Rotating and charged holes, and where this course stops

Idea. Everything is built from two lengths, \(r_g\) and \(a\), and one identity, \(2r_gr_+ = r_+^2+a^2\).

1. \(r_g = GM/c^2 = 1.32712\times10^{21}/8.98755\times10^{16} = 14\,766\) m and \(a = 0.9r_g = 13\,290\) m. With \(\sqrt{1-0.81} = 0.435890\),

\[r_+ = 1.435890\,r_g = 21\,203\ \mathrm m = 21.20\ \mathrm{km},\qquad r_- = 0.564110\,r_g = 8\,330\ \mathrm m = 8.33\ \mathrm{km}.\]

For the ergosurface, \(r_{\mathrm E}(\theta) = r_g\big(1 + \sqrt{1 - a_*^2\cos^2\theta}\big)\):

• \(\theta = 0\): \(\sqrt{0.19} = 0.43589\), \(r_{\mathrm E} = 1.4359\,r_g = 21.20\) km — the horizon itself.
• \(\theta = 30°\): \(\cos^2 = 0.75\), \(\sqrt{1-0.6075} = 0.62650\), \(r_{\mathrm E} = 1.6265\,r_g = 24.02\) km.
• \(\theta = 60°\): \(\cos^2 = 0.25\), \(\sqrt{1-0.2025} = 0.89303\), \(r_{\mathrm E} = 1.8930\,r_g = 27.95\) km.
• \(\theta = 90°\): \(\cos = 0\), \(r_{\mathrm E} = 2r_g = 29.53\) km.

2. \(A = 4\pi(r_+^2+a^2) = 8\pi r_gr_+ = 8\pi(14766)(21203) = 7.869\times10^{9}\ \mathrm{m^2}\). A Euclidean sphere of that area has radius \(\sqrt{A/4\pi} = 25\,023\) m \(= 25.02\) km, which is larger than \(r_+ = 21.20\) km. No contradiction: \(r\) is Boyer–Lindquist's radial coordinate, not an areal radius. In Schwarzschild the two coincided; in Kerr they do not, because \(g_{\theta\theta} = \Sigma\) varies with \(\theta\) and the horizon is oblate.

3. \(r_+^2+a^2 = 2r_gr_+ = 6.2617\times10^{8}\ \mathrm{m^2}\), so

\[\Omega_{\mathrm H} = \frac{ac}{r_+^2+a^2} = \frac{(1.3290\times10^{4})(2.99792\times10^{8})}{6.2617\times10^{8}} = 6363\ \mathrm{rad\,s^{-1}},\] which is \(6363/2\pi = 1013\) Hz. And with \(r_+ - r_- = 1.2873\times10^{4}\) m,

\[\kappa = \frac{c^2(r_+-r_-)}{2(r_+^2+a^2)} = \frac{(8.98755\times10^{16})(1.2873\times10^{4})}{1.25234\times10^{9}} = 9.238\times10^{11}\ \mathrm{m\,s^{-2}} .\]

4. On \(r = r_+\), \(\theta = \pi/2\): \(\Delta = 0\) and \(\Sigma = r_+^2\), so the preamble's identity gives \(g_{\varphi\varphi} = (r_+^2+a^2)^2/r_+^2\) and

\[\ell = \int_0^{2\pi}\sqrt{g_{\varphi\varphi}}\,\mathrm d\varphi = 2\pi\,\frac{r_+^2+a^2}{r_+} = 2\pi\,\frac{2r_gr_+}{r_+} = 4\pi r_g ,\] with \(a\) gone. Numerically \(4\pi(14766) = 1.8556\times10^{5}\) m \(= 185.6\) km.

Check. Dimensions: \(\Omega_{\mathrm H}\) is \(\mathrm{m\cdot m\,s^{-1}/m^2} = \mathrm{s^{-1}}\) ✓; \(\kappa\) is \(\mathrm{m^2s^{-2}\cdot m/m^2} = \mathrm{m\,s^{-2}}\) ✓. Limit \(a\to0\): \(r_+\to2r_g = r_s = 29.53\) km, \(r_-\to0\), \(r_{\mathrm E}\to r_s\) at every \(\theta\), \(A\to4\pi r_s^2 = 1.096\times10^{10}\ \mathrm{m^2}\), \(\Omega_{\mathrm H}\to0\), and \(\kappa\to c^4/4GM = 1.522\times10^{12}\ \mathrm{m\,s^{-2}}\) ✓ — and the equatorial circumference \(4\pi r_g = 2\pi r_s\) ✓, as it must be for a round sphere. Special case: at \(\theta = 0\) the ergosurface coincides with \(r_+\), so the ergosphere has zero thickness on the axis ✓.

Another way

For part 2, avoid the area integral: the induced metric on the horizon is \(\Sigma\,\mathrm d\theta^2 + (r_+^2+a^2)^2\sin^2\theta\,\mathrm d\varphi^2/\Sigma\), whose determinant is \((r_+^2+a^2)^2\sin^2\theta\) with the \(\Sigma\) cancelling exactly. The area element is therefore \((r_+^2+a^2)\sin\theta\,\mathrm d\theta\,\mathrm d\varphi\), and the integral is immediate. That the \(\Sigma\) cancels is the reason \(A\) is so simple, and it is worth noticing rather than grinding through.

Idea. Every part is the timelike condition on \(\partial_t + \Omega\partial_\varphi\), with \(\Omega = 0\) in part 1 and \(\Omega\) free in parts 2 and 3.

1. A curve with \(r,\theta,\varphi\) constant has tangent proportional to \(\partial_t\), so it is timelike iff \(g(\partial_t,\partial_t) = g_{tt} \lt 0\). From the metric,

\[g_{tt} = -\Big(1 - \frac{2r_gr}{\Sigma}\Big)c^2 = -\frac{c^2}{\Sigma}\big(\Sigma - 2r_gr\big) = -\frac{c^2}{\Sigma}\big(r^2 - 2r_gr + a^2\cos^2\theta\big),\] and \(\Sigma \gt 0\) away from the ring, so \(g_{tt} \lt 0 \iff r^2 - 2r_gr + a^2\cos^2\theta \gt 0\). The quadratic in \(r\) has roots \(r_g \pm \sqrt{r_g^2 - a^2\cos^2\theta}\) and is positive outside them; the larger root is \(r_{\mathrm E}(\theta)\).

2. \(g(\xi_\Omega,\xi_\Omega) = g_{tt} + 2\Omega g_{t\varphi} + \Omega^2g_{\varphi\varphi}\). Since \(g_{\varphi\varphi} \gt 0\) outside the horizon, this parabola opens upwards and is negative exactly strictly between its roots,

\[\Omega_\pm = \frac{-g_{t\varphi}\pm\sqrt{g_{t\varphi}^2 - g_{tt}g_{\varphi\varphi}}}{g_{\varphi\varphi}} = \omega \pm \frac{c\sqrt\Delta\,\sin\theta}{g_{\varphi\varphi}}, \qquad \omega := -\frac{g_{t\varphi}}{g_{\varphi\varphi}},\] using the preamble's discriminant identity. Real roots need \(\Delta \ge 0\), i.e. \(r \ge r_+\), and the window closes at the horizon.

