Charged black-bounce spacetimes

Our first task consists in showing that our spacetime is indeed everywhere regular. Conveniently, given that our spacetime is static, examination of the Kretschmann scalar $K=R^{\mu\nu\rho\sigma}\,R_{\mu\nu\rho\sigma}$ will be sufficient to accomplish this task [4].

Indeed, a simple computation shows that the Kretschmann scalar is quartic in $Q$ and given by

	$\displaystyle K$	$\displaystyle=\frac{4}{(r^{2}+\ell^{2})^{6}}\Bigg{\{}\left(3\ell^{4}-10\ell^{2}r^{2}+14r^{4}\right)Q^{4}$		(2.5)
		$\displaystyle+\sqrt{r^{2}+\ell^{2}}\left[2\ell^{2}(2\ell^{2}-3r^{2})\sqrt{r^{2}+\ell^{2}}-2m(5\ell^{4}-11\ell^{2}r^{2}+12r^{4})\right]Q^{2}$
		$\displaystyle+3\ell^{8}+(9m^{2}+6r^{2})\ell^{6}+3r^{2}(r^{2}-m^{2})\ell^{4}-8m\ell^{2}(\ell^{4}-r^{4})\sqrt{r^{2}+\ell^{2}}+12m^{2}r^{6}\Bigg{\}}\ .$

In the limit as $r\rightarrow 0$ we have the finite result

$$\lim_{r\rightarrow 0}K=\frac{12}{\ell^{8}}\,Q^{4}+\frac{8(2\ell-5m)}{\ell^{7}}\,Q^{2}+\frac{4\left(3\ell^{2}-8m\ell+9m^{2}\right)}{\ell^{6}}\ .$$

(2.6)

So in this manifestly static situation we are then guaranteed that all of the orthonormal curvature tensor components are automatically finite [4] (nonetheless for completeness we provide in Appendix A expressions for other curvature invariants). We may then conclude that the geometry is indeed globally regular.

In order to complete our investigation on curvature, we display here the explicit forms for the various non-zero curvature components. This is mostly conveniently done in the above introduced orthonormal basis. The orthonormal components of the Riemann curvature tensor are given by

	$\displaystyle R^{\hat{t}\hat{r}}{}_{\hat{t}\hat{r}}$	$\displaystyle=\frac{\ell^{2}-3r^{2}}{(r^{2}+\ell^{2})^{3}}Q^{2}-\frac{m(\ell^{2}-2r^{2})}{(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.7a)
	$\displaystyle R^{\hat{t}\hat{\theta}}{}_{\hat{t}\hat{\theta}}=R^{\hat{t}\hat{\phi}}{}_{\hat{t}\hat{\phi}}$	$\displaystyle=\frac{r^{2}}{(r^{2}+\ell^{2})^{3}}Q^{2}-\frac{mr^{2}}{(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.7b)
	$\displaystyle R^{\hat{r}\hat{\theta}}{}_{\hat{r}\hat{\theta}}=R^{\hat{r}\hat{\phi}}{}_{\hat{r}\hat{\phi}}$	$\displaystyle=\frac{r^{2}-\ell^{2}}{(r^{2}+\ell^{2})^{3}}Q^{2}+\frac{m(2\ell^{2}-r^{2})-\ell^{2}\sqrt{r^{2}+\ell^{2}}}{(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.7c)
	$\displaystyle R^{\hat{\theta}\hat{\phi}}{}_{\hat{\theta}\hat{\phi}}$	$\displaystyle=-\frac{r^{2}}{(r^{2}+\ell^{2})^{3}}Q^{2}+\frac{2mr^{2}+\ell^{2}\sqrt{r^{2}+\ell^{2}}}{(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.7d)

and in the limit as $r\rightarrow 0$ we find

$$R^{\hat{t}\hat{r}}{}_{\hat{t}\hat{r}}\rightarrow\frac{Q^{2}-m\ell}{\ell^{4}}\ ,\quad R^{\hat{t}\hat{\theta}}{}_{\hat{t}\hat{\theta}}\rightarrow 0\ ,\quad R^{\hat{r}\hat{\theta}}{}_{\hat{r}\hat{\theta}}\rightarrow-\frac{Q^{2}+\ell^{2}-2m\ell}{\ell^{4}}\ ,\quad R^{\hat{\theta}\hat{\phi}}{}_{\hat{\theta}\hat{\phi}}\rightarrow\frac{1}{\ell^{2}}\ .$$

