Part F · Einstein · branch

Rotating and charged holes, and where this course stops

Every hole that has ever formed rotates. Kerr's solution has an ergosphere you can mine, a hidden symmetry nobody expected, and — at the end — a temperature this course cannot derive.

needs 36, 15~3 h slides + 6 h problemsproblems: ./mr problems 40

Unit 36 solved the vacuum equations for a round, static star, and unit 37 read three measurements off the answer. But nothing round and static has ever existed. Every star rotates, so every hole that forms from one rotates, and the rotating solution was not found until 1963 — forty-eight years after Schwarzschild's, by Roy Kerr, who was looking for algebraically special metrics rather than for a black hole. It is not obtainable by the method of unit 36: no ansatz with two unknown functions of one variable contains it.

This unit has two jobs, and they pull in different directions. The first is to do the Kerr geometry honestly — the ergosphere, the two horizons, the energy you can take out, the fourth constant of motion that makes its orbits integrable. The second is to say where this course's methods stop. That is not a rhetorical flourish: there is a precise line, and it is crossed on the slide where the horizon acquires a temperature. Everything before that line is classical differential geometry and can be derived here. Everything after needs quantum field theory in curved spacetime, which this course has deliberately not taught, and so is stated with its source named.

How to read this deck

One object runs through every section: a Kerr hole of ten solar masses spun to $a_* = 0.9$ — unit 36's hole, set turning. Section 1 writes down the metric and locates its ergosphere and its two horizons. Section 2 defines the Killing horizon and its surface gravity, which is the quantity the last section turns into a temperature. Section 3 takes energy out of the hole and proves the ceiling. Section 4 adds charge and states the no-hair theorems. Section 5 separates the Hamilton–Jacobi equation and finds Carter's constant, which is unit 15's method paying its largest dividend. Section 6 is how a hole forms, and the theorem that says it must contain a singularity. Section 7 is the four laws, the temperature, and the edge.

Conventions

Unit 36's, unchanged: signature $(-,+,+,+)$, factors of $c$ kept, $t$ in seconds so $g_{tt}$ carries a $c^2$, $\mathrm d\Omega^2 = \mathrm d\theta^2 + \sin^2\theta\,\mathrm d\varphi^2$. Two abbreviations do all the work: $$r_g := \frac{GM}{c^2}\ (\text{a length}),\qquad a := \frac{J}{Mc}\ (\text{also a length}),\qquad a_* := \frac{a}{r_g}\ (\text{dimensionless}),$$ so $r_s = 2r_g$. Boyer–Lindquist coordinates $(t, r, \theta, \varphi)$ throughout, with $$\Sigma := r^2 + a^2\cos^2\theta,\qquad \Delta := r^2 - 2r_gr + a^2 .$$ Momenta are covector components, $p_\mu = \partial S/\partial x^\mu$, so $E := -p_t$ is an energy in joules and $L := p_\varphi$ an angular momentum in $\mathrm{J\,s}$.

Constants for every number: $c = 299\,792\,458$ m/s, $G = 6.67430\times10^{-11}\ \mathrm{m^3kg^{-1}s^{-2}}$, $GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3s^{-2}}$, $\hbar = 1.054572\times10^{-34}$ J s, $k_{\mathrm B} = 1.380649\times10^{-23}$ J/K.

40 · overview

One hole, turning

The running example: unit 36's ten-solar-mass hole, given the spin a real collapsed star would have. $r_g = 14.77$ km, $a_* = 0.9$.

outer horizon $r_+$	21.20 km
inner horizon $r_-$	8.33 km
ergosphere, equator	29.53 km
horizon's spin rate $\Omega_{\mathrm H}/2\pi$	1013 Hz
surface gravity $\kappa$	$9.24\times10^{11}$ m/s²
extractable	15.27% of $Mc^2$
Hawking temperature	3.75 nK

Each is computed in the section that introduces it; the last from a formula this course cannot prove.

the arc

1 the metric · 2 Killing horizons and $\kappa$ · 3 the Penrose process and its 29.3% ceiling · 4 charge, and no hair · 5 Carter's constant · 6 collapse, trapped surfaces, singularity theorems · 7 the four laws, and the edge

getting around

Dotted underlines are glossary terms: click for a definition, G for the whole list. N opens the notes, where the longer proofs, the history and the citations live. A solid underline — the Kerr metric — shows a piece of the running example on hover.

proved, or cited

Kerr's metric is verified, not derived. The no-hair theorems, the zeroth law and Hawking's temperature are cited, with sources. Everything else here is proved.

Why $a_* = 0.9$ and not $a_* = 1$. Extremal holes are the clean limiting case and the place the 29.3% ceiling is attained, but they are not attainable: Thorne showed in 1974 that photon capture from a thin accretion disc caps the spin-up at $a_* \approx 0.998$, and measured spins of stellar-mass holes scatter over the whole range (GRS 1915+105 sits near the top, Cygnus X-1 at $a_* \gt 0.95$, others near zero). $0.9$ is a real value, and keeping $a_* \lt 1$ means every square root in this deck is of a positive number, which extremal geometry quietly stops being true of.

Why a branch and not a spine unit. Nothing later in this course needs Kerr. It is here because it is the only other exact solution of comparable importance, because its hidden symmetry is the single best advertisement for unit 15's Hamilton–Jacobi machinery, and because the honest place to stop a general-relativity course is at the point where the classical theory starts predicting things it cannot explain.

section 1

Rotation: the Kerr metric

A metric with a cross term. What the cross term does to space, and the two surfaces it creates that Schwarzschild has only one of.

40 · the metric

Kerr's solution, written down

the Kerr metric, in Boyer–Lindquist coordinates $$\begin{aligned}\mathrm ds^2 = &-\Big(1 - \frac{2r_gr}{\Sigma}\Big)c^2\mathrm dt^2 \;-\; \frac{4r_gra\sin^2\theta}{\Sigma}\,c\,\mathrm dt\,\mathrm d\varphi\\ &+ \frac{\Sigma}{\Delta}\,\mathrm dr^2 + \Sigma\,\mathrm d\theta^2 + \Big(r^2 + a^2 + \frac{2r_gra^2\sin^2\theta}{\Sigma}\Big)\sin^2\theta\,\mathrm d\varphi^2 ,\end{aligned}$$

$\Sigma = r^2 + a^2\cos^2\theta$, $\Delta = r^2 - 2r_gr + a^2$. It solves $R_{\mu\nu} = 0$ for every $M$ and every $a$. Cited: Kerr 1963; the verification is mechanical and long, and the derivation needs machinery this course has not taught.

dimensions

$r_g$, $a$, $r$ are lengths, so $\Sigma$ and $\Delta$ are areas and every term is an area. $c\,\mathrm dt$ is a length; $g_{t\varphi}$ carries one factor of $c$, $g_{tt}$ two.

limit $a \to 0$

$\Sigma \to r^2$, $\Delta \to r^2(1 - r_s/r)$, the cross term vanishes, and every component collapses to unit 36's metric exactly.

limit $M \to 0$

$\mathrm ds^2 \to -c^2\mathrm dt^2 + \frac{\Sigma}{r^2+a^2}\mathrm dr^2 + \Sigma\mathrm d\theta^2 + (r^2+a^2)\sin^2\theta\,\mathrm d\varphi^2$: flat, in oblate spheroidal coordinates.

Two new features, both from the cross term: $g_{t\varphi} \ne 0$ means the geometry distinguishes $\varphi$ from $-\varphi$, and $g_{tt}$ vanishes somewhere other than $\Delta = 0$.

The $M \to 0$ check, in full. Set $x = \sqrt{r^2+a^2}\sin\theta\cos\varphi$, $y = \sqrt{r^2+a^2}\sin\theta\sin\varphi$, $z = r\cos\theta$. Differentiating and summing $\mathrm dx^2 + \mathrm dy^2 + \mathrm dz^2$ gives, after the cross terms cancel, $$\frac{r^2+a^2\cos^2\theta}{r^2+a^2}\,\mathrm dr^2 + (r^2+a^2\cos^2\theta)\,\mathrm d\theta^2 + (r^2+a^2)\sin^2\theta\,\mathrm d\varphi^2,$$ which is exactly the spatial part above. The surfaces of constant $r$ are oblate spheroids of semi-axes $\sqrt{r^2+a^2}$ (equatorial) and $r$ (polar); $r = 0$ is not a point but the disc $x^2 + y^2 \le a^2$, $z = 0$, whose rim $x^2+y^2 = a^2$ is $r = 0, \theta = \pi/2$. This is worth holding on to: it is why $r = 0$ in Kerr is a ring and not a point.

What "Boyer–Lindquist" buys. Kerr wrote his solution in what are now called Kerr–Schild coordinates, $g_{\mu\nu} = \eta_{\mu\nu} + H\ell_\mu\ell_\nu$ with $\ell$ null — elegant, and regular across the horizon, but with three off-diagonal terms. Boyer and Lindquist (1967) found the chart above, in which only $g_{t\varphi}$ survives and the $a \to 0$ limit is manifestly Schwarzschild. The price is the same one unit 36 paid: the chart breaks down at $\Delta = 0$, and crossing the horizon needs a different one.

Why no ansatz finds it. Unit 36's method worked because sphericity plus staticity left two functions of one variable. Kerr is stationary and axisymmetric, which leaves four functions of two variables, and $R_{\mu\nu} = 0$ for that ansatz is a coupled nonlinear PDE system (the Ernst equations, 1968) with no elementary route. Kerr found the metric by imposing an algebraic condition on the Weyl tensor — that the spacetime be of Petrov type D, meaning its Weyl tensor has two repeated principal null directions — which reduces the PDE system to something solvable. That the resulting metric described a rotating hole was not obvious and was not what he was looking for.

The ring singularity. $\Sigma = 0$ requires $r = 0$ and $\theta = \pi/2$ together. The Kretschmann scalar there is $K = 48r_g^2(r^2-a^2\cos^2\theta)\big[\Sigma^2 - 16r^2a^2\cos^2\theta\big]/\Sigma^6$ (cited; Chandrasekhar, The Mathematical Theory of Black Holes, §58), which reduces to unit 36's $48r_g^2/r^6 = 12r_s^2/r^6$ at $a = 0$ and diverges as $\Sigma \to 0$. So the ring is a genuine curvature singularity. Unlike Schwarzschild's, it is timelike: it can be avoided, and the maximal extension of Kerr lets a geodesic pass through the disc to a region with $r \lt 0$, where closed timelike curves exist near the ring. That region is almost certainly not physical, for reasons the collapse section gives.

40 · the metric

What the cross term does: space is dragged

Drop a particle from rest at large $r$ with $L = p_\varphi = 0$: no angular momentum at all. Its angular velocity is

$$\omega(r,\theta) := \frac{\mathrm d\varphi}{\mathrm dt} = \frac{g^{t\varphi}}{g^{tt}} = -\,\frac{g_{t\varphi}}{g_{\varphi\varphi}} = \frac{2r_gra\,c}{(r^2+a^2)^2 - \Delta a^2\sin^2\theta} ,$$

and it is not zero. A zero-angular-momentum observer is nonetheless carried around the hole: this is frame dragging.

far field

The denominator $\to r^4$ as $r\to\infty$, so

$$\omega \;\longrightarrow\; \frac{2r_gac}{r^3} \;=\; \frac{2GJ}{c^2r^3} .$$

Mass alone never produces this; only $J$ does. It is Lense–Thirring precession, 1918.

checks

Dimensions: $GJ/c^2r^3$ is $\mathrm{(m^3kg^{-1}s^{-2})(kg\,m^2s^{-1})/(m^2s^{-2}\cdot m^3)} = \mathrm{s^{-1}}$ ✓. Limit $J \to 0$: $\omega \to 0$, and Schwarzschild has no preferred sense of rotation ✓. Sign: $a \gt 0$ gives $\omega \gt 0$, dragging with the hole ✓.

the Earth does it too

$J_\oplus = 5.85\times10^{33}\ \mathrm{kg\,m^2s^{-1}}$. A gyroscope in a polar circular orbit at $r = 7013$ km precesses at $GJ_\oplus/2c^2r^3 = 6.29\times10^{-15}$ rad/s, which is 41 milliarcseconds per year.

