or How Matter and Light Can Escape from a Rotating Black Hole

Abstract

We show how to construct models for rotating black holes with comoving matter inside, and how they canc be interpreted as models for rotating universes. The metric along the polar axis of such a black hole is fundamentally different from the metric elsewhere inside the black hole. We show that this creates the possibility that matter or light can escape through the poles. As a corollary, we show how the Lemaitre-Tolman metric can be obtained from the Schwarzschild metric.

1. Introduction

The motive for this paper is the statement in Dejonghe [1] that matter and light may be able to escape a rotating black hole through the poles. The rationale behind that statement was that in a nonrotating nonempty black hole (i.e., a black hole containing a distributed mass density, not only containing a central point mass), the accompanying metric allows this phenomenon under certain conditions. These cannot realistically be met globally in a spherical symmetry, however, which was the framework [1] was written in. However, there is some similarity between the Schwarzschild metric and the Kerr metric, the latter taken along the polar axis. Since the polar axis is one-dimensional, the conditions mentioned below could be met in a Kerr metric along the polar axis. This statement was but a conjecture in Ref. [1]. The trigger of this paper was to prove it.

Sections 2 and 3 set the stage and do not contain new material. The core of the paper is to be found in sections 4 and 5. Section 6 recapitulates [1], but here from a different perspective. Section 7 contains the framework for the construction of models for a nonempty rotating black hole, that can also be seen as cosmological models for rotating universes in the presence of a cosmological constant. Section 8 reconnects with Ref. [1] and proves that matter and light can escape from a rotating nonempty black hole through the poles.

2. The Kerr–de Sitter metric

The Kerr–de Sitter solution (hereafter KdS) of Einstein’s equations represents a rotating black hole with mass $\mathcal{M}$ in a world with a cosmological constant $\mathrm{\Lambda }$. It is the generalization of the Kerr geometry, which has two parameters: the geometrical mass of the black hole $M=G\mathcal{M}/c^2$ and a rotation parameter $a=L_z/\left(\mathcal{M}c\right)$, with $L_z$ an angular momentum. Both parameters have the dimension of a length. We follow here largely Hackmann et al. [2] and denote as usual

$${\rho }^2\left(r,\vartheta \right)=r^2+a^2{cos}^2\vartheta =r^2+a^2-a^2{sin}^2\vartheta ,$$E1

$${\mathrm{\Delta }}_r\left(r\right)=r^2-2\mathit{Mr}+a^2-\frac{\mathrm{\Lambda }}{3}{r}^2\left(r^2+a^2\right)\text{and}{\mathrm{\Delta }}_{\vartheta }\left(\vartheta \right)=1+\frac{\mathrm{\Lambda }}{3}{a}^2{cos}^2\vartheta .$$E2

The metric reads

$${\mathit{ds}}^2=g_{00}{\mathit{dt}}^2+g_{11}{\mathit{dr}}^2+g_{22}d{\vartheta }^2+g_{33}d{\phi }^2+2g_{03}\mathit{\text{dtdφ}},$$E3

with

$$g_{00}=\frac{1}{{\rho }^2}\left({\mathrm{\Delta }}_r-a^2{sin}^2\vartheta {\mathrm{\Delta }}_{\vartheta }\right),g_{11}=-\frac{{\rho }^2}{{\mathrm{\Delta }}_r},g_{22}=-\frac{{\rho }^2}{{\mathrm{\Delta }}_{\vartheta }},$$E4

$$g_{33}=\frac{{sin}^2\vartheta }{{\rho }^2}\left[{\mathrm{\Delta }}_r{a}^2{sin}^2\vartheta -{\mathrm{\Delta }}_{\vartheta }{\left(r^2+a^2\right)}^2\right]\text{and}{g}_{03}=\frac{a{sin}^2\vartheta }{{\rho }^2}\left[{\mathrm{\Delta }}_{\vartheta }\left(r^2+a^2\right)-{\mathrm{\Delta }}_r\right].$$E5

The coordinate ‘time’ $t$ has the dimension of a length. We note that the parameters $M$ and $\mathrm{\Lambda }$ are tucked away in the mutually independent functions ${\mathrm{\Delta }}_r\left(r\right)$ and ${\mathrm{\Delta }}_{\vartheta }\left(\vartheta \right)$.

In the sequel, we will also use the following shorthand notation, inspired on (1):

$${{\rho }^{\prime }}^2\left(r,\vartheta \right)=r^2+a^2+a^2{sin}^2\vartheta .$$E6

3. The orbits

3.1 The general case $a\ne 0$

It is essential for this paper that the Hamilton–Jacobi equation for geodesic motion, which reads (see also Chandrasekhar [3])

$$2{\partial }_{\tau }S=g^{\mathit{\mu \nu }}{\partial }_{x^{\mu }}S{\partial }_{x^{\nu }}S$$E7

with ${\partial }_{x^{\mu }}$ the partial derivative with respect to variable $x^{\mu }$, $\tau$ the proper time and $S$ Hamilton’s principal function, is separable in de KdS spacetime. Its solution is

$$S=\frac{c^2}{2}\mathit{\delta \tau }-\mathit{cEt}+{\mathit{cL}}_z\phi +{\epsilon }_{r}c\int \frac{\sqrt{R\left(r^{\prime }\right)}}{{\mathrm{\Delta }}_r\left(r^{\prime }\right)}d{r}^{\prime }+{\epsilon }_{\vartheta }c\int \frac{\sqrt{\mathrm{\Theta }\left({\vartheta }^{\prime }\right)}}{{\mathrm{\Delta }}_{\vartheta }\left({\vartheta }^{\prime }\right)}d{\vartheta }^{\prime }.$$E8

In this equation $\delta =0$ for light and $\delta =1$ for material particles, in which case $E$ is the energy normalized to the rest energy $m_0{c}^2$ (and thus dimensionless) and $L_z$ is the angular momentum normalized with $m_{0}c$ (and thus with the dimension of a length), while, also here, $t$ has the dimension of a length. The integrals stand for primitives of the integrands. The sign constants ${\epsilon }_r$ and ${\epsilon }_{\vartheta }$ equal $\pm 1$, independent of each other. Their signs equal the signs of the derivatives of $r$ and $\vartheta$ (as functions of $\tau$). The functions $R\left(r\right)$ and $\mathrm{\Theta }\left(\vartheta \right)$ are given by

$$R\left(r\right)={\left[E\left(r^2+a^2\right)-{\mathit{aL}}_z\right]}^2-\left(\delta r^2+a^{2}K\right){\mathrm{\Delta }}_r\left(r\right)$$E9

and

$$\mathrm{\Theta }\left(\vartheta \right)=-\frac{1}{{sin}^2\left(\vartheta \right)}{\left[\mathit{aE}{sin}^2\vartheta -L_z\right]}^2+a^2\left(K-\delta {cos}^2\vartheta \right){\mathrm{\Delta }}_{\vartheta }\left(\vartheta \right).$$E10

In both these equations appears, apart from $E$ and $L_z$, the integration constant $a^{2}K$ which follows from the separability. Note that our definition of $K$ differs from the usual one by a factor $a^2$, but was chosen here in order that $K$ has the same dimension as $E$, and thus dimensionless in our units¹ ($a^2{\mathit{Km}}_0{c}^2$ has the dimension of a square of an action).

In this paper we will only be dealing with orbits that have $L_z=0$. Since $\phi$ does not enter the equations, what we call here an orbit is in fact a filled ring centered on the polar axis. We denote an orbit in the ring with $\left(\overline{t}\left(\tau \right),\overline{r}\left(\tau \right),\overline{\vartheta }\left(\tau \right),\overline{\phi }\left(\tau \right)\right)$, here with omitted reference to the integration constants $E$ and $K$.

Our formulae will be explicit for infalling orbits (hence ${\epsilon }_r=-1$). For outgoing orbits (for as far as this makes sense) $\sqrt{R}$ has to be changed to $-\sqrt{R}$ (hence ${\epsilon }_r=+1$). The quantity $\sqrt{\mathrm{\Theta }}$ can have both signs, depending, for infalling orbits, on whether the orbit is ‘above’ (${\epsilon }_{\vartheta }=+1$, $0\le \vartheta \le \pi /2$) or ‘below’ (${\epsilon }_{\vartheta }=-1$, $\pi /2\le \vartheta \le \pi$) the equatorial plane (and vice versa for outgoing orbits). Here we choose ${\epsilon }_{\vartheta }=+1$.

We will need the equations for these orbits in differential form as well as in quadrature form. The differential equations are (for completeness the terms in $L_z\ne 0$ are also shown between brackets):

$$\begin{array}{l}{\rho }^2\dot{r}={\rho }^2{r}_{\tau }=-c\sqrt{R}\\ {\rho }^2\dot{\vartheta }={\rho }^2{\vartheta }_{\tau }=c\sqrt{\mathrm{\Theta }}\\ {\rho }^2\dot{\phi }={\rho }^2{\phi }_{\tau }=\mathit{cEa}\left[\frac{r^2+a^2}{{\mathrm{\Delta }}_r}-\frac{1}{{\mathrm{\Delta }}_{\vartheta }}\right]\left(+\frac{L_z}{{sin}^2\vartheta }\left[\frac{1}{{\mathrm{\Delta }}_{\vartheta }}-\frac{a^2{sin}^2\vartheta }{{\mathrm{\Delta }}_r}\right]\right)\\ {\rho }^2\dot{t}={\rho }^2{t}_{\tau }=\mathit{cE}\left[\frac{{\left(r^2+a^2\right)}^2}{{\mathrm{\Delta }}_r}-\frac{a^2{sin}^2\vartheta }{{\mathrm{\Delta }}_{\vartheta }}\right]\left(+{\mathit{aL}}_z\left[\frac{1}{{\mathrm{\Delta }}_{\vartheta }}-\frac{r^2+a^2}{{\mathrm{\Delta }}_r}\right]\right),\end{array}$$E11

with the dot differentiation with respect to $\tau$. The functions $R\left(r\right)$ and $\mathrm{\Theta }\left(\vartheta \right)$ now ($L_z=0$) read:

$$R\left(r\right)=E^2{\left(r^2+a^2\right)}^2-\left(r^2+a^{2}K\right){\mathrm{\Delta }}_r\left(r\right)\ge 0$$E12

and

$$\mathrm{\Theta }\left(\vartheta \right)=-a^2{E}^2{sin}^2\vartheta +a^2\left(K-{cos}^2\vartheta \right){\mathrm{\Delta }}_{\vartheta }\left(\vartheta \right)\ge 0.$$E13

The quadratures are obtained by differentiation of $S$ with respect to the constants of the integration. For infalling orbits above the equatorial plane, Hamilton’s principal function can be written as

$$S=\frac{c^2}{2}\delta \left(\tau -{\tau }_0\right)-\mathit{cE}\left(\overline{t}-{\overline{t}}_0\right)+{\mathit{cL}}_z\left(\overline{\phi }-{\overline{\phi }}_0\right)+c{\int }_{\overline{r}}^{{\overline{r}}_0}\frac{\sqrt{R\left(r\right)}}{{\mathrm{\Delta }}_r\left(r\right)}\mathit{dr}+c{\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}}{{\mathrm{\Delta }}_{\vartheta }\left(\vartheta \right)}\mathit{d\vartheta },$$E14

with ${\overline{t}}_0=\overline{t}\left({\tau }_0\right)$, ${\overline{r}}_0=\overline{r}\left({\tau }_0\right)$, ${\overline{\vartheta }}_0=\overline{\vartheta }\left({\tau }_0\right)$ and ${\overline{\phi }}_0=\overline{\phi }\left({\tau }_0\right)$ the coordinates of any point along an orbit. Upon partial differentiation of $S$ with respect to $K$, $\delta$, $E$ and $L_z$ we obtain respectively

$$a^2{\int }_{\overline{r}}^{{\overline{r}}_0}\frac{\mathit{dr}}{\sqrt{R\left(r\right)}}=a^2{\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{\mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}}$$E15

$$c\left(\tau -{\tau }_0\right)={\int }_{\overline{r}}^{{\overline{r}}_0}\frac{r^2\mathit{dr}}{\sqrt{R\left(r\right)}}+a^2{\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{{cos}^2\vartheta \mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}}$$E16

$$\overline{t}-{\overline{t}}_0=E{\int }_{\overline{r}}^{{\overline{r}}_0}\frac{{\left(r^2+a^2\right)}^2\mathit{dr}}{\sqrt{R\left(r\right)}{\mathrm{\Delta }}_r\left(r\right)}-a^{2}E{\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{{sin}^2\vartheta \mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}{\mathrm{\Delta }}_{\vartheta }\left(\vartheta \right)}$$E17

$$\overline{\phi }-{\overline{\phi }}_0=\mathit{aE}{\int }_{\overline{r}}^{{\overline{r}}_0}\frac{\left(r^2+a^2\right)\mathit{dr}}{\sqrt{R\left(r\right)}{\mathrm{\Delta }}_r\left(r\right)}-\mathit{aE}{\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{1}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}{\mathrm{\Delta }}_{\vartheta }\left(\vartheta \right)}\mathit{d\vartheta }$$E18

The first equation suffices to determine $\overline{\vartheta }\left(r\right)$, a function depending on choices for $E$, $K$ and $\left(r_0,{\vartheta }_0\right)$ (apart from also depending on $a$, $M$ and $\mathrm{\Lambda }$, of course, but we will omit explicit reference to these latter dependencies). The second will in addition yield, be it implicitly, $\overline{r}\left(\tau ,E,K\right)$ and $\overline{\vartheta }\left(\tau ,E,K\right)$ after choosing ${\tau }_0$.