(2.8)

For the orthonormal components of the Ricci tensor we find

	$\displaystyle R^{\hat{t}}{}_{\hat{t}}$	$\displaystyle=\frac{\ell^{2}-r^{2}}{(r^{2}+\ell^{2})^{3}}Q^{2}-\frac{m\ell^{2}}{(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.9a)
	$\displaystyle R^{\hat{r}}{}_{\hat{r}}$	$\displaystyle=-\frac{Q^{2}}{(r^{2}+\ell^{2})^{2}}+\frac{\ell^{2}(3m-2\sqrt{r^{2}+\ell^{2}})}{(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.9b)
	$\displaystyle R^{\hat{\theta}}{}_{\hat{\theta}}$	$\displaystyle=R^{\hat{\phi}}{}_{\hat{\phi}}=\frac{r^{2}-\ell^{2}}{(r^{2}+\ell^{2})^{3}}Q^{2}+\frac{2m\ell^{2}}{(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.9c)

and in the limit as $r\rightarrow 0$

$$R^{\hat{t}}{}_{\hat{t}}\rightarrow\frac{Q^{2}-m\ell}{\ell^{4}}\ ;\quad R^{\hat{r}}{}_{\hat{r}}\rightarrow\frac{3m\ell-2\ell^{2}-Q^{2}}{\ell^{4}}\ ;\quad R^{\hat{\theta}}{}_{\hat{\theta}}=R^{\hat{\phi}}{}_{\hat{\phi}}\rightarrow\frac{2m\ell-Q^{2}}{\ell^{4}}\ .$$

(2.10)

Finally, for the Weyl tensor we obtain

	$\displaystyle C^{\hat{t}\hat{r}}{}_{\hat{t}\hat{r}}$	$\displaystyle=-2C^{\hat{t}\hat{\theta}}{}_{\hat{t}\hat{\theta}}=-2C^{\hat{t}\hat{\phi}}{}_{\hat{t}\hat{\phi}}=-2C^{\hat{r}\hat{\theta}}{}_{\hat{r}\hat{\theta}}=-2C^{\hat{r}\hat{\phi}}{}_{\hat{r}\hat{\phi}}=C^{\hat{\theta}\hat{\phi}}{}_{\hat{\theta}\hat{\phi}}$
		$\displaystyle=\frac{2(\ell^{2}-3r^{2})}{3(r^{2}+\ell^{2})^{3}}Q^{2}+\frac{2\ell^{2}\sqrt{r^{2}+\ell^{2}}-3m(\ell^{2}-2r^{2})}{3(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.11)

and in the limit as $r\rightarrow 0$

$$C^{\hat{t}\hat{r}}{}_{\hat{t}\hat{r}}\rightarrow\frac{2Q^{2}+2\ell^{2}-3m\ell}{3\ell^{4}}\ .$$

(2.12)

Now that the regularity of spacetime is out of question, we shall proceed by discussing its geometrical properties, i.e. horizons and characteristic orbits.