Gravity Probe B measured $37.2 \pm 7.2$ mas/yr against a full prediction of $39.2$ (cited: Everitt et al., PRL 106, 221101, 2011).

Where $\omega = -g_{t\varphi}/g_{\varphi\varphi}$ comes from. "Zero angular momentum" means $p_\varphi = 0$. Raise the index: $\dot\varphi = p^\varphi/m = (g^{\varphi t}p_t + g^{\varphi\varphi}p_\varphi)/m = g^{\varphi t}p_t/m$, and likewise $\dot t = g^{tt}p_t/m$. Hence $\mathrm d\varphi/\mathrm dt = g^{\varphi t}/g^{tt}$. For a metric whose only off-diagonal entry in the $(t,\varphi)$ block is $g_{t\varphi}$, inverting the $2\times2$ block gives $g^{t\varphi} = -g_{t\varphi}/D$ and $g^{tt} = g_{\varphi\varphi}/D$ with $D = g_{tt}g_{\varphi\varphi} - g_{t\varphi}^2$, so the ratio is $-g_{t\varphi}/g_{\varphi\varphi}$. The closed form uses the identity $g_{\varphi\varphi} = \big[(r^2+a^2)^2 - \Delta a^2\sin^2\theta\big]\sin^2\theta/\Sigma$, which is two lines of algebra from the metric and is worth keeping: it makes the $\sin^2\theta$ cancel, so $\omega$ has no $\theta$ in it at all once written this way, and it makes the horizon value obvious — set $\Delta = 0$ and use $2r_gr_+ = r_+^2+a^2$ to get $ac/(r_+^2+a^2)$. The single factor of $c$ is inherited from $g_{t\varphi}$, which carries one because the metric is written with $c\,\mathrm dt$ while $t$ is measured in seconds.

The Earth number, honestly. The Lense–Thirring precession rate of a gyroscope at position $\vec r$ is $\vec\Omega = (G/c^2r^3)\big[3(\vec J\cdot\hat r)\hat r - \vec J\big]$ (cited; it follows from the gravitomagnetic form of unit 38's linearised metric). Averaging over a circular polar orbit, $\langle(\hat J\cdot\hat r)\hat r\rangle = \tfrac12\hat J$, so $\langle\vec\Omega\rangle = G\vec J/2c^2r^3$ — the formula used in the box. With $I_\oplus = 0.3307M_\oplus R_\oplus^2$ and GP-B's 642 km altitude that is 41 mas/yr, about 5% above the mission's 39.2, the difference being the Earth's real moment of inertia distribution and the orbit's small eccentricity. The much larger geodetic (de Sitter) precession, $\tfrac32(GM_\oplus/c^2r)\sqrt{GM_\oplus/r^3}$, comes out at 6638 mas/yr against the published 6606.1 — the same kind of agreement. Both are unit 32's parallel transport, measured.

Why this is not a force. Nothing pushes the particle round. Its worldline is a geodesic; $\varphi$ increases because the metric's notion of "not rotating" is itself rotating relative to the stars. This is Mach's objection made quantitative, and it is the reason "inertial frame" is a local statement in general relativity and a global one in Newton's.

40 · the metric

Worked example 1 — the ergosphere, and the end of standing still

where can an observer hold $r$, $\theta$, $\varphi$ fixed?

Step 1 — what "static" is. A worldline with $r, \theta, \varphi$ constant has tangent $u\propto\partial_t$, so it is an observer's worldline iff $g_{tt} \lt 0$.

Step 2 — read off $g_{tt}$.

$$g_{tt} = -\Big(1 - \frac{2r_gr}{\Sigma}\Big)c^2 = -\frac{c^2}{\Sigma}\big(\Sigma - 2r_gr\big).$$

Step 3 — substitute $\Sigma$. $\Sigma - 2r_gr = r^2 - 2r_gr + a^2\cos^2\theta$, so $g_{tt} \lt 0 \iff$ that quadratic is positive.

Step 4 — solve the quadratic. $r^2 - 2r_gr + a^2\cos^2\theta$ opens upwards, so it is positive outside its larger root, which is

$$\boxed{\;r_{\mathrm E}(\theta) = r_g + \sqrt{r_g^2 - a^2\cos^2\theta}\;}$$

and negative just inside it, where $g_{tt} \gt 0$. Outside $r_{\mathrm E}$ static observers exist; inside, none do.

The surface depends on $\theta$: the $a^2\cos^2\theta$ term vanishes on the equator and is largest on the axis. The next slide puts in the running hole's numbers and draws it.

Step 1, said carefully. "Static" here means: the observer's spatial Boyer–Lindquist coordinates never change. That is a statement relative to the coordinates, but not an empty one, because $\varphi$ is fixed by the axial Killing field and $r,\theta$ by the geometry of the orbit spheroids — a static observer is one at rest with respect to the distant stars. The condition for such a curve to be a possible worldline is that its tangent be timelike, and its tangent is proportional to $\partial_t$ because $\dot r = \dot\theta = \dot\varphi = 0$.

The smaller root. The quadratic's other root is $r_g - \sqrt{r_g^2 - a^2\cos^2\theta}$, which lies inside $r_-$ and bounds an inner ergoregion around the ring. It matters for the maximal extension and not for anything an infalling observer from outside will meet before the Cauchy horizon, so this deck does not use it.

40 · the metric · example

The ergosphere, drawn: thick at the equator, gone at the poles

worked example 1, continued

Step 5 — the running hole. Put $a = 0.9\,r_g$ into the boxed $r_{\mathrm E}(\theta)$:

$\theta = 0$ (pole)	$1.436\,r_g = 21.20$ km
$\theta = 45°$	$1.771\,r_g = 26.16$ km
$\theta = 90°$ (equator)	$2\,r_g = 29.53$ km

At the pole $r_{\mathrm E} = r_g + \sqrt{r_g^2 - a^2}$, which is the outer horizon $r_+$ of the next slide: the two surfaces touch there.

checks

Lengths throughout ✓. At $a \to 0$, $r_{\mathrm E} \to r_s$, where standing still does become impossible ✓. At the equator $r_{\mathrm E} = 2r_g$ for every $a$ ✓.

The shaded crescent is the ergosphere: thickest on the equator, pinched to nothing on the axis, where $r_{\mathrm E}$ meets the horizon. The dots are the three rows of the table. Boyer–Lindquist $r$ is drawn as distance from the centre, so this is a coordinate picture, not a scale model.

Reading the figure. Each point is plotted at distance $r$ from the centre and angle $\theta$ from the rotation axis, so the outer horizon $r = r_+$ and the inner one $r = r_-$ are circles and the ergosurface is the curve $r_{\mathrm E}(\theta)$. The proper distances are not these: the spatial metric is not $\mathrm dr^2 + r^2\mathrm d\theta^2$, and in the flat-space coordinates of the Kerr metric slide's notes the horizon itself would look oblate. What the picture gets right is the topology and the ordering — which surface lies inside which, and where they touch — and that is what the argument uses. The arrow at the top is the sense of rotation, the direction of $J$.

The surface $r_{\mathrm E}$ has several names. Static limit (Penrose's), ergosurface, or the outer boundary of the ergoregion. "Ergo" is Greek for work, and the name was chosen because of what section 3 does there. The region between $r_+$ and $r_{\mathrm E}$ is the ergosphere, and despite the name it is not a sphere: it is the volume between a spheroid and the horizon, pinched to zero thickness at the poles.

Why the poles are special. At $\theta = 0$, $\cos^2\theta = 1$ and $r_{\mathrm E} = r_g + \sqrt{r_g^2 - a^2} = r_+$: the ergosurface meets the horizon. Physically, the dragging is a rotational effect and the rotation axis is the one place it has no direction to drag you in. Mathematically, $g_{tt}$ and $\Delta$ vanish together there because $\sin\theta = 0$ kills the difference $\Sigma - 2r_gr$ versus $\Delta$, which is $a^2\cos^2\theta - a^2 = -a^2\sin^2\theta$.

What is not true. Being unable to stand still is not the same as being unable to escape. The ergosphere is outside the horizon; a rocket in it can leave, and must simply also be moving in $\varphi$ while it does. Confusing the static limit with the horizon is the single most common error about Kerr, and the slide Inside the ergosphere: a window of allowed rotations is the repair: inside the ergosphere the set of allowed angular velocities is a nonempty interval that does not contain zero.

40 · the metric

Two horizons, and a bound on spin

$g_{rr} = \Sigma/\Delta$ blows up at the roots of $\Delta = r^2 - 2r_gr + a^2$:

$$r_\pm = r_g \pm \sqrt{r_g^2 - a^2} .$$

Both are coordinate singularities — $\Sigma$, and every curvature invariant, is finite there — and $r_+$ is a null hypersurface, hence an event horizon, by exactly unit 36's argument.

the Kerr bound

If $a \gt r_g$, i.e. $|J| \gt GM^2/c$, then $\Delta \gt 0$ everywhere: no horizon, and the ring singularity is naked and visible from infinity.

$a = r_g$ is extremal: the two horizons merge. Our hole has $a_* = 0.9$, so $r_+ = 1.436\,r_g = 21.20$ km and $r_- = 0.564\,r_g = 8.33$ km.

$\Delta$ is a parabola in $r$ whose minimum, $a^2 - r_g^2$, rises with spin. At $a_* = 0.9$ it dips below zero between the two horizons; at $a_* = 1$ it just touches, and the horizons merge; past that it never reaches zero, and nothing hides the centre.

Reading the figure. In units of $r_g$, $\Delta/r_g^2 = x^2 - 2x + a_*^2$ with $x = r/r_g$: the same parabola shifted up by $a_*^2$. Its vertex is always at $x = 1$, so the two roots sit symmetrically about $r_g$ — that is $r_+ + r_- = 2r_g$ — and their product is $a^2$. Raising the spin lifts the parabola, which pulls the roots together until they meet at $a = r_g$. Between the roots $\Delta \lt 0$, so $g_{rr} = \Sigma/\Delta$ is negative there: inside the outer horizon $r$ is a time coordinate, exactly as inside unit 36's $r_s$.

Why $\Delta = 0$ is a null surface. The surface $r = r_+$ has normal covector $n_\mu = \partial_\mu(r - r_+) = \delta^r_\mu$, so $n\cdot n = g^{rr} = \Delta/\Sigma$, which vanishes exactly on it. A hypersurface whose normal is null is itself null, and unit 36 showed that a null hypersurface is one-way: its generators are null geodesics lying in it, and no causal curve crosses it outwards. The inverse metric component $g^{rr} = \Delta/\Sigma$ is the honest object to use here, since $g_{rr} = \Sigma/\Delta$ is the one that diverges and divergence of a component proves nothing (unit 36's whole point).

A chart across the horizon. As in unit 36, define a Kerr analogue of the ingoing Eddington–Finkelstein coordinates by $\mathrm dv = c\,\mathrm dt + (r^2+a^2)\mathrm dr/\Delta$ and $\mathrm d\tilde\varphi = \mathrm d\varphi + a\,\mathrm dr/\Delta$. In $(v, r, \theta, \tilde\varphi)$ the metric is smooth and non-degenerate at $\Delta = 0$; the twist in $\tilde\varphi$ is the new feature, and it is the coordinate statement that an infaller is dragged.

40 · the metric

Behind the inner horizon, and past the bound

the inner horizon is not benign

$r_-$ is a Cauchy horizon: past it, data on a spacelike slice outside no longer determine the future. Infalling perturbations are infinitely blueshifted against it, and the backreaction ("mass inflation", Poisson–Israel 1990) is believed to turn it into a curvature singularity.

checks

$a \to 0$: $r_+ \to 2r_g = r_s$ and $r_- \to 0$ ✓, one horizon, as Schwarzschild has. Dimensions: lengths ✓. Product $r_+r_- = a^2$ and sum $r_+ + r_- = 2r_g$, straight from the quadratic — both useful later.

Cosmic censorship (Penrose 1969, a conjecture): generic collapse from regular data never produces a naked singularity.

Every attempt to spin a hole past $a_* = 1$ by feeding it angular momentum has failed; the hole refuses the last particles. In the previous slide's picture: the particles that would lift the extremal parabola clear of the axis are exactly the ones that are turned away.