The other two have right hand sides that tend to $+\infty$ at the horizons ${\mathrm{\Delta }}_r=0$ or ${\mathrm{\Delta }}_{\vartheta }=0$, making $t$ and $\phi$ unsuitable as coordinates that describe the orbits there. In our final results, however, $\overline{t}$ and $\overline{\phi }$ will have disappeared altogether. Eqs. (14)–(18) can easily be modified to accommodate infalling orbits below the equatorial plane and outgoing orbits above and below the equatorial plane.

In the sequel, we will frequently need the partial derivatives with respect to $E$ and $K$ of all functions $\overline{t}\left(\tau \right)$, $\overline{r}\left(\tau \right)$, $\overline{\vartheta }\left(\tau \right)$ and $\overline{\phi }\left(\tau \right)$. We will denote these derivatives by the coordinate with added subscript (but no bar, for simplicity). We will not need explicit expressions for $r_E$, $r_K$, ${\vartheta }_E$ and ${\vartheta }_K$. It can be shown that $r_E$ and $r_K$ change sign with ${\epsilon }_r$, and likewise ${\vartheta }_E$ and ${\vartheta }_K$ with ${\epsilon }_{\vartheta }$. The expressions for ${\phi }_E$ and ${\phi }_K$ are obtained from (17):

$${\phi }_E=-\frac{\mathit{aE}\left({\overline{r}}^2+a^2\right)}{\sqrt{\overline{R}}{\overline{\mathrm{\Delta }}}_r}{r}_E-\frac{\mathit{aE}}{\sqrt{\overline{\mathrm{\Theta }}}{\overline{\mathrm{\Delta }}}_{\vartheta }}{\vartheta }_E+w_E\left(\overline{r},\overline{\vartheta }\right)$$E19

and

$${\phi }_K=-\frac{\mathit{aE}\left({\overline{r}}^2+a^2\right)}{\sqrt{\overline{R}}{\overline{\mathrm{\Delta }}}_r}{r}_K-\frac{\mathit{aE}}{\sqrt{\overline{\mathrm{\Theta }}}{\overline{\mathrm{\Delta }}}_{\vartheta }}{\vartheta }_K+w_K\left(\overline{r},\overline{\vartheta }\right)$$E20

with

$$\begin{array}{l}{w}_E\left(\overline{r}\right)=-a{\int }_{\overline{r}}^{{\overline{r}}_0}\frac{\left(r^2+a^2\right)\left(r^2+a^{2}K\right)}{R^{3/2}\left(r\right)}\mathit{dr}-a^3{\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{K-{cos}^2\left(\vartheta \right)}{{\mathrm{\Theta }}^{3/2}\left(\vartheta \right)}\mathit{d\vartheta }\\ w_K\left(\overline{r}\right)=\frac{a^{3}E}{2}{\int }_{\overline{r}}^{{\overline{r}}_0}\frac{r^2+a^2}{R^{3/2}\left(r\right)}\mathit{dr}+\frac{a^{3}E}{2}{\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{1}{{\mathrm{\Theta }}^{3/2}\left(\vartheta \right)}\mathit{d\vartheta }.\end{array}$$E21

The expressions for $t_E$ and $t_K$ are calculated with the aid of (14):

$${\mathit{Et}}_E=-\frac{\sqrt{\overline{R}}}{{\overline{\mathrm{\Delta }}}_r}{r}_E+\frac{\sqrt{\overline{\mathrm{\Theta }}}}{{\overline{\mathrm{\Delta }}}_{\vartheta }}{\vartheta }_E\text{and}{\mathit{Et}}_K=-\frac{\sqrt{\overline{R}}}{{\overline{\mathrm{\Delta }}}_r}{r}_K+\frac{\sqrt{\overline{\mathrm{\Theta }}}}{{\overline{\mathrm{\Delta }}}_{\vartheta }}{\vartheta }_K.$$E22

We note that also ${\phi }_E$, ${\phi }_K$ and $t_E$, $t_K$ are independent of the signs ${\epsilon }_r$ or ${\epsilon }_{\vartheta }$.

The problem of finding the analytic solutions of geodesic motion in the KdS spacetime has a long history and was fully solved in Ref. [2], with many references therein. In this paper, we will only touch on this difficult and specialist topic when we consider nonempty rotating black holes (or universes) with $\mathrm{\Lambda }=0$, this for the purpose of illustrating the procedure of actually constructing a model for a nonempty rotating black hole or universe. We will solve the equations of motion in that case in terms of elliptic functions and Carlson elliptic integrals (Appendix D).

3.2 The case $a=0$

It follows from (13) that $\mathrm{\Theta }=0$, hence, with (11), $\vartheta$ and $\phi$ are constants (see also (18)). The orbits are radial $\overline{r}\left(\tau ,E,\vartheta \right)$, with $\vartheta$ taking on the role of $K$. The latter is not undefined, since it follows from (32) that $K={cos}^2\vartheta +{sin}^2\vartheta {\mathrm{\Delta }}_r\left(r_a\right)/r_a^2=$ ${cos}^2\vartheta +{sin}^2\vartheta \mathrm{\Delta }\left(r_a\right)$. In addition we find, with (2) and (12):

$$\begin{array}{c}{\mathrm{\Delta }}_r=r^2\left(1-\frac{2M}{r}-\frac{\mathrm{\Lambda }}{3}{r}^2\right)\equiv r^2\mathrm{\Delta },{\mathrm{\Delta }}_{\vartheta }=1,\\ R=r^2\left(E^2{r}^2-{\mathrm{\Delta }}_r\right)=r^4\left(E^2-\mathrm{\Delta }\right).\end{array}$$E23

The equations of the motion are

$$\dot{r}=r_{\tau }=-c\sqrt{E^2-\mathrm{\Delta }}\text{and}\dot{t}=t_{\tau }=\frac{\mathit{cE}}{\mathrm{\Delta }}.$$E24

Eq. (15) does not exist, since it reduces to $0=0$. All the orbits can reach $r=0$. With that choice for $r_0$, we can recast (16) in the form

$$c\left(\tau -{\tau }_0\right)=-{\int }_0^{\overline{r}}\frac{\sqrt{r}\mathit{dr}}{\sqrt{2M-\left(1-E^2\right)r+\mathrm{\Lambda }{r}^3/3}}=-{\int }_0^{\overline{r}}\frac{\mathit{dr}}{\sqrt{E^2-\mathrm{\Delta }}},$$E25

which is the same as the defining equation for the downward branch of the function cyc in Ref. [1], Appendix 1. Of all the partial derivatives, only $r_E$ and $t_E$ are left in a nontrivial way. We will need (22):

$${\mathit{Et}}_E=-\frac{\sqrt{E^2-\overline{\mathrm{\Delta }}}}{\overline{\mathrm{\Delta }}}.$$E26

As for the other partial derivatives, $r_K$ and $t_K$ are to be replaced by $r_{\vartheta }=0$ and $t_{\vartheta }=0$, ${\vartheta }_E=0$ and ${\vartheta }_K$ has to be replaced by ${\vartheta }_{\vartheta }=1$. The functions $w_E=w_K=0$.

3.3 The case $\vartheta =0$

Now the integrals over $\vartheta$ are indeterminate since $\mathrm{\Theta }=0$, but can all be evaluated using (15). We obtain

$$c\left(\tau -{\tau }_0\right)={\int }_{\overline{r}}^{r_0}\frac{r^2+a^2}{\sqrt{R\left(r\right)}}\mathit{dr}\text{and}\overline{t}-{\overline{t}}_0=E{\int }_{\overline{r}}^{r_0}\frac{{\left(r^2+a^2\right)}^2}{\sqrt{R\left(r\right)}{\mathrm{\Delta }}_r\left(r\right)}\mathit{dr}.$$E27

The equation for $\overline{\phi }$ is, as the above integral, a proper one for ${\mathrm{\Delta }}_r>0$ and ${\mathrm{\Delta }}_{\vartheta }>0$:

$$\overline{\phi }-{\overline{\phi }}_0=\mathit{aE}{\int }_{\overline{r}}^{r_0}\left(\frac{r^2+a^2}{{\mathrm{\Delta }}_r\left(r\right)}-\frac{1}{{\mathrm{\Delta }}_{\vartheta }}\right)\frac{\mathit{dr}}{\sqrt{R\left(r\right)}}.$$E28

Between the brackets, we recognize an expression that is proportional to $\dot{\phi }$, but the meaning of this last integral is a limit $\vartheta \to 0$, since $\phi$ is not defined on the polar axis. We will not need it anyway.

The orbits are radial $\overline{r}\left(\tau ,E\right)$. Regarding the partial derivatives, only $r_E$ and $t_E$ are left. The others do not exist, since $\vartheta \left(=0\right)$ is not a coordinate anymore and neither is $K\left(=1\right)$. We will not need $w_E$ nor $w_K$.

4. The travelator metric

4.1 The general case $a\ne 0$

We now pass from the coordinates $\left(x^{\alpha }\right)=\left(t,r,\vartheta ,\phi \right)$ to new coordinates $\left({\overline{x}}^{\alpha }\right)=\left(\tau ,E,K,\psi \right)$. The angle $\psi$ is the same as $\phi$ and is additive to it, i.e. $\phi =\overline{\phi }\left(E,K,\tau \right)+\psi$. Such a metric is called a synchronous metric (see also Landau and Lifshitz [4]). This metric is the generalization of the Novikov metric, much as the synchronous metric of Khatsymovski [5] is the generalization of the Lemaitre metric.

Each of the old coordinates is now function of $E$, $K$, $\tau$, and , and we adopt the transformation matrix

$$C=\left(C_{\mathit{\alpha \beta }}\right)=\left(\begin{array}{cccc}{t}_{\tau }& r_{\tau }& {\vartheta }_{\tau }& {\phi }_{\tau }\\ t_E& r_E& {\vartheta }_E& {\phi }_E\\ t_K& r_K& {\vartheta }_K& {\phi }_K\\ 0& 0& 0& 1\end{array}\right)=\left(\frac{\partial x^{\beta }}{\partial {\overline{x}}^{\alpha }}\right),$$E29

where $t_{\tau }$, $r_{\tau }$, ${\vartheta }_{\tau }$, and ${\phi }_{\tau }$ are given by (11).² The indices $\alpha$ and $\beta$ are simply matrix indices. The metric transforms according to ${\overline{g}}_{\mathit{\alpha \beta }}=\frac{\partial x^{\rho }}{\partial {\overline{x}}^{\alpha }}\frac{\partial x^{\sigma }}{\partial {\overline{x}}^{\beta }}{g}_{\mathit{\rho \sigma }}$, or in matrix form

$$\left(\overline{g}\right)=\mathcal{T}\left(g\right){\mathcal{T}}^T,$$E30

with $\mathcal{T}=C$. The matrices $\left(g\right)$ and $\left(\overline{g}\right)$ are therefore congruent.

The determinant of $C$ is given by a remarkably simple expression:

$$det\left(C\right)=\frac{c}{E}\left(r_E{\vartheta }_K-r_K{\vartheta }_E\right).$$E31

It does not contain singularities connected with the event horizons ${\mathrm{\Delta }}_r=0$ or ${\mathrm{\Delta }}_{\vartheta }=0$. We recognize in the last factor the Jacobian of the transformation from $\left(E,K\right)$ to $\left(\overline{r},\overline{\vartheta }\right)$, with $\tau$ as a parameter.

Since the determinant of the KdS metric $\left(g\right)$ equals $-{\rho }^4{sin}^2\vartheta$, we obtain

$$det\left(\overline{g}\right)=-\frac{c^2{\rho }^4}{E^2}{\left(r_E{\vartheta }_K-r_K{\vartheta }_E\right)}^2{sin}^2\vartheta .$$E32

From the concept of Gaussian coordinate systems, as developed, e.g., in Ref. [6], we know that a coordinate system based on geodesics that starts off with zero velocity has a metric with ${\overline{g}}_{0i}={\overline{g}}_{i0}=0,i=\mathrm{1,2,3}$ for as long as the metric is valid. Hence, our consideration of orbits with $L_z=0$ only. That this is true in our case can be verified with (30) and $\mathcal{T}=C$: inserting the general equations of motion (with $L_z\ne 0$) into the expression for $2{\overline{g}}_{03}$ (factor of the term in $\mathit{d\tau }\mathit{d\psi }$), we find indeed that ${\overline{g}}_{03}=0$ globally only if $L_z=0$. For the coefficients $2{\overline{g}}_{01}$ (factor of the term in $\mathit{d\tau }\mathit{dE}$) and $2{\overline{g}}_{02}$ (factor of the term in $\mathit{d\tau }\mathit{dK}$) we need to make use of (22).

