Darboux diagonalization of the spatial 3-metric in Kerr spacetime

Abstract

The astrophysical importance of the Kerr spacetime cannot be overstated. Of the currently known exact solutions to the Einstein field equations, the Kerr spacetime stands out in terms of its direct applicability to describing astronomical black hole candidates. In counterpoint, purely mathematically, there is an old classical result of differential geometry, due to Darboux, that all 3-manifolds can have their metrics recast into diagonal form. In the case of the Kerr spacetime the Boyer–Lindquist coordinates provide an explicit example of a diagonal spatial 3-metric. Unfortunately, as we demonstrate herein, Darboux diagonalization of the spatial 3-slices of the Kerr spacetime is incompatible with simultaneously putting the Kerr metric into unit-lapse form while retaining manifest axial symmetry. This no-go theorem is somewhat reminiscent of the no-go theorem to the effect that the spatial 3-slices of the Kerr spacetime cannot be chosen to be conformally flat.

A Appendix: some explicit metrics

In this appendix we list some of the explicit metrics utilized in the text.

1.1 A.1 Unit-lapse metrics

The general form for a unit-lapse metric is [17]:

$$\begin{aligned} \mathrm{d}s^2 = - dt^2 + h_{ij} (dx^i - v^i dt) (dx^j-v^j dt). \end{aligned}$$

(A.1)

Then

$$\begin{aligned} g_{ab}= & {} \left[ \begin{array}{c|c} -1 + (h^{ij} v_i v_j) &{} -v_j \\ \hline -v_i &{} h_{ij} \end{array}\right] _{ab}; \nonumber \\ \qquad \qquad g^{ab}= & {} \left[ \begin{array}{c|c} -1 &{} -v^j \\ \hline -v^i &{} h^{ij} - v^i v^j\end{array}\right] ^{ab}. \end{aligned}$$

(A.2)

Note $\det (g_{ab})=-\det (h_{ij})$ and $g^{tt}=-1$.

1.2 A.2 Lense–Thirring (Painlevé–Gullstrand version)

For the Painlevé–Gullstrand version of Lense–Thirring [22]:

$$\begin{aligned} \mathrm{d}s^2= & {} - d t^2 +\left( d r+\sqrt{2m\over r} dt\right) ^2\nonumber \\&+\, r^2 \left( {{d}}\theta ^2+\sin ^2\theta \left( {{d}}\phi - {2J\over r^3} dt\right) ^2\right) , \end{aligned}$$

(A.3)

Then

$$\begin{aligned} g_{ab} = \left[ \begin{array}{c|ccc} -1+{2m\over r} + {4 J^2\sin ^2\theta \over r^4} &{} \sqrt{2m\over r} &{} 0 &{} -{2J\sin ^2\theta \over r}\\ \hline \sqrt{2m\over r} &{} 1 &{} 0 &{} 0\\ 0&{} 0 &{} r^2 &{} 0\\ -{2J\sin ^2\theta \over r} &{} 0 &{} 0 &{} r^2\sin ^2\theta \\ \end{array} \right] _{ab}. \end{aligned}$$

(A.4)

Here $h_{ij}$ is not just diagonal, it is flat 3-space.

$$\begin{aligned} g^{ab} = \left[ \begin{array}{c|ccc} -1&{} \sqrt{2m\over r} &{} 0 &{} -{2J\over r^3}\\ \hline \sqrt{2m\over r} &{} 1-{2m\over r} &{} 0 &{} \sqrt{2m\over r} {2J\over r^3}\\ 0&{} 0 &{} {1\over r^2} &{} 0\\ -{2J\over r^3} &{} \sqrt{2m\over r} {2J\over r^3} &{} 0 &{} {1\over r^2\sin ^2\theta } - {4J^2\over r^6}\\ \end{array} \right] ^{ab}. \end{aligned}$$

(A.5)

Note this is unit lapse, $g^{tt}=-1$.