The over-spinning question, and what actually happens. Take an extremal hole and throw in a particle with enough angular momentum to push $a$ past $r_g$. Wald showed in 1974 that the particles that would do it are precisely the ones the hole will not absorb: the centrifugal and spin-curvature repulsion turns them away. The calculation has been sharpened many times since — for near-extremal holes it needs second-order self-force terms to work out — and no counterexample has survived. This is evidence for cosmic censorship, not a proof of it, and the conjecture remains open.

Mass inflation, briefly. An observer crossing $r_-$ sees the entire infinite future history of the external universe compressed into finite proper time, so any radiation tail falling in is blueshifted without bound. The Ori model shows the resulting mass function diverging while tidal distortion stays integrable — a "weak" singularity a body might survive crossing. Whether that counts as censorship holding is a live question; recent work on Reissner–Nordström–de Sitter has produced parameter ranges where it appears to fail.

40 · the metric

Inside the ergosphere: a window of allowed rotations

A stationary observer is one with constant $r,\theta$ and $\varphi = \varphi_0 + \Omega t$, so $u \propto \xi_\Omega := \partial_t + \Omega\,\partial_\varphi$. It is a worldline iff

$$g(\xi_\Omega,\xi_\Omega) = g_{tt} + 2\Omega\,g_{t\varphi} + \Omega^2 g_{\varphi\varphi} \;\lt\; 0 .$$

A quadratic in $\Omega$ opening upwards ($g_{\varphi\varphi} \gt 0$), so it is negative exactly between its roots:

$$\Omega_\pm = \tfrac{-g_{t\varphi} \pm \sqrt{g_{t\varphi}^2 - g_{tt}g_{\varphi\varphi}}}{g_{\varphi\varphi}} = \omega \pm \sqrt{\omega^2 - g_{tt}/g_{\varphi\varphi}} .$$

three regimes

Outside $r_{\mathrm E}$: $g_{tt} \lt 0$, so $\Omega_- \lt 0 \lt \Omega_+$ and $\Omega = 0$ is allowed.
In the ergosphere: $g_{tt} \gt 0$, so $0 \lt \Omega_- \lt \Omega_+$ — you must co-rotate.
On $r_+$: the discriminant vanishes, the window shuts, and $\Omega_- = \Omega_+ = \Omega_{\mathrm H}$.

The running hole's equator. Far out the band of allowed $\Omega$ straddles zero; at $r_{\mathrm E} = 2r_g$ its lower edge reaches zero, and inside it every observer must co-rotate; at $r_+$ the band has shrunk to the single value $\Omega_{\mathrm H}$.

Reading the figure. On the equator $\Sigma = r^2$ and the metric components are $g_{tt} = -(1 - 2r_g/r)c^2$, $g_{t\varphi} = -2r_gac/r$ and $g_{\varphi\varphi} = r^2 + a^2 + 2r_ga^2/r$; the curves are $\Omega_\pm$ computed from them for $a_* = 0.9$, with $c/r_g = 2.030\times10^4\ \mathrm{s^{-1}}$ converting to rad/s. At $r = 3r_g$ the window runs from $-2.65$ to $+5.01\times10^3$ rad/s; at $r = 2r_g$ from exactly $0$ to $6.50\times10^3$; at $r_+$ it is the single point $6363$ rad/s. The dashed middle curve is $\omega$, and it does not vanish anywhere outside the horizon.

Reading $\omega$ again. The midpoint of the window is $\omega = -g_{t\varphi}/g_{\varphi\varphi}$, the ZAMO angular velocity from the frame-dragging slide. So the allowed window is "the dragging rate, plus or minus what your rocket can do against the local light cone", and it closes when the light cone has tipped so far that its whole width is on one side of $\Omega = 0$ — the ergosphere — and then to a single ray — the horizon. The next slide draws that tipping.

40 · the metric

The window shuts: the horizon turns rigidly

the angular velocity of the horizon $$\Omega_{\mathrm H} = \frac{ac}{r_+^2 + a^2} ,$$

independent of $\theta$: the horizon turns rigidly. For the hole, $6363\ \mathrm{rad\,s^{-1}}$ — 1013 revolutions a second, a 0.988 ms period.

checks

$\mathrm{m}\cdot\mathrm{m\,s^{-1}}/\mathrm{m^2} = \mathrm{s^{-1}}$ ✓. $a \to 0$: $\Omega_{\mathrm H}\to 0$, and a Schwarzschild horizon does not turn ✓. Extremal $a = r_g$: $\Omega_{\mathrm H} = c/2r_g$, the largest it can be ✓.

Every stationary observer on the horizon turns at the same rate, whatever its $\theta$. Section 2's rigidity theorem says no stationary rotating horizon has a choice; section 7's zeroth law rests on it.

The light cone cut by the $t$–$\varphi$ plane, on the equator. Up is $c\,\mathrm dt$, across is proper distance round the hole; the shaded wedge holds the stationary directions $\partial_t + \Omega\partial_\varphi$ that are timelike. Dragging tips it: past $r_{\mathrm E}$ it no longer contains $\partial_t$, and at $r_+$ it closes to one ray.

Reading the figure. In each panel the vertical is one unit of $c\,\mathrm dt$ and the horizontal is $\sqrt{g_{\varphi\varphi}}\,\Omega\,\mathrm dt$, the proper distance a stationary observer moves round the hole in that time. The two solid edges are the null directions $\Omega = \Omega_\pm$ of the previous slide and the dotted ray is the ZAMO, $\Omega = \omega$. In these units the edges sit at $(-0.61, +0.82)$ at $r = 4r_g$, at $(+0.24, +0.75)$ at $r = 1.7r_g$, and both at $0.63$ on the horizon — computed from the equatorial metric components, not sketched. The dashed vertical is $\partial_t$: inside the wedge outside $r_{\mathrm E}$, outside it within.

Why $\Omega_{\mathrm H}$ has no $\theta$ in it — the computation. On $r = r_+$ we have $r_+^2 - 2r_gr_+ + a^2 = 0$, i.e. $2r_gr_+ = r_+^2 + a^2$. Then $$-g_{t\varphi}\big|_{r_+} = \frac{2r_gr_+a\sin^2\theta}{\Sigma}c = \frac{(r_+^2+a^2)a\sin^2\theta}{\Sigma}c,$$ and $$g_{\varphi\varphi}\big|_{r_+} = \Big(r_+^2+a^2 + \frac{(r_+^2+a^2)a^2\sin^2\theta}{\Sigma}\Big)\sin^2\theta = (r_+^2+a^2)\frac{\Sigma + a^2\sin^2\theta}{\Sigma}\sin^2\theta = \frac{(r_+^2+a^2)^2\sin^2\theta}{\Sigma},$$ using $\Sigma + a^2\sin^2\theta = r_+^2 + a^2$. Dividing, every $\theta$ cancels and $\Omega_{\mathrm H} = ac/(r_+^2+a^2)$. A Newtonian fluid body cannot rotate rigidly and stay in equilibrium; a horizon has no choice.

The discriminant on the horizon. $g_{t\varphi}^2 - g_{tt}g_{\varphi\varphi} = \Delta\sin^2\theta\,c^2$ for the Kerr metric — a one-line computation from the two displayed expressions plus $g_{tt} = -c^2(\Delta - a^2\sin^2\theta)/\Sigma$. So the window's width is $2c\sqrt{\Delta}\sin\theta/g_{\varphi\varphi}$, which closes exactly at $\Delta = 0$. That is the precise sense in which the horizon is the last surface on which anything can be stationary at all, and it forces every stationary observer there to share one angular velocity.

section 2

Killing horizons, and the one number a horizon has

A definition that makes no reference to coordinates, and the constant it attaches to every stationary horizon — the constant that turns into a temperature in section 7.

40 · killing horizons

Killing horizon: the definition

Killing horizon

Let $\xi$ be a Killing vector field (unit 31). A null hypersurface $\mathcal N$ is a Killing horizon of $\xi$ if $\xi$ is null and non-vanishing on $\mathcal N$.

On a null hypersurface the normal is also tangent, so $\xi$ generates $\mathcal N$. No chart appears anywhere in that.

Kerr's horizon is one

$r = r_+$ is a Killing horizon of

$$\xi_{\mathrm H} := \partial_t + \Omega_{\mathrm H}\,\partial_\varphi .$$

proof

$g(\xi_\Omega,\xi_\Omega) = g_{\varphi\varphi}(\Omega - \Omega_-)(\Omega - \Omega_+)$, and on $r_+$ the two roots coincide at $\Omega_{\mathrm H}$ (The window shuts), so the value at $\Omega = \Omega_{\mathrm H}$ is zero. $\xi_{\mathrm H}$ is a Killing field as a constant-coefficient combination of two Killing fields. $\square$

event horizon $\ne$ Killing horizon, in general

An event horizon is defined globally — the boundary of the past of future null infinity — and deciding where it is needs the whole future history. A Killing horizon is defined locally, from a symmetry. They coincide here because of a theorem, not a definition.

Hawking's rigidity theorem (cited)

In a stationary, analytic, asymptotically flat spacetime obeying the dominant energy condition, the event horizon is a Killing horizon; if not static, its Killing field is $\partial_t + \Omega_{\mathrm H}\partial_\varphi$ and the spacetime is axisymmetric. Hawking 1972; Hawking & Ellis §9.3.

So a rotating stationary hole must be axisymmetric and its horizon must turn rigidly. Both were checked by hand for Kerr; there was never a choice.

Why a null hypersurface's normal is tangent to it. Let $\mathcal N = \{f = 0\}$ with $n_\mu = \partial_\mu f$. Null means $g^{\mu\nu}n_\mu n_\nu = 0$, i.e. $n\cdot n = 0$, i.e. $n$ is orthogonal to itself. A tangent vector to $\mathcal N$ is exactly one orthogonal to $n$; so $n$ is tangent. This is the peculiarity of Lorentzian signature that makes horizons possible at all, and unit 36 used it to argue that $r = r_s$ is one-way.

Where the analyticity in the rigidity theorem hurts. Hawking's original proof assumes the metric is real-analytic, which is a much stronger hypothesis than smoothness and has no physical justification — an analytic function is determined everywhere by its behaviour in any open set, so assuming it is close to assuming the answer. Removing it is a substantial research programme (Alexakis–Ionescu–Klainerman have removed it in the near-Kerr regime); the general case is open. This is typical of black-hole uniqueness results: physically compelling, technically fragile.

The bifurcation surface. Where $\xi_{\mathrm H}$ vanishes — the two-sphere at the crossing point of the two branches of the horizon in the maximally extended spacetime, analogous to the origin of unit 36's Kruskal diagram — the definition above excludes the surface, since $\xi$ must be non-vanishing. A horizon containing such a surface is called a bifurcate Killing horizon, and on one the zeroth law is a theorem of pure differential geometry (Rácz–Wald 1996) rather than a consequence of the field equations. A hole formed by collapse has no bifurcation surface, which is why the zeroth law for real holes is stated with an energy condition.

40 · killing horizons

Surface gravity

On $\mathcal N$, $\xi\cdot\xi = 0$; just off it, $\xi\cdot\xi \ne 0$. So $\nabla^\mu(\xi\cdot\xi)$ is normal to $\mathcal N$ — and on a null hypersurface, normal means proportional to $\xi$.

surface gravity

$\kappa$ is the function on $\mathcal N$ defined by

$$\nabla^\mu(\xi_\nu\xi^\nu) = -\frac{2\kappa}{c}\,\xi^\mu \quad\text{on }\mathcal N .$$

equivalently: the generators are geodesic, badly parametrised $$\xi^\nu\nabla_\nu\xi^\mu = \frac{\kappa}{c}\,\xi^\mu .$$

proof, one line

Killing means $\nabla_\mu\xi_\nu = -\nabla_\nu\xi_\mu$. So $\nabla^\mu(\xi^\nu\xi_\nu) = 2\xi^\nu\nabla^\mu\xi_\nu = -2\xi^\nu\nabla_\nu\xi^\mu$. Compare with the definition. $\square$

$\kappa$ depends on a normalisation

$\xi \mapsto \lambda\xi$ sends $\kappa\mapsto\lambda\kappa$. The convention is fixed at infinity: $\xi_{\mathrm H} = \partial_t + \Omega_{\mathrm H}\partial_\varphi$ with $g(\partial_t,\partial_t)\to-c^2$ as $r\to\infty$. Without that sentence $\kappa$ is meaningless. The $1/c$ is the price of keeping $t$ in seconds; it makes $\kappa$ an acceleration rather than an inverse time.

physical reading

$\kappa$ is the redshifted proper acceleration of a stationary observer in the limit $r\to r_+$: the force per unit mass that must be applied at infinity, through an ideal rope, to hold a unit mass just above the horizon. The local force diverges; the force at the far end does not.