A zero ‘start’ velocity also means a zero velocity in the meridional plane. The radius at which the radial velocity is zero is an apocenter radius and we denote it with a subscript $a$, hence $r_a$. The polar angle $\vartheta$ at which the tangential velocity in the meridional plane is zero we similarly denote by ${\vartheta }_a$. When the expressions for $R\left(r\right)$ and $\mathrm{\Theta }\left(\vartheta \right)$, as given by (12) and (13), are equated to zero (as is the case for $r=r_a$ and $\vartheta ={\vartheta }_a$), they can easily be solved for $E^2$ and $K$:

$$\begin{array}{c}-g_{33}\left(r_a,{\vartheta }_a\right){\rho }^2\left(r_a,{\vartheta }_a\right)E^2={sin}^2{\vartheta }_a{\rho }^2\left(r_a,{\vartheta }_a\right){\mathrm{\Delta }}_r\left(r_a\right){\mathrm{\Delta }}_{\vartheta }\left({\vartheta }_a\right)\\ -g_{33}\left(r_a,{\vartheta }_a\right){\rho }^2\left(r_a,{\vartheta }_a\right)K={sin}^2{\vartheta }_a\left[{\left(r_a^2+a^2\right)}^2{cos}^2{\vartheta }_a{\mathrm{\Delta }}_{\vartheta }\left({\vartheta }_a\right)+r_a^2{sin}^2{\vartheta }_a{\mathrm{\Delta }}_r\left(r_a\right)\right].\end{array}$$E33

Clearly $E^2$ and $K$ are positive for all initial conditions $r_a$ and ${\vartheta }_a$ outside the black hole. Better still, $r_a$ and ${\vartheta }_a$ are valid substitutes for $E$ and $K$, with the advantage that they are intuitive (we assume $r_a<+\infty$).

The inequality (12) can also be written as $E^2\ge V_{\mathrm{eff}}\left(r\right)$ (with)

$$V_{\mathrm{eff}}\left(r\right)=\frac{\left(r^2+a^{2}K\right){\mathrm{\Delta }}_r\left(r\right)}{{\left(r^2+a^2\right)}^2}.$$E34

A qualitative inspection of this curve shows that the orbits we consider here can also attain a radius where the radial velocity is again zero (hence a turning point) which we call the pericenter radius; we denote that radius by ${\overline{r}}_p$. In addition, there are orbits that can attain $r=0$ with nonzero velocity. They are called polar orbits.

As for the ‘orbital’ part of the metric ${\overline{g}}_{\mathit{ij}}\left(\tau ,E,K\right),i=\mathrm{1,2,3}$, we thus assume that at $t=\tau =0$ all orbits start off with zero velocity. In that case, all the partial derivatives that appear in the transformation matrix (29) are regular in the vicinity of $\left(r_a,{\vartheta }_a\right)$, and hence the metric ${\overline{g}}_{\mathit{ij}}\left(\tau ,E,K\right)$, as calculated in (30) with $\mathcal{T}=C$, is regular there. We also note that we deal here with very special orbits for which $\left(r_a,{\vartheta }_a\right)$ is a point of the orbit. In general $r_a$ and ${\vartheta }_a$ will be reached at different times, since solving (33) does not imply that ${\vartheta }_a\left(\tau \right)=r_a\left(\tau \right)$ for some $\tau$.

Now that we have established regularity of the metric ${\overline{g}}_{\alpha ,\beta }$ for $\tau$ small, it still remains to be seen what happens at later times, since there is no theorem that guarantees regularity, and thus validity. Therefore, we have to go through the chore of actually calculating the ${\overline{g}}_{\mathit{ij}}\left(\tau ,E,K\right),i=\mathrm{1,2,3}$. Since we are now in the process of getting rid of the singular coordinates $t$ and $\phi$, we can expect to also get rid of the singularities in the original KdS metric. The coefficients in the third row and/or column are

$${\overline{g}}_{13}={\overline{g}}_{31}=a_1{r}_E+b_1{\vartheta }_E+g_{33}{w}_E$$E35

$${\overline{g}}_{23}={\overline{g}}_{32}=a_1{r}_K+b_1{\vartheta }_K+g_{33}{w}_K$$E36

$${\overline{g}}_{33}=g_{33}$$E37

with

$$\begin{array}{l}{a}_1=\frac{1}{E\sqrt{\overline{R}}}\left[\left({\overline{r}}^2+a^2\right){\mathit{aE}}^2{sin}^2\overline{\vartheta }+\left({\overline{r}}^2+a^{2}K\right)g_{03}\right]\\ b_1=\frac{1}{E\sqrt{\overline{\mathrm{\Theta }}}}\left[\left({\overline{r}}^2+a^2\right){\mathit{aE}}^2{sin}^2\overline{\vartheta }+a^2\left(K-{cos}^2\vartheta \right)g_{03}\right].\end{array}$$E38

The remaining coefficients are

$$\begin{array}{l}{\overline{g}}_{21}={\overline{g}}_{12}=a_2{r}_E{r}_K+b_2{\vartheta }_E{\vartheta }_K+c_2\left(r_E{\vartheta }_K+r_K{\vartheta }_E\right)+\\ +a_1\left(r_E{w}_K+r_K{w}_E\right)+b_1\left({\vartheta }_E{w}_K+{\vartheta }_K{w}_E\right)+g_{33}{w}_E{w}_K\end{array}$$E39

$${\overline{g}}_{11}=a_2{r}_E^2+b_2{\vartheta }_E^2+2c_2{r}_E{\vartheta }_E+2a_1{r}_E{w}_E+2b_1{\vartheta }_E{w}_E+g_{33}{w}_E^2$$E40

$${\overline{g}}_{22}=a_2{r}_K^2+b_2{\vartheta }_K^2+2c_2{r}_K{\vartheta }_K+2a_1{r}_K{w}_K+2b_1{\vartheta }_K{w}_K+g_{33}{w}_K^2$$E41

with

$$\begin{array}{l}{a}_2=\frac{{\overline{r}}^2+a^{2}K}{E^2\overline{R}}\left[\left({\overline{r}}^2+a^{2}K\right)g_{00}-E^2{{\overline{\rho }}^{\prime }}^2\right]\\ b_2=\frac{a^2\left(K-{cos}^2\overline{\vartheta }\right)}{E^2\overline{\mathrm{\Theta }}}\left[a^2\left(K-{cos}^2\overline{\vartheta }\right)g_{00}-E^2{{\overline{\rho }}^{\prime }}^2\right]\\ c_2=\frac{a^2}{E^2\sqrt{\overline{R}}\sqrt{\overline{\mathrm{\Theta }}}}\left[\left({\overline{r}}^2+a^{2}K\right)\left(K-{cos}^2\overline{\vartheta }\right)g_{00}-E^2\left(2{\overline{r}}^2{sin}^2\overline{\vartheta }-{\overline{\rho }}^2+K{{\overline{\rho }}^{\prime }}^2\right)\right].\end{array}$$E42

In the above expressions (35)–(39),³ it is, of course, understood that $g_{00}\left(\overline{r},\overline{\vartheta }\right)$, $g_{03}\left(\overline{r},\overline{\vartheta }\right)$, and $g_{33}\left(\overline{r},\overline{\vartheta }\right)$. We also note that the metric is independent of ${\epsilon }_r$ and ${\epsilon }_{\vartheta }$.

Clearly, all singularities associated with ${\mathrm{\Delta }}_r$ and ${\mathrm{\Delta }}_{\vartheta }$ are gone, but an $\overline{R}$ or a $\overline{\mathrm{\Theta }}$ in a denominator causes a singularity whenever $r_a$, $r_p$ or ${\vartheta }_a$ is reached. Technically, this is a consequence of having to split up a regular limit into two singular terms in the expressions for ${\phi }_E$ and ${\phi }_K$. Hence, we have again two complementary representations of ${\overline{g}}_{\mathit{\alpha \beta }}$: the one in (29) with $\mathcal{T}=C$ that is valid in the vicinity of $\tau =0$ and the one that is valid for each orbit $\left(E,K\right)$ up to the time ${\tau }_p$. Since (1) the orbits are continuously replenished from far away and (2) the singularities associated with the $R\left(\overline{r}\left(0,E,K\right)\right)$ and $\mathrm{\Theta }\left(\overline{\vartheta }\left(0,E,K\right)\right)$ are removable, the metric given by Eqs. (35)–(39) suffices as the only metric.

The KdS metric is a regular one outside the ergosphere $r=M+\sqrt{M^2-a^2{cos}^2\vartheta }$. The synchronous metric can be seen there as a kind of travelator or conveyor belt that one can ‘step on’ (this requires a boost to match the velocity of the travelator ring that passes by) to enter this valid metric that carries one through the ergosphere and the region ${\mathrm{\Delta }}_r\left(r\right)<0$ between the event horizons $R_{\pm }=M\pm \sqrt{M^2-a^2}$. The ergosphere has disappeared, illustrating the fact that the presence of an ergosphere is a property of the velocity of the observer rather than an intrinsic feature of the rotating black hole. The travelator consists of rings labeled by $\left(E,K\right)$, and these are centered on the polar axis $\vartheta =0$. The travelator metric can be freely traveled as long as one stays clear of the ${\tau }_p$ of the rings that one is standing on. Since all ${\overline{r}}_p$ are located in the inner KdS spacetime (defined here as the spacetime inside the inner event horizon $R_{-}$), it is always possible to ‘step off’ the travelator ‘in time’ after having passed $R_{-}$, again by applying a boost to match the velocity of objects that are at rest in another valid metric.

As explained above, we need not bother with the singularities associated with $R$ and $\mathrm{\Theta }$, in the KdS case, that is. However, in the next section, we will consider nonempty rotating black holes, and we will have to deal with these singularities. Inspection of the expressions (35)–(42) and those for $w_E$ and $w_K$, in combination with the appropriate differentials $\mathit{dE}$ and $\mathit{dK}$ of the travelator metric, reveals that all singularities appear as

$$\frac{r_E}{\sqrt{R}}\mathit{dE}=2\frac{d\sqrt{R}}{{\partial }_{r}R},\frac{{\vartheta }_E}{\sqrt{\mathrm{\Theta }}}\mathit{dE}=2\frac{d\sqrt{\mathrm{\Theta }}}{{\partial }_{\vartheta }\mathrm{\Theta }}$$E43

and similarly for the differentials in $K$. Since the zeroes of $R$ and $\mathrm{\Theta }$ are simple zeroes, ${\partial }_{r}R$ and ${\partial }_{\vartheta }\mathrm{\Theta }$ are nonzero there. Hence, if we pass, locally, from $\left(E,K\right)$ to $\left(\sqrt{R},\sqrt{\mathrm{\Theta }}\right)$, the metric in these coordinates is regular and Lorentzian.

It may seem remarkable that the travelator metric (35)–(42) has such a simple determinant (32). However, this can be verified using the common techniques of determinant evaluation, not without effort.

4.2 The case $a=0$

The transformation matrix from the coordinates $\left(x^{\alpha }\right)=\left(t,r,\vartheta ,\phi \right)$ to the new coordinates $\left({\overline{x}}^{\alpha }\right)=\left(\tau ,E,\vartheta ,\phi \right)$ is now simply

$$\left(C_{\mathit{\alpha \beta }}\right)=\left(\begin{array}{cccc}{t}_{\tau }& r_{\tau }& 0& 0\\ t_E& r_E& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\frac{\partial x^{\beta }}{\partial {\overline{x}}^{\alpha }}\right),det\left(C\right)=\frac{c{r}_E}{E}.$$E44

Since $\vartheta$ and $\phi$ need no transformation, we obtain immediately from the KdS metric and the expression for $a_2$

$${\mathit{ds}}^2=c^{2}d{\tau }^2-\frac{r_E^2}{E^2}{\mathit{dE}}^2-{\overline{r}}^{2}d{\vartheta }^2-{\overline{r}}^2{sin}^2\mathit{\vartheta d}{\psi }^2.$$E45

Note that in this metric $\overline{r}\left(E,\tau \right)$ as well as $r_E\left(E,\tau \right)$. This is the travelator version of the Schwarzschild-$\mathrm{\Lambda }$ metric (also called the Schwarzschild-de Sitter metric). In Ref. [1], this metric is called the Novikov metric.