3. At \(r = 2r_g\), \(\theta = \pi/2\): \(\Sigma = 4r_g^2\), \(\Delta = 4r_g^2 - 4r_g^2 + a^2 = a^2\), and \(g_{tt} = 0\) (this is the ergosurface). Then

\[g_{t\varphi} = -\frac{2r_g(2r_g)a}{4r_g^2}c = -ac,\qquad g_{\varphi\varphi} = \frac{(4r_g^2+a^2)^2 - a^4}{4r_g^2} .\] With \(a = 0.9r_g\): \((4.81)^2 - 0.6561 = 23.1361 - 0.6561 = 22.480\), divided by \(4\), gives \(g_{\varphi\varphi} = 5.6200\,r_g^2\). Because \(g_{tt} = 0\) the quadratic factorises, \(\Omega(2g_{t\varphi} + \Omega g_{\varphi\varphi})\), so

\[\Omega_- = 0,\qquad \Omega_+ = -\frac{2g_{t\varphi}}{g_{\varphi\varphi}} = 2\omega = \frac{2(0.9)r_gc}{5.6200\,r_g^2} = 0.32028\,\frac{c}{r_g}.\] With \(c/r_g = 2.99792\times10^{8}/14766 = 2.0303\times10^{4}\ \mathrm{s^{-1}}\), that is \(\Omega_+ = 6.50\times10^{3}\ \mathrm{rad\,s^{-1}}\), and \(\omega = 3251\ \mathrm{rad\,s^{-1}}\). The lower value \(\Omega_- = 0\) is the static observer, who survives — just — precisely on the ergosurface and nowhere inside it.

4. True: inside the ergosphere every observer is dragged, \(\Omega \gt 0\), and no one can stand still relative to the distant stars. False: the ergosphere is outside the event horizon, and a rocket in it can leave and reach infinity. The static limit and the horizon are different surfaces, and they touch only at the poles.

Check. Dimensions: \(\Omega_\pm\) come from \(\sqrt{g_{t\varphi}^2}/g_{\varphi\varphi} = \mathrm{(m^2s^{-1})/m^2} = \mathrm{s^{-1}}\) ✓. Limit \(a\to0\): \(\omega\to0\) and \(\Omega_\pm \to \pm c\sqrt{f}/(r\sin\theta)\), symmetric about zero, so a static observer exists wherever \(f \gt 0\) ✓. Consistency with P1: at \(r = r_+\) the window shuts at \(\Omega_{\mathrm H} = 6363\ \mathrm{rad\,s^{-1}}\), which lies inside \((0, 6503)\) — the allowed band narrows as one descends, as it must ✓.

Idea. One substitution, \(2r_gr = r^2+a^2-\Delta\), does all three algebraic parts. It is just \(\Delta\)'s definition rearranged, and it is the only identity worth memorising about Kerr.

1. From the metric,

\[\frac{\Sigma\,g_{\varphi\varphi}}{\sin^2\theta} = (r^2+a^2)\Sigma + 2r_gra^2\sin^2\theta .\] Substitute \(\Sigma = r^2+a^2-a^2\sin^2\theta\) in the first term and \(2r_gr = r^2+a^2-\Delta\) in the second:

\[= (r^2+a^2)^2 - (r^2+a^2)a^2\sin^2\theta + (r^2+a^2)a^2\sin^2\theta - \Delta a^2\sin^2\theta = (r^2+a^2)^2 - \Delta a^2\sin^2\theta .\] The two middle terms cancel exactly, which is the whole content of the identity. \(\square\)

2. \(g_{t\varphi} = -2r_gra\sin^2\theta\,c/\Sigma\), so

\[\omega = -\frac{g_{t\varphi}}{g_{\varphi\varphi}} = \frac{2r_gra\sin^2\theta\,c/\Sigma}{\big[(r^2+a^2)^2-\Delta a^2\sin^2\theta\big]\sin^2\theta/\Sigma} = \frac{2r_gra\,c}{(r^2+a^2)^2 - \Delta a^2\sin^2\theta},\] with both \(\Sigma\) and \(\sin^2\theta\) cancelling — so \(\omega\) has no explicit \(\theta\) in it once written this way, only through \(r\). On \(r = r_+\), \(\Delta = 0\), so the denominator is \((r_+^2+a^2)^2\) and the numerator is \(2r_gr_+ac = (r_+^2+a^2)ac\), giving

\[\omega\big|_{r_+} = \frac{(r_+^2+a^2)ac}{(r_+^2+a^2)^2} = \frac{ac}{r_+^2+a^2} = \Omega_{\mathrm H},\] independent of \(\theta\). \(\square\)

3. \(g(\xi_\Omega,\xi_\Omega)\) is a quadratic in \(\Omega\) with roots \(\Omega_\pm = \omega \pm c\sqrt\Delta\sin\theta/g_{\varphi\varphi}\), so it factorises as \(g_{\varphi\varphi}(\Omega-\Omega_-)(\Omega-\Omega_+)\). At \(\Delta = 0\) the discriminant identity gives \(g_{t\varphi}^2 - g_{tt}g_{\varphi\varphi} = 0\), so the roots coincide at \(\Omega_- = \Omega_+ = \omega|_{r_+} = \Omega_{\mathrm H}\). Evaluating the quadratic at \(\Omega = \Omega_{\mathrm H}\) therefore gives \(g_{\varphi\varphi}(\Omega_{\mathrm H}-\Omega_{\mathrm H})^2 = 0\): \(\xi_{\mathrm H}\) is null. \(\square\) It is non-vanishing there (its \(\partial_t\) component is 1), so \(r = r_+\) is a Killing horizon of it.

4. A self-gravitating fluid in equilibrium cannot rotate rigidly unless it is specially prepared: differential rotation is the generic state, because different shells feel different effective gravity, and the Sun does exactly that. A horizon has no such freedom — part 2 shows every generator shares one \(\Omega_{\mathrm H}\). Hawking's rigidity theorem says this was never a property of Kerr in particular: any stationary, non-static, asymptotically flat horizon obeying the dominant energy condition is a Killing horizon of \(\partial_t + \Omega_{\mathrm H}\partial_\varphi\) with \(\Omega_{\mathrm H}\) constant.

Check. Dimensions: \(\omega\) is \(\mathrm{m\cdot m\cdot m\cdot m\,s^{-1}/m^4} = \mathrm{s^{-1}}\) ✓. Limit \(a\to0\): the numerator vanishes, \(\omega \equiv 0\), and Schwarzschild has no preferred sense of rotation ✓. Special case \(\theta = \pi/2\), \(r\) large: the denominator is \(\approx r^4\) and \(\omega \approx 2r_gac/r^3 = 2GJ/c^2r^3\), the Lense–Thirring rate ✓. Number: at \(r_+\) for the running hole the formula returns \(6363\ \mathrm{rad\,s^{-1}}\), as in P1 ✓.

Idea. \(\kappa\) is a limit of "local force times redshift", and the two definitions differ only by Killing's equation.

1. By the product rule and metric compatibility, \(\nabla^\mu(\xi^\nu\xi_\nu) = 2\xi^\nu\nabla^\mu\xi_\nu\). Killing's equation is \(\nabla_\mu\xi_\nu + \nabla_\nu\xi_\mu = 0\), so \(\nabla^\mu\xi_\nu = -\nabla_\nu\xi^\mu\) after raising and lowering. Hence \(\nabla^\mu(\xi\cdot\xi) = -2\xi^\nu\nabla_\nu\xi^\mu\), and setting this equal to \(-(2\kappa/c)\xi^\mu\) gives \(\xi^\nu\nabla_\nu\xi^\mu = (\kappa/c)\xi^\mu\). \(\square\) The second form says the generators are geodesics run with a non-affine parameter, the non-affinity being \(\kappa\).