1.3 A.3 Boyer–Lindquist

The standard form for the Boyer–Lindquist version of Kerr is:

$$\begin{aligned} (g_{ab})_{BL} = {\left[ \begin{array}{c|cc|c} - 1+{2mr\over \rho ^2} &{} 0 &{} 0 &{} -{2mar\sin ^2\theta \over \rho ^2}\\ \hline 0&{}{\rho ^2\over \Delta } &{}0&{}0\\ 0&{}0&{}{\rho ^2}&{}0\\ \hline - {2mar\sin ^2\theta \over \rho ^2}&{}0&{} 0&{} \Sigma \sin ^2\theta \end{array}\right] }_{ab}. \end{aligned}$$

(A.6)

Note $h_{ij}$ is diagonal.

$$\begin{aligned} (g^{ab})_{BL} = {\left[ \begin{array}{c|cc|c} - 1-{2mr(r^2+a^2)\over \rho ^2\Delta } &{} 0 &{} 0 &{} -{2mar\over \rho ^2\Delta }\\ \hline 0&{}{\Delta \over \rho ^2} &{}0&{}0\\ 0&{}0&{}{1\over \rho ^2}&{}0\\ \hline - {2mar\over \rho ^2\Delta }&{}0&{} 0&{} {1-2mr/\rho ^2\over \Delta \sin ^2\theta } \end{array}\right] }^{ab}. \end{aligned}$$

(A.7)

Here $\rho ^2=r^2+a^2\cos ^2\theta$, and $\Delta =r^2+a^2-2mr$, while $\Sigma = r^2+a^2 + {2mra^2\over \rho ^2} \sin ^2\theta$.

Note this is not unit lapse, $g^{tt}\ne -1$.

1.4 A.4 Boyer–Lindquist-rain

For the Boyer–Lindquist-rain version of the Kerr metric $(g_{ab})_{\hbox {BL-rain}}$ we have [17]:

$$\begin{aligned} {\left[ \begin{array}{c|cc|c} - 1+{2mr\over \rho ^2} &{} \left( 1-{2mr\over \rho ^2}\right) {\sqrt{2mr(r^2+a^2)}\over \Delta } &{} 0 &{} -{2mar\sin ^2\theta \over \rho ^2}\\ \hline \left( 1-{2mr\over \rho ^2}\right) {\sqrt{2mr(r^2+a^2)}\over \Delta }&{} {\rho ^2\over \Delta } - \left( 1-{2mr\over \rho ^2}\right) {2mr(r^2+a^2)\over \Delta ^2} &{}0&{}{2mar\sin ^2\theta \over \rho ^2} {\sqrt{2mr(r^2+a^2)}\over \Delta } \\ 0&{}0&{}{\rho ^2}&{}0\\ \hline - {2mar\sin ^2\theta \over \rho ^2}&{}{2mar\sin ^2\theta \over \rho ^2} {\sqrt{2mr(r^2+a^2)}\over \Delta } &{} 0&{} \Sigma \sin ^2\theta \end{array}\right] }_{ab} \end{aligned}$$

(A.8)

Note $h_{ij}$ is not diagonal.

For the inverse metric

$$\begin{aligned} (g^{ab})_{\hbox {BL-rain}}= {\left[ \begin{array}{c|cc|c} - 1 &{} {\sqrt{2mr(r^2+a^2)}\over \rho ^2} &{} 0 &{} -{2mar\over \rho ^2\Delta }\\ \hline {\sqrt{2mr(r^2+a^2)}\over \rho ^2} &{}{\Delta \over \rho ^2} &{}0&{}0\\ 0&{}0&{}{1\over \rho ^2}&{}0\\ \hline - {2mar\over \rho ^2\Delta }&{}0&{} 0&{} {1-2mr/\rho ^2\over \Delta \sin ^2\theta } \end{array}\right] }^{ab}. \end{aligned}$$

(A.9)

Note this is unit lapse, $g^{tt}=-1$. Oddly the spatial part of the inverse metric $g^{ij}$ is diagonal, but this is not the same as saying $h_{ij}$ is diagonal.