4.3 The case $\vartheta =0$

Here we are dealing with a subset of the orbits, those that are confined to the polar axis and that are therefore dynamically decoupled from the rest. The transformation matrix from the coordinates $\left(x^{\alpha }\right)=\left(t,r\right)$ to the new coordinates $\left({\overline{x}}^{\alpha }\right)=\left(\tau ,E\right)$ is the lower dimensional version of (44):

$$\left(C_{\mathit{\alpha \beta }}\right)=\left(\begin{array}{cc}{t}_{\tau }& r_{\tau }\\ t_E& r_E\end{array}\right)=\left(\frac{\partial x^{\beta }}{\partial {\overline{x}}^{\alpha }}\right),det\left(C\right)=\frac{c{r}_E}{E}.$$E46

The coordinate $\phi$ is not present, and $K=1$. Hence we obtain

$${\mathit{ds}}^2=c^{2}d{\tau }^2-\frac{r_E^2}{E^2}{\mathit{dE}}^2.$$E47

Again $r_E\left(E,\tau \right)$. The similarity with the previous case is only formal and qualitative, since here $a$ is not necessarily zero and, therefore, the orbits are different. The result (47) can, of course, also be obtained from (35)–(43) with $\mathit{d\vartheta }=\mathit{d\psi }=0$.

5. The Einstein equations

The KdS metric satisfies the Einstein equations

$$R^{\mathit{\alpha \beta }}-\frac{1}{2}{\mathit{Rg}}^{\mathit{\alpha \beta }}-\mathrm{\Lambda }{g}^{\mathit{\alpha \beta }}=0$$E48

with $R^{\mathit{\alpha \beta }}$ the Ricci tensor and $R$ the Riemann scalar,⁴ and that is, of course, no different for the travelator metric. From now on, we will lift the restriction that all orbits on which the travelator metric is built start off with zero velocity at a given time, say $\tau =0$. The only consequence is that the travelator metric will also be singular at $\tau =0$, but, as already indicated, these singularities are removable. Hence, any family of orbits that satisfies (15) and (16) gives rise to a travelator metric.

What we will need for the Einstein equations in the KdS metric can be found in Appendix A. The contravariant Ricci tensor $R^{\mathit{\alpha \beta }}$ in the KdS metric transforms into the travelator metric according to ${\overline{R}}^{\mathit{\alpha \beta }}=\frac{\partial {\overline{x}}^{\alpha }}{\partial x^{\rho }}\frac{\partial {\overline{x}}^{\beta }}{\partial x^{\sigma }}{R}^{\mathit{\rho \sigma }}$, or in matrix form

$$\left(\overline{R}\right)={\mathcal{T}}^T\left(R\right)\mathcal{T}$$E49

with $\mathcal{T}=D=C^{-1}$. The matrix $D$ is given in Appendix B. For as far as the contravariant metric tensor in the travelator metric is concerned, we will here only need the rather trivial ${\overline{g}}^{00}=1/c^2$ and the ${\overline{R}}^{00}$ component of the contravariant Ricci tensor in Appendix C.

One could set out to verify (48) analytically for both the KdS and the travelator metric, but that is, again, a chore because all the derivatives of $g_{00}$, in addition to ${\mathrm{\Delta }}_r$ and ${\mathrm{\Delta }}_{\vartheta }$, must be exposed explicitly in terms of $M$, $a$ and $\mathrm{\Lambda }$.⁵ For this paper, a numerical check has been performed, which is comparatively easy, and zeroes have been obtained up to machine accuracy. This latter fact shows, by the way, that the zeroes are not trivial but a result of all kinds of cancelations. Moreover, the identity (48) is satisfied irrespective of whatever values for $r_E$, ${\vartheta }_E$, $r_K$, ${\vartheta }_K$, $w_E$, and are assumed.

Next, we note that in the KdS metric the constants $M$ and $a$ are independent. Therefore, we can choose two new parameters, say $\lambda$ and $\mu$, unrelated to the KdS coordinates, and consider.⁶

$$M=m\left(\lambda ,\mu \right)\text{and}a=A\left(\lambda ,\mu \right).$$E50

We write the KdS metric with these functions instead of $M$ and $a$. Looking now at the equations for the orbits (15)–(18), we see that all four constants $M$, $a$, $E$, and $K$ can be freely chosen, which is a direct consequence of the separability of the equations of the motion. Hence we can also consider $E\left(\lambda ,\mu \right)$ and $K\left(\lambda ,\mu \right)$. We now transform the KdS metric, this time in its $\left(\lambda ,\mu \right)$ livery, to the travelator metric (35)–(42), thus in terms of $m\left(\lambda ,\mu \right)$, $A\left(\lambda ,\mu \right)$, $\tau$, $E\left(\lambda ,\mu \right)$, $K\left(\lambda ,\mu \right)$, $\psi$ and the differentials $\mathit{d\tau }$, $\mathit{dE}$, $\mathit{dK}$, and $\mathit{d\psi }$. Next follows a transformation from $\left(E,K\right)$ to the new variables $\left(\lambda ,\mu \right)$, using the transformation (30) with

$$\mathcal{T}=U=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& E_{\lambda }& K_{\lambda }& 0\\ 0& E_{\mu }& K_{\mu }& 0\\ 0& 0& 0& 1\end{array}\right)$$E51

with obvious notations. That metric $\left(\overline{\overline{g}}\right)$ (1) still satisfies the Einstein Eqs. (48) because of covariance, (2) features the variables $\tau$, $\lambda$, $\mu$, $\psi$ and the differentials $\mathit{d\tau }$, $\mathit{d\lambda }$ (replacing $\mathit{dE}$), $\mathit{d\mu }$ (replacing $\mathit{dK}$) and $\mathit{d\psi }$ and (3) is the same as (34)–(41) with the subscripts $E$ and $K$ replaced by $\lambda$ and $\mu$.

We note that, since $E$ and $K$ are attributes of a comoving ring, $\lambda$ and $\mu$ replace them as attributes of a comoving ring. We thus obtain a two-dimensional manifold of comoving rings, the motion of which is determined by (15) and (16), equations that feature $m\left(\lambda ,\mu \right)$, $A\left(\lambda ,\mu \right)$, $E\left(\lambda ,\mu \right)$, and $K\left(\lambda ,\mu \right)$.

If $a$ and $M$ are not functions of $\lambda$ and $\mu$, but just constants, then we have produced a travelator metric that is free of the KdS singularities and satisfies the Einstein equations for empty space (48). In that case, we have merely replaced the constants of the motion $E$ and $K$ by other constants of the motion $\lambda$ and $\mu$. But $A\left(\lambda ,\mu \right)$ and $m\left(\lambda ,\mu \right)$ need not be constants, and we now investigate what this means for the Einstein equations for a nonempty rotating black hole:

$${\overline{\overline{R}}}^{\mathit{\alpha \beta }}-\frac{1}{2}{\overline{\overline{R}}\overline{\overline{g}}}^{\mathit{\alpha \beta }}-\mathrm{\Lambda }{\overline{\overline{g}}}^{\mathit{\alpha \beta }}=\frac{8\mathit{\pi G}}{c^4}{\overline{\overline{T}}}^{\mathit{\alpha \beta }},$$E52

with ${\overline{\overline{T}}}^{\mathit{\alpha \beta }}$ the stress-energy-momentum tensor.

We consider comoving matter:

$${\overline{\overline{u}}}^{\alpha }=\left(\frac{\mathit{d\tau }}{\mathit{d\tau }},\frac{\mathit{d\lambda }}{\mathit{d\tau }},\frac{\mathit{d\mu }}{\mathit{d\tau }},\frac{\mathit{d\psi }}{\mathit{d\tau }}\right)=\left(\mathrm{1,0,0,0}\right).$$E53

The Ricci tensor ${\overline{\overline{R}}}^{\mathit{\alpha \beta }}$ contains the derivatives ${\partial }_r^n{\mathrm{\Delta }}_r\left(r,M,a^2\right)$, ${\partial }_{\vartheta }^n{\mathrm{\Delta }}_{\vartheta }\left(\vartheta ,a^2\right)$ with $n=1,2$. When taken as such, i.e., with the partial derivative only pertaining to the first argument, all ${\overline{\overline{T}}}^{\mathit{\alpha \beta }}$ are zero. However, since $\lambda$ and $\mu$ are now coordinates, the inverse functions $\lambda \left(\tau ,r,\vartheta \right)$ and $\mu \left(\tau ,r,\vartheta \right)$ also make sense. Indeed, the functions $\overline{r}\left(\tau ,\lambda ,\mu \right)$ and $\overline{\vartheta }\left(\tau ,\lambda ,\mu \right)$ from Section 3 tell us at which $\left(r,\vartheta \right)$ the ring with label $\left(\lambda ,\mu \right)$ intersects the meridian plane at time $\tau$. Hence, the inverse functions $\lambda \left(\tau ,r,\vartheta \right)$ and $\mu \left(\tau ,r,\vartheta \right)$ tell us which ring with label $\left(\lambda ,\mu \right)$ happens to be at $\left(r,\vartheta \right)$ at time $\tau$. We can, therefore, consider as alternative for ${\partial }_r{\mathrm{\Delta }}_r$ the expression

$$\begin{array}{ll}{\partial }_r{\mathrm{\Delta }}_r\left(r,m\left(\lambda ,\mu \right),A^2\left(\lambda ,\mu \right)\right)& ={\partial }_r{\mathrm{\Delta }}_r\left(r,M,a^2\right)+m_{\overline{r}}{\partial }_m{\mathrm{\Delta }}_r\left(r,m,A^2\right)+A_{\overline{r}}^2{\partial }_{A^2}{\mathrm{\Delta }}_r\left(r,m,A^2\right)\\ & ={\partial }_r{\mathrm{\Delta }}_r\left(r,M,a^2\right)-2{\mathit{rm}}_{\overline{r}}+\left(1-\frac{\mathrm{\Lambda }}{3}{r}^2\right)A_{\overline{r}}^2\\ & \equiv {\partial }_r{\mathrm{\Delta }}_r\left(r,M,a^2\right)+\overline{{\partial }_r{\mathrm{\Delta }}_r}\end{array}$$E54

and similarly for ${\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }$

$$\begin{array}{ll}{\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }\left(\vartheta ,A^2\left(\lambda ,\mu \right)\right)& ={\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }\left(\vartheta ,a^2\right)+A_{\overline{\vartheta }}^2{\partial }_{A^2}{\mathrm{\Delta }}_{\vartheta }\left(\vartheta ,A^2\right)\\ & ={\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }\left(\vartheta ,a^2\right)+\frac{1}{3}\mathrm{\Lambda }{cos}^2\vartheta A_{\overline{\vartheta }}^2\\ & \equiv {\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }\left(\vartheta ,a^2\right)+\overline{{\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }}.\end{array}$$E55

The symbols with subscripts $\overline{r}$ and $\overline{\vartheta }$ stand for the expressions

$$\begin{array}{l}{f}_{\overline{r}}\left(\tau ,r,\vartheta \right)={\lambda }_r{\partial }_{\lambda }f+{\mu }_r{\partial }_{\mu }f,f=m,A^2\\ A_{\overline{\vartheta }}^2\left(\tau ,r,\vartheta \right)={\lambda }_{\vartheta }{\partial }_{\lambda }{A}^2+{\mu }_{\vartheta }{\partial }_{\mu }{A}^2\end{array}$$E56

and ${\lambda }_r$, ${\lambda }_{\vartheta }$, ${\mu }_r$, and ${\mu }_{\vartheta }$ stand for partial derivatives of the functions $\lambda \left(\tau ,r,\vartheta \right)$ and $\mu \left(\tau ,r,\vartheta \right)$ in our usual notations. Since we rather have $\overline{r}\left(\tau ,\lambda ,\mu \right)$ and $\overline{\vartheta }\left(\tau ,\lambda ,\mu \right)$ at our disposition, these partial derivatives are calculated using

$$\left(\begin{array}{cc}{\lambda }_r& {\mu }_r\\ {\lambda }_{\vartheta }& {\mu }_{\vartheta }\end{array}\right)\left(\tau ,\lambda ,\mu \right)=\frac{1}{{\overline{r}}_{\lambda }{\overline{\vartheta }}_{\mu }-{\overline{\vartheta }}_{\lambda }{\overline{r}}_{\mu }}\left(\begin{array}{cc}{\overline{\vartheta }}_{\mu }& -{\overline{\vartheta }}_{\lambda }\\ -{\overline{r}}_{\mu }& {\overline{r}}_{\lambda }\end{array}\right)\left(\tau ,\lambda ,\mu \right).$$E57

Expressions for the second derivatives are obtained analogously to (54) and (55). We write

$$\begin{array}{ll}{\partial }_r^2{\mathrm{\Delta }}_r\left(r,m\left(\lambda ,\mu \right),A^2\left(\lambda ,\mu \right)\right)& ={\partial }_r^2{\mathrm{\Delta }}_r\left(r,M,a^2\right)+\overline{{\partial }_r^2{\mathrm{\Delta }}_r}\\ \hfill {\partial }_{\vartheta }^2{\mathrm{\Delta }}_{\vartheta }\left(\vartheta ,A^2\left(\lambda ,\mu \right)\right)& ={\partial }_{\vartheta }^2{\mathrm{\Delta }}_{\vartheta }\left(\vartheta ,a^2\right)+\overline{{\partial }_{\vartheta }^2{\mathrm{\Delta }}_{\vartheta }}\end{array}$$E58

with again

$$\begin{array}{ll}\overline{{\partial }_r^2{\mathrm{\Delta }}_r}& =\left({\lambda }_r{\partial }_{\lambda }+{\mu }_r{\partial }_{\mu }\right)\left[{\left.{\partial }_r{\mathrm{\Delta }}_r\right|}_{M=m,a^2=A^2}+\overline{{\partial }_r{\mathrm{\Delta }}_r}\right]\\ \overline{{\partial }_{\vartheta }^2{\mathrm{\Delta }}_{\vartheta }}& =\left({\lambda }_{\vartheta }{\partial }_{\lambda }+{\mu }_{\vartheta }{\partial }_{\mu }\right)\left[{\left.{\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }\right|}_{a^2=A^2}+\overline{{\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }}\right].\end{array}$$E59

Whether one holds on to ‘the original’ ${\partial }_r^n{\mathrm{\Delta }}_r$ and ${\partial }_{\vartheta }^n{\mathrm{\Delta }}_{\vartheta }$ (in which $M$ and $a^2$ are constants) or adopts the alternatives $\overline{{\partial }_r^n{\mathrm{\Delta }}_r}$ and $\overline{{\partial }_{\vartheta }^n{\mathrm{\Delta }}_{\vartheta }}$ (in which $m$ and $A^2$ are functions of $\lambda$ and $\mu$), is a matter of choice for every equation $\left(\alpha ,\beta \right)$ in (52) separately.