3 (do this first; it is the concrete route). A static observer has \(u^\mu = (u^t,0,0,0)\) with \(g_{tt}(u^t)^2 = -c^2\), so \(u^t = f^{-1/2}\). Its acceleration is

\[a^\mu = u^\nu\nabla_\nu u^\mu \Rightarrow a^r = \Gamma^r_{tt}(u^t)^2 = \tfrac12ff'c^2\cdot\frac1f = \tfrac12f'c^2 ,\] with the other components zero (nothing depends on \(t\), and \(u^r = u^\theta = u^\varphi = 0\)). Its magnitude is

\[|a| = \sqrt{g_{rr}(a^r)^2} = \frac{1}{\sqrt f}\cdot\frac{f'c^2}{2} = \frac{c^2r_s}{2r^2\sqrt{f}} \xrightarrow{\ r\to r_s\ } \infty .\] Multiplying by the redshift factor \(\sqrt f\) — the ratio of force felt at infinity to force felt locally, unit 36 — removes the divergence:

\[\sqrt f\,|a| = \frac{f'c^2}{2} = \frac{c^2r_s}{2r^2} \xrightarrow{\ r\to r_s\ } \frac{c^2}{2r_s} = \frac{c^4}{4GM} .\]

2. The scalar form, first. On a Killing horizon \(\xi\) is hypersurface orthogonal, so Frobenius' theorem (unit 31) gives \(\xi_{[\mu}\nabla_\nu\xi_{\lambda]} = 0\). Writing \(A_{\nu\lambda} := \nabla_\nu\xi_\lambda\), which is antisymmetric by Killing's equation, that expands to \(\xi_\mu A_{\nu\lambda} = \xi_\nu A_{\mu\lambda} - \xi_\lambda A_{\mu\nu}\). Contract with \(A^{\nu\lambda}\): the two right-hand terms are equal, so

\[\xi_\mu A_{\nu\lambda}A^{\nu\lambda} = 2\,\xi_\nu A^{\nu\lambda}A_{\mu\lambda} = 2\frac{\kappa}{c}\,\xi^\lambda\nabla_\mu\xi_\lambda = -2\frac{\kappa}{c}\,\xi^\lambda\nabla_\lambda\xi_\mu = -2\frac{\kappa^2}{c^2}\,\xi_\mu ,\]

using \(\xi_\nu\nabla^\nu\xi^\lambda = (\kappa/c)\xi^\lambda\) twice and the antisymmetry of \(A\) once. Dividing by \(\xi_\mu\), \((\nabla^\mu\xi^\nu)(\nabla_\mu\xi_\nu) = -2\kappa^2/c^2\), that is \(\kappa^2 = -\tfrac12c^2(\nabla^\mu\xi^\nu)(\nabla_\mu\xi_\nu)\). \(\square\)

Now evaluate it. With \(\xi = \partial_t\), the covector is \(\xi_\nu = g_{\nu t}\), whose only component is \(\xi_t = -fc^2\). Then

\[\nabla_r\xi_t = \partial_r\xi_t - \Gamma^\lambda_{rt}\xi_\lambda = -f'c^2 - \Gamma^t_{rt}(-fc^2) = -f'c^2 + \frac{f'}{2f}fc^2 = -\tfrac12f'c^2 ,\]

and by antisymmetry \(\nabla_t\xi_r = +\tfrac12f'c^2\), with every other component zero. Hence

\[(\nabla^\mu\xi^\nu)(\nabla_\mu\xi_\nu) = 2\,g^{rr}g^{tt}(\nabla_r\xi_t)^2 = 2\,f\Big({-}\frac{1}{fc^2}\Big)\frac{f'^2c^4}{4} = -\frac{f'^2c^2}{2},\]

so \(\kappa^2 = -\tfrac12c^2\big({-}f'^2c^2/2\big) = \tfrac14f'^2c^4\), that is \(\kappa = \tfrac12f'c^2\) — at every \(r\), not only on the horizon, and identical to part 3's redshifted acceleration. On \(r = r_s\), \(f' = 1/r_s\) and \(\kappa = c^2/2r_s = c^4/4GM\). \(\square\) The scalar form is the standard one (Wald, eq. (12.5.2), where it reads \(\kappa^2 = -\tfrac12(\nabla^a\xi^b)(\nabla_a\xi_b)\) in geometrised units). The \(c^2\) is pure bookkeeping: the geometrised Killing field is \(\xi/c\), so \((\nabla\xi)^2\) here is \(c^2\) times the geometrised one, while \(\kappa\) as an acceleration is \(c^2\) times the geometrised \(\kappa\) — and \(c^4 = c^2\cdot c^2\) balances.

4. \(c^4 = 8.0776\times10^{33}\) and \(4GM = 4\times1.32712\times10^{21} = 5.3085\times10^{21}\), so \(\kappa = 1.5216\times10^{12}\ \mathrm{m\,s^{-2}}\). A winch at infinity lowering a one-kilogram mass to just above the horizon feels a tension tending to \(1.52\times10^{12}\) N — large, but finite, whereas the mass itself is being crushed by an unboundedly large local acceleration.

Check. Dimensions: \(f'c^2\) is \(\mathrm{m^{-1}\cdot m^2s^{-2}} = \mathrm{m\,s^{-2}}\) ✓. Limit: as \(M\to\infty\), \(\kappa\to0\) — a supermassive hole has a gentle horizon, which is why an infaller into Sgr A* is not torn apart at crossing (unit 36) ✓. Consistency with the Kerr formula: \(a\to0\) gives \(\kappa = c^2(r_+-r_-)/2(r_+^2+a^2) = c^2r_s/2r_s^2 = c^2/2r_s\) ✓. Number: \(1.52\times10^{12}\) is \(1.55\times10^{11}\) times Earth's \(g\) ✓.

Idea. The ansatz \(W = -Et + L\varphi + W_r + W_\theta\) uses the two Killing vectors for \(t\) and \(\varphi\) and then guesses that \(r\) and \(\theta\) also separate. There was no reason to expect it; that it works is the discovery.

1. Geodesics are the integral curves of the Hamiltonian flow of \(H = \tfrac12g^{\mu\nu}p_\mu p_\nu\) on \(T^*\mathcal M\) (unit 14), with \(p_\mu = \mu u_\mu\), so \(H = \tfrac12\mu^2 g^{\mu\nu}u_\mu u_\nu = -\tfrac12\mu^2c^2\), a constant. The Hamilton–Jacobi equation (unit 15) is \(\partial_\lambda S + H(x, \partial S/\partial x) = 0\). Since \(H\) carries no explicit \(\lambda\), \(\partial_\lambda S\) is constant and equals \(\tfrac12\mu^2c^2\), so \(S = \tfrac12\mu^2c^2\lambda + W(x)\) with

\[g^{\mu\nu}\partial_\mu W\,\partial_\nu W + \mu^2c^2 = 0 .\]

2. With \(\partial_tW = -E\), \(\partial_\varphi W = L\), \(\partial_rW = W_r'\), \(\partial_\theta W = W_\theta'\), multiply through by \(\Sigma\):

\[\Delta W_r'^2 + W_\theta'^2 - \frac{\big[(r^2+a^2)^2-\Delta a^2\sin^2\theta\big]E^2}{\Delta c^2} + \frac{4r_graEL}{\Delta c} + \frac{(\Delta-a^2\sin^2\theta)L^2}{\Delta\sin^2\theta} + \mu^2c^2\Sigma = 0,\] the \(t\varphi\) term appearing twice and with \(\partial_tW\,\partial_\varphi W = -EL\). Collect the three \(E,L\) terms over \(\Delta\), writing \(\mathcal E = E/c\) and \(2r_gr = r^2+a^2-\Delta\):

\[-\frac{(r^2+a^2)^2\mathcal E^2}{\Delta} + a^2\mathcal E^2\sin^2\theta + \frac{2(r^2+a^2-\Delta)a\mathcal EL}{\Delta} + \frac{L^2}{\sin^2\theta} - \frac{a^2L^2}{\Delta} .\] The terms with \(1/\Delta\) assemble into \(-\big[(r^2+a^2)\mathcal E - aL\big]^2/\Delta\), and the rest into \(a^2\mathcal E^2\sin^2\theta - 2a\mathcal EL + L^2/\sin^2\theta = (L - a\mathcal E\sin^2\theta)^2/\sin^2\theta\). Rearranged, that is the displayed equation. \(\square\)