Dimensions: an acceleration, $\mathrm{m\,s^{-2}}$.

Full proof that $\nabla^\mu(\xi\cdot\xi)$ is normal. $\xi\cdot\xi$ is constant (zero) along $\mathcal N$, so its gradient annihilates every vector tangent to $\mathcal N$; a covector annihilating the tangent space of a hypersurface is a multiple of its normal. On a null hypersurface the normal direction is spanned by $\xi$ itself, whence the proportionality. The factor $-2$ in the definition is chosen so that the second, more useful, form comes out with no factor.

The Schwarzschild value, derived. Take a static observer at radius $r$ in unit 36's metric, $f = 1 - r_s/r$. Its four-velocity is $u^\mu = (f^{-1/2}, 0,0,0)$ — normalised by $g_{tt}(u^t)^2 = -fc^2/f = -c^2$. Its proper acceleration is $$a^r = u^\nu\nabla_\nu u^r = \Gamma^r_{tt}(u^t)^2 = \frac{ff'c^2}{2}\cdot\frac1f = \frac{f'c^2}{2},$$ using unit 36's $\Gamma^r_{tt} = ff'c^2/2$. Its magnitude is $|a| = \sqrt{g_{rr}}\,|a^r| = f^{-1/2}f'c^2/2$, which diverges as $r\to r_s$. Multiplying by the redshift factor $\sqrt f$ — the ratio of the force felt at infinity to the force felt locally, unit 36's gravitational redshift — gives $$\sqrt f\,|a| = \frac{f'c^2}{2} = \frac{c^2r_s}{2r^2}\ \xrightarrow{\ r\to r_s\ }\ \frac{c^2}{2r_s} = \frac{c^4}{4GM},$$ finite. That limit is $\kappa$. The geometric definition agrees: for a Killing field the scalar $\kappa^2 = -\tfrac12c^2(\nabla^\mu\xi^\nu)(\nabla_\mu\xi_\nu)$ follows from the boxed relation, and for $\xi = \partial_t$ in Schwarzschild the only independent component is $\nabla_r\xi_t = -\tfrac12f'c^2$, giving $\kappa = \tfrac12f'c^2$ at every $r$ — which is the redshifted acceleration above, not merely its limit. Problem P4 does that computation.

Why the rope is the right picture. Lower a unit mass on a massless rope from infinity. The tension at the bottom is the local proper acceleration and grows without bound; the tension at the top is redshifted by $\sqrt f$ and tends to $\kappa$. The work recovered at the top as the mass is lowered to the horizon is therefore finite — in fact the whole rest energy, which is the classical reason a black hole is the most efficient engine there is, and the thought experiment Bekenstein was analysing when he proposed that horizons carry entropy.

40 · killing horizons

Worked example 2 — $\kappa$ and $\Omega_{\mathrm H}$ for the running hole

ten solar masses, $a_* = 0.9$: $r_g = 14.766$ km, $a = 13.290$ km

Step 1 — the horizons. $\sqrt{1 - a_*^2} = \sqrt{0.19} = 0.43589$, so

$$r_\pm = r_g(1 \pm 0.43589) = 21.203,\ 8.330\ \text{km}.$$

Step 2 — the combination that keeps appearing. $r_+^2 + a^2 = 2r_gr_+ = 2(14766)(21203) = 6.2617\times10^{8}\ \mathrm{m^2}$.

Step 3 — $\Omega_{\mathrm H}$.

$$\Omega_{\mathrm H} = \frac{ac}{r_+^2+a^2} = \frac{1.3290\times10^{4}\cdot 2.9979\times10^{8}}{6.2617\times10^{8}}$$

$= 6363\ \mathrm{rad\,s^{-1}}$, i.e. $1013$ Hz, a period of $0.988$ ms.

Step 4 — $\kappa$. For a Killing horizon at a root of $\Delta$,

$$\kappa = \frac{c^2\,\Delta'(r_+)}{2(r_+^2+a^2)} = \frac{c^2\,(r_+ - r_-)}{2(r_+^2+a^2)},$$

using $\Delta'(r) = 2r - 2r_g$ and $r_+ + r_- = 2r_g$. Numerically

$$\kappa = \frac{8.9875\times10^{16}\cdot 1.2873\times10^{4}}{1.2523\times10^{9}}$$

$= 9.238\times10^{11}\ \mathrm{m\,s^{-2}}$.

checks

Dimensions: $\mathrm{m^2s^{-2}}\cdot\mathrm m/\mathrm{m^2} = \mathrm{m\,s^{-2}}$ ✓. $a\to0$: $\Delta = r(r-r_s)$, $\Delta'(r_s) = r_s$, $r_+^2+a^2 = r_s^2$, so $\kappa\to c^4/4GM = 1.522\times10^{12}$ — spinning up has cut $\kappa$ by 39% ✓. $a\to r_g$: $r_+ = r_-$, $\kappa\to0$ ✓ — the third law, in advance.

Step 4, where the formula comes from. Two routes. (i) Direct: compute $\xi_{\mathrm H}^\nu\nabla_\nu\xi_{\mathrm H}^\mu$ in Boyer–Lindquist coordinates on $r = r_+$; the only surviving component is radial, and the algebra collapses using $2r_gr_+ = r_+^2+a^2$. (ii) Slicker: in the ingoing Kerr chart $(v, r, \theta, \tilde\varphi)$ the near-horizon metric has the form of a two-dimensional Rindler wedge times a sphere, with $g(\xi_{\mathrm H},\xi_{\mathrm H}) = -\Delta\Sigma/[(r^2+a^2)^2 - \Delta a^2\sin^2\theta]$, and $\kappa^2 = -\tfrac12c^2(\nabla^\mu\xi^\nu)(\nabla_\mu\xi_\nu)$ evaluated at $\Delta = 0$ gives $\Delta'(r_+)^2c^4/4(r_+^2+a^2)^2$. Both are mechanical; neither is illuminating, so the formula is given and checked rather than derived on the slide. (Cited: Wald, General Relativity, §12.5.)

The sign of the trend. $\kappa$ falls as the hole spins up, from $c^4/4GM$ at $a_*=0$ to zero at $a_*=1$. In the thermodynamic dictionary of section 7 that reads: a rapidly spinning hole is cold. An extremal hole has zero temperature, which is the whole content of the third law and the reason extremality is unreachable.

Scale, for calibration. $9.2\times10^{11}\ \mathrm{m\,s^{-2}}$ is $9.4\times10^{10}$ times the Earth's $g$. It is also, by the rope reading, the tension per kilogram that a winch at infinity would feel — which is why "you could mine a black hole" is a real statement and not a joke.

section 3

The Penrose process: taking energy out

Energy is what a Killing field measures. Inside the ergosphere that Killing field is spacelike, and the measurement can come out negative.

40 · penrose process

Negative energy, and why it is not a paradox

For a particle with four-momentum $p$, the conserved energy measured at infinity is

$$E = -g(\partial_t, p) = -p_t .$$

That is a pairing with a Killing field, not a local energy: every observer measures $-g(u_{\text{obs}}, p) \gt 0$, because $u_{\text{obs}}$ is timelike.

in the ergosphere, $E$ can be negative

Inside the ergosphere, $\partial_t$ is spacelike. There exist future-directed timelike $p$ at every such point with $E \lt 0$.

proof

Write $X = \partial_t$, $g(X,X) = s^2 \gt 0$. Pick a future-directed unit timelike $u$ with $g(u,X) = 0$ (possible: $X^\perp$ is a Lorentzian 3-space). Put $p = m(u + \beta X/s)$ with $0 \lt \beta \lt c$. Then $g(p,p) = m^2(-c^2 + \beta^2) \lt 0$, so $p$ is timelike and future-directed, and $E = -g(X,p) = -m\beta s \lt 0$. $\square$

what is not happening

No local observer sees a negative energy; the particle's rest mass and locally measured energy are positive. $E \lt 0$ means: the work needed to carry it to infinity exceeds what it has. It never gets there — such orbits are confined to the ergosphere and must cross the horizon.

sanity check against Newton

A bound Newtonian orbit has negative total energy too. The difference: here the rest energy $\mu c^2$ is part of $E$, so $E \lt 0$ means the binding exceeds the entire rest mass — impossible outside the ergosphere, routine inside.

The consequence: a particle that falls in with $E_2 \lt 0$ reduces the hole's mass.

Why $X^\perp$ is Lorentzian. If $X$ is spacelike, the restriction of $g$ to the three-dimensional subspace $X^\perp$ has signature $(-,+,+)$: the orthogonal complement of a positive-norm vector in a $(-,+,+,+)$ space keeps the one negative direction. So $X^\perp$ contains timelike vectors, and we may pick a unit one and orient it to the future. If $X$ were timelike, as it is outside the ergosphere, $X^\perp$ would be spacelike and no such $u$ would exist — which is exactly why the construction fails there, and why $E \gt 0$ outside.

The upper bound on how negative. Not arbitrarily: the construction needs $|\beta| \lt c$, so $|E| \lt mcs = mc\sqrt{g_{tt}}$, and $g_{tt}$ is bounded on the ergosphere. Its maximum over the ergosphere of a Kerr hole is attained at the equator on the horizon and is finite, which is ultimately why the extractable fraction has a ceiling at all.

Penrose's own telling. Penrose proposed the process in 1969 in Rivista del Nuovo Cimento, in the paper that also proposed cosmic censorship, and told it as a parable about an advanced civilisation living on a scaffold around a Kerr hole, lowering rubbish into the ergosphere and arranging for it to split so that the useful half comes back up hot. It is the cleanest demonstration that "the energy of a black hole" is not all locked away — that a hole has a mass and, separately, a rotational energy that is on the table.

40 · penrose process

The process, and the inequality that caps it

the Penrose process

Send a particle with $E_0 \gt 0$ into the ergosphere and split it: $p_0 = p_1 + p_2$, so $E_0 = E_1 + E_2$. Arrange $E_2 \lt 0$ and let fragment 2 fall in; then $E_1 = E_0 - E_2 \gt E_0$, and the hole absorbs $\delta(Mc^2) = E_2 \lt 0$, $\delta J = L_2 \lt 0$: it loses mass and spin.

where the negative energy must go

Fragment 2 cannot escape or orbit: an $E \lt 0$ curve stays inside the ergosphere, and every timelike curve confined there falls in. The bookkeeping closes because the hole's mass is defined at infinity too — same Killing field, same pairing.

What caps the gain is a statement about what the horizon will accept, and it is the next slide.

The equatorial plane from above, the two circles at the running hole's radii. Both fragments move with the rotation, as every worldline in the ergosphere must; the one sent in with $E_2 \lt 0$ is swallowed, and its partner leaves with more energy than the particle brought. Paths schematic.

Reading the figure. The two circles are drawn to scale: the horizon at $r_+ = 21.20$ km and the ergosurface at $r_{\mathrm E} = 2r_g = 29.53$ km on the equator. The orbits are not computed; what is faithful is that the split happens between the circles, that both fragments head round in the sense of rotation (inside $r_{\mathrm E}$ nothing can do otherwise), and that only the outgoing fragment ever reaches the region where a distant observer could catch it. The bookkeeping is the whole point: $E_1 = E_0 - E_2$ exceeds $E_0$ by $|E_2|$, and the hole's mass falls by the same $|E_2|/c^2$.