6. The Lemaitre–Tolman metric (the case $a=0$)

We now illustrate the procedure that is outlined in the previous section for the simplest possible case $a=0$. We need only the coordinate $\lambda$.

We first revisit the metric (45) in $\left(\tau ,\lambda ,\vartheta ,\psi \right)$ (coordinates)

$${\mathit{ds}}^2=c^{2}d{\tau }^2-\frac{{\overline{r}}_{\lambda }^2}{E^2\left(\lambda \right)}d{\lambda }^2-{\overline{r}}^{2}d{\vartheta }^2-{\overline{r}}^2{sin}^2\mathit{\vartheta d}{\psi }^2,$$E60

with, again ${\overline{r}}_{\lambda }={\partial }_{\lambda }\overline{r}$.⁷ The metric (60) is exactly what has been called in Ref. [1] the Lemaitre-Tolman metric (hereafter L-T metric) if we set

$$E^2\left(\lambda \right)=1-2e\left(\lambda \right)$$E61

in the notations of Ref. [1]. What we call here $\lambda$ is mostly known as $r$ in the L-T models, which was called the shell label in Ref. [1]. The only parameter $M$, which does not appear in the metric but solely in the equation of the motion (26), is now a function of $\lambda$. As before, we write $m\left(\lambda \right)$, and in this simple case we know that $m\left(\lambda \right)$ is the mass inside the shell with label $\lambda$. We assume $0\le \lambda \le 1$ with the $\lambda =1$ shell the shell beyond which there is no mass anymore.

As for the Einstein equations, we must first transform the contravariant tensors in the appendices which are in the $\left(\tau ,E,\vartheta ,\psi \right)$ coordinates to the $\left(\tau ,\lambda ,\vartheta ,\psi \right)$ coordinates with the transformation (49) using the transformation matrix $V=U^{-1}$. This transformation produces a contravariant tensor with the same coefficients, except for these that carry just one index 2 which are divided by $E_{\lambda }$, and the 22-index, which is divided by $E_{\lambda }^2$. Since the Einstein equations feature only diagonal coefficients, this transformation has no effect on the final equations.

The stress-energy-momentum tensor reads, in spherical geometry with anisotropic pressures (see, e.g., Doneva & Yazadjiev [7]):

$${\overline{\overline{T}}}^{\mathit{\alpha \beta }}={\rho }_0{\overline{\overline{u}}}^{\alpha }{\overline{\overline{u}}}^{\beta }+p_r{e}_r^{\alpha }{e}_r^{\beta }+p_t\left({\overline{\overline{u}}}^{\alpha }{\overline{\overline{u}}}^{\beta }-e_r^{\alpha }{e}_r^{\beta }-g^{\mathit{\alpha \beta }}\right),$$E62

since the $\lambda$-direction is the radial direction. The tensor that an arbitrary tangential pressure $p_t$ (i.e., a pressure orthogonal to the radial direction) is multiplied by in (63), is orthogonal to both ${\overline{\overline{u}}}^{\alpha }$ and $e_r^{\alpha }$.

We find for the 00-equation, with the expressions in Appendices A and C.2.

$$-\frac{1}{c^2}\left(\frac{\mathrm{\Delta }-1}{r^2}+\frac{{\partial }_r\mathrm{\Delta }}{r}+\mathrm{\Lambda }\right)=\frac{8\mathit{\pi G}}{c^4}{\overline{\overline{T}}}^{00}.$$E63

When $M$ is a constant, this expression reduces to $0=0$. When $m\left(\lambda \right)$, we obtain

$$-\frac{\overline{{\partial }_r{\mathrm{\Delta }}_r}}{{\overline{r}}^3}=\frac{8\mathit{\pi G}}{c^2}{\overline{\overline{T}}}^{00},$$E64

or, with (54)

$$2\frac{{\lambda }_r{\partial }_{\lambda }m}{{\overline{r}}^2}=\frac{8\mathit{\pi G}}{c^2}{\overline{\overline{T}}}^{00}.$$E65

For a comoving medium we have ${\overline{\overline{T}}}^{00}={\overline{T}}^{00}={\rho }_0$ and in this simple case ${\lambda }_r^{-1}={\overline{r}}_{\lambda }$, thus

$$\frac{{\partial }_{\lambda }m\left(\lambda \right)}{{\overline{r}}^2\left(\lambda ,\tau \right){\overline{r}}_{\lambda }\left(\lambda ,\tau \right)}=\frac{4\mathit{\pi G}}{c^2}{\rho }_0.$$E66

This is exactly the same equation (mutatis mutandis) as the one valid for the L–T models, as, e.g., given in Eq. (6) of Ref. [1]. It is the general-relativistic form of Poisson’s equation.

The 11-equation is almost identical to the 00-equation and yields again either $0=0$, (or)

$$\frac{{\partial }_{\lambda }m}{{\overline{r}}^2{\overline{r}}_{\lambda }}=-\frac{{\overline{r}}_{\lambda }^2}{E^2}\frac{4\mathit{\pi G}}{c^4}{\overline{\overline{T}}}^{11}.$$E67

Since ${\overline{\overline{T}}}^{11}=-p_r{\overline{\overline{g}}}^{11}$ is a positive quantity or zero, this equation is not compatible with (66), unless we choose for this equation ${\partial }_{\lambda }m=0$. This shows that we are free to make this choice for each equation, since we obtained already the L-T solution, and this choice confirms this freedom. We conclude that the flux of the radial flow through any sphere is zero, confirming that spheres are comoving surfaces.

The 22-equation equals

$$\frac{1}{2r^2}{\partial }_r^2\mathrm{\Delta }+\frac{1}{r^3}{\partial }_r\mathrm{\Delta }+\frac{1}{r^2}\mathrm{\Lambda }=\frac{8\mathit{\pi G}}{c^4}{\overline{\overline{T}}}^{22}$$E68

Here ${\overline{\overline{T}}}^{22}=T^{22}=-p_t{\overline{\overline{g}}}^{22}$. Again, ${\overline{\overline{T}}}^{22}=0$ is the solution if ${\partial }_{\lambda }m=0$. If ${\partial }_{\lambda }m\ne 0$ we obtain

$$-\frac{1}{{\overline{r}\overline{r}}_{\lambda }^2}{\partial }_{\lambda }^{2}m=\frac{8\mathit{\pi G}}{c^4}{\overline{\overline{T}}}^{22}.$$E69

We know from Ref. [1] that regular solutions in the center behave like $m\left(\lambda \right)\sim {\lambda }^3$ for $\lambda \to 0$. Hence we conclude that a ${\overline{\overline{T}}}^{22}\ge 0$ is impossible, unless we allow for a singularity in the center.

The 33-equation adds no further information, since ${sin}^2\vartheta {\overline{\overline{R}}}^{33}={\overline{\overline{R}}}^{22}$, ${sin}^2\vartheta {\overline{\overline{g}}}^{33}={\overline{\overline{g}}}^{22}$ and ${sin}^2\vartheta {\overline{\overline{T}}}^{33}={\overline{\overline{T}}}^{22}$.

In this simple case, we have three independent equations, yielding two structurally different solutions: (1) empty space with central singularity, i.e., the Schwarzschild-$\mathrm{\Lambda }$ solution and (2) the L-T solution.

In order to make the L-T solution explicit, we still need to prescribe $0\le e\left(\lambda \right)\le \frac{1}{2}$ and $0\le m\left(\lambda \right)\le M=m\left(1\right)$. The latter inequality follows from the requirement that the outermost massive shell $\lambda =1$ must match the Schwarzschild-$\mathrm{\Lambda }$ solution at maximum expansion. The functions $e\left(\lambda \right)$ and $m\left(\lambda \right)$ must be such that no shell $\lambda <1$ has a maximum expansion that is larger than the outermost $\lambda =1$ shell. We also need to assign, for a given cosmic time $\tau$ and for every shell $\lambda$, a function $r\left(\lambda \right)$, together with the specification whether the shell is expanding or contracting. In the parlance of Ref. [1] this means the specification of a phase function.

Out of the great variety of models that come with these choices, there is a special class of models for which all shells attain their maximum extent at the same cosmic time. These shells will have in general different radial periods which will cause eventually shell collisions. The radial period of shell $\lambda$ is given by (24):

$$2T_r=2{\int }_0^{{\overline{r}}_0}\frac{\sqrt{r}\mathit{dr}}{\sqrt{2m\left(\lambda \right)-2e\left(\lambda \right)r+\mathrm{\Lambda }{r}^3/3}},$$E70

with ${\overline{r}}_0$ the smallest positive root of the radicand. Writing, as in Ref. [1],

$$\overline{r}=p\left(\lambda \right)\mathrm{cyc}\left(e,m,\mathrm{\Lambda },\tau \right)$$E71

we obtain

$$\frac{p^{3/2}}{\sqrt{2m}}{\int }_0^{w_0}\frac{\sqrt{\mathrm{cyc}}d\mathrm{cyc}}{\sqrt{1-\left(\mathit{ep}/m\right)\mathrm{cyc}+\left(\mathrm{\Lambda }/6\right)\left(p^3/m\right){\mathrm{cyc}}^3/3}},$$E72

with $w_0$ the smallest real positive root for cyc of the radicand. If that radial period is to be the same for all shells, which means that shells will not cross, the radicand should not depend on $\lambda$. Assuming $p\left(\lambda \right)=2\mathit{M\lambda }$, we find that $e\left(\lambda \right)=\frac{1}{2}{\lambda }^2$ and $m\left(\lambda \right)=M{\lambda }^3$. This is the spherical version of the Robertson–Walker metric for a closed universe.

Finally, we remark that we did obtain the L-T solution starting from the Schwarzschild-$\mathrm{\Lambda }$ solution (since $a=0$), ‘simply’ by choosing another coordinate system that is comoving. This shows that the L–T solution can be derived from the Schwarzschild-$\mathrm{\Lambda }$ solution by coordinate transformation, but that could only be accomplished because the equations of motion are separable.

7. Dusty models in the case $a\ne 0$

7.1 The general case

We now include a (pressureless) dusty medium with comoving proper-density ${\rho }_0$ in the travelator metric; the case $a=0$ has learned us that pressures are unlikely to be realistic within this paradigm. The stress-energy-momentum tensor, therefore, has the simplest nonzero form possible: ${\overline{\overline{T}}}^{00}={\rho }_0$ and all other components are zero. We note that, while in many cases, the Eq. (52) could also be taken in its covariant form, this is not the case; because the metric ${\overline{\overline{g}}}^{\mathit{\alpha \beta }}$ is not diagonal, it is imperative to use the contravariant form for the stress-energy-momentum tensor. All 10 equations of (52) remain obviously satisfied if ${\rho }_0=0$. We will now investigate which special form ${\rho }_0$ has to assume in order to satisfy

$${\overline{\overline{R}}}^{00}-\frac{1}{2}{\overline{\overline{R}}\overline{\overline{g}}}^{00}-\mathrm{\Lambda }{\overline{\overline{g}}}^{00}=\frac{8\mathit{\pi G}}{c^4}{\rho }_0$$E73

if ${\rho }_0\ne 0$.