In view of the diagonal metric environment, horizon locations are characterised by

$\displaystyle g_{tt}$

$\displaystyle=0\quad\Longrightarrow\quad r_{H}=S_{1}\sqrt{\left(m+S_{2}\sqrt{m^{2}-Q^{2}}\right)^{2}-\ell^{2}}\ .$

(2.13)

Here $S_{1},S_{2}=\pm 1$, and choice of sign for $S_{1}$ dictates which universe we are in, whilst the choice of sign on $S_{2}$ corresponds to an outer/inner horizon respectively. For horizons to exist we need both $|Q|\leq m$ and $\ell\leq m\pm\sqrt{m^{2}-Q^{2}}$. The case $|Q|=m$ while $\ell\leq m$ leads to extremal horizons at $r_{H}=\pm\sqrt{m^{2}-\ell^{2}}$. When $|Q|>m$ there are no horizons, and the geometry is that of a traversable wormhole. Finally when $|Q|\leq m$ but $\ell$ is large enough, $\ell>m\pm\sqrt{m^{2}-Q^{2}}$, first the inner horizons vanish and then the outer horizons vanish.

The structure of the maximally-extended spacetime can be visualised with the aid of the Penrose diagrams in figures 1 and 2 for the two qualitatively different regular black hole cases. Both diagrams will be relevant for the rotating generalisation of the next section, too; they are analogous to those of [7] (both) and [1] (the first one).

It is straightforward to calculate the surface gravity at the event horizon in our universe for the black-bounce–Reissner–Nordström spacetime. Keeping in mind that we are still working in curvature coordinates, the surface gravity $\kappa_{H}$ reduces to [42]

$$\kappa_{H}=\lim_{r\rightarrow r_{H}}\frac{1}{2}\frac{\partial_{r}g_{tt}}{\sqrt{-g_{tt}g_{rr}}}\ .$$

(2.14)

For our metric we have $g_{tt}=-1/g_{rr}$ so this simplifies to

$\displaystyle\kappa_{H}=\left.\frac{1}{2}\partial_{r}g_{tt}\right|_{r_{H}}=\frac{\sqrt{(m+\sqrt{m^{2}-Q^{2}})^{2}-\ell^{2}}\;\sqrt{m^{2}-Q^{2}}}{(m+\sqrt{m^{2}-Q^{2}})^{3}}=\kappa_{H}^{\text{RN}}\sqrt{\frac{r_{H}^{2}}{r_{H}^{2}+\ell^{2}}}\ ,$

(2.15)

where $\kappa_{H}^{\text{RN}}$ is the surface gravity of a Reissner–Nordström black hole of the same mass and charge, and as usual the associated Hawking temperature is given by $k_{B}T_{H}=\frac{\hbar}{2\pi}\kappa_{H}$ [42]. (This gives the usual Reissner–Nordström result as $\ell\to 0$.)

Let us briefly examine the coordinate locations of the notable orbits for our candidate spacetime, the innermost stable circular orbit (ISCO) and photon sphere [43, 44, 45]. Firstly we define the object

$$\epsilon=\begin{cases}-1&\quad\mbox{massive particle, \emph{i.e.}\ timelike worldline}\\ 0&\quad\mbox{massless particle, \emph{i.e.}\ null worldline}.\end{cases}$$

(2.16)

Considering the affinely parameterised tangent vector to the worldline of a massive or massless particle, and fixing $\theta=\pi/2$ in view of spherical symmetry, we obtain the reduced equatorial problem

$$\frac{{\mathrm{d}}s^{2}}{{\mathrm{d}}\lambda^{2}}=-g_{tt}\left(\frac{{\mathrm{d}}t}{{\mathrm{d}}\lambda}\right)^{2}+g_{rr}\left(\frac{{\mathrm{d}}r}{{\mathrm{d}}\lambda}\right)^{2}+(r^{2}+\ell^{2})\left(\frac{{\mathrm{d}}\phi}{{\mathrm{d}}\lambda}\right)^{2}=\epsilon\ .$$

(2.17)

The Killing symmetries yield the following expressions for the conserved energy $E$, and angular momentum per unit mass $L$

$$\left(1-\frac{2m}{\sqrt{r^{2}+\ell^{2}}}+\frac{Q^{2}}{r^{2}+\ell^{2}}\right)\left(\frac{{\mathrm{d}}t}{{\mathrm{d}}\lambda}\right)=E\ ;\quad(r^{2}+\ell^{2})\left(\frac{{\mathrm{d}}\phi}{{\mathrm{d}}\lambda}\right)=L\ ,$$