The split is ordinary four-momentum conservation. At a single event, conservation of four-momentum in a local inertial frame is unit 27's law, unchanged: general relativity has no global momentum conservation, but it has a local one at each point, and that is all the argument uses. $E = -p_t$ is then conserved separately along each fragment's geodesic by unit 36's Killing argument — which is unit 13's Noether theorem, again.

Why the process is not an engine anyone will build. Bardeen, Press and Teukolsky showed in 1972 that the split must be relativistic: the fragments need a relative speed above about $c/2$ to put one of them on a negative-energy orbit. A chemical or nuclear explosion is nowhere near that. The astrophysically relevant versions do not use particles at all — they use fields, which is Fields do it better: superradiance.

40 · penrose process

The cap: what the horizon will accept

the horizon inequality

Every causal, future-directed $p$ crossing $r_+$ satisfies

$$\delta(Mc^2) \;\ge\; \Omega_{\mathrm H}\,\delta J .$$

proof, three lines

$\xi_{\mathrm H}$ is future-directed null there and $p$ future-directed causal, so $-g(\xi_{\mathrm H}, p) \ge 0$. Expand: it is $E - \Omega_{\mathrm H}L$. The hole's increments are $\delta(Mc^2) = E$, $\delta J = L$. $\square$

checks

$\Omega_{\mathrm H}\delta J$ is $\mathrm{s^{-1}\cdot kg\,m^2s^{-1}} = \mathrm J$ ✓. At $a = 0$ the inequality is $\delta M \ge 0$: a Schwarzschild hole only grows, and has no ergosphere to mine ✓.

In one line: what is taken from the hole is charged against its spin, at the rate $\Omega_{\mathrm H}$.

40 · penrose process

Irreducible mass: the part you cannot have

The horizon's area is $A = 4\pi(r_+^2+a^2) = 8\pi r_gr_+$ (the $\sin^2\theta$ in $g_{\varphi\varphi}$ and the $\Sigma$'s cancel — see the notes).

irreducible mass $$M_{\mathrm{irr}} := \frac{c^2}{G}\sqrt{\frac{A}{16\pi}} \;=\; M\sqrt{\frac{r_+}{2r_g}} .$$

Christodoulou mass formula $$M^2 = M_{\mathrm{irr}}^2 + \Big(\frac{cJ}{2GM_{\mathrm{irr}}}\Big)^{\!2} .$$

The hole's mass splits into an irreducible part and a rotational energy $\;(M - M_{\mathrm{irr}})c^2$.

The Christodoulou formula is Pythagoras: for the running hole, $10^2 = 8.47^2 + 5.31^2$ in solar masses. The best possible extraction holds the horizontal leg fixed and shrinks the vertical one to zero, so the hypotenuse falls to $M_{\mathrm{irr}}$ — the $1.53\,M_\odot$ of worked example 3.

Reading the figure. The legs are $M_{\mathrm{irr}} = M\sqrt{r_+/2r_g} = 10 \times 0.84732 = 8.473\,M_\odot$ and $cJ/2GM_{\mathrm{irr}} = \sqrt{M^2 - M_{\mathrm{irr}}^2} = 5.311\,M_\odot$, drawn to scale. Every process obeying the horizon inequality keeps the horizontal leg from shrinking (the next slide proves it), so the hypotenuse can come down at most to the horizontal leg's length — and it does so only if the leg stays exactly fixed while the spin is drained. For an extremal hole the triangle is isosceles, both legs $M/\sqrt2$, which is where the $29.3\%$ ceiling comes from.

The area, computed. The horizon is $r = r_+$, $t$ constant. Its induced metric is $\Sigma\,\mathrm d\theta^2 + g_{\varphi\varphi}\mathrm d\varphi^2$ with $g_{\varphi\varphi}|_{r_+} = (r_+^2+a^2)^2\sin^2\theta/\Sigma$ (computed in the notes to The window shuts: the horizon turns rigidly), so the area element is $\sqrt{\Sigma\cdot(r_+^2+a^2)^2\sin^2\theta/\Sigma} = (r_+^2+a^2)\sin\theta$ — the $\Sigma$ cancels exactly. Integrating, $A = 2\pi\cdot2\cdot(r_+^2+a^2) = 4\pi(r_+^2+a^2)$. Then $2r_gr_+ = r_+^2+a^2$ gives the second form. Note that the horizon's intrinsic geometry is not that of a round sphere: for $a_* \gt \sqrt3/2 \approx 0.866$ its Gaussian curvature is negative near the poles, and it cannot be isometrically embedded in $\mathbb R^3$ (Smarr 1973). Our $a_* = 0.9$ hole is just past that threshold.

The formula's shape. $M^2 = M_{\mathrm{irr}}^2 + (J/2M_{\mathrm{irr}})^2$ (geometrised) is a Pythagorean relation, and the two legs are "what is stored in the horizon" and "what is stored in the rotation". Its differential is exactly what section 7 calls the first law: $\delta M = (\kappa/8\pi)\delta A + \Omega_{\mathrm H}\delta J$. One can go either way; we derived the inequality first because it is the physical constraint, and the equality is what the inequality saturates for a reversible process.

40 · penrose process

The irreducible mass never falls, and nor does the area

the area can only grow, from the horizon inequality

$\delta(Mc^2) \ge \Omega_{\mathrm H}\delta J \iff \delta M_{\mathrm{irr}} \ge 0 \iff \delta A \ge 0$.

Section 7 will call the last of these the second law of black-hole mechanics.

checks

$J = 0$: $M_{\mathrm{irr}} = M$, nothing extractable ✓. Extremal $J = GM^2/c$: $r_+ = r_g$, $M_{\mathrm{irr}} = M/\sqrt2$ ✓, and the formula returns $M^2/2 + M^2/2 = M^2$ ✓. Dimensions: $cJ/GM_{\mathrm{irr}}$ is $\mathrm{(m\,s^{-1})(kg\,m^2s^{-1})/(m^3kg^{-1}s^{-2}\,kg)} = \mathrm{kg}$ ✓.

proof of the equivalence

Work in $G = c = 1$ for the algebra, where $M_{\mathrm{irr}}^2 = \tfrac12\big(M^2 + \sqrt{M^4 - J^2}\big)$. Writing $x = \sqrt{M^4-J^2}$ and differentiating,

$$2M_{\mathrm{irr}}\,\delta M_{\mathrm{irr}} = \frac{M(M^2+x)\,\delta M - \tfrac12 J\,\delta J}{x} ,$$

so $\delta M_{\mathrm{irr}}\ge0 \iff \delta M \ge \dfrac{J\,\delta J}{2M(M^2+x)}$. And $r_+ = (M^2+x)/M$, so $\Omega_{\mathrm H} = a/(2Mr_+) = J/\big(2M(M^2+x)\big)$ — the same coefficient. $\square$

40 · penrose process

Worked example 3 — the ceiling: 29.3 per cent

how much of $Mc^2$ can be extracted?

Step 1 — what is conserved. $M_{\mathrm{irr}}$ never decreases. Best case: it stays exactly constant (a reversible extraction, $\delta M = \Omega_{\mathrm H}\delta J$).

Step 2 — the endpoint. Spin cannot be extracted forever: at $J = 0$ the ergosphere is gone. So the final state is a Schwarzschild hole of mass $M_{\mathrm f} = M_{\mathrm{irr}}$.

Step 3 — the fraction.

$$\eta = 1 - \frac{M_{\mathrm{irr}}}{M} = 1 - \sqrt{\tfrac12\big(1+\sqrt{1-a_*^2}\,\big)} .$$

Step 4 — maximise. $\eta$ increases with $a_*$; at $a_* = 1$, $r_+ = r_g$ and

$$\eta_{\max} = 1 - \tfrac1{\sqrt2} = 0.29289 = \mathbf{29.3\%} .$$

Is the ceiling attained? Only in an idealised limit: reversible extraction needs each step to saturate $\delta M = \Omega_{\mathrm H}\delta J$, meaning fragments that reach the horizon with exactly zero "wasted" energy, which requires infinitely many infinitely gentle steps. Any real process increases $A$ and so falls short. This is precisely the structure of a Carnot cycle, and the parallel is not an accident — section 7 says why.

The whole spin, or not. Note that nothing forces the extraction to stop at $J = 0$; one could overshoot and spin the hole the other way, but that costs energy rather than yielding it. The maximum is at $J = 0$ because $\eta$ is computed against a fixed $M_{\mathrm{irr}}$ and $M = \sqrt{M_{\mathrm{irr}}^2 + (J/2M_{\mathrm{irr}})^2}$ is minimised there.

40 · penrose process · example

The extractable fraction, across every spin

worked example 3, continued

Step 5 — the running hole. $a_* = 0.9$: $r_+/2r_g = 1.43589/2 = 0.717945$, $\sqrt{\ } = 0.84732$,

$$\eta = 15.27\%,\qquad \eta Mc^2 = 2.73\times10^{47}\ \mathrm J .$$

$a_*$	$\eta$
0.1	0.13%
0.5	3.41%
0.9	15.27%
0.998 (Thorne limit)	27.09%
1 (extremal)	29.29%

checks

$a_*\to0$: $\eta\to0$ ✓. $\eta$ is dimensionless ✓. Cross-check: $2.7286\times10^{47}\,\mathrm J/c^2 = 3.04\times10^{30}$ kg $= 1.53\,M_\odot$, and $M - M_{\mathrm{irr}} = 0.15268\times10\,M_\odot = 1.53\,M_\odot$ ✓.

$\eta = 1 - \sqrt{\tfrac12(1+\sqrt{1-a_*^2})}$. Almost nothing is on the table until the hole spins fast: $\eta$ grows like $a_*^2/8$ at first, and half the ceiling is only reached at $a_* \approx 0.89$. The last tenth of spin from $0.9$ is worth nearly as much as everything before it.

Reading the figure. For small spin $\sqrt{1-a_*^2} \approx 1 - a_*^2/2$, so the bracket is $1 - a_*^2/4$ and its root $1 - a_*^2/8$: $\eta \approx a_*^2/8$, which gives $0.125\%$ at $a_* = 0.1$, the table's first row. Near $a_* = 1$ the inner root has infinite slope, which is why the curve shoots up at the right — and why the difference between Thorne's $0.998$ and exact extremality, two parts in a thousand of spin, is still more than two per cent of the rest energy.

The arithmetic, spelled out. $M = 10M_\odot = 1.9884\times10^{31}$ kg, so $Mc^2 = 1.7871\times10^{48}$ J and $\eta Mc^2 = 0.152684\times1.7871\times10^{48} = 2.7286\times10^{47}$ J. For scale, the Sun radiates $3.828\times10^{26}$ W, so this is $2.26\times10^{13}$ years of solar output — about $1600$ times the age of the universe, from one stellar-mass hole's spin.

How this compares to everything else. Hydrogen fusion converts $0.7\%$ of rest mass. Accretion onto a Schwarzschild hole down to the innermost stable circular orbit releases $1 - \sqrt{8/9} = 5.72\%$ (unit 37). Accretion onto an extremal Kerr hole, prograde, releases $1 - 1/\sqrt3 = 42.3\%$ — more than the Penrose ceiling, because that is energy extracted from the infalling matter, not from the hole. The $29.3\%$ here is different in kind: it is energy taken out of the hole itself, leaving it lighter than it was.

40 · penrose process

Fields do it better: superradiance

Replace the particle by a classical wave. A mode $\propto e^{-i\omega t + im\varphi}$ carries energy and angular momentum into the horizon in the fixed ratio $\mathrm dE/\mathrm dL = \omega/m$ — a statement about unit 29's stress tensor, and nothing quantum.

superradiance

Put that ratio into the horizon inequality $\mathrm dE - \Omega_{\mathrm H}\mathrm dL \ge 0$:

$$\mathrm dL\Big(\frac{\omega}{m} - \Omega_{\mathrm H}\Big) \ge 0 .$$

So if $\;\omega \lt m\,\Omega_{\mathrm H}$, then $\mathrm dL \lt 0$ and $\mathrm dE \lt 0$: the hole loses energy, and the reflected wave comes back amplified.

Zel'dovich predicted it in 1971 for a rotating absorbing cylinder, before anyone connected it to holes.

checks

$\omega/m$ is a rad/s, like $\Omega_{\mathrm H}$ ✓. An axisymmetric mode ($m = 0$) would need $\omega \lt 0$: impossible ✓. Schwarzschild ($\Omega_{\mathrm H} = 0$): never ✓.