With the aid of Appendix C we find

$$\begin{array}{l}\frac{a^2}{2{\overline{\rho }}^4}\left(K-{cos}^2\overline{\vartheta }\right){\partial }_{\overline{r}}^2{\mathrm{\Delta }}_r-\frac{\overline{r}}{{\overline{\rho }}^6}\left[{\overline{\rho }}^2+2a^2\left(K-{cos}^2\overline{\vartheta }\right)\right]{\partial }_{\overline{r}}{\mathrm{\Delta }}_r-12{\overline{\rho }}^4\left({\overline{r}}^2+a^{2}K\right){\partial }_{\overline{\vartheta }}^2{\mathrm{\Delta }}_{\vartheta }-\\ -\left\{\frac{a^2\left(K-{cos}^2\overline{\vartheta }\right)}{{\overline{\rho }}^4}\left[3\left({\overline{r}}^2+a^2\right)+a^2{sin}^2\overline{\vartheta }\right]+\frac{2a^2{sin}^2\overline{\vartheta }}{{\overline{\rho }}^2}+3\right\}\frac{cos\overline{\vartheta }}{2{\overline{\rho }}^2}\frac{{\partial }_{\overline{\vartheta }}{\mathrm{\Delta }}_{\vartheta }}{sin\overline{\vartheta }}-\\ -\frac{g_{00}}{{\overline{\rho }}^4}\left[\left({\overline{r}}^2+a^{2}K\right)\left(1-\frac{4{\overline{r}}^2}{{\overline{\rho }}^2}\right)+2{\overline{r}}^2\right]+{\mathrm{\Delta }}_{\vartheta }{\overline{\rho }}^2+\\ +\frac{a^2\left(g_{00}-{\mathrm{\Delta }}_{\vartheta }\right)}{{\overline{\rho }}^4}\left[\left(K-{cos}^2\overline{\vartheta }\right)\left(\frac{4a^2{cos}^2\overline{\vartheta }}{{\overline{\rho }}^2}-1\right)+2{cos}^2\overline{\vartheta }\right]-\mathrm{\Lambda }=\frac{8\mathit{\pi G}}{c^2}{\rho }_0.\end{array}$$

The right-hand side of this expression evaluates to 0 if $M$ and $a^2$ are constants. If, on the other hand, we consider the alternatives (54), (55), and (58), the additional terms that these dependencies entail are those featuring the derivatives of the $\mathrm{\Delta }$‘s:

$$\begin{array}{lll}\frac{8\pi G}{c^2}{\overline{\rho }}_0& =& \frac{A^2}{2{\rho }^4}\left(K-{\cos }^2\vartheta \right)\overline{{\partial }_r^2{\text{Δ}}_r}r{\rho }^6\left[{\rho }^2+2A^2\left(K-{\cos }^2\vartheta \right)\right]\overline{{\partial }_r{\text{Δ}}_r}12{\rho }^4\left(r^2+A^{2}K\right)\overline{{\partial }_{\vartheta }^2{\text{Δ}}_{\vartheta }}+\\ & & +\left\{\frac{A^2\left(1-K\right)}{{\rho }^4}\left[3\left(r^2+A^2\right)+A^2{\sin }^2\vartheta \right]-A^2{\sin }^2\vartheta {\rho }^4\left[5r^2+4A^2+A^2{\cos }^2\vartheta \right]-3\right\}\frac{\cos \vartheta }{2{\rho }^2}\frac{\overline{{\partial }_{\vartheta }{\text{Δ}}_{\vartheta }}}{\sin \vartheta },\end{array}$$

resulting in an expression for the comoving density ${\rho }_0$. It is interesting to note that the 00 equation is the only Einstein equation that does not feature any of the $r_E$, $r_K$, ${\vartheta }_E$, ${\vartheta }_K$, $w_E$ or $w_K$.

7.2 The case $\mathrm{\Lambda }=0$

We will now illustrate the analysis for the case $\mathrm{\Lambda }=0$. The expression for the comoving density simplifies greatly:

$$\frac{8\mathit{\pi G}}{c^2}{\overline{\rho }}_0=\frac{A^2}{2{\rho }^4}\left(K-{cos}^2\vartheta \right)\overline{{\partial }_r^2{\mathrm{\Delta }}_r}-\frac{r}{{\rho }^6}\left[{\rho }^2+2A^2\left(K-{cos}^2\vartheta \right)\right]\overline{{\partial }_r{\mathrm{\Delta }}_r}$$E74

In order to make this solution more explicit, we need to prescribe $m\left(\lambda ,\mu \right)$ and $A\left(\lambda ,\mu \right)$ on the one hand, and $E\left(\lambda ,\mu \right)$ and $K\left(\lambda ,\mu \right)$ on the other hand. It is convenient to replace $\left(E,K\right)$ by $\left(r_a,{\vartheta }_a\right)$, with now, of course, $r_a\left(\lambda ,\mu \right)$ and ${\vartheta }_a\left(\lambda ,\mu \right)$. The constants $E$ and $K$ are then given by (33). We will assume $0\le \lambda \le 1$ and , with no constituent matter ‘beyond’ the $\lambda =1$ rings. We recall that the comoving rings form a two-dimensional manifold, each ring being labeled by a pair $\left(\lambda ,\mu \right)$, or a pair $\left(E,K\right)$. From the case, $a=0$ we also learned that each shell there moves as if it were in a Schwarzschild-$\mathrm{\Lambda }$ metric with mass $m\left(\lambda \right)$, independent of time. Likewise, in this case, each ring moves as if it were in a KdS metric with parameters $m\left(\lambda ,\mu \right)$ and $A\left(\lambda ,\mu \right)$, again, independent of time.

Since $\mathrm{\Theta }\left(\vartheta \right)\ge 0$, every orbit is located in the region ${\vartheta }_a\le \vartheta \le \pi -{\vartheta }_a$, with the aperture ${\vartheta }_a$ given by $\mathrm{\Theta }=-E^2{sin}^2{\vartheta }_a+K-{cos}^2{\vartheta }_a=0$. Since $R\left(r\right)\ge 0$, every orbit is located between $r_a$ and ${\overline{r}}_p$, ${\overline{r}}_p\le r\le r_a$. There are no restrictions on $0\le {\vartheta }_a\le \pi /2$.

As an example, we now assume ${sin}^2{\vartheta }_a=\mu$. In order to avoid the KdS singularity at $r=0$ and $\vartheta =\pi /2$, we assume $A\left(0,\mu \right)=0$. The simplest choice seems to be

$$\begin{array}{ll}{A}^2\left(\lambda ,\mu \right)& =\left({\lambda }^2\mu \right)a^2\text{for}\lambda \to 0\\ & =a^2-a_b\left(\mu \right){\left(1-\lambda \right)}^{\alpha }\text{for}\lambda \to 1,\end{array}$$E75

with $a_b\left(\mu \right)$ a function and $\alpha$ a constant. The factor ${\lambda }^2$ ensures that $A^2\to 0$ for $\lambda \to 0$ sufficiently smoothly.

As for $m\left(\lambda ,\mu \right)$, we follow Ref. [1] and propose a monotonically increasing function such that

$$\begin{array}{ll}m\left(\lambda ,\mu \right)& =m_c{\lambda }^3\text{for}\lambda \to 0\\ & =M-m_b{\left(1-\lambda \right)}^{\beta }\text{for}\lambda \to 1,\end{array}$$E76

with , $m_b$ and $\beta$ real constants. We know from the $a=0$ case that the shell $E=0$ must reach the Schwarzschild-$\mathrm{\Lambda }$ horizon $M$, and all other shells must satisfy $r_a\le M$. Likewise here, in order to realize the match between the inner and the outer KdS solution, at least one ring with $E=0$ must have the apocenter radius $R_{+}=M+\sqrt{M^2-a^2}$. We will assume for simplicity that all rings $E=0$ do.

We are still left with the specification of $r_a\left(\lambda ,\mu \right)$. The apocenter radius $r_a$ plays the same role (but is not the same) as the quantity $p\left(r\right)$ in Ref. [1]. We already indicated that $r_a\left(1,\mu \right)=R_{+}$. For simplicity, we assume

$$\begin{array}{ll}{r}_a\left(\lambda ,\mu \right)& =r_{a,c}\lambda \text{for}\lambda \to 0\\ & =R_{+}-r_{a,b}{\left(1-\lambda \right)}^{\gamma }\text{for}\lambda \to 1,\end{array}$$E77

with $r_{a,c}$, $r_{a,b}$ and $\gamma$ constants. The functions ${\vartheta }_a\left(\mu \right)$ and $r_a\left(\lambda \right)$ now being known, the energy follows from (33). We find

$$\begin{array}{l}1-E^2\equiv 2e\to \frac{2m_c}{r_{a,c}}{\lambda }^2\text{for}\lambda \to 0\\ \to 1-\frac{m_b}{M}{\left(1-\lambda \right)}^{\beta }+\frac{r_{a,b}}{{\mathit{MR}}_{+}}\sqrt{M-a^2}{\left(1-\lambda \right)}^{\gamma }\text{for}\lambda \to 1,\end{array}$$E78

in agreement with Ref. [1] for $\lambda \to 0$.

In order to calculate $\rho \left(\tau ,\lambda ,\mu \right)$, we still have to choose initial conditions. This is no different from the case $a=0$, where we have seen that the classical cosmological models (in their spherical form) also require very special initial conditions in order to avoid collisions of shells. Out of the orbits discussed in Appendix D we can choose those that all attain their apocenter radius (i.e., maximum expansion) at the same time $\tau$. Numerical experiments show that it is indeed possible to choose initial conditions such that ${\rho }_0>0$ for subsequent times up to maximum expansion. This statement should be interpreted as no more than a proof of concept: i.e. a model for a rotating nonempty black hole or a rotating cosmological model with a travelator metric (35)–(42). A more detailed exploration of these models is beyond the scope of this paper.

8. Mass ejection or light emission

In Ref. [1] it has been shown that in general the $g_{11}$ coefficient in the metric (69) will prevent not comoving matter and light to leave the black hole. More precisely, when an orbit (matter or light) attains the shell for which $E\left(\lambda \right)=1-2e\left(\lambda \right)=0$, the coefficient $g_{11}$ will become in general singular, causing an outgoing orbit to turn back.⁸ It is important that the condition $E\left(\lambda \right)=0$ is independent of $\tau$, hence it is the property of one particular shell to act as a barrier. That shell is the outermost shell that carries some constituent matter, and it is the only shell that can reach the Schwarzschild radius from within. Only when, at the very same time that the orbit reaches that shell $E=0$, also (1) ${\partial }_{\lambda }\overline{r}\left(\lambda ,\tau \right)=0$ for that shell, which has been shown in Ref. [1] to be the condition that shells collide (in the continuum limit), and (2) that shell reaches the Schwarzschild radius, is it possible to create a ‘neck’, i.e., a nonzero $0/0$ in $g_{11}$, through which matter or light can escape. It was noted in Ref. [1] that in spherical symmetry, these conditions are very unlikely to be globally satisfied, but that along the polar axis of a rotating black hole, there might be a possibility due to the similarity of the Kerr metric and the Schwarzschild metric along the polar axis. But that statement was a conjecture.

Now, we are in the position to be more precise with the help of the travelator metric (35)–(42). A $\left(\lambda ,\mu \right)$ ring of constituent comoving matter orbits in a region bounded by 2 radii, ${\overline{r}}_p$ and $r_a$ determined by the condition $R=0$, and a cone ${\vartheta }_a\le \vartheta \le \pi -{\vartheta }_a$, with ${\vartheta }_a$ determined by the condition $\mathrm{\Theta }=0$. Every now and then, the ring will touch these boundaries, which act as impenetrable walls. At such an event, all coefficients of the orbital part of the travelator metric become singular, but, as already indicated, the singularities are removable locally. We note that at all other times, the comoving ring produces a regular metric.

Any particle, matter or light, that is not comoving will continuously visit different rings in $\left(\lambda ,\mu \right)$ space. It might visit a comoving ring that just happens to touch its boundary. Hence the particle will be forced to visit a ring for which $R>0$ of $\mathrm{\Theta }>0$, that is, it remains in the interior of the black hole since all comoving rings remain in the interior of the black hole.

The metric along the polar axis (47), on the other hand, is fundamentally different, since no singularities associated with $R=0$ or are present. That metric is formally identical to the L-T metric, and the mechanism as indicated above can take place. Hence, if the conditions are right, matter or light can leave the rotating black hole through the poles.

9. Concluding remark

The paradigm presented here can also be used to construct models for rotating dust balls, or rotating universes that are not the interior of a black hole. It suffices that all rings have $0<E\left(\lambda ,\mu \right)<1$ and judicious choices for $A^2\left(\lambda ,\mu \right)$ and $m\left(\lambda ,\mu \right)$ are made. Since the mechanism outlined above for outflow along the polar axis is independent of the presence of a horizon, it opens up the possibility of bipolar outflows in fast-rotating dust balls, such as the vicinity of very young stars.