(2.18)

yielding the “effective potential” for geodesic orbits

$$V_{\epsilon}(r)=\left(1-\frac{2m}{\sqrt{r^{2}+\ell^{2}}}+\frac{Q^{2}}{r^{2}+\ell^{2}}\right)\left[-\epsilon+\frac{L^{2}}{r^{2}+\ell^{2}}\right]\ .$$

(2.19)

Starting from a given geometry, the discussion of the associated stress-energy tensor necessarily requires us to fix the geometrodynamics. In the following we shall assume that this is everywhere described by general relativity. While this might seem a crude assumption for geometries associated to a possible regularisation of singularities by quantum gravity, we do expect it — once the regular spacetime settles down in an equilibrium state after the collapse — to be a good approximation everywhere, if the final regularisation scale $\ell$ is much larger than the Planck scale, or at least sufficiently far away from the core region otherwise.

Given the above assumption, the determination of the stress-energy tensor associated to our spacetime is easily accomplished by computing the non-zero components of the Einstein tensor. This leads to the following decomposition for the total stress-energy tensor $T^{\hat{\mu}}{}_{\hat{\nu}}$ valid outside the outer horizon (and inside the inner horizon):

$$\frac{1}{8\pi}\,G^{\hat{\mu}}{}_{\hat{\nu}}=T^{\hat{\mu}}{}_{\hat{\nu}}=\left[T_{\text{bb}}\right]^{\hat{\mu}}{}_{\hat{\nu}}+\left[T_{Q}\right]^{\hat{\mu}}{}_{\hat{\nu}}=\operatorname{diag}\left(-\varepsilon,p_{r},p_{t},p_{t}\right)\ .$$

(2.27)

In contrast, between the inner and outer horizons we have:

$$\frac{1}{8\pi}\,G^{\hat{\mu}}{}_{\hat{\nu}}=T^{\hat{\mu}}{}_{\hat{\nu}}=\left[T_{\text{bb}}\right]^{\hat{\mu}}{}_{\hat{\nu}}+\left[T_{Q}\right]^{\hat{\mu}}{}_{\hat{\nu}}=\operatorname{diag}\left(p_{r},-\varepsilon,p_{t},p_{t}\right)\ .$$

(2.28)

Here $\left[T_{\text{bb}}\right]^{\hat{\mu}}{}_{\hat{\nu}}$ is the stress-energy tensor for the original electrically neutral black-bounce spacetime from ref. [1], and $\left[T_{Q}\right]^{\hat{\mu}}{}_{\hat{\nu}}$ is the charge-dependent contribution to the stress-energy. Examination of the non-zero components of the Einstein tensor, outside the outer horizon (and inside the inner horizon), yields the following:

	$\displaystyle\varepsilon$	$\displaystyle=\frac{Q^{2}(r^{2}-2\ell^{2})}{8\pi(r^{2}+\ell^{2})^{3}}+\frac{\ell^{2}\left(4m-\sqrt{r^{2}+\ell^{2}}\right)}{8\pi(r^{2}+\ell^{2})^{5/2}}\ ,$		(2.29a)
	$\displaystyle-p_{r}$	$\displaystyle=\frac{Q^{2}r^{2}}{8\pi(r^{2}+\ell^{2})^{3}}+\frac{\ell^{2}}{8\pi(r^{2}+\ell^{2})^{2}}\ ,$		(2.29b)
	$\displaystyle p_{t}$	$\displaystyle=\frac{Q^{2}r^{2}}{8\pi(r^{2}+\ell^{2})^{3}}+\frac{\ell^{2}\left(\sqrt{r^{2}+\ell^{2}}-m\right)}{8\pi(r^{2}+\ell^{2})^{5/2}}\ .$		(2.29c)

For the radial null energy condition (NEC) [16, 17, 18, 19, 20, 24, 23, 22, 21, 25, 26], outside (inside) the outer (inner) horizon, we have:

