Kerr black hole

The Kerr black hole is the asymptotically flat stationary axisymmetric vacuum black hole with mass $M$ and angular momentum $J=Ma$.

A Kerr black hole is the Kerr metric with mass $M>0$ and spin parameter satisfying $|a|\le M$ in geometrized units; its outer horizon is $r_{+}=M+\sqrt{M^2-a^2}$.

Lense-Thirring precession is the rotation of orbital angular momentum caused by the gravitational field of a spinning mass. In the secular weak-field picture, the precession vector is parallel to the central spin. Thus an annulus has torque proportional to ${\mathbf{J}}_h\times \delta {\mathbf{J}}_d$, perpendicular to ${\mathbf{J}}_h$. Summing such torques preserves this orthogonality even if their radial coefficients differ. Dissipation and the internal response of a warped accretion disk can supplement precession with an alignment torque.

For a weakly gravitating thin spherical shell of mass $M$, radius $R$ and uniform angular velocity $\Omega$, its interior linearized gravity metric has constant scalar potentials $A=-GM/R$, $C=GM/R$ and $g_{0i}=\omega (\hat{\mathbf{z}}\times \mathbf{x}{)}_i$. The scalar mass shell is a monopole. Its mass current is a dipole: regular interior and decaying exterior radial solutions are proportional to $r$ and $r^{-2}$, and their derivative jump fixes the factor $4/3$. Local inertial axes co-rotate with the shell at $\omega$ relative to asymptotic nonrotating axes. This is the leading interior Lense-Thirring precession, neglecting quadratic rotation and higher gravitational orders.

At linear order in $ϵ=GM/R$ and the leading rotation term, this coordinate change takes the interior shell metric to Minkowski spacetime. Consequently the retained interior Riemann curvature tensor vanishes although inertial axes rotate relative to distant reference axes. A freely moving particle has $\mathbf{X}={\mathbf{X}}_0+\mathbf{V}T$, while its original coordinates obey $d^2\mathbf{x}/d{t}^2=2\mathbf{\omega }\times d\mathbf{x}/dt$ to first rotation order. Terms of order ${\omega }^2$ would include centrifugal acceleration and are outside this approximation. A gyroscope at rest to the retained order has its parallel-transported spin precessing at $\mathbf{\omega }$ in the original coordinates.

For the opposite internal torques on a black hole and its accretion disk, the total angular momentum ${\mathbf{J}}_t$ and the black-hole spin magnitude $J_h$ are constant. If $A=K_2|{\mathbf{J}}_h\times {\mathbf{J}}_d{|}^2$, then $d({\mathbf{J}}_h\cdot {\mathbf{J}}_t)/dt=A$ and $d{J}_d^2/dt=-2A$. The $K_1$ term produces precession; the positive $K_2$ term increases the alignment of the hole with the total vector while decreasing the disk angular-momentum magnitude. The exact direction equation is $\dot{c}=K_2{J}_h{J}_t(1-c^2)$ for $c=\cos \angle ({\mathbf{J}}_h,{\mathbf{J}}_t)$.

After persistent dissipative alignment of the hole with the conserved total angular momentum, the final vectors are ${\mathbf{J}}_{h,f}=J_h{\hat{\mathbf{J}}}_t$ and ${\mathbf{J}}_{d,f}=(J_t-J_h){\hat{\mathbf{J}}}_t$. The disk is therefore antiparallel to the hole if $J_t<J_h$. Squaring $J_t^2=J_h^2+J_{d,0}^2+2J_h{J}_{d,0}\cos {\theta }_0$ gives the displayed initial-angle criterion. The disk and hole need not finish aligned with each other even though the hole aligns with the total. Equality $J_t=J_h$ leaves zero final disk angular momentum rather than a defined disk spin direction.