A. The Ricci tensor and Rieman scalar in the KdS spacetime

We adopt here the definition $R_{\mathit{\alpha \beta }}={\mathrm{\Gamma }}_{\mathit{\alpha \beta },\rho }^{\rho }-{\mathrm{\Gamma }}_{\mathit{\alpha \rho },\beta }^{\rho }+{\mathrm{\Gamma }}_{\mathit{\sigma \rho }}^{\rho }{\mathrm{\Gamma }}_{\mathit{\alpha \beta }}^{\sigma }-{\mathrm{\Gamma }}_{\mathit{\sigma \beta }}^{\rho }{\mathrm{\Gamma }}_{\mathit{\alpha \rho }}^{\sigma }$. The tables that follow are to be read as follows: the quantity in the first column is the sum of terms consisting of the product of the second column with the third.

$R$	${\partial }_r^2{\mathrm{\Delta }}_r$	$\frac{1}{{\rho }^2}$
	${\partial }_{\vartheta }^2{\mathrm{\Delta }}_{\vartheta }$	$\frac{1}{{\rho }^2}$
	$\frac{{\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }}{sin\vartheta }$	$\frac{3cos\vartheta }{{\rho }^2}$
	${\mathrm{\Delta }}_{\vartheta }$	$-\frac{2}{{\rho }^2}$

$R^{00}$	${\partial }_r^2{g}_{00}\frac{1}{2{\rho }^2{\mathrm{\Delta }}_r}$	${\left(r^2+a^2\right)}^2$
	${\partial }_{\vartheta }^2{g}_{00}\frac{1}{2{\rho }^2{\mathrm{\Delta }}_{\vartheta }}$	$1$
	${\partial }_r{g}_{00}$	$-\frac{r}{{\rho }^2{sin}^2\vartheta }\frac{1}{{\mathrm{\Delta }}_r{\mathrm{\Delta }}_{\vartheta }}{g}_{33}$
	${\partial }_{\vartheta }{g}_{00}\frac{cot\vartheta }{2{\rho }^4}$	$\frac{2{\left(r^2+a^2\right)}^2}{{\mathrm{\Delta }}_r}-\frac{{{\rho }^{\prime }}^2}{{\mathrm{\Delta }}_{\vartheta }}$

B. The transformation matrix $D$

All $r$ and $\vartheta$ in this section should be read as $\overline{r}$ and $\overline{\vartheta }$.

B.1 The case $a\ne 0$

$$\begin{array}{ll}{D}_{00}& =\frac{E}{c}\\ D_{01}& =\frac{E}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\left[\sqrt{R}{\vartheta }_K+\sqrt{\mathrm{\Theta }}{r}_K\right]\\ D_{02}& =-\frac{E}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\left[\sqrt{R}{\vartheta }_E+\sqrt{\mathrm{\Theta }}{r}_E\right]\\ D_{03}& =\frac{E}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\left[\sqrt{R}\left({\vartheta }_E{w}_K-{\vartheta }_K{w}_E\right)+\sqrt{\mathrm{\Theta }}\left(r_E{w}_K-r_K{w}_E\right)\right]\end{array}$$E79

$$\begin{array}{l}{D}_{10}=\frac{1}{c}\sqrt{R}{\mathrm{\Delta }}_r\\ D_{11}=\frac{1}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\left[\frac{\sqrt{R}}{{\mathrm{\Delta }}_r}\left(\sqrt{R}{\vartheta }_K+\sqrt{\mathrm{\Theta }}{r}_K\right)+{\rho }^2{\vartheta }_K\right]\\ D_{12}=-\frac{1}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\left[\frac{\sqrt{R}}{{\mathrm{\Delta }}_r}\left(\sqrt{R}{\vartheta }_E+\sqrt{\mathrm{\Theta }}{r}_E\right)+{\rho }^2{\vartheta }_E\right]\\ D_{13}=\frac{\mathit{aE}\left(r^2+a^2\right)}{\sqrt{R}{\mathrm{\Delta }}_r}+\left[\frac{R}{{\mathrm{\Delta }}_r}+{\rho }^2\right]\frac{{\vartheta }_E{w}_K-{\vartheta }_K{w}_E}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}+\frac{\sqrt{R\mathrm{\Theta }}}{{\mathrm{\Delta }}_r}\frac{r_E{w}_K-r_K{w}_E}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\end{array}$$E80

$$\begin{array}{l}{D}_{20}=-\frac{1}{c}\sqrt{\mathrm{\Theta }}{\mathrm{\Delta }}_{\vartheta }\\ D_{21}=-\frac{1}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\left[\frac{\sqrt{\mathrm{\Theta }}}{{\mathrm{\Delta }}_{\vartheta }}\left(\sqrt{R}{\vartheta }_K+\sqrt{\mathrm{\Theta }}{r}_K\right)+{\rho }^2{r}_K\right]\\ D_{22}=\frac{1}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\left[\frac{\sqrt{\mathrm{\Theta }}}{{\mathrm{\Delta }}_{\vartheta }}\left(\sqrt{R}{\vartheta }_E+\sqrt{\mathrm{\Theta }}{r}_E\right)+{\rho }^2{r}_E\right]\\ D_{23}=\frac{\mathit{aE}}{\sqrt{\mathrm{\Theta }}{\mathrm{\Delta }}_{\vartheta }}-\left[\frac{\mathrm{\Theta }}{{\mathrm{\Delta }}_{\vartheta }}+{\rho }^2\right]\frac{r_E{w}_K-r_K{w}_E}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}-\frac{\sqrt{R\mathrm{\Theta }}}{{\mathrm{\Delta }}_{\vartheta }}\frac{{\vartheta }_E{w}_K-{\vartheta }_K{w}_E}{{\rho }^2\left(r_E{\vartheta }_K-{\vartheta }_E{r}_K\right)}\end{array}$$E81

$$D_{30}=D_{31}=D_{32}=0,D_{33}=1.$$E82

C. The contravariant Ricci tensor in the travelator spacetime

C.1 The case $a\ne 0$

All $r$ and $\vartheta$ in this section should be read as $\overline{r}$ and $\overline{\vartheta }$.

${\overline{R}}^{00}$	$\frac{1}{c^2}\frac{a^2}{2{\rho }^4}$	${\partial }_{\vartheta }^2{\mathrm{\Delta }}_{\vartheta }$	$-\left(K-{cos}^2\vartheta \right)$
		$\frac{{\partial }_{\vartheta }{\mathrm{\Delta }}_{\vartheta }}{sin\vartheta }\frac{cos\vartheta }{{\rho }^2}$	$-\left(K-{cos}^2\vartheta \right)\left[3\left(r^2+a^2\right)+a^2{sin}^2\vartheta \right]-2{\rho }^2{sin}^2\vartheta$
		$2\left(g_{00}-{\mathrm{\Delta }}_{\vartheta }\right)$	$\left(K-{cos}^2\vartheta \right)\left(\frac{4a^2}{{\rho }^2}{cos}^2\vartheta -1\right)+2{cos}^2\vartheta$
	$\frac{1}{c^2}\frac{1}{{\rho }^4}$	$\frac{1}{2}{\partial }_r^2{\mathrm{\Delta }}_r$	$r^2+a^{2}K$
		${\partial }_r{\mathrm{\Delta }}_r\frac{r}{{\rho }^2}$	$-r^2+a^2{cos}^2\vartheta -2a^{2}K$
		$g_{00}$	$\left(r^2+a^{2}K\right)\left(\frac{4r^2}{{\rho }^2}-1\right)-2r^2$

C.2 The case $a=0$

Taking account of

$$\begin{array}{ll}{\mathrm{\Delta }}_{\vartheta }& ={\vartheta }_K=1\\ R& =r^2\left(E^2{r}^2-{\mathrm{\Delta }}_r\right)=r^4\left(E^2{r}^2-\mathrm{\Delta }\right)\\ r_K& ={\vartheta }_E=w_E=w_K=\mathrm{\Theta }=0\\ g_{00}& =\mathrm{\Delta }\end{array}$$

we find

$${\overline{R}}^{00}=\frac{1}{c^2}\left(\frac{1}{2}{\partial }_r^2\mathrm{\Delta }+\frac{1}{r}{\partial }_r\mathrm{\Delta }\right)$$E83

D. The orbits in the case $\mathrm{\Lambda }=0$

D.1 The solution for $\overline{\vartheta }\left(r\right)$

In this appendix we first solve (15) explicitly. We denote

$$I_r\left(\overline{r}\right)={\int }_{\overline{r}}^{{\overline{r}}_0}\frac{\mathit{dr}}{\sqrt{R\left(r\right)}}$$E84

and we need to solve

$$I_r\left(\overline{r}\right)=I_{\vartheta }\left(\overline{\vartheta }\right)={\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{\mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}}=\frac{1}{a}{\int }_{{\overline{\vartheta }}_0}^{\overline{\vartheta }}\frac{\mathit{d\vartheta }}{\sqrt{K-E^2-\left(1-E^2\right){cos}^2\vartheta }}.$$E85

In this expression ${\overline{\vartheta }}_0$ is any attainable polar angle of the orbit. Be $0\le {\vartheta }_a\le \pi /2$ the root of the radicand:

$$cos{\vartheta }_a=k,k^2=\frac{K-E^2}{1-E^2}\le 1.$$E86

We now choose ${\overline{\vartheta }}_0={\vartheta }_a$, and thus $\left({\overline{r}}_0,{\vartheta }_a\right)$ is a point along the orbit where the tangential velocity in the meridional plane is zero. We consider first the branch for increasing $\vartheta$:

$$I_{\vartheta ,1}\left(\overline{\vartheta }\right)={\int }_{{\vartheta }_a}^{\overline{\vartheta }}\frac{\mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}}=\frac{1}{a}{\int }_{{\vartheta }_a}^{\overline{\vartheta }}\frac{\mathit{d\vartheta }}{\sqrt{K-E^2-\left(1-E^2\right){cos}^2\vartheta }},{\vartheta }_a\le \overline{\vartheta }\le \pi -{\vartheta }_a.$$E87

The change of variable

$$\nu =\frac{1}{k}cos\vartheta ,-1\le \nu \le 1$$E88

leads to the equation

$$a\sqrt{K-E^2}{I}_{\vartheta ,1}\left(\overline{\nu }\right)=k{\int }_{\overline{\nu }}^1\frac{\mathit{d\nu }}{\sqrt{1-{\nu }^2}\sqrt{1-k^2{\nu }^2}}.$$E89

This can be rewritten as

$$\frac{{\mathrm{sn}}^{-1}}{}\left(\overline{\nu }\right)={\int }_0^{\overline{\nu }}\mathit{d\nu }\sqrt{1-{\nu }^2}\sqrt{1-k^2{\nu }^2}=\mathit{K}\left(k\right)-\frac{a}{k}\sqrt{K-E^2}{I}_{\vartheta ,1}\left(\overline{\nu }\right),$$E90

with $\mathit{K}\left(k\right)$ the complete elliptic integral of the first kind. Inversion produces

$$cos\overline{\vartheta }=k\mathrm{sn}\left(\mathit{K}\left(k\right)-\frac{a}{k}\sqrt{K-E^2}{I}_r\left(\overline{r}\right)\right).$$E91

Clearly, the angular period $2\mathrm{\Delta }{I}_{\vartheta }=2{\int }_{{\vartheta }_a}^{\pi -{\vartheta }_a}$, defined as the increment of the argument of the elliptic sine after which sn repeats itself, is $4\mathit{K}\left(k\right)$. When $I_r\left(\overline{r}\right)\to 0$ we obtain $cos\overline{\vartheta }=k\mathrm{sn}\left(\mathit{K}\left(k\right)\right)=k=cos{\vartheta }_a$.

As for the second branch of the integral in $\vartheta$, for decreasing $\vartheta$, we write

$$I_{\vartheta ,2}\left(\overline{\vartheta }\right)={\int }_{{\vartheta }_a}^{\overline{\vartheta }}\frac{\mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}}={\int }_{{\vartheta }_a}^{\pi -{\vartheta }_a}\frac{\mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}}+{\int }_{\pi -{\vartheta }_a}^{\overline{\vartheta }}\frac{\mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}},$$E92

which can be written as

$$\frac{a}{k}\sqrt{K-E^2}{I}_{\vartheta ,2}\left(\overline{\nu }\right)=3\mathit{K}\left(k\right)+{\int }_0^{\overline{\nu }}\frac{\mathit{d\nu }}{\sqrt{1-{\nu }^2}\sqrt{1-k^2{\nu }^2}}.$$E93

Inversion yields

$$cos\overline{\vartheta }=k\mathrm{sn}\left(\frac{a}{k}\sqrt{K-E^2}{I}_r\left(\overline{r}\right)-3\mathit{K}\left(k\right)\right).$$E94

As for $I_r\left(\overline{r}\right)$, we again need to consider two cases. When $\overline{r}$ decreases, we write ${\int }_{\overline{r}}^{{\overline{r}}_0}={\int }_{\overline{r}}^{r_a}-{\int }_{{\overline{r}}_0}^{r_a}$. When $\overline{r}$ subsequently increases, we write ${\int }_{\overline{r}}^{{\overline{r}}_0}={\int }_{{\overline{r}}_p}^{{\overline{r}}_0}+{\int }_{{\overline{r}}_p}^{\overline{r}}=\mathrm{\Delta }{I}_r+{\int }_{{\overline{r}}_p}^{\overline{r}}$, and so on, such that $I_r\left(\overline{r}\right)$ always increases. Hence, we are left with the integrals ${\int }_{\overline{r}}^{r_a}$ and ${\int }_{{\overline{r}}_p}^{\overline{r}}$.