3. Expand the \(\theta\) fraction and add \(\mu^2c^2a^2\cos^2\theta\) from \(\mu^2c^2\Sigma\):

\[\frac{L^2}{\sin^2\theta} - 2a\mathcal EL + a^2\mathcal E^2\sin^2\theta + \mu^2c^2a^2\cos^2\theta = \cos^2\theta\Big[a^2\big(\mu^2c^2-\mathcal E^2\big) + \frac{L^2}{\sin^2\theta}\Big] + \big(L - a\mathcal E\big)^2 .\] So the equation reads \(F(r) + G(\theta) = 0\) with

\[G(\theta) = W_\theta'^2 + \cos^2\theta\Big[a^2\big(\mu^2c^2-\mathcal E^2\big) + \frac{L^2}{\sin^2\theta}\Big] + (L-a\mathcal E)^2,\] \[F(r) = \Delta W_r'^2 + \mu^2c^2r^2 - \frac{\big[(r^2+a^2)\mathcal E - aL\big]^2}{\Delta} .\] Each must be constant. Defining Carter's constant as the \(\theta\)-part without its \((L-a\mathcal E)^2\) tail,

\[\boxed{\;Q := W_\theta'^2 + \cos^2\theta\Big[a^2\big(\mu^2c^2-\mathcal E^2\big) + \frac{L^2}{\sin^2\theta}\Big]\;}\] the two ordinary differential equations are

\[W_\theta'^2 = Q - \cos^2\theta\Big[a^2\big(\mu^2c^2-\mathcal E^2\big) + \frac{L^2}{\sin^2\theta}\Big], \qquad \Delta W_r'^2 = \frac{\big[(r^2+a^2)\mathcal E - aL\big]^2}{\Delta} - \mu^2c^2r^2 - Q - (L-a\mathcal E)^2 .\]

4. At \(a = 0\), \(Q = p_\theta^2 + L^2\cot^2\theta\). On a sphere the total angular momentum satisfies \(L_{\text{tot}}^2 = p_\theta^2 + p_\varphi^2/\sin^2\theta\) and \(L_z = p_\varphi = L\), so \(L_{\text{tot}}^2 - L_z^2 = p_\theta^2 + L^2(\sin^{-2}\theta - 1) = p_\theta^2 + L^2\cot^2\theta = Q\). For the sign: if \(E \lt \mu c^2\) then \(\mu^2c^2 - \mathcal E^2 = \mu^2c^2 - E^2/c^2 \gt 0\), so every term in \(Q\) is non-negative and \(Q \ge 0\). Equality forces \(W_\theta' = p_\theta = 0\) and \(\cos\theta = 0\) at every \(\lambda\) — that is, the orbit sits in the equatorial plane with no \(\theta\)-momentum. Conversely such an orbit has \(Q = 0\). \(\square\)

5. Unit 13's theorem produces a conserved quantity from a one-parameter group of transformations of configuration space, and the associated constant is linear in the momenta. \(Q\) is quadratic in the momenta and comes from a Killing tensor — a symmetric \((0,2)\)-tensor with \(\nabla_{(\lambda}K_{\mu\nu)} = 0\) — which generates a symmetry of the geodesic flow on \(T^*\mathcal M\) but no isometry of \(\mathcal M\) at all.

Check. Dimensions: \(W_\theta'\) is \(\mathrm{J\,s}\) and so is \(L\); \(a\mathcal E\) is \(\mathrm{m\cdot kg\,m\,s^{-1}} = \mathrm{J\,s}\) ✓, so \(Q\) is \((\mathrm{J\,s})^2\) throughout. Limit \(a\to0\) is part 4 ✓. Special case: an equatorial orbit has \(Q = 0\), and the radial equation then reduces to \(\Delta W_r'^2 = \big[(r^2+a^2)\mathcal E - aL\big]^2/\Delta - \mu^2c^2r^2 - (L-a\mathcal E)^2\), which at \(a = 0\) is unit 36's radial equation \(\dot r^2 = \mathcal E^2/\mu^2 - f(c^2 + L^2/\mu^2r^2)\) after dividing by \(\mu^2\) ✓.

Idea. Integrability is a counting statement with three conditions attached, and the value of the problem is in checking all three rather than the count alone.

1. Liouville–Arnold: let \((\mathcal P, \omega)\) be a \(2n\)-dimensional symplectic manifold and \(f_1 = H, f_2, \dots, f_n\) smooth functions with \(\{f_i, f_j\} = 0\) for all \(i, j\) and \(\mathrm df_1, \dots, \mathrm df_n\) linearly independent on a level set \(M_c = \{f_i = c_i\}\). If \(M_c\) is compact and connected it is diffeomorphic to an \(n\)-torus; the flow of \(H\) on it is linear in suitable coordinates, and there exist action–angle variables in a neighbourhood. For geodesics on a 4-manifold the phase space is \(T^*\mathcal M\), of dimension \(8 = 2n\), so \(n = 4\) and four constants are needed.

2. In canonical coordinates \((x^\mu, p_\mu)\), \(\{f, g\} = \partial_{x^\mu}f\,\partial_{p_\mu}g - \partial_{p_\mu}f\,\partial_{x^\mu}g\). Take \(f = E = -p_t\): then \(\partial_{x^\mu}E = 0\) and \(\partial_{p_\mu}E = -\delta^\mu_t\), so \(\{E, g\} = \partial_tg\). Hence \(\{E, H\} = \partial_tH = \tfrac12(\partial_tg^{\mu\nu})p_\mu p_\nu = 0\), because Boyer–Lindquist components are \(t\)-independent; and \(\{E, L\} = \partial_tp_\varphi = 0\). Identically with \(\varphi\): \(\{L, g\} = -\partial_\varphi g\), so \(\{L, H\} = 0\) by axisymmetry. \(\square\)

3. \(Q = K_{\mu\nu}p^\mu p^\nu\) up to the constant combination \((L - a\mathcal E)^2\), which commutes with everything in sight since \(E\) and \(L\) do. By the same computation as part 2, \(\{E, Q\} = \partial_tQ = (\partial_tK_{\mu\nu})p^\mu p^\nu\), which vanishes because \(\mathcal L_{\partial_t}K = 0\); likewise \(\{L, Q\} = 0\). And \(\{H, Q\} = 0\) is exactly the statement that \(Q\) is constant along the flow of \(H\), which is the Killing-tensor theorem. \(\square\)

4. \(\mathrm dE\) and \(\mathrm dL\) involve only \(p_t, p_\varphi\) and are always independent of each other. \(\mathrm dH\) and \(\mathrm dQ\) both involve \(p_\theta\), but only \(H\) involves \(p_r\), so a dependence needs the \(p_\theta\)-parts to align — which happens where \(p_\theta = 0\) and \(\cos\theta = 0\) simultaneously, that is on the equatorial orbits, \(Q = 0\). That is a measure-zero set, and nothing is lost: equatorial Kerr geodesics live on the invariant submanifold \(\theta = \pi/2\), \(p_\theta = 0\), which is symplectic of dimension \(6 = 2n\) with \(n = 3\), and there \(H, E, L\) are three independent constants in involution — exactly the number Liouville–Arnold asks for — so those orbits are solved by quadrature directly, as unit 37 did for Schwarzschild.