For nonzero $J_h,J_t$ and a noncollinear initial configuration, the alignment equation integrates to $\mathrm{artanh}c(t)=\mathrm{artanh}c(0)+J_h{J}_t{\int }_0^t{K}_2(s)ds$. Hence the positive dissipative torque aligns the hole with the total vector if its integrated strength diverges. A fixed positive coefficient is sufficient. Pointwise positivity alone need not suffice for time-dependent coefficients: with $J_h=J_t=1$, $K_1=0$, $K_2=e^{-t}$ and $c(0)=0$, the limiting cosine is $\tanh 1<1$. Exactly antiparallel initial spin and total angular momentum form a stationary, nongeneric exception.

The two-dimensional symmetry-axis restriction of the Kerr metric has simple horizons at $r_{\pm }$ for $0<|a|<M$, but is regular at $r=0$ because the ring is off-axis. Its static region below the inner horizon continues to an asymptotically flat end at negative infinity. The horizon blocks repeat in the maximal analytic extension. This ideal continuation is distinct from a claim of physical stability at the Cauchy horizon.

Conserved stationary energy and unit timelike normalization give this equation in the Kerr axial analytic extension. For $a>0$, the negative-sheet maximum of the potential is $1+M/a$ at $r=-a$. Energy squared above this maximum permits an inward orbit to pass through regular $r=0$ and continue to negative infinity without a turning point. At energy squared one, zero is a turning point. For negative spin the threshold is expressed with $|a|$.

For nonzero spin, the curvature singularity of the Kerr metric is the ring $x^2+y^2=a^2$, $z=0$ in oblate Cartesian coordinates. The rest of the $r=0$ disk is regular and allows continuation to negative $r$. The entire disk must not be mistaken for a curvature singularity, and axial trajectories do not hit the ring.

For the massless Klein-Gordon equation on a Kerr black hole, the covariant wave operator separates into radial and angular ordinary differential equations. With $K=(r^2+a^2)\omega -am$ and angular separation constant $\Lambda$, the radial equation is $(\Delta R^{\prime }{)}^{\prime }+[K^2/\Delta -a^2{\omega }^2+2am\omega -\Lambda ]R=0$. The angular equation is $(\sin \theta {\Theta }^{\prime }{)}^{\prime }/\sin \theta +[a^2{\omega }^2{\cos }^2\theta -m^2/{\sin }^2\theta +\Lambda ]\Theta =0$. No extra radial factor is included in the displayed ansatz.

A scalar spheroidal harmonic is a regular angular solution of $(\sin \theta S^{\prime }{)}^{\prime }/\sin \theta +[c^2{\cos }^2\theta -m^2/{\sin }^2\theta +\Lambda ]S=0$; with $e^{im\varphi }$ it is smooth at both axes. For real $c$, regularity gives discrete eigenvalues ${\Lambda }_{\ell m}(c)$. At $c=0$ it reduces to the angular part of a spherical harmonic, with $\Lambda =\ell (\ell +1)$. The sign of the $c^2$ term distinguishes the convention used here.

The outer Killing horizon of a Kerr black hole is generated by $k+{\Omega }_{H}m$, with the stationary Killing vector field $k$ normalized at infinity and the axial generator $m$ having $2\pi$-periodic orbits. In ingoing Kerr coordinates its generators satisfy $d\chi /dv={\Omega }_H$. This is the horizon's angular velocity relative to the nonrotating asymptotic stationary frame.

Ingoing Kerr coordinates cancel the explicit inverse-$\Delta$ singularities of the Boyer-Lindquist coordinates metric. They give a smooth, nondegenerate Kerr metric at the future outer horizon when $M>|a|$. Its stationary and axial Killing vector fields are ${\partial }_v$ and ${\partial }_{\chi }$. The advanced chart need not cover every horizon branch of the maximal extension.

The irreducible mass of a four-dimensional uncharged Kerr black hole is $M_{\mathrm{irr}}=\sqrt{A/(16\pi )}$. Its mass satisfies $M^2=M_{\mathrm{irr}}^2+J^2/(4M_{\mathrm{irr}}^2)$. Hawking's area theorem prevents a classical process from decreasing $M_{\mathrm{irr}}$ under the theorem's energy and global hypotheses.

