Extremal limits and Kerr spacetime

Abstract

The fact that one must evaluate the near-extremal and near-horizon limits of Kerr spacetime in a specific order, is shown to lead to discontinuity in the extremal limit, such that this limiting spacetime differs nontrivially from the precisely extremal spacetime. This is established by first showing a discontinuity in the extremal limit of the maximal analytic extension of the Kerr geometry, given by Carter. Next, we examine the ISCO of the exactly extremal Kerr geometry and show that on the event horizon of the extremal Kerr black hole, it coincides with the principal null geodesic generator of the horizon, having vanishing energy and angular momentum. We find that there is no such ISCO in the near-extremal geometry, thus garnering additional support for our primary contention. We relate this disparity between the two geometries to the lack of a trapping horizon in the extremal situation.

Appendix: Maximal analytical extension of Kerr spacetime (off axis of symmetry)

The maximal analytic extension of the non-extremal Kerr spacetime along the off axis of symmetry was first reported by Carter [8]. As we showed for symmetry axis the analytic extension is not continuous at the extremal limit r ₊→r ₋ in Sect. 3, here we implemented for the off axis symmetry and only derive the extremal case. For non-extremal case see Carter’s [8] paper.

The complete analytic extension of the extremal Kerr spacetime thus cannot be obtained as a limiting case of the non-extremal geometry as previously, one needs to treat the extremal case separately [8]. Defining the double null coordinates and angular coordinates we have

$$ \begin{array}{@{}l} \displaystyle du+dv=2\frac{r^2+M^2}{\varDelta }dr,\\ \displaystyle d\phi+d\tilde{\phi}=2\frac{M}{\varDelta }\,dr. \end{array} $$

(A.1)

Define ignorable angle coordinates ϕ ^⋆ given by

$$ 2\,d\phi^{\star}=d\phi-d\tilde{\phi}-\frac{(du-dv)}{2M}, $$

(A.2)

which are constant on the null generator of the horizon at r=M. The metric is thus given by

$$\begin{aligned} ds^2 =& \frac{\varDelta }{8\rho^2} \biggl[\frac{\rho^2}{r^2+M^2}+ \frac {\rho_{\star}^2}{2M^2} \biggr] \frac{(r^2-M^2)\sin^2\theta}{(r^2+M^2)} \\&{}\times \bigl(du^2+dv^2\bigr)+\rho^2 \, d \theta^2 \\&{}+\frac{\varDelta }{2\rho^2} \biggl[\frac{\rho^4}{(r^2+M^2)^2}+\frac {\rho_{\star}^4}{4M^4} \biggr]\,du \,dv \\&{}-\frac{\varDelta M\sin^2\theta}{\rho^2} \biggl[M\sin^2\theta \,d\phi^{\star }- \frac{\rho_{\star}^2}{2M^2}(du-dv) \biggr]\,d\phi^{\star} \\&{}+\frac{\sin^2\theta}{\rho^2} \biggl[ \frac {(M^2-r^2)}{4M}(du-dv) \\&{}-\bigl(r^2+M^2 \bigr)\,d\phi^{\star} \biggr]^2, \end{aligned}$$

(A.3)

where $\rho_{\star}^{2}=M^{2}(1+\cos^{2}\theta)$, Δ=(r−M)², ρ ²=(r ²+M ²cos² θ). r is now defined implicitly as a function of u and v by

$$ F(r)=u+v=2r^\ast, $$

(A.4)

where r ^∗ is called the ‘tortoise’ coordinate, determined by

$$ dr^\ast=\frac{(r^2+M^2)\,dr}{\varDelta }=\frac{(r^2+M^2)\,dr}{(r-M)^2}. $$

(A.5)

Integrating (A.5) yields

$$ r^\ast=\int\frac{(r^2+M^2)\,dr}{(r-M)^2}=r+ 2M \biggl[ \ln\vert r-M \vert - {M \over2(r-M)} \biggr]. $$

(A.6)

Near the horizon r=M this has a leading pole-type singularity

$$ r^\ast\approx\frac{M^2}{(r-M)} $$

(A.7)

instead of a logarithmic one. To locate the event horizon at a finite region in the coordinate chart, we follow Ref. [8] and introduce null coordinates U, V such that

$$\begin{aligned} u = \tan U,\qquad v = \cot V. \end{aligned}$$

(A.8)

This implies that

$$\begin{aligned} \tan U + \cot V = 2 r^{\ast}(U,V). \end{aligned}$$

(A.9)

Therefore the complete extremal Kerr metric in (U,V,θ,ϕ ^⋆) is given by

$$\begin{aligned} ds^2 =& \frac{\varDelta }{8\rho^2} \biggl[\frac{\rho^2}{r^2+M^2}+ \frac {\rho_{\star}^2}{2M^2} \biggr] \frac{(r^2-M^2)\sin^2\theta}{(r^2+M^2)} \\&{}\times \bigl(\sec^4U dU^2+\csc ^4V \,dV^2 \bigr) +\rho^2 \, d\theta^2 \\&{}-\frac{\varDelta }{2\rho^2} \biggl[\frac{\rho^4}{(r^2+M^2)^2}+\frac{\rho _{\star}^4}{4M^4} \biggr] \sec^2U \csc^2V \, dU\,dV \\&{}-\frac{\varDelta M\sin^2\theta}{\rho^2} \biggl[M\sin^2\theta \,d\phi^{\star} \\&{}-\frac{\rho_{\star}^2}{2M^2}\bigl(\sec^2U\,dU+\csc^2VdV\bigr) \biggr] \, d\phi ^{\star} \\&{} +\frac{\sin^2\theta}{\rho^2} \biggl[\frac{(M^2-r^2)}{4M} \bigl(\sec^2U\, dU \\&{}+\csc^2V \,dV \bigr) -\bigl(r^2+M^2 \bigr)\,d\phi^{\star} \biggr]^2. \end{aligned}$$

(A.10)

It can easily be checked that in the limit θ=0 we obtain the metric (18) for the symmetry axis.