5. First: put a spin on the test body. The Mathisson–Papapetrou equations are not integrable and their numerics show unit 17's islands and chaotic sea. Second: the two-body problem. Two comparable holes have no known constants beyond the ten Poincaré ones, radiate away their energy, and are solved only by numerical relativity — which is why unit 38's templates are computed rather than written down.

Check. Count: \(2n = 8\), four constants, four commuting flows, level sets of dimension \(8 - 4 = 4\) ✓ — the right dimension for a 4-torus. Limit: at \(a = 0\), Schwarzschild has \(H, E, L_x, L_y, L_z\), five constants, of which only four can be in involution (\(\{L_x, L_y\} = L_z \ne 0\)); the system is superintegrable, and superintegrability is why the orbit is planar. Kerr loses that and keeps bare integrability ✓.

Idea. \(Q\) is fixed by the orbit's highest latitude, and at \(a = 0\) that is exactly the Newtonian statement \(L_z = L\cos i\).

1. With \(p_\theta = 0\) at \(\theta_{\min} = 50°\),

\[\hat Q = \cos^2 50°\Big[a_*^2r_g^2(1 - \hat E^2) + \frac{\hat L^2}{\sin^2 50°}\Big] .\] Now \(1 - \hat E^2 = 1 - 0.8836 = 0.1164\), so \(a_*^2(1-\hat E^2) = 0.81\times0.1164 = 0.094284\); and \(\hat L^2/\sin^2 50° = 16/0.586824 = 27.265423\). The bracket is \(27.359707\,r_g^2\), and

\[\hat Q = 0.413176\times27.359707\,r_g^2 = 11.3044\,r_g^2 .\] With \(r_g = 14766\) m, \(r_g^2 = 2.18034\times10^{8}\ \mathrm{m^2}\), so \(\hat Q = 2.4647\times10^{9}\ \mathrm{m^2}\).

2. \(\hat L_{\text{tot}} = \sqrt{11.3044 + 16} = \sqrt{27.3044} = 5.22536\,r_g\), so \(\cos i = 4.0/5.22536 = 0.765495\) and \(i = 40.049°\).

3. At \(a_* = 0\) the bracket loses its first term: \(\hat Q_0 = 0.413176\times27.265423 = 11.2654\,r_g^2\), \(\hat L_{\text{tot}} = \sqrt{27.2654} = 5.22162\,r_g\), \(\cos i = 4.0/5.22162 = 0.766043\), \(i = 40.000°\) — exactly \(90° - 50°\). Proof. With \(a = 0\), \(\hat Q_0 = \hat L^2\cos^2\theta_{\min}/\sin^2\theta_{\min}\), so

\[\hat Q_0 + \hat L^2 = \hat L^2\Big(\frac{\cos^2\theta_{\min}}{\sin^2\theta_{\min}} + 1\Big) = \frac{\hat L^2}{\sin^2\theta_{\min}},\qquad \cos i = \frac{\hat L}{\hat L/\sin\theta_{\min}} = \sin\theta_{\min} = \cos(90° - \theta_{\min}).\ \square\]

4. The shift is \(40.049° - 40.000° = 0.049°\), about three arcminutes. \(\hat Q\) is larger for the spinning hole, by \(0.039\,r_g^2\), because for a bound orbit \(1 - \hat E^2 \gt 0\) and the term \(a^2\cos^2\theta(1-\hat E^2)\) it adds is positive. For an unbound orbit (\(\hat E \gt 1\)) that term changes sign and \(\hat Q\) can even be negative — those are the "vortical" geodesics that never reach the equatorial plane, and they exist only when \(a \ne 0\).

5. \(Q\) depends on \(\theta\) only through \(\cos^2\theta\), so the \(\theta\)-motion is symmetric about \(\theta = 90°\): the turning points are \(\theta = 50°\) and \(\theta = 130°\), and the orbit oscillates between them, crossing the equator twice per polar period.

Check. Dimensions: \(\hat Q\), \(\hat L^2\) and \(a^2\) are all areas, and \(1 - \hat E^2\) is dimensionless ✓. Limit \(\theta_{\min}\to90°\): \(\cos^2\theta_{\min}\to0\), so \(\hat Q\to0\) and \(i\to0\) — an equatorial orbit ✓. Limit \(\theta_{\min}\to0\): \(\hat Q\to\infty\) and \(i\to90°\), a polar orbit, which indeed requires \(\hat L\to0\) at fixed \(\hat L_{\text{tot}}\) ✓. Number: at \(a_* = 0\) the inclination came out at \(40.000°\), against \(90° - \theta_{\min} = 40°\) ✓.

Idea. Every condition is a contraction of \(T_{\mu\nu}\) with one or two vectors, and for a perfect fluid the only two scalars available are \(u\cdot v\) and \(v\cdot v\).

1. NEC. For null \(k\), \(g_{\mu\nu}k^\mu k^\nu = 0\), so \(T_{\mu\nu}k^\mu k^\nu = (\rho + p/c^2)(u\cdot k)^2\). Since \((u\cdot k)^2 \gt 0\) for timelike \(u\) and non-zero null \(k\), the condition is \(\rho + p/c^2 \ge 0\), i.e. \(\rho c^2 + p \ge 0\).

WEC. Let \(u_{\text{obs}}\) be timelike with \(u\cdot u_{\text{obs}} = -\gamma c^2\), \(\gamma \ge 1\), and \(u_{\text{obs}}\cdot u_{\text{obs}} = -c^2\). Then

\[T_{\mu\nu}u^\mu_{\text{obs}}u^\nu_{\text{obs}} = (\rho + p/c^2)\gamma^2c^4 - pc^2 = c^4\Big[\gamma^2\rho + (\gamma^2-1)\frac{p}{c^2}\Big].\] At \(\gamma = 1\) this is \(\rho c^4\), so \(\rho \ge 0\) is necessary. As \(\gamma\to\infty\) the bracket is dominated by \(\gamma^2(\rho + p/c^2)\), so \(\rho c^2 + p \ge 0\) is necessary. Both together make the bracket a sum of non-negative terms, hence sufficient.

SEC. \(T = g^{\mu\nu}T_{\mu\nu} = (\rho+p/c^2)(-c^2) + 4p = -\rho c^2 + 3p\), and \(T_{\mu\nu}u^\mu u^\nu = (\rho+p/c^2)c^4 - pc^2 = \rho c^4\). So

\[\big(T_{\mu\nu}-\tfrac12Tg_{\mu\nu}\big)u^\mu u^\nu = \rho c^4 - \tfrac12(-\rho c^2+3p)(-c^2) = \tfrac{c^2}{2}\big(\rho c^2 + 3p\big),\] so the condition is \(\rho c^2 + 3p \ge 0\), together with \(\rho c^2 + p \ge 0\) from boosted observers (the same \(\gamma\to\infty\) argument).

DEC. The flux measured by \(u\) is \(-T^\mu{}_\nu u^\nu\); in the comoving frame it is \((\rho c^2, \vec0)\) and in a boosted frame it acquires a spatial part proportional to \((\rho c^2 + p)v\). Requiring it to be causal gives \(|p| \le \rho c^2\), in addition to the WEC.

2.