1.5 A.5 Eddington–Finklestein-rain

In Eddington–Finkelstein-rain coordinates the covariant metric is given by [17]:

$$\begin{aligned} (g_{ab})_{\hbox {EF-rain}}= {\left[ \begin{array}{c|cc|c} - 1+{2mr\over \rho ^2} &{} g_{tr} &{} 0 &{}- {2mar\over \rho ^2}\sin ^2\theta \\ \hline g_{tr} &{}g_{rr} &{}0&{}g_{r\phi }\\ 0&{}0&{}\rho ^2&{}0\\ \hline -{2mar\over \rho ^2}\sin ^2\theta &{}g_{r\phi }&{}0&{} \Sigma \sin ^2\theta \end{array}\right] }_{ab} \end{aligned}$$

(A.10)

subject to the relatively messy results that

$$\begin{aligned} g_{rr}= & {} 1+ {a^2\sin ^2\theta (2mr/\rho ^2) \over (r^2+a^2)(1+ \sqrt{2mr/(r^2+a^2)})^2}; \end{aligned}$$

(A.11)

$$\begin{aligned} g_{tr}= & {} { {2mr/\rho ^2} + \sqrt{2mr/(r^2+a^2)} \over 1+ \sqrt{2mr/(r^2+a^2)}}; \end{aligned}$$

(A.12)

$$\begin{aligned} g_{r\phi }= & {} -a \sin ^2\theta \left( 1 +{2mr/\rho ^2} + \sqrt{2mr/(r^2+a^2)} \over 1 + \sqrt{2mr/( r^2+a^2)}\right) . \end{aligned}$$

(A.13)

Note $h_{ij}$ is not diagonal. The inverse metric is much simpler

$$\begin{aligned} (g^{ab})_{\hbox {EF-rain}}= {\left[ \begin{array}{c|cc|c} - 1&{}{\sqrt{2mr(r^2+a^2)}\over \rho ^2}&{} 0 &{} {\sqrt{2mra^2/(r^2+a^2)}\over \rho ^2 (1+\sqrt{2mr/(r^2+a^2)})}\\ \hline {\sqrt{2mr(r^2+a^2)}\over \rho ^2}&{} {\Delta \over \rho ^2} &{}0&{} {a\over \rho ^2}\\ 0&{}0&{}{1\over \rho ^2}&{}0\\ \hline {\sqrt{2mra^2/(r^2+a^2)}\over \rho ^2 (1+\sqrt{2mr/(r^2+a^2)})}&{} {a\over \rho ^2}&{} 0&{} {1\over \rho ^2\sin ^2\theta } \end{array}\right] }^{ab}. \end{aligned}$$

(A.14)

Note this is unit lapse, $g^{tt}=-1$.

1.6 A.6 Doran

The Doran metric is [15, 17]

$$\begin{aligned} (g_{ab})_{{\mathrm {Doran}}} = \left[ \begin{array}{c|cc|c} -1 +{2mr\over \rho ^2} &{} \sqrt{2mr\over a^2+r^2} &{} 0 &{} -{2mar\sin ^2\theta \over \rho ^2}\\ \hline \sqrt{2mr\over a^2+r^2} &{} {\rho ^2\over r^2+a^2}&{} 0 &{} -a\sqrt{2mr\over a^2+r^2} \sin ^2\theta \\ 0 &{}0 &{} \rho ^2&{} 0\\ \hline -{2mar\sin ^2\theta \over \rho ^2} &{} -a\sqrt{2mr\over a^2+r^2} \sin ^2\theta &{} 0 &{} \Sigma \sin ^2\theta \end{array}\right] _{ab}. \end{aligned}$$

(A.15)

Note $h_{ij}$ is not diagonal.

For the inverse metric

$$\begin{aligned} (g^{ab})_{{\mathrm {Doran}}} = \left[ \begin{array}{c|cc|c} -1 &{} {\sqrt{2mr(a^2+r^2)}\over \rho ^2} &{} 0 &{} 0\\ \hline {\sqrt{2mr(a^2+r^2)}\over \rho ^2} &{} {\Delta \over \rho ^2}&{} 0 &{} {a\sqrt{2mr\over a^2+r^2} \over \rho ^2}\\ 0 &{}0 &{} {1\over \rho ^2} &{} 0\\ \hline 0 &{} {a\sqrt{2mr\over a^2+r^2} \over \rho ^2} &{} 0 &{} {1\over (a^2+r^2) \sin ^2\theta } \end{array}\right] ^{ab}. \end{aligned}$$

(A.16)

Note this is unit lapse, $g^{tt}=-1$.