Kerr Black Holes & Rotating Spacetimes
Kerr Black Holes & Rotating Spacetimes
The Kerr metric (1963) is the unique stationary, axisymmetric vacuum solution, describing a rotating black hole characterized by mass \(M\) and angular momentum \(J\). It predicts frame-dragging, the ergosphere, and the Penrose process — extracting rotational energy from a spinning black hole.
Kerr Metric (Boyer-Lindquist Coordinates)
With \(a = J/(Mc)\), \(\rho^2 = r^2 + a^2\cos^2\theta\), \(\Delta = r^2 - r_s r + a^2\):
\(ds^2 = -\!\left(1-\frac{r_s r}{\rho^2}\right)c^2 dt^2 - \frac{2r_s r a\sin^2\theta}{\rho^2}c\,dt\,d\phi + \frac{\rho^2}{\Delta}dr^2 + \rho^2 d\theta^2 + \frac{\Sigma^2\sin^2\theta}{\rho^2}d\phi^2\), \(\Sigma^2 = (r^2+a^2)^2 - a^2\Delta\sin^2\theta\).
Ergosphere & Penrose Process
The static limit (ergosphere boundary) is at \(r_\text{erg} = \frac{r_s + \sqrt{r_s^2 - 4a^2\cos^2\theta}}{2}\). Between the ergosphere and event horizon \(r_+ = (r_s+\sqrt{r_s^2-4a^2})/2\), \(\partial_t\) is spacelike. A decaying particle can eject one fragment with negative energy, extracting up to \(29\%\) of the BH mass (maximal spin).
Example 1: Horizon Radii
Find outer and inner horizons for \(a = 0.9\,GM/c^2\).
Solution: \(r_\pm = \frac{r_s}{2} \pm \sqrt{\left(\frac{r_s}{2}\right)^2 - a^2}= GM/c^2(1\pm\sqrt{1-0.81})\). Thus \(r_+ \approx 1.436\,GM/c^2\), \(r_- \approx 0.564\,GM/c^2\).
Example 2: Frame Dragging (ZAMO)
Find the angular velocity \(\Omega\) of a zero-angular-momentum observer (ZAMO) far from the BH.
Solution: \(\Omega = -g_{t\phi}/g_{\phi\phi} \approx \frac{2GJ}{c^2 r^3}\) for \(r\gg r_s\). This Lense-Thirring frame dragging was measured by Gravity Probe B to \(\sim 19\%\) accuracy.
Practice
- Show the Kerr metric reduces to Schwarzschild when \(a=0\).
- Derive the condition \(g_{tt}=0\) that defines the ergosphere boundary.
- Compute the maximum energy extracted per unit rest mass via the Penrose process.
- What is the ISCO radius for an extremal Kerr BH? Compare to Schwarzschild.
Show Answer Key
1. Kerr metric in Boyer-Lindquist: $ds^2 = -(1-\frac{r_s r}{\Sigma})c^2dt^2 - \frac{2r_s ra\sin^2\theta}{\Sigma}cdtd\phi + \frac{\Sigma}{\Delta}dr^2 + \Sigma d\theta^2 + (r^2+a^2+\frac{r_s ra^2\sin^2\theta}{\Sigma})\sin^2\theta d\phi^2$, where $\Sigma = r^2+a^2\cos^2\theta$, $\Delta = r^2-r_sr+a^2$. Setting $a=0$: $\Sigma=r^2$, $\Delta=r^2-r_sr$, cross term vanishes, recovering Schwarzschild. ✓
2. Ergosphere boundary: $g_{tt}=0$, i.e., $1-r_sr/\Sigma=0$, giving $r_{\text{ergo}} = \frac{r_s}{2}+\sqrt{(r_s/2)^2-a^2\cos^2\theta}$. At the equator ($\theta=\pi/2$): $r_{\text{ergo}}=r_s$, the Schwarzschild radius. At the poles ($\theta=0$): $r_{\text{ergo}}=r_+$ (event horizon). Between $r_+$ and $r_{\text{ergo}}$, no static observer can exist — all must co-rotate with the black hole (frame dragging).
3. Penrose process: a particle enters the ergosphere and splits into two. One piece falls into the BH with negative energy (possible in the ergosphere where $g_{tt}>0$, allowing negative-energy orbits). The other escapes with more energy than the original. Maximum efficiency: $\Delta E/E = 1 - \sqrt{(1+\sqrt{1-a_*^2})/2}$ where $a_* = a/(r_s/2)$. For extremal Kerr ($a_*=1$): $\eta_{\max} = 1-1/\sqrt{2} \approx 29\%$.
4. ISCO for Kerr: depends on spin. For prograde orbits around extremal Kerr ($a=M$): $r_{\text{ISCO}} = GM/c^2$ (= $r_s/2$, at the horizon). For Schwarzschild ($a=0$): $r_{\text{ISCO}} = 3r_s = 6GM/c^2$. For retrograde orbits around extremal Kerr: $r_{\text{ISCO}} = 9GM/c^2$. The prograde ISCO being at the horizon means up to 42% of rest mass energy can be radiated during inspiral (vs. 5.7% for Schwarzschild).