Blandford–Znajek process

The astrophysical version (1977): a magnetic field anchored in an accretion disc threads the horizon, the same inequality applies to its angular-momentum flux, and the hole drives a Poynting-flux jet along its axis. The leading model for the jets of active galactic nuclei — the Penrose process, run with fields rather than rubbish.

For the running hole, the superradiant band is $\omega \lt m\times6363\ \mathrm{rad\,s^{-1}}$.

Why $\mathrm dE/\mathrm dL = \omega/m$. For a field with time dependence $e^{-i\omega t}$ and angular dependence $e^{im\varphi}$, the stress-energy fluxes through the horizon are $\mathrm dE/\mathrm dt = -\oint T^r{}_t$ and $\mathrm dL/\mathrm dt = \oint T^r{}_\varphi$, and for a single mode the integrands are proportional with ratio $\omega/m$, because the only $t$- and $\varphi$-dependence is in the phase $-\omega t + m\varphi$ and $T$ is quadratic in the field. Unit 29 derived $T^{\mu\nu}$ for a field by varying the action with respect to the metric; this is that tensor, evaluated on a mode.

The black-hole bomb. Press and Teukolsky (1972) noticed that surrounding a Kerr hole with a reflecting mirror turns superradiance into an exponential instability: the wave is amplified, reflected, amplified again. A massive field supplies its own mirror — modes with $\omega$ below the mass gap cannot escape to infinity — so a light boson of the right mass would make astrophysical Kerr holes unstable and spin them down. Measured black-hole spins therefore exclude ranges of ultralight boson masses, which is one of the stranger particle-physics constraints in existence and a live line of work with LIGO data.

Where the quantum version would go. Replace the classical incident wave by the vacuum, and the same amplification acts on vacuum fluctuations: that is spontaneous superradiance, and it is the rotational part of Hawking radiation. The step from "a wave is amplified" to "the vacuum radiates" is exactly the step this course cannot take, and section 7 says why.

section 4

Charge, and the theorem that there is nothing else

A second parameter, a second pair of horizons, and a uniqueness theorem saying that mass, spin and charge exhaust the list.

40 · charge

Reissner–Nordström

the Reissner–Nordström metric $$\mathrm ds^2 = -f(r)c^2\mathrm dt^2 + \frac{\mathrm dr^2}{f(r)} + r^2\mathrm d\Omega^2,$$ $$f(r) = 1 - \frac{r_s}{r} + \frac{r_Q^2}{r^2},\qquad r_Q^2 := \frac{GQ^2}{4\pi\varepsilon_0c^4} .$$

Sourced by a point charge: unit 36's derivation with $R_{\mu\nu} = \tfrac{8\pi G}{c^4}T_{\mu\nu}$ and unit 29's Maxwell stress tensor. No trace reversal, since $T^\mu{}_\mu = 0$ in four dimensions.

checks

$[r_Q^2] = \mathrm{(m^3kg^{-1}s^{-2})(J\,m)/(m^4s^{-4})} = \mathrm{m^2}$ ✓. $Q\to0$: Schwarzschild ✓. The $1/r^2$ term is the field energy outside $r$, gravitating — hence the $+$ sign, and it reduces the attraction ✓.

Two horizons again, at the roots of $f$: $\;r_\pm = r_g \pm \sqrt{r_g^2 - r_Q^2}$, merging when $r_Q = r_g$, i.e. $|Q| = M\sqrt{4\pi\varepsilon_0 G}$. Past that, a naked singularity. Exactly Kerr's structure, with $r_Q$ where $a$ was.

why RN is studied at all

Nothing astrophysical is charged. RN is the laboratory for the inner horizon: spherically symmetric, so mass inflation can be analysed in closed form there and then believed about Kerr.

Kerr–Newman metric

Charge and spin together: Kerr with $\Delta = r^2 - 2r_gr + a^2 + r_Q^2$, plus $A = -\dfrac{Qr\,(c\,\mathrm dt - a\sin^2\theta\,\mathrm d\varphi)}{4\pi\varepsilon_0\Sigma}$. Newman et al., 1965.

The source, in full. The electromagnetic field of the charge is $F = \mathrm dA$ with $A = -\dfrac{Q}{4\pi\varepsilon_0 r}\,\mathrm dt$, so the only nonzero component is $F_{tr} = -Q/(4\pi\varepsilon_0r^2)$ — Coulomb, unchanged, because unit 34's Maxwell equations $\mathrm dF = 0$, $\mathrm d\star F = 0$ are conformally invariant in four dimensions and the spherical ansatz does the rest. Unit 29's stress tensor $T_{\mu\nu} = \varepsilon_0(F_{\mu\alpha}F_\nu{}^\alpha - \tfrac14g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta})$ then has $T^t{}_t = T^r{}_r = -T^\theta{}_\theta = -T^\varphi{}_\varphi$, and feeding that into unit 36's $R_{\mu\nu}$ computation changes one line of the algebra: instead of $(rA)' = 1$ one gets $(rA)' = 1 - r_Q^2/r^2$, integrating to $A = 1 - r_s/r + r_Q^2/r^2$.

Tracelessness, and why it matters twice. $T^\mu{}_\mu = \varepsilon_0(F^{\mu\alpha}F_{\mu\alpha} - \tfrac44F_{\alpha\beta}F^{\alpha\beta}) = 0$ only because the spacetime dimension is 4 — the $\tfrac14$ in the definition and the $\delta^\mu_\mu = 4$ must cancel. It is the same fact that makes electromagnetism conformally invariant in four dimensions, and it is why $R = 0$ for RN and why the trace-reversed field equation of unit 35 collapses to $R_{\mu\nu} \propto T_{\mu\nu}$.

The singularity is still spacelike-free. Like Kerr's ring and unlike Schwarzschild's, RN's $r=0$ is timelike: the $+r_Q^2/r^2$ term dominates at small $r$ and makes $f \to +\infty$, so an infaller is repelled rather than crushed, and can miss the singularity and re-emerge through a white hole into another universe in the maximal extension. The Cauchy horizon at $r_-$ is exactly where that picture stops being trustworthy.

40 · charge

Why no real hole is charged

two numbers that settle it

1. An electron is over-extremal by a factor $2\times10^{21}$. The ratio for a body of charge $Q$ and mass $m$ is

$$\frac{r_Q}{r_g} = \frac{Q}{m\sqrt{4\pi\varepsilon_0G}} .$$

For an electron, $\sqrt{4\pi\varepsilon_0G} = 8.617\times10^{-11}\ \mathrm{C\,kg^{-1}}$ and

$$\frac{r_Q}{r_g} = \frac{1.602\times10^{-19}}{9.109\times10^{-31}\cdot 8.617\times10^{-11}} = 2.04\times10^{21}.$$

2. Extremality for the running hole needs $Q_{\max} = M\sqrt{4\pi\varepsilon_0G} = 1.71\times10^{21}$ C. At one part in $10^{18}$ of that — $1.5\times10^{3}$ C — the electrostatic repulsion of a proton already equals the gravitational attraction, at every radius.

the conclusion

A charged hole preferentially accretes the opposite sign from the plasma around it. Neutralisation is immediate; the equilibrium charge is zero to eighteen figures.

checks

$[\sqrt{4\pi\varepsilon_0G}] = \mathrm{C\,kg^{-1}}$ ✓. The ratio is the electron–electron force ratio, square-rooted: $(2.04\times10^{21})^2 = 4.2\times10^{42}$ ✓.

So the astrophysical family has one parameter, or two. $Q$ survives in the theory because the uniqueness theorem needs it, not because anything has it.

Where $\sqrt{4\pi\varepsilon_0G}$ comes from. Extremality is $r_Q = r_g$, i.e. $GQ^2/4\pi\varepsilon_0c^4 = G^2M^2/c^4$, i.e. $Q^2 = 4\pi\varepsilon_0GM^2$. The combination $\sqrt{4\pi\varepsilon_0 G} = 8.6174\times10^{-11}$ C/kg is the charge-to-mass ratio at which a body is its own extremal black hole, and comparing anything's $Q/m$ to it is the whole question.

The second number, computed. Equal forces on a proton at radius $r$: $Q e/4\pi\varepsilon_0r^2 = GMm_p/r^2$, so $Q = 4\pi\varepsilon_0GMm_p/e = 1542$ C, independent of $r$. Against $Q_{\max} = 1.714\times10^{21}$ C that is $9.0\times10^{-19}$. So a hole charged to one part in $10^{18}$ of extremality already repels protons as hard as it attracts them; beyond that it accretes electrons preferentially and discharges. The estimate is crude — it ignores the plasma's Debye screening and its temperature — and it does not need to be better, because the answer is off by eighteen orders of magnitude.

The one place charge is taken seriously. Extremal and near-extremal charged holes are central to string theory's account of black-hole entropy: Strominger and Vafa counted the microstates of a supersymmetric extremal charged hole in 1996 and reproduced $A/4$ exactly. They are also the setting for the weak gravity conjecture, which says a consistent quantum theory must contain some particle with $Q/m$ above the extremal ratio — the electron obliges, by $10^{21}$.

40 · charge

No hair: three numbers, and that is the whole list

the no-hair theorem (cited)

Let $(\mathcal M, g, F)$ be a four-dimensional, asymptotically flat, stationary solution of the Einstein–Maxwell equations, containing a black hole whose event horizon is connected and non-degenerate, with the exterior regular. Then it is a Kerr–Newman solution, determined by $(M, J, Q)$ alone.

Israel 1967/68 (static case); Carter 1971 and Robinson 1975 (the rotating vacuum case); Hawking 1972 (rigidity and horizon topology). Wald §12.3; Heusler, Black Hole Uniqueness Theorems.

Everything else the collapsing star had — its shape, composition, magnetic field, baryon number, the colour of its light — is radiated away or swallowed. Unit 38's ringdown is that radiation.

four caveats, all real

Analyticity. The rigidity step assumes it, and nobody can justify it.
Electrovac only. With Yang–Mills fields there are hairy solutions (Bartnik–McKinnon 1988); with a conformally coupled scalar too. Hair is forbidden for ordinary matter, not for all matter.
Four dimensions. In five, black rings (Emparan–Reall 2002) have horizon topology $S^1\times S^2$ and the theorem is simply false.
Degenerate horizons — the extremal case — are excluded and need separate work.

test

If the theorem holds, every multipole moment of a hole is fixed by $M$ and $J$: $M_\ell + iS_\ell = M(ia)^\ell$. The quadrupole is then $-Ma^2$, and LIGO's ringdown spectroscopy and the Event Horizon Telescope both test it. So far it holds.

Wheeler's phrase and Bekenstein's objection. "Black holes have no hair" is Wheeler's, from around 1970; the story is that Feynman found it indecent. The physical content is an information puzzle from the start: a star has an enormous number of possible internal configurations, and the hole it collapses to has three parameters. Where did the rest go? Bekenstein's answer — the horizon area is an entropy counting them — is the subject of section 7, and the question of whether the information is really gone is the one the last slide leaves open.

Why the multipole relation is the sharp test. Geroch and Hansen showed that a stationary asymptotically flat vacuum spacetime is determined by its multipole moments $M_\ell$ (mass) and $S_\ell$ (current). For Kerr they collapse onto a single line: $M_\ell + iS_\ell = M(ia)^\ell$, so $M_0 = M$, $S_1 = Ma = J/c$, $M_2 = -Ma^2$, and every higher moment follows. Measuring $M$, $J$ and $M_2$ independently and finding $M_2 \ne -J^2/M c^2$ would falsify the Kerr hypothesis in one measurement. Extreme-mass-ratio inspirals seen by LISA are expected to do this to a part in $10^{4}$; they are also why the next section matters, since modelling those orbits needs Kerr geodesics to be integrable.