For initially well-separated Kerr black holes with no orbital angular momentum, exact axisymmetry gives $J_f=J_1+J_2$. Put $X=(A_1+A_2)/(8\pi )$. The black-hole area theorem and the irreducible mass identity implyIndeed, $X\ge |J_1|+|J_2|\ge |J_f|$, and $\frac{1}{2}(Y+J_f^2/Y)$ is nondecreasing for $Y\ge X$. Opposite spin signs lower $J_f^2$ at fixed initial areas and hence raise the allowed radiated-energy bound; this need not be achievable in an actual merger.

Boyer-Lindquist coordinates adapt the Kerr metric to its stationary and axial symmetries. The horizon radii are the roots $r_{\pm }=M\pm \sqrt{M^2-a^2}$ of $\Delta =r^2-2Mr+a^2$.

The Kerr ergoregion is the region where the asymptotically timelike stationary Killing field becomes spacelike. Its outer boundary is $r=M+\sqrt{M^2-a^2{\cos }^2\theta }$.

The Kerr realization of an ergosphere lies outside the outer horizon where the stationary Killing vector is spacelike; its outer boundary is $r=M+\sqrt{M^2-a^2{\cos }^2\theta }$.

For subextreme Kerr spacetime, the outer stationary-limit surface is $s=r-r_e(\theta )=0$, where $r_e=M+\sqrt{M^2-a^2{\cos }^2\theta }$. Away from the rotation axis, $r_e>r_{+}$ and $\Delta (r_e)=a^2{\sin }^2\theta >0$. Its normal is therefore spacelike and the surface timelike. Although the stationary Killing vector field has zero norm on it, it is not normal to it, and the surface is not a Killing horizon. The actual outer Killing horizon is $r=r_{+}$, generated by $K+{\Omega }_H\Phi$.

Superradiance is amplification of a scattered wave by extracting rotational or other available conserved energy from a background. For a rotating Kerr black hole, a bosonic mode of frequency $\omega$ and azimuthal integer $m$ can have negative horizon Killing-energy flux when $0<\omega <m{\Omega }_H$. The reflected energy can then exceed the incident energy. A stationary Killing flow need not define a globally positive particle energy.

Outside a subextremal Kerr black hole, regular causal curves obey $(-g_{t\varphi }-\sqrt{D})/g_{\varphi \varphi }\le d\varphi /dt\le (-g_{t\varphi }+\sqrt{D})/g_{\varphi \varphi }$, where $D=g_{t\varphi }^2-g_{tt}{g}_{\varphi \varphi }>0$. Positive radial and polar metric terms establish this necessary bound. In the open Kerr ergoregion both endpoints are positive for $a>0$, forcing co-rotation.

The Penrose process extracts rotational energy by allowing a fragment with negative stationary Killing energy to enter the black hole while another fragment escapes with more energy than the original body carried.

For a future-directed causal four-momentum $p$ crossing a future Killing horizon with generator $\xi =t+{\Omega }_H\phi$, the Lorentzian inner product obeys $p\cdot \xi \le 0$. With $E=-p\cdot t$ and $L=p\cdot \phi$, this is $E\ge {\Omega }_{H}L$. A massive particle has strict inequality. The same bound follows in the nonextremal quasistationary regime from the Physical-process first law for a rotating black hole and the second law of black-hole mechanics. Negative Killing energy in a Penrose process is allowed, but its angular momentum must satisfy this bound.

On the Kerr symmetry axis, the radial metric has $F=\Delta /(r^2+a^2)$ and the tortoise coordinate satisfies $d{r}_{\ast }/dr=F^{-1}=(r^2+a^2)/\Delta$.

Writing $s=\sqrt{1-J^2/M^4}=\sqrt{1-a^2/M^2}$, the heat capacity of a Kerr black hole at fixed angular momentum isIt is positive for $\sqrt{2\sqrt{3}-3}<|a|/M<1$, diverges at the lower endpoint, and tends to zero in the extremal limit.