$\displaystyle\varepsilon+p_{r}=-\frac{\ell^{2}\left[r^{2}+\ell^{2}-2m\sqrt{r^{2}+\ell^{2}}+Q^{2}\right]}{4\pi(r^{2}+\ell^{2})^{3}}=-\frac{\ell^{2}f(r)}{4\pi(r^{2}+\ell^{2})^{2}}\ .$

(2.30)

It is then clear that outside the outer horizon (or inside the inner one), where $f(r)>0$, we will have violation of the radial NEC. At the horizons, where $f(r)=0$, we always have $\left.\left(\varepsilon+p_{r}\right)\right|_{H}=0$. (This on-horizon marginal satisfaction of the NEC is a quite generic phenomenon [46, 42, 47, 48].)

We can trivially extract the explicit form for $\left[T_{Q}\right]^{\hat{\mu}}{}_{\hat{\nu}}$:

	$\displaystyle\left[T_{Q}\right]^{\hat{\mu}}{}_{\hat{\nu}}$	$\displaystyle=\frac{Q^{2}r^{2}}{8\pi(r^{2}+\ell^{2})^{3}}\operatorname{diag}\left(\frac{2\ell^{2}}{r^{2}}-1,-1,1,1\right)$
		$\displaystyle=\frac{Q^{2}r^{2}}{8\pi(r^{2}+\ell^{2})^{3}}\left[\operatorname{diag}\left(-1,-1,1,1\right)+\operatorname{diag}\left(\frac{2\ell^{2}}{r^{2}},0,0,0\right)\right]\ .$		(2.31)

In the current situation, the first term above can be interpreted as the usual Maxwell stress-energy tensor [49]

$$\left[T_{\text{Maxwell}}\right]^{\hat{\mu}}{}_{\hat{\nu}}=\frac{1}{4\pi}\left[-F^{\hat{\mu}}{}_{\hat{\alpha}}F^{\hat{\alpha}}{}_{\hat{\nu}}-\frac{1}{4}\delta^{\hat{\mu}}{}_{\hat{\nu}}F^{2}\right]\ ,$$

(2.32)

while the second term can be interpreted as the stress-energy of “charged dust”, with the density of the dust involving both the bounce parameter $\ell$ and the total charge $Q$. Overall we have

$$\left[T_{Q}\right]^{\hat{\mu}}{}_{\hat{\nu}}=\left[T_{\text{Maxwell}}\right]^{\hat{\mu}}{}_{\hat{\nu}}+\Xi\;V^{\hat{\mu}}V_{\hat{\nu}}\ .$$

(2.33)

The vector $V^{\hat{\mu}}$ is the normalised unit timelike eigenvector of the stress-energy, which in the current situation reduces to the normalised time-translation Killing vector, while the dust density $\Xi$ has to be determined. We obtain

$$\left[T_{Q}\right]^{\hat{t}}{}_{\hat{t}}=-\varepsilon_{\text{em}}=\left[T_{\text{Maxwell}}\right]^{\hat{t}}{}_{\hat{t}}-\Xi=-\frac{1}{8\pi}E^{2}-\Xi\ .$$

(2.34)

Comparing with eq. (2.5), we find for the electric field strength $E$

$\displaystyle E$

$\displaystyle=\frac{Qr}{(r^{2}+\ell^{2})^{3/2}}=E_{\text{RN}}\left[\frac{r^{3}}{(r^{2}+\ell^{2})^{3/2}}\right]\ ,$

(2.35)

where $E_{\text{RN}}$ is the electric field strength of a Reissner–Nordström black hole. For the density of the dust $\Xi$ we find

$$\Xi=-\frac{1}{4\pi}\frac{Q^{2}\ell^{2}}{(r^{2}+\ell^{2})^{3}}\ .$$

(2.36)