"Non-degenerate" is not a technicality. Extremal holes have $\kappa = 0$, no bifurcation surface, and an infinite throat; the uniqueness proofs' boundary analysis fails on them. They are also where the entropy counting of string theory works, so the excluded case is the interesting one — a recurring pattern.

section 5

Carter's constant: a symmetry with no group behind it

Kerr has two Killing vectors and needs four constants. Unit 15's Hamilton–Jacobi equation separates anyway, and the reason is an object unit 16's definition of a tensor already makes room for.

40 · carter

Two Killing vectors are one short

Geodesic motion on a 4-manifold is a Hamiltonian system on $T^*\mathcal M$, dimension $8 = 2n$ with $n = 4$ (unit 14). Unit 17's Liouville–Arnold theorem asks for four independent constants in involution.

constant	from	Schwarzschild	Kerr
$H = \tfrac12g^{\mu\nu}p_\mu p_\nu$	the metric	✓	✓
$E = -p_t$	$\partial_t$	✓	✓
$L = p_\varphi$	$\partial_\varphi$	✓	✓
$L_x, L_y$	$SO(3)$	✓	—

Schwarzschild has $SO(3)$, so orbits are planar and one may set $\theta \equiv \pi/2$ — unit 36 did exactly that. Kerr's isometry group is only $\mathbb R\times SO(2)$: orbits are not planar, $\theta$ genuinely moves, and three constants is one short.

what being one short would mean

With three constants on an eight-dimensional phase space the reduced motion is on a three-dimensional level set with a two-dimensional flow — enough room for chaos, in the precise sense of unit 17: Poincaré sections with island chains and a chaotic sea, positive Lyapunov exponents, no closed-form orbits.

the empirical fact

Kerr geodesics are not chaotic. Numerical integrations show clean quasi-periodic motion, and the orbits are described by three frequencies. Something is conserving a fourth quantity, and it is not any Killing vector, because Kerr has no more.

Carter found it in 1968 by the one method available: try to separate the Hamilton–Jacobi equation, and see what falls out.

Why $SO(3)$ collapses the problem and $SO(2)$ does not. Spherical symmetry gives three Killing vectors whose conserved quantities are the three components of $\vec L$; the vector $\vec L$ is constant, and $\vec r\cdot\vec L = 0$ confines the orbit to a plane. Axial symmetry gives one, so only the projection $L_z$ is conserved, and nothing stops the orbit wandering in $\theta$. Unit 08's central-force reduction used precisely the planarity that Kerr destroys.

"In involution" is the extra condition. Unit 17's theorem needs the $n$ constants to Poisson-commute, not merely to be independent. $\{H, E\} = \{H, L\} = \{E, L\} = 0$ here because $E$ and $L$ generate commuting flows (translations in $t$ and rotations in $\varphi$) that preserve $H$. Whether the fourth constant is in involution with the other three is a thing to check, and it is — see the Liouville–Arnold slide.

Kerr geodesics are integrable; Kerr is not integrable for everything. Two caveats worth holding. First, the statement is about geodesics: add a spin to the test body (the Mathisson–Papapetrou equations) and integrability is lost, with genuine chaos appearing in the numerics. Second, the two-body problem in general relativity is not integrable at all, which is why unit 38's waveforms need numerical relativity. Integrability is a property of this particular one-body problem, and it is a gift.

40 · carter

Killing tensors: the object that does it

Killing tensor

A Killing tensor of rank 2 on $(\mathcal M, g)$ is a symmetric $(0,2)$-tensor field $K$ — that is, a smooth field of symmetric bilinear maps $T_p\mathcal M\times T_p\mathcal M\to\mathbb R$, unit 16's definition — satisfying

$$\nabla_{(\lambda}K_{\mu\nu)} = 0 ,$$

the round brackets denoting the symmetrisation over all three indices.

what it conserves

$\;\mathcal K := K_{\mu\nu}p^\mu p^\nu$ is constant along every geodesic.

proof, one line

$\dfrac{\mathrm d\mathcal K}{\mathrm d\lambda} = p^\lambda\nabla_\lambda(K_{\mu\nu}p^\mu p^\nu) = (\nabla_\lambda K_{\mu\nu})p^\lambda p^\mu p^\nu = (\nabla_{(\lambda}K_{\mu\nu)})p^\lambda p^\mu p^\nu = 0$, the second step because $p^\lambda\nabla_\lambda p^\mu = 0$ and the third because $p^\lambda p^\mu p^\nu$ is totally symmetric. $\square$

rank 1 is a Killing vector

$\nabla_{(\mu}\xi_{\nu)} = 0$ is Killing's equation (unit 31), and it conserves $\xi_\mu p^\mu$ — unit 36's $E$ and $L$. The metric itself is always a Killing tensor, since $\nabla g = 0$; it conserves $g_{\mu\nu}p^\mu p^\nu = -\mu^2c^2$.

no group behind it

A Killing vector generates a one-parameter group of isometries, and its constant is unit 13's Noether charge for that symmetry. A Killing tensor generates nothing on $\mathcal M$: it is a symmetry of the geodesic flow on $T^*\mathcal M$ that is invisible in the isometry group. Hence the name hidden symmetry.

Kerr has one (Walker–Penrose 1970), built from two special null directions $\ell, n$: notes.

The symmetrisation, written out. $\nabla_{(\lambda}K_{\mu\nu)} = \tfrac13(\nabla_\lambda K_{\mu\nu} + \nabla_\mu K_{\nu\lambda} + \nabla_\nu K_{\lambda\mu})$, using that $K$ is already symmetric in its two indices so only three of the six permutations are distinct. The condition is weaker than $\nabla K = 0$ (a covariantly constant tensor, which forces reducible holonomy and is far too strong) and is exactly what the one-line proof needs.

What "Petrov type D" means. At each point the Weyl tensor singles out at most four null directions $k$, its principal null directions, those with $k_{[\epsilon}C_{\alpha]\beta\gamma[\delta}k_{\zeta]}k^\beta k^\gamma = 0$. Generically there are four distinct ones; "type D" means there are only two, each counted twice (each then satisfies the stronger $C_{\alpha\beta\gamma[\delta}k_{\zeta]}k^\beta k^\gamma = 0$). Schwarzschild and Kerr are both type D, and it is the condition Kerr imposed to find his metric (notes to Kerr's solution, written down). It is cited here, not developed: only the two directions themselves are used below.

Kerr's Killing tensor, explicitly. Let $\ell$ and $n$ be the two principal null directions, $$\ell^\mu = \Big(\frac{r^2+a^2}{\Delta},\,1,\,0,\,\frac{a}{\Delta}\Big),\qquad n^\mu = \frac{1}{2\Sigma}\Big(r^2+a^2,\,-\Delta,\,0,\,a\Big)$$ in Boyer–Lindquist $(t,r,\theta,\varphi)$ up to the usual factors of $c$, normalised so $\ell\cdot n = -1$. Then $K_{\mu\nu} = 2\Sigma\,\ell_{(\mu}n_{\nu)} + r^2g_{\mu\nu}$ is a Killing tensor, and $K_{\mu\nu}p^\mu p^\nu = Q + (L - aE/c)^2$ — Carter's constant, up to the piece built from the Killing vectors. Verifying $\nabla_{(\lambda}K_{\mu\nu)} = 0$ is a page of Christoffel symbols and is left to the reference (Chandrasekhar §62; Walker & Penrose, Commun. Math. Phys. 18, 265).

The Noether reading, for the learner who wants one. There is a Noether theorem here, but not on configuration space. On $T^*\mathcal M$, the function $\mathcal K$ generates a Hamiltonian flow that commutes with the geodesic flow; that is the symmetry. It is not induced by any diffeomorphism of $\mathcal M$, because it moves points of $T^*\mathcal M$ in a way that depends quadratically on the momentum. Unit 13's theorem was for point transformations; this is the canonical-transformation generalisation, and it is the reason "symmetry" in Hamiltonian mechanics is a larger notion than "symmetry of the space".

Where else hidden symmetries show up. The Runge–Lenz vector of the Kepler problem is the other famous one: it is conserved, it is not the Noether charge of any point symmetry of configuration space, and it is what makes the Kepler ellipse close. Kerr's Carter constant is its closest relative in this course.

40 · carter

Worked example 4 — separating the Kerr Hamilton–Jacobi equation

unit 15's method on unit 14's geodesic Hamiltonian. Write $\mathcal E := E/c$ and $P := (r^2+a^2)\mathcal E - aL$

Step 1 — the Hamiltonian. Geodesics are the flow of $H = \tfrac12g^{\mu\nu}p_\mu p_\nu$, with $p_\mu = \mu u_\mu$ and $H = -\tfrac12\mu^2c^2$ fixed.

Step 2 — the HJ equation. $\partial_\lambda S + H = 0$ with $H$ free of $\lambda$, so $S = \tfrac12\mu^2c^2\lambda + W(x)$ and

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

Step 3 — use the two Killing vectors. $t$ and $\varphi$ are cyclic, so

$$W = -E\,t + L\,\varphi + W_r(r) + W_\theta(\theta) .$$

Step 4 — multiply by $\Sigma$. With the inverse Kerr metric (notes),

$$\Delta W_r'^2 + W_\theta'^2 + \mu^2c^2\Sigma = \frac{P^2}{\Delta} - \frac{N^2}{\sin^2\theta},$$

where $N := L - a\mathcal E\sin^2\theta$.

Step 5 — expand the $\theta$ terms. The second fraction plus $\mu^2c^2a^2\cos^2\theta$ is

$$\Theta\cos^2\theta + (L - a\mathcal E)^{2},$$

where $\Theta := a^2(\mu^2c^2 - \mathcal E^2) + L^2/\sin^2\theta$.

Step 6 — it splits. What is left is a function of $r$ plus a function of $\theta$, so each is constant:

Carter's constant $$Q := W_\theta'^2 + \Theta\cos^2\theta$$

and then, automatically,

$$\Delta W_r'^2 = \frac{P^2}{\Delta} - \mu^2c^2r^2 - Q - (L - a\mathcal E)^{2}.$$

Step 4, the inverse metric. Inverting the $(t,\varphi)$ block and the two diagonal entries gives $$\Sigma g^{tt} = -\frac{(r^2+a^2)^2 - \Delta a^2\sin^2\theta}{\Delta c^2},\quad \Sigma g^{t\varphi} = -\frac{2r_gra}{\Delta c},\quad \Sigma g^{rr} = \Delta,\quad \Sigma g^{\theta\theta} = 1,$$ $$\Sigma g^{\varphi\varphi} = \frac{\Delta - a^2\sin^2\theta}{\Delta\sin^2\theta} .$$ Substituting, collecting and completing the square in $E$ and $L$ gives the displayed line exactly; the identity was checked symbolically and the check is the one the learner should repeat, since it is the only step in this example that is not two lines.

Step 5, in full. The $\theta$-dependent pieces of step 4 are $$\frac{(L - a\mathcal E\sin^2\theta)^2}{\sin^2\theta} + \mu^2c^2a^2\cos^2\theta = \frac{L^2}{\sin^2\theta} - 2a\mathcal EL + a^2\mathcal E^2\sin^2\theta + \mu^2c^2a^2\cos^2\theta .$$ Write $\sin^2\theta = 1 - \cos^2\theta$ in the third term and $L^2/\sin^2\theta = L^2 + L^2\cos^2\theta/\sin^2\theta$ in the first: $$= \cos^2\theta\Big[a^2\mu^2c^2 - a^2\mathcal E^2 + \frac{L^2}{\sin^2\theta}\Big] + L^2 - 2a\mathcal EL + a^2\mathcal E^2,$$ and the last three terms are $(L - a\mathcal E)^2$. Everything left depends on $\theta$ only through the bracket, and the constant tail joins the $r$ side. That is the whole separation.

Step 6, why "each must be constant". This is unit 15's separation argument, unchanged: if $f(r) + g(\theta) = 0$ for all $r$ and $\theta$ in an open set, then $\partial_r f = 0$ and $\partial_\theta g = 0$, so each is a constant and they are negatives of each other. Carter's contribution was not this step but the discovery that the ansatz of step 3 works at all — there was no reason to expect a metric with a $t\varphi$ cross term and no third Killing vector to separate, and the $r$ and $\theta$ dependence conspires in a way that looks like a coincidence until the Killing tensor explains it.