• Dust, \(p = 0\), \(\rho \gt 0\): \(\rho c^2 \ge 0\), \(\rho c^2 + 3p = \rho c^2 \ge 0\), \(\rho c^2 \ge 0 = |p|\). All four, none marginal.
• Radiation, \(p = \rho c^2/3\): NEC \(\tfrac43\rho c^2\), SEC \(2\rho c^2\), DEC \(\rho c^2 \ge \rho c^2/3\). All four.
• Stiff fluid, \(p = \rho c^2\): NEC \(2\rho c^2\), SEC \(4\rho c^2\), DEC \(\rho c^2 \ge \rho c^2\) — satisfied marginally. All four.
• \(\Lambda \gt 0\), \(\rho_\Lambda c^2 = \Lambda c^4/8\pi G \gt 0\), \(p = -\rho_\Lambda c^2\): NEC \(\rho c^2 + p = 0\), marginal ✓; WEC \(\rho \gt 0\) and the same zero ✓; DEC \(\rho c^2 = |p|\), marginal ✓; SEC \(\rho c^2 + 3p = -2\rho_\Lambda c^2 \lt 0\), violated.

3. Penrose (1965). If \((\mathcal M, g)\) is globally hyperbolic with a non-compact Cauchy surface, if \(R_{\mu\nu}k^\mu k^\nu \ge 0\) for every null \(k\) — the NEC, through Einstein's equation — and if \(\mathcal M\) contains a closed trapped surface, then \(\mathcal M\) is future null geodesically incomplete. \(\Lambda \gt 0\) satisfies the NEC (marginally), so this theorem is untouched: black holes in an accelerating universe still have singularities.

Hawking (1967). If \(R_{\mu\nu}u^\mu u^\nu \ge 0\) for every timelike \(u\) — the SEC — if there are no closed timelike curves, and if there is a compact spacelike hypersurface with expansion \(\theta \le -C \lt 0\) everywhere, then every past-directed timelike geodesic is incomplete. \(\Lambda \gt 0\) violates the SEC exactly, so this theorem's energy condition hypothesis fails and its conclusion cannot be drawn.

4. De Sitter space fails the strong energy condition and nothing else: it is a vacuum solution with \(\Lambda \gt 0\), has compact Cauchy surfaces in the global slicing, and has no closed timelike curves. Since it is geodesically complete, the SEC hypothesis cannot be dropped or weakened to the NEC — the theorem's conclusion is genuinely false without it. A hypothesis with an explicit counterexample is not a technicality.

5. \(\rho c^2 + p = \big(\tfrac12\dot\phi^2 + V\big) + \big(\tfrac12\dot\phi^2 - V\big) = \dot\phi^2 \ge 0\): the NEC always holds. And \(\rho c^2 + 3p = \tfrac12\dot\phi^2 + V + \tfrac32\dot\phi^2 - 3V = 2\dot\phi^2 - 2V\), which is negative exactly when \(V \gt \dot\phi^2\) — the slow-roll condition. So inflation is a SEC-violating epoch by construction. \(\square\)

Check. Dimensions: every entry is an energy density, \(\mathrm{J\,m^{-3}}\) ✓. Ordering: DEC \(\Rightarrow\) WEC \(\Rightarrow\) NEC is visible in the derived inequalities, and SEC sits outside that chain — it implies the NEC but says nothing about \(\rho \ge 0\) ✓. Number: with unit 39's \(\Omega_\Lambda = 0.685\) and \(\Omega_{\mathrm m} = 0.315\), today's mix has \(\rho c^2 + 3p = (0.315 - 2\times0.685)\rho_{\text{crit}}c^2 = -1.055\,\rho_{\text{crit}}c^2 \lt 0\): the SEC is violated by the actual universe, not merely by a hypothetical fluid ✓.

Idea. The nonlinear term \(-\tfrac12\theta^2\) does all the work, and it is a Riccati term: the substitution \(u = 1/\theta\) linearises it.

1. With \(\omega = 0\), \(\sigma^2 \ge 0\) and \(R_{\mu\nu}k^\mu k^\nu \ge 0\),

\[\frac{\mathrm d\theta}{\mathrm d\lambda} \le -\tfrac12\theta^2 .\] Suppose \(\theta\) stays finite and negative on \([0, \lambda]\). Put \(u = 1/\theta \lt 0\). Then

\[\frac{\mathrm du}{\mathrm d\lambda} = -\frac{1}{\theta^2}\frac{\mathrm d\theta}{\mathrm d\lambda} \ge \frac{1}{\theta^2}\cdot\tfrac12\theta^2 = \tfrac12 ,\] so \(u(\lambda) \ge u(0) + \lambda/2 = 1/\theta_0 + \lambda/2\). The right-hand side reaches \(0\) at \(\lambda = -2/\theta_0 = 2/|\theta_0|\). But \(u\) is negative wherever \(\theta\) is finite and negative, so the assumption fails before then: \(u\to0^-\) and \(\theta\to-\infty\) at some \(\lambda \le 2/|\theta_0|\). \(\square\) (\(\theta\) cannot become positive first: it would have to pass through \(0\), and \(\theta' \le -\tfrac12\theta^2 \le 0\) forbids increase.)

2. The twist vanishes exactly when the congruence is hypersurface orthogonal, by Frobenius' theorem (unit 31): a distribution is integrable iff it is involutive, and the vanishing of \(\omega\) is the involutivity of the screen distribution. The null geodesics orthogonal to a closed 2-surface \(S\) are orthogonal to the family of surfaces obtained by dragging \(S\) along them, so they are hypersurface orthogonal by construction and \(\omega \equiv 0\). Choosing this congruence is what makes Penrose's proof work, since a rotating congruence could defocus.

3. In flat space the ingoing null normals of a sphere of radius \(r\) converge on the centre. Parametrise by affine \(\lambda\) with \(k = \partial_t - \partial_r\) in suitable units, so the cross-sectional area at parameter \(\lambda\) is \(\mathcal A \propto (r-\lambda)^2\) and

\[\theta = \frac{\dot{\mathcal A}}{\mathcal A} = \frac{-2}{r-\lambda},\qquad \theta_0 = -\frac2r .\] The bound gives \(\lambda \le 2/|\theta_0| = r\), and indeed \(\theta\to-\infty\) at \(\lambda = r\) exactly. The bound is attained because the inequality is an equality here: \(\sigma = \omega = 0\) and \(R_{\mu\nu} = 0\), so \(\theta' = -\tfrac12\theta^2\) exactly. The caustic is the centre of the sphere, where all the rays meet — an entirely ordinary event.

4. Penrose applies part 1 to the two null congruences orthogonal to a trapped surface, where \(\theta_0 \lt 0\) by definition: focusing forces caustics within finite affine parameter, the generators of \(\partial J^+(S)\) therefore all terminate, and \(\partial J^+(S)\) is compact — which contradicts its having to inject into a non-compact Cauchy surface, unless some generator is inextendible short of its caustic. A caustic is not a singularity: nothing diverges there but the congruence's own cross-section, as part 3 shows with the centre of an ordinary sphere in flat space; the singularity appears only from the global contradiction.