The radicand $R$ has four roots. One of them is $r_0=r_a$. In case the other three are real we choose the largest of them as $r_1={\overline{r}}_p$, and the others as $r_2$ and $r_3$. In case two of them are complex conjugate, we choose the other real root as $r_1={\overline{r}}_p$, and $r_2$ and $r_3$ are the complex conjugate ones. The first integral we consider reads

$$I_{r,1}\left(\overline{r}\right)={\int }_{\overline{r}}^{r_a}\frac{\mathit{dr}}{\sqrt{R\left(r\right)}}=\frac{1}{\sqrt{1-E^2}}{\int }_{\overline{r}}^{r_0}\frac{\mathit{dr}}{\sqrt{\left(r_0-r\right)\left(r-r_1\right)\left(r-r_2\right)\left(r-r_3\right)}}.$$E95

The transformation $t=\left(r-\overline{r}\right)/\left(r_0-r\right)$ brings this integral in the form

$$I_{r,1}\left(\overline{r}\right)=\frac{2}{\sqrt{1-E^2}}\frac{\sqrt{r_0-\overline{r}}}{\sqrt{\left(r_0-r_1\right)\left(r_0-r_2\right)\left(r_0-r_3\right)}}{R}_F\left(\frac{\overline{r}-r_1}{r_0-r_1},\frac{\overline{r}-r_2}{r_0-r_2},\frac{\overline{r}-r_3}{r_0-r_3}\right)$$E96

with $R_F$ a Carlson elliptic integral.

The second integral reads

$$I_{r,2}\left(\overline{r}\right)={\int }_{{\overline{r}}_p}^{\overline{r}}\frac{\mathit{dr}}{\sqrt{R\left(r\right)}}=\frac{1}{\sqrt{1-E^2}}{\int }_{r_1}^{\overline{r}}\frac{\mathit{dr}}{\sqrt{\left(r_0-r\right)\left(r-r_1\right)\left(r-r_2\right)\left(r-r_3\right)}}.$$E97

The transformation $t=\left(\overline{r}-r\right)/\left(r-r_1\right)$ brings this integral in the form

$$I_{r,2}\left(\overline{r}\right)=\frac{2}{\sqrt{1-E^2}}\frac{\sqrt{\overline{r}-r_1}}{\sqrt{\left(r_0-r_1\right)\left(r_1-r_2\right)\left(r_1-r_3\right)}}{R}_F\left(\frac{r_0-\overline{r}}{r_0-r_1},\frac{\overline{r}-r_2}{r_1-r_2},\frac{\overline{r}-r_3}{r_1-r_3}\right).$$E98

If we substitute $\overline{r}=r_1$ in $I_{r,1}\left(\overline{r}\right)$ and $\overline{r}=r_0$ in $I_{r,2}\left(\overline{r}\right)$, we should get identical expressions, amounting to $\mathrm{\Delta }{I}_r$. Inspection of these expressions shows that the following equality should hold:

$$R_F\left(0,y,z\right)=\frac{1}{\sqrt{y}}\frac{1}{\sqrt{z}}{R}_F\left(0,\frac{1}{y},\frac{1}{z}\right).$$E99

This can be proven by writing down the integrals, and changing one of the integral variables, say $t$, to $1/t$. Clearly $I_{r,2}\left(\overline{r}\right)=\mathrm{\Delta }{I}_r-I_{r,1}\left(\overline{r}\right)$.

We can now summarize the construction of $\overline{\vartheta }\left(r\right)$. The position on the orbit is defined by (1) the independent variable $r$, ${\overline{r}}_p\le r\le r_a$, and (2) the number of full radial half cycles $m$ (i.e., from $r_a$ to ${\overline{r}}_p$ and vice versa) that the orbit has already gone through since its ‘start’ at $\left(r_0,{\vartheta }_a\right)$. Then

$$\begin{array}{ll}{I}_r\left(\overline{r}\right)& =m\mathrm{\Delta }{I}_r+I_{r,1}\left(\overline{r}\right),m\text{even}\\ & =m\mathrm{\Delta }{I}_r+I_{r,2}\left(\overline{r}\right),m\mathrm{odd}.\end{array}$$E100

To this must be added either $I_{r,1}\left(r_0\right)$ or $I_{r,2}\left(r_0\right)$, depending on the case. The angle $\overline{\vartheta }$ is determined by $n=int\left(I_r\left(\overline{r}\right)/\mathrm{\Delta }{I}_{\vartheta }\right)$ and

$$\begin{array}{l}cos\overline{\vartheta }\left(\overline{r}\right)=k\overline{\nu }\left(\overline{r}\right)=k\mathrm{sn}\left(\mathit{K}\left(k\right)-\frac{a}{k}\sqrt{K-E^2}\left[I_r\left(\overline{r}\right)-n\mathrm{\Delta }{I}_{\vartheta }\right]\right),n\text{even}\\ =k\mathrm{sn}\left(\frac{a}{k}\sqrt{K-E^2}\left[I_r\left(\overline{r}\right)-\left(n-1\right)\mathrm{\Delta }{I}_{\vartheta }\right]-3\mathit{K}\left(k\right)\right),n\mathrm{odd}.\end{array}$$E101

A little remark may be added at this point. In the travelator metric, we can construct the travelator with orbits that start off at large apocenter radii $r_a$, and wait for a sufficiently long time until the orbits arrive ‘in the vicinity’ of the black hole. From (4.5) we find in that case

$$E^2\to 1-\frac{2M}{r_a},K\to 1-\frac{2M}{r_a}{sin}^2{\vartheta }_a\text{and}\frac{K-E^2}{1-E^2}\to {cos}^2{\vartheta }_a.$$E102

Hence $\mathrm{\Theta }\to 0$, and the orbits tend toward straight lines of constant $\vartheta ={\vartheta }_a$ in the meridional plane.

D.2 The solution for $\overline{r}\left(\tau \right)$

We next turn to the integrals (3.10) in the expression for $\tau$. Using integral tables [8], we find

$$\begin{array}{l}-2{\tau }_{r,1}\left(\overline{r}\right)=-2{\int }_{\overline{r}}^{r_a}\frac{r^2\mathit{dr}}{\sqrt{R}}=\left(r_0+r_1+r_2+r_3\right)I_3^{\prime }\left(\overline{r}\right)+\left(r_3-r_1\right)\left(r_3-r_2\right)I_2\left(\overline{r}\right)-\\ -\left[2r_0^2-\left(r_1-r_0\right)\left(r_2-r_0\right)\right]I_1\left(\overline{r}\right)-2\frac{\sqrt{\left(r_0-\overline{r}\right)\left(\overline{r}-r_1\right)\left(\overline{r}-r_2\right)}}{\sqrt{\overline{r}-r_3}},\end{array}$$E103

with

$$\begin{array}{l}{I}_1\left(\overline{r}\right)=2\frac{\sqrt{r_0-\overline{r}}}{\sqrt{\left(r_0-r_1\right)\left(r_0-r_2\right)\left(r_0-r_3\right)}}{R}_F\left(\frac{\overline{r}-r_1}{r_0-r_1},\frac{\overline{r}-r_2}{r_0-r_2},\frac{\overline{r}-r_3}{r_0-r_3}\right)\\ I_2\left(\overline{r}\right)=\frac{2}{3}\frac{{\left(r_0-\overline{r}\right)}^{3/2}}{\sqrt{\left(r_0-r_1\right)\left(r_0-r_2\right)}{\left(r_0-r_3\right)}^{3/2}}{R}_D\left(\frac{\overline{r}-r_1}{r_0-r_1},\frac{\overline{r}-r_2}{r_0-r_2},\frac{\overline{r}-r_3}{r_0-r_3}\right)\\ I_3^{\prime }\left(\overline{r}\right)=\frac{2}{3}\frac{{\left(r_0-\overline{r}\right)}^{3/2}}{\sqrt{\left(r_0-r_1\right)\left(r_0-r_2\right)\left(r_0-r_3\right)}}{R}_J\left(\frac{\overline{r}-r_1}{r_0-r_1},\frac{\overline{r}-r_2}{r_0-r_2},\frac{\overline{r}-r_3}{r_0-r_3},1\right).\end{array}$$E104

After $m$ radial half cycles we obtain

$$\begin{array}{ll}{\tau }_r\left(\overline{r}\right)& =m{\tau }_r\left({\overline{r}}_p\right)+{\tau }_{r,1}\left(\overline{r}\right),m\text{even}\\ & =\left(m+1\right){\tau }_r\left({\overline{r}}_p\right)-{\tau }_r\left(\overline{r}\right)\equiv m{\tau }_r\left({\overline{r}}_p\right)+{\tau }_{r,2}\left(\overline{r}\right),m\mathrm{odd}.\end{array}$$E105

To this must be added either ${\tau }_{r,1}\left(r_0\right)$ or ${\tau }_{r,2}\left(r_0\right)$, depending on the case.

The integral in $\vartheta$ (reads)

$${\tau }_{\vartheta }\left(\overline{\vartheta }\right)=a^2{\int }_{{\vartheta }_a}^{\overline{\vartheta }}\frac{{cos}^2\vartheta \mathit{d\vartheta }}{\sqrt{\mathrm{\Theta }\left(\vartheta \right)}}=\frac{a}{\sqrt{K-E^2}}{k}^3{\int }_{\overline{\nu }}^1\frac{{\nu }^2\mathit{d\nu }}{\sqrt{1-{\nu }^2}\sqrt{1-k^2{\nu }^2}}.$$E106

When ${\vartheta }_a\le \overline{\vartheta }\le \pi -{\vartheta }_a$ we obtain

$${\tau }_{\vartheta ,1}\left(\overline{\nu }\right)=\frac{\mathit{ak}}{\sqrt{K-E^2}}\left[\mathit{K}\left(k\right)-\mathit{E}\left(k\right)-F\left(arcsin\nu ,k\right)+E\left(arcsin\nu ,k\right)\right],$$E107

or

$${\tau }_{\vartheta ,1}\left(\overline{\nu }\right)=\frac{a}{\sqrt{K-E^2}}\frac{k^3}{3}\left[R_D\left(0,1-k^2,1\right)-{\overline{\nu }}^3{R}_D\left(1-{\overline{\nu }}^2,1-k^2{\overline{\nu }}^2,1\right)\right],$$E108

while for the second branch, for decreasing $\overline{\vartheta }$, we obtain

$${\tau }_{\vartheta ,2}\left(\overline{\nu }\right)=\frac{\mathit{ak}}{\sqrt{K-E^2}}\left[3\left[\mathit{K}\left(k\right)-\mathit{E}\left(k\right)\right]+F(arcsin\nu ,k)-E(arcsin\nu ,k)\right],$$E109

or

$${\tau }_{\vartheta ,2}\left(\overline{\nu }\right)=\frac{a}{\sqrt{K-E^2}}{k}^3\left[R_D\left(0,1-k^2,1\right)+\frac{{\overline{\nu }}^3}{3}{R}_D\left(1-{\overline{\nu }}^2,1-k^2{\overline{\nu }}^2,1\right)\right].$$E110

We recall that the angle $\overline{\vartheta }$ is determined by (D18), and thus

$$\begin{array}{l}{\tau }_{\vartheta }\left(\overline{\nu }\right)=n{\tau }_{\vartheta ,1}\left(-1\right)+{\tau }_{\vartheta ,1}\left(\overline{\nu }\right),n\text{even}\\ {\tau }_{\vartheta }\left(\overline{\nu }\right)=\left(n-1\right){\tau }_{\vartheta ,1}\left(-1\right)+{\tau }_{\vartheta ,2}\left(\overline{\nu }\right),n\mathrm{odd}.\end{array}$$E111

Finally,

$$c\left[\tau \left(\overline{r},\overline{\vartheta }\right)-{\tau }_0\right]={\tau }_r\left(\overline{r}\right)+{\tau }_{\vartheta }\left(\overline{\nu }\right).$$E112

Since $\overline{\nu }\left(\overline{r}\right)$ according to (D18), this expression can be inverted to yield $\overline{r}\left(\tau \right)$.