All told, we have the following form for the electromagnetic stress-energy tensor for our regularised Reissner–Nordström spacetime

$$\left[T_{Q}\right]^{\hat{\mu}}{}_{\hat{\nu}}=\frac{1}{4\pi}\left[-F^{\hat{\mu}}{}_{\hat{\alpha}}F^{\hat{\alpha}}{}_{\hat{\nu}}-\frac{1}{4}\delta^{\hat{\mu}}{}_{\hat{\nu}}F^{2}\right]-\frac{1}{4\pi}\frac{Q^{2}\ell^{2}}{(r^{2}+\ell^{2})^{3}}V^{\hat{\mu}}V_{\hat{\nu}}\ .$$

(2.37)

Finally, the electromagnetic potential is easily extracted via integrating eq. (2.35), and in view of asymptotic flatness we may set the constant of integration to zero, yielding

$$A_{\mu}=(\Phi_{\text{em}}(r),0,0,0)=-\frac{Q}{\sqrt{r^{2}+\ell^{2}}}\;(1,0,0,0).$$

(2.38)

Note that this really is simply the electromagnetic potential from standard Reissner–Nordström spacetime, $-Q/r$, under the map $r\to\sqrt{r^{2}+\ell^{2}}$.

It is easy to verify that the electromagnetic field-strength tensor $F_{\mu\nu}=\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}$ satisfies $F_{[\mu\nu,\sigma]}=0$. The inhomogeneous Maxwell equation is, using $z=\sqrt{r^{2}+\ell^{2}}$:

$\displaystyle\nabla^{\hat{\mu}}F_{\hat{\mu}\hat{\nu}}$

$\displaystyle={\frac{Q\ell^{2}}{z^{6}}\;\sqrt{z^{2}-2mz+Q^{2}}}\;\left(1,0,0,0\right).$

(2.39)

(We warn the reader that the situation will get somewhat messier once we add rotation.)

Again using $z:=\sqrt{r^{2}+\ell^{2}}$, where this shorthand now stands for the equatorial value of the parameter $\rho$, the Ricci scalar is given by

$$R=2\ell^{2}\frac{m\left(\rho^{4}-2z^{4}\right)+Q^{2}z^{3}+z^{3}\left(\rho^{2}-2\Delta\right)}{\rho^{6}z^{3}}.$$

(3.5)

It is clearly finite in the limit $r\to 0$, i.e. $z\to\ell$. So are the Kretschmann scalar and the invariants $R^{\mu\nu}R_{\mu\nu}$ and $C^{\mu\nu\rho\sigma}C_{\mu\nu\rho\sigma}$. The only potentially dangerous behaviour arises from the denominators, which, in the $r\to 0$ limit, take the form

$$\frac{1}{\ell^{2}(\ell^{2}+a^{2}\cos^{2}\theta)^{6}}.$$

(3.6)

As long as $\ell\neq 0$, therefore, these quantities are never infinite.

We now turn our attention to the Einstein and Ricci tensors. We note that the (mixed) Einstein tensor $G^{\mu}_{\ \nu}$ can be diagonalised over the real numbers: its four eigenvectors $\{e_{\hat{\mu}}\}_{\mu=t,r,\theta,\phi}$ form a globally defined tetrad [52] and have explicit Boyer–Lindquist components

	$\displaystyle\big{(}e_{\hat{t}}\big{)}^{\mu}$	$\displaystyle=\frac{1}{\sqrt{\rho^{2}\absolutevalue{\Delta}}}\left(r^{2}+\ell^{2}+a^{2},0,0,a\right),\quad\big{(}e_{\hat{r}}\big{)}^{\mu}=\sqrt{\frac{\absolutevalue{\Delta}}{\rho^{2}}}\left(0,1,0,0\right),$
	$\displaystyle\big{(}e_{\hat{\theta}}\big{)}^{\mu}$	$\displaystyle=\frac{1}{\sqrt{\rho^{2}}}\left(0,0,1,0\right),\quad\big{(}e_{\hat{\phi}}\big{)}^{\mu}=\frac{1}{\sin\theta\sqrt{\rho^{2}}}\left(a\sin^{2}\theta,0,0,1\right).$		(3.7)

Eigenvectors are defined up to multiplicative, possibly dimensionful constants. This choice of normalisation ensures that the tetrad $\{e_{\hat{\mu}}\}$ is orthonormal and reduces to eq. (2.3) in the limit $a\to 0$. We will therefore use the tetrad (3.7), along with the coordinate basis, to express components of tensors.