5. With \(\theta_0 = 0\) and \(R_{\mu\nu}k^\mu k^\nu \ge \epsilon\), the equation gives \(\theta' \le -\epsilon\) while \(\theta\) is near zero, so \(\theta(\lambda) \le -\epsilon\lambda\). At \(\lambda_1\) the expansion is at most \(-\epsilon\lambda_1\); applying part 1 from there costs a further \(2/(\epsilon\lambda_1)\). The total is at most \(g(\lambda_1) = \lambda_1 + 2/(\epsilon\lambda_1)\), minimised at \(\lambda_1 = \sqrt{2/\epsilon}\), giving

\[\lambda_{\text{caustic}} \le 2\sqrt{\frac{2}{\epsilon}} .\]

Check. Dimensions: \(\theta\) is \(\mathrm m^{-1}\), so \(2/|\theta_0|\) is a length ✓; \(\epsilon\) is \(\mathrm m^{-2}\), so \(2\sqrt{2/\epsilon}\) is a length ✓. Limit \(\epsilon\to0\) in part 5: the bound diverges, correctly — with no matter and no initial convergence, parallel rays stay parallel forever ✓. Number: for the ingoing normals of a sphere of radius 1 m in flat space, the caustic is at \(\lambda = 1\) m ✓, and the bound is \(2/2 = 1\) m, attained ✓.

Idea. There is one constant in the problem, \(\sqrt{4\pi\varepsilon_0G} = 8.617\times10^{-11}\ \mathrm{C\,kg^{-1}}\), and everything is a comparison against it.

1. \([Q^2/4\pi\varepsilon_0] = \mathrm{N\,m^2} = \mathrm{kg\,m^3s^{-2}}\), so \([GQ^2/4\pi\varepsilon_0] = \mathrm{(m^3kg^{-1}s^{-2})(kg\,m^3s^{-2})} = \mathrm{m^6s^{-4}}\), and dividing by \(c^4 = \mathrm{m^4s^{-4}}\) leaves \(\mathrm{m^2}\) ✓. Extremality:

\[\frac{GQ^2}{4\pi\varepsilon_0c^4} = \frac{G^2M^2}{c^4} \iff Q^2 = 4\pi\varepsilon_0GM^2 \iff |Q| = M\sqrt{4\pi\varepsilon_0G}.\] Numerically \(4\pi\varepsilon_0 = 1.112650\times10^{-10}\), times \(G = 6.67430\times10^{-11}\), gives \(7.42629\times10^{-21}\), whose square root is \(8.6176\times10^{-11}\ \mathrm{C\,kg^{-1}}\).

2. \(r_Q/r_g = Q/(m\sqrt{4\pi\varepsilon_0G})\). For an electron,

\[\frac{r_Q}{r_g} = \frac{1.6021766\times10^{-19}} {(9.1093837\times10^{-31})(8.6176\times10^{-11})} = 2.041\times10^{21}.\] Over-extremal, by twenty-one orders of magnitude. (An electron is not a black hole for other reasons — its Compton wavelength dwarfs both lengths — but taken at face value the Reissner–Nordström solution with its parameters has no horizon.)

3. \(M = 10M_\odot = 1.98841\times10^{31}\) kg gives \(Q_{\max} = 1.98841\times10^{31}\times8.6176\times10^{-11} = 1.714\times10^{21}\) C. For the Earth, \(M_\oplus = 5.972\times10^{24}\) kg gives \(Q_{\max} = 5.146\times10^{14}\) C.

4. Balance on a proton at radius \(r\): \(Qe/4\pi\varepsilon_0r^2 = GMm_{\mathrm p}/r^2\), and \(r\) cancels, leaving

\[Q_{\text{bal}} = \frac{4\pi\varepsilon_0GMm_{\mathrm p}}{e} = \frac{(1.11265\times10^{-10})(1.32712\times10^{21})(1.6726219\times10^{-27})}{1.6021766\times10^{-19}} = 1.542\times10^{3}\ \mathrm C .\] As a fraction, \(1.542\times10^{3}/1.714\times10^{21} = 9.0\times10^{-19}\).

5. A hole charged to more than a part in \(10^{18}\) of extremality repels protons harder than it attracts them, so it preferentially swallows electrons from the surrounding plasma and discharges within moments. Charge is therefore an exact parameter of the theory with no astrophysical realisation, and every real hole is described by \(M\) and \(J\) alone.

Check. Dimensions: \(4\pi\varepsilon_0GMm_{\mathrm p}/e\) is \(\mathrm{(C^2N^{-1}m^{-2})(m^3s^{-2})(kg)/C} = \mathrm C\) ✓ once \(GM\) is used as a single factor. Limit: \(Q\to0\) gives Schwarzschild in every formula ✓. Cross-check on part 2: the electron–electron electrostatic-to-gravitational force ratio is \(e^2/4\pi\varepsilon_0Gm_{\mathrm e}^2 = (r_Q/r_g)^2 = 4.17\times10^{42}\), which is the standard \(\sim10^{42}\) ✓.

Idea. Everything follows from one causality fact about the horizon. The algebra of part 3 is what converts an inequality on increments into a monotone function, and a monotone function is what gives a ceiling.

1. The lemma. Let \(k\) be future-directed null and \(p\) future-directed causal. In an orthonormal frame choose axes so that \(k = \alpha(1, 1, 0, 0)\) with \(\alpha \gt 0\), and write \(p = (p^0, \vec p)\) with \(p^0 \ge |\vec p|\) (causal, future-directed). Then

\[g(k, p) = \alpha\big(-p^0 + p^1\big) \le \alpha\big(-p^0 + |\vec p|\big) \le 0 . \]

The inequality. A particle crossing the horizon has future-directed causal \(p\), and \(\xi_{\mathrm H}\) is future-directed null there (P3, part 3). So \(-g(\xi_{\mathrm H}, p) \ge 0\). Expand:

\[-g(\xi_{\mathrm H}, p) = -g(\partial_t, p) - \Omega_{\mathrm H}\,g(\partial_\varphi, p) = E - \Omega_{\mathrm H}L .\] The hole's mass and spin increase by \(\delta(Mc^2) = E\) and \(\delta J = L\), so \(\delta(Mc^2) \ge \Omega_{\mathrm H}\delta J\). \(\square\)

2. With \(G = c = 1\), \(r_+ = M + \sqrt{M^2-a^2}\) and \(a = J/M\), so \(\sqrt{M^2 - J^2/M^2} = x/M\) with \(x := \sqrt{M^4-J^2}\), and \(r_+ = (M^2+x)/M\). Hence

\[M_{\mathrm{irr}}^2 = \frac{Mr_+}{2} = \frac{M^2+x}{2} = \tfrac12\big(M^2 + \sqrt{M^4-J^2}\big).\] Rearranging, \(2M_{\mathrm{irr}}^2 - M^2 = x\); squaring, \(4M_{\mathrm{irr}}^4 - 4M_{\mathrm{irr}}^2M^2 + M^4 = M^4 - J^2\), so \(4M_{\mathrm{irr}}^2(M_{\mathrm{irr}}^2 - M^2) = -J^2\) and

\[M^2 = M_{\mathrm{irr}}^2 + \frac{J^2}{4M_{\mathrm{irr}}^2}. \ \square\]

3. Differentiate \(M_{\mathrm{irr}}^2 = \tfrac12(M^2 + x)\) with \(x = \sqrt{M^4-J^2}\), so \(\delta x = (4M^3\delta M - 2J\delta J)/2x = (2M^3\delta M - J\delta J)/x\):

\[2M_{\mathrm{irr}}\,\delta M_{\mathrm{irr}} = M\,\delta M + \frac{2M^3\delta M - J\delta J}{2x} = \frac{M(2x + 2M^2)\,\delta M - J\,\delta J}{2x} = \frac{M(M^2+x)\,\delta M - \tfrac12J\,\delta J}{x}.\] Since \(M_{\mathrm{irr}}, x \gt 0\) for a non-extremal hole,

\[\delta M_{\mathrm{irr}} \ge 0 \iff \delta M \ge \frac{J\,\delta J}{2M(M^2+x)} .\] But \(\Omega_{\mathrm H} = a/(r_+^2+a^2) = a/2Mr_+\) and \(r_+ = (M^2+x)/M\), \(a = J/M\), so

\[\Omega_{\mathrm H} = \frac{J/M}{2M(M^2+x)/M} = \frac{J}{2M(M^2+x)},\] the very same coefficient. Hence \(\delta M_{\mathrm{irr}} \ge 0 \iff \delta M \ge \Omega_{\mathrm H}\delta J\). \(\square\) Since \(A = 16\pi M_{\mathrm{irr}}^2\), this is also \(\delta A \ge 0\).