What the two ODEs are for. Differentiating and using $p_r = W_r'$, $p_\theta = W_\theta'$ turns them into $\mu^2\Sigma^2\dot r^2 = R(r)$ and $\mu^2\Sigma^2\dot\theta^2 = \mathcal T(\theta)$, where $R(r) := P^2 - \Delta\big[\mu^2c^2r^2 + Q + (L-a\mathcal E)^2\big]$ and $\mathcal T(\theta) := Q - \Theta\cos^2\theta$ — two decoupled first-order equations, each reducible to quadrature. That is unit 00's programme, four units of machinery later, on the hardest spacetime in the course.

40 · carter

Reading $Q$: what the fourth constant measures

the $a \to 0$ limit

$Q$ $\to p_\theta^2 + L^2\cot^2\theta = L_{\text{tot}}^2 - L_z^2$: the angular momentum off the axis. So $Q$ generalises "how tilted is the orbit".

sign and equatoriality

For a bound orbit ($E \lt \mu c^2$) the bracket is positive, so $Q \ge 0$, with $Q = 0$ if and only if the orbit lies in the equatorial plane ($\cos\theta \equiv 0$ and $p_\theta \equiv 0$).

a number for the hole

Per unit mass, $\hat E = E/\mu c^2 = 0.95$ and $\hat L = L/\mu c = 3.5\,r_g$; the orbit turns at $\theta = 60°$, where $p_\theta = 0$. Then

$$\hat Q = \cos^2 60°\Big[a_*^2r_g^2(1 - \hat E^2) + \frac{\hat L^2}{\sin^2 60°}\Big]$$ $$= 0.25\big[0.0790 + 16.3333\big]r_g^2 = 4.103\,r_g^2 .$$

checks

$\hat Q$ is a squared length, matching $\hat L^2$ ✓. At $a_* = 0$: $\hat Q = 4.0833\,r_g^2$, $\hat L_{\text{tot}} = \sqrt{\hat Q + \hat L^2} = 4.0415\,r_g$, $\cos i = 3.5/4.0415 = 0.86603 = \cos 30°$ — inclination exactly $90° - \theta_{\min}$, as Newtonian geometry demands ✓. At $a_* = 0.9$, $i = 30.06°$ ✓.

$Q$ is not "total angular momentum"

It reduces to $L_{\text{tot}}^2 - L_z^2$ only at $a = 0$. For $a \ne 0$ it mixes in $E$ and $\mu$, and no vector $\vec L$ exists whose components it is. The name is a mnemonic, not a statement.

Orbits with $Q \gt 0$ oscillate in $\theta$ between $\theta_{\min}$ and $\pi-\theta_{\min}$ while precessing in $\varphi$: the three-frequency motion the next slide names.

The $a\to0$ limit, derived. At $a = 0$, $Q = p_\theta^2 + L^2\cot^2\theta$. In Schwarzschild the total angular momentum satisfies $L_{\text{tot}}^2 = p_\theta^2 + p_\varphi^2/\sin^2\theta$ (the standard decomposition on a sphere, unit 08's $L^2$ in spherical coordinates). Subtract $L_z^2 = p_\varphi^2 = L^2$: $L_{\text{tot}}^2 - L_z^2 = p_\theta^2 + L^2(\sin^{-2}\theta - 1) = p_\theta^2 + L^2\cot^2\theta$ ✓.

The sign claim, proved. $Q = p_\theta^2 + \cos^2\theta[a^2(\mu^2c^2 - E^2/c^2) + L^2/\sin^2\theta]$. For a bound orbit $E \lt \mu c^2$, so $\mu^2c^2 - E^2/c^2 \gt 0$ and every term is non-negative; $Q = 0$ then forces $p_\theta = 0$ and $\cos\theta = 0$ simultaneously and for all $\lambda$. Conversely an equatorial orbit has both. For unbound orbits with $E \gt \mu c^2$ the first bracket term turns negative and $Q \lt 0$ becomes possible — those are the "vortical" geodesics that never cross the equatorial plane, and they exist only for $a \ne 0$.

The arithmetic of the example. $a_*^2(1 - \hat E^2) = 0.81\times(1 - 0.9025) = 0.81\times 0.0975 = 0.078975$. $\hat L^2/\sin^2 60° = 12.25/0.75 = 16.33333$. Sum $16.41231$, times $\cos^2 60° = 0.25$, gives $4.10308\,r_g^2$. With $r_g = 14766$ m that is $8.946\times10^{8}\ \mathrm{m^2}$ per unit $\mu^2c^2$. The inclination check: $\hat L_{\text{tot}} = \sqrt{4.10308 + 12.25} = 4.04389$, $\cos i = 3.5/4.04389 = 0.865502$, $i = 30.0599°$.

Why anyone computes $Q$ in practice. Extreme-mass-ratio inspirals — a stellar-mass hole spiralling into a supermassive one — are LISA's flagship source, and their waveforms are built by evolving $(E, L, Q)$ slowly under radiation reaction while the fast motion is solved exactly by the separated equations above. Without separability the templates would have to come from numerical relativity at a cost nobody can pay, for a signal that stays in band for years. Carter's 1968 paper is load-bearing for a mission launching in the 2030s.

40 · carter

Four constants: Liouville–Arnold, and no chaos

Kerr geodesics are completely integrable

On $T^*\mathcal M$, $\dim = 8$, the four functions

$$H = \tfrac12g^{\mu\nu}p_\mu p_\nu,\quad E = -p_t,\quad L = p_\varphi,\quad Q$$

are independent and pairwise in involution. By unit 17's Liouville–Arnold theorem, the compact level sets are 4-tori carrying action–angle variables, and the motion is quasi-periodic.

why they commute

$\{H,E\} = \{H,L\} = 0$: Killing vectors preserve $H$. $\{E,L\} = 0$: $\partial_t$ and $\partial_\varphi$ commute. $\{H,Q\} = 0$: the Killing-tensor theorem. $\{E,Q\} = \{L,Q\} = 0$: $\mathcal L_{\partial_t}K = \mathcal L_{\partial_\varphi}K = 0$, since $K$ is built from $\Sigma$, $r$, $g$ and the principal null directions. $\square$

what it buys

Three physical frequencies $\Omega_r$, $\Omega_\theta$, $\Omega_\varphi$. Perihelion precession is $\Omega_\varphi - \Omega_r$, unit 37's $43''$ generalised; nodal precession is $\Omega_\varphi - \Omega_\theta$, zero exactly when $a = 0$.

the honest caveat

"No chaos" concerns compact level sets. Plunging and scattering orbits are not compact, and the theorem says nothing about them — but nor is there anything chaotic to find, since the separated equations solve those by quadrature too.

This is unit 15's payoff: a method introduced for its elegance is the only way anyone has found the fourth constant.

Independence, checked. The four differentials $\mathrm dH$, $\mathrm dE$, $\mathrm dL$, $\mathrm dQ$ must be linearly independent on a dense subset of $T^*\mathcal M$. $\mathrm dE$ and $\mathrm dL$ involve only $p_t$ and $p_\varphi$; $\mathrm dH$ and $\mathrm dQ$ both involve $p_r$ and $p_\theta$, and they are independent because $H$ contains $\Delta p_r^2/\Sigma$ while $Q$ contains no $p_r$ at all. They degenerate exactly on the equatorial orbits ($Q = 0$, $p_\theta = 0$), which is a measure-zero set and the standard place Liouville–Arnold tori degenerate.

Contrast with unit 17. Hénon–Heiles has two degrees of freedom and one constant too few, and the Poincaré section shows islands drowning in a chaotic sea as the energy rises. Kerr has four degrees of freedom and exactly four constants, so the section is a clean set of curves at every energy and spin. The difference is one function, and it exists here for a reason — the Killing tensor — that has no analogue in Hénon–Heiles. Unit 17's lesson was that integrability is exceptional; Kerr is one of the exceptions, and nobody predicted it.

Three frequencies, measured. For a slightly eccentric, slightly inclined orbit far from the hole, $\Omega_\varphi - \Omega_r \approx 3(GM)^{3/2}/c^2r^{5/2}$ — unit 37's Mercury formula — and $\Omega_\varphi - \Omega_\theta \approx 2GJ/c^2r^3$, the Lense–Thirring rate from section 1. That the same two combinations appear here, out of an exact integrable system rather than a perturbation expansion, is the cleanest check that the machinery is consistent.

section 6

How holes form, and why they must be singular

One exact collapse solution, then the theorem that makes the singularity generic — stated with its energy condition in the open, because that is where it can fail.

40 · collapse

Oppenheimer–Snyder: collapse, solved exactly

A uniform ball of pressureless dust, at rest at radius $R_0$, released. Two exact pieces, glued:

inside and outside

Outside: Schwarzschild, by Birkhoff (unit 36).
Inside: homogeneous and isotropic, so by unit 39 a piece of a closed FLRW dust universe, run backwards.

Israel's junction conditions (cited) fix the curvature constant from $R_0$ and $M$, and unit 39's cycloid is the answer, with $\tau_{\text{tot}} = \tfrac\pi2\sqrt{R_0^3/2GM}$ from rest to $r=0$:

$$R(\eta) = \tfrac{R_0}{2}(1 + \cos\eta),\quad \tau = \sqrt{\tfrac{R_0^3}{8GM}}\,(\eta + \sin\eta) .$$

The surface's radius on its own clock, for any mass and starting radius. It starts like a body under a constant pull (dashed), then runs away: the first half of the radius takes $82\%$ of the time, the second half the last $18\%$.

Reading the figure. Dividing $R(\eta)$ by $R_0$ and $\tau(\eta)$ by $\tau_{\text{tot}} = \pi\sqrt{R_0^3/8GM}$ removes every parameter: $R/R_0 = \tfrac12(1+\cos\eta)$ against $\tau/\tau_{\text{tot}} = (\eta + \sin\eta)/\pi$, one curve for all collapses. $R = R_0/2$ at $\eta = \pi/2$, where $\tau/\tau_{\text{tot}} = (\pi/2 + 1)/\pi = 0.818$. The dashed curve is $R_0 - \tfrac12(GM/R_0^2)\tau^2$, which in these units is $1 - (\pi^2/16)(\tau/\tau_{\text{tot}})^2$: the cycloid's first two Taylor terms. The pull grows as $1/R^2$, so the real fall pulls ahead of it.

Why the interior is FLRW. Unit 39 proved that homogeneity and isotropy of the spatial slices force the metric into FLRW form, with the curvature constant $k$ the only choice. A uniform ball of dust, in the frame comoving with the dust, is homogeneous and isotropic within the ball; the argument is local and does not care that the slices are bounded. The ball is collapsing, so it is the contracting half of a closed dust universe — literally unit 39's recollapsing model, cut off at a comoving radius. Oppenheimer and Snyder published this in Physical Review in September 1939, in the same issue as the outbreak of the war, and it was largely ignored for twenty-five years.

The matching conditions, in one sentence. Israel's conditions require the induced metric and the extrinsic curvature to agree from both sides across the surface. The first makes the areal radius continuous; the second, with no surface layer of matter, makes $R_0$ and $M$ satisfy $2GM/c^2 = R_0\sin^2\chi_0$ for the comoving radius $\chi_0$ of the surface in the closed interior. The full statement is Israel 1966; the cleanest exposition is Poisson, A Relativist's Toolkit, §3.7 and §3.8.

40 · collapse

Half an hour on the star's clock, for ever on ours

a number

A solar mass from the Sun's radius, $R_0 = 6.957\times10^{8}$ m. Then $R_0^3/2GM_\odot = 1.2687\times10^{6}\,\mathrm{s^2}$ and $\tau_{\text{tot}} = \tfrac\pi2(1126.4) = 1769$ s — 29.5 minutes on the surface's own clock.

checks

$\sqrt{\mathrm{m^3/(m^3s^{-2})}} = \mathrm s$ ✓. It is exactly Newton's free-fall time from $R_0$ — as it must be, since by Birkhoff the surface follows a radial Schwarzschild geodesic ✓.

A distant observer sees none of this. The surface never crosses $r_s$ in coordinate time, but its light redshifts and its luminosity decays as $e^{-tc^3/3\sqrt3\,GM}$ (cited: MTW §32.4) — for the hole, an e-folding time