In particular, the components of the Einstein tensor are

	$\displaystyle G_{\hat{t}\hat{t}}$	$\displaystyle=\operatorname{sign}\Delta\frac{\ell^{2}\left[2m\left(z^{2}-\rho^{2}\right)+z\left(\rho^{2}-2\Delta\right)\right]+Q^{2}z\left(\rho^{2}-\ell^{2}\right)}{\rho^{6}z}\ ,$		(3.8a)
	$\displaystyle G_{\hat{r}\hat{r}}$	$\displaystyle=-\operatorname{sign}\Delta\frac{\ell^{2}\left[2m\left(z^{2}-\rho^{2}\right)+\rho^{2}z\right]+Q^{2}z\left(\rho^{2}-\ell^{2}\right)}{\rho^{6}z}\ ,$		(3.8b)
	$\displaystyle G_{\hat{\theta}\hat{\theta}}$	$\displaystyle=\frac{\ell^{2}\left[m\left(-\rho^{4}-2\rho^{2}z^{2}+2z^{4}\right)+\rho^{2}z^{3}\right]+Q^{2}z^{3}\left(\rho^{2}-\ell^{2}\right)}{\rho^{6}z^{3}}\ ,$		(3.8c)
	$\displaystyle G_{\hat{\phi}\hat{\phi}}$	$\displaystyle=\frac{\ell^{2}\left[m\left(-\rho^{4}-2\rho^{2}z^{2}+6z^{4}\right)+z^{3}\left(2\Delta-\rho^{2}\right)\right]+Q^{2}z^{3}\left(\rho^{2}-3\ell^{2}\right)}{\rho^{6}z^{3}}\ .$		(3.8d)

We note that

$$G_{\hat{t}\hat{t}}+G_{\hat{r}\hat{r}}=-\operatorname{sign}\Delta\;\frac{2\ell^{2}\Delta}{\rho^{6}}$$

(3.9)

which is well-behaved at $\Delta=0$.

The Ricci tensor is clearly diagonal, too:

	$\displaystyle R_{\hat{t}\hat{t}}$	$\displaystyle=\operatorname{sign}\Delta\frac{m\ell^{2}\left(-\rho^{4}-2\rho^{2}z^{2}+4z^{4}\right)+Q^{2}z^{3}\left(\rho^{2}-2\ell^{2}\right)}{\rho^{6}z^{3}}\ ,$		(3.10a)
	$\displaystyle R_{\hat{r}\hat{r}}$	$\displaystyle=\operatorname{sign}\Delta\frac{\ell^{2}\left(m\left(\rho^{4}+2\rho^{2}z^{2}-4z^{4}\right)-2\Delta z^{3}\right)+Q^{2}z^{3}\left(2\ell^{2}-\rho^{2}\right)}{\rho^{6}z^{3}}\ ,$		(3.10b)
	$\displaystyle R_{\hat{\theta}\hat{\theta}}$	$\displaystyle=\frac{Q^{2}}{\rho^{4}}-\frac{2\ell^{2}}{\rho^{6}}\left[\Delta+\frac{\rho^{2}(m-z)}{z}\right],$		(3.10c)
	$\displaystyle R_{\hat{\phi}\hat{\phi}}$	$\displaystyle=\frac{2m\ell^{2}\left(2z^{2}-\rho^{2}\right)+Q^{2}z\left(\rho^{2}-2\ell^{2}\right)}{\rho^{6}z}\ .$		(3.10d)

Similarly, we note that

$$R_{\hat{t}\hat{t}}+R_{\hat{r}\hat{r}}=-\operatorname{sign}\Delta\;\frac{2\ell^{2}\Delta}{\rho^{6}}$$

(3.11)

which is well-behaved at $\Delta=0$.