4. \(M_{\mathrm{irr}}\) never decreases, and the best case is that it stays constant. The extraction stops at \(J = 0\), where the hole is Schwarzschild with mass \(M_{\mathrm{irr}}\). So

\[\eta = 1 - \frac{M_{\mathrm{irr}}}{M} = 1 - \sqrt{\frac{r_+}{2r_g}} = 1 - \sqrt{\tfrac12\big(1 + \sqrt{1-a_*^2}\big)} .\]

• \(a_* = 0.9\): \(\sqrt{1-0.81} = 0.435890\), \(\tfrac12(1.435890) = 0.717945\), \(\sqrt{\ } = 0.847316\), \(\eta = 0.15268 = \mathbf{15.27\%}\).
• \(a_* = 0.998\): \(\sqrt{1-0.996004} = 0.063214\), \(\tfrac12(1.063214) = 0.531607\), \(\sqrt{\ } = 0.729114\), \(\eta = \mathbf{27.09\%}\).
• \(a_* = 1\): \(\eta = 1 - 1/\sqrt2 = 0.292893 = \mathbf{29.29\%}\).

5. \(Mc^2 = 1.98841\times10^{31}\times8.98755\times10^{16} = 1.7871\times10^{48}\) J, so \(\eta Mc^2 = 0.152684\times1.7871\times10^{48} = 2.729\times10^{47}\) J. That is more than the \(5.72\%\) of unit 37 — but the two are not competing, because unit 37's efficiency is the fraction of the infalling matter's rest energy released before it crosses the horizon, whereas \(\eta\) is a fraction of the hole's own mass, taken out and leaving the hole lighter. Accretion makes a hole heavier while radiating; the Penrose process makes it lighter.

Check. Dimensions: \(\eta\) is dimensionless ✓; \(\eta Mc^2\) is joules ✓. Limits: \(a_*\to0\) gives \(\eta\to0\), nothing to extract from a Schwarzschild hole ✓; \(J = 0\) in the Christodoulou relation gives \(M = M_{\mathrm{irr}}\) ✓; extremal \(J = M^2\) gives \(M_{\mathrm{irr}} = M/\sqrt2\) and the relation returns \(M^2/2 + M^2/2 = M^2\) ✓. Cross-check on part 5: \(2.729\times10^{47}\,\mathrm J/c^2 = 3.04\times10^{30}\,\mathrm{kg} = 1.53\,M_\odot\), and \((1 - 0.847316)\times10\,M_\odot = 1.53\,M_\odot\) ✓.

Idea. Area is not mass. Mass adds and can be radiated away; area adds and cannot decrease, and the merger is the place where the two come apart visibly.

1. \(A = 4\pi(r_+^2+a^2) = 8\pi r_gr_+\) and \(r_+ = r_g(1+\sqrt{1-a_*^2})\), so

\[A = 8\pi r_g^2\big(1 + \sqrt{1-a_*^2}\big),\qquad r_g = \frac{GM}{c^2}.\] At \(a_* = 0\), \(A = 16\pi r_g^2 = 16\pi G^2M^2/c^4\) ✓, which is \(4\pi r_s^2\) as it must be.

2. Work in units of \(r_{g,\odot}^2\) with \(r_{g,\odot} = GM_\odot/c^2 = 1.32712\times10^{20}/8.98755\times10^{16} = 1476.6\) m, so \(r_{g,\odot}^2 = 2.18034\times10^{6}\ \mathrm{m^2}\).

• Initial: \(A_{\mathrm i}/r_{g,\odot}^2 = 16\pi(36^2 + 29^2) = 16\pi(1296 + 841) = 16\pi(2137) = 1.07417\times10^{5}\), so \(A_{\mathrm i} = 2.342\times10^{11}\ \mathrm{m^2}\).
• Final: \(\sqrt{1 - 0.67^2} = \sqrt{0.5511} = 0.742361\), so \(A_{\mathrm f}/r_{g,\odot}^2 = 8\pi(62)^2(1.742361) = 8\pi(3844)(1.742361) = 1.68330\times10^{5}\), so \(A_{\mathrm f} = 3.670\times10^{11}\ \mathrm{m^2}\).

3. The area rose by \(A_{\mathrm f}/A_{\mathrm i} = 1.5671\), that is \(\mathbf{+56.7\%}\). The mass fell from \(65\,M_\odot\) to \(62\,M_\odot\), that is \(\mathbf{-4.6\%}\). Entirely compatible: the second law constrains \(A\), not \(M\), and the merger is the cleanest demonstration that the two are independent. Three solar masses of rest energy left as radiation while the horizon grew by more than half.

4. \(S/k_{\mathrm B} = c^3A/4G\hbar\) with \(c^3/4G\hbar = 2.69440\times10^{25}/(2.81549\times10^{-44}) = 9.5702\times10^{68}\ \mathrm{m^{-2}}\):

\[\frac{S_{\mathrm i}}{k_{\mathrm B}} = 2.242\times10^{80},\qquad \frac{S_{\mathrm f}}{k_{\mathrm B}} = 3.513\times10^{80}.\] The comparison tests the second law — Hawking's area theorem, in its thermodynamic reading. It is a measurement, not a calculation: LIGO published a direct test of the area law from the inspiral and ringdown portions of this event in 2021 and found it satisfied at \(95\%\) confidence.

5. Set \(A_{\mathrm f} = A_{\mathrm i}\): \(8\pi(3844)(1+\sqrt{1-a_*^2}) = 16\pi(2137)\), so \(1 + \sqrt{1-a_*^2} = 2(2137)/3844 = 1.11186\) and \(\sqrt{1-a_*^2} = 0.11186\), giving \(a_* = 0.99372\). Any \(a_*\) above that would mean a decrease. Such a spin is allowed by the Kerr bound (\(a_* \le 1\)) but is not what a merger of two non-spinning holes produces: the final spin comes from the orbital angular momentum at plunge and sits near \(0.69\) for equal masses. So the area theorem is not violated, and — importantly — it could have been, had nature chosen differently. The test has teeth.

Check. Dimensions: \(A\) is \(\mathrm m^2\) ✓, \(c^3/G\hbar\) is \(\mathrm m^{-2}\) ✓, \(S/k_{\mathrm B}\) dimensionless ✓. Limit: if the final hole were non-spinning, \(A_{\mathrm f}/r_{g,\odot}^2 = 16\pi(3844) = 1.932\times10^{5}\), a rise of \(80\%\) — larger, as expected, since spin shrinks the horizon at fixed mass ✓. Sanity: \(M_{\mathrm{irr,f}} = 62\sqrt{1.742361/2} = 57.9\,M_\odot\), against \(\sqrt{36^2+29^2} = 46.2\,M_\odot\) for the initial pair combined in quadrature — and quadrature is the right combination, because area adds and \(A \propto M_{\mathrm{irr}}^2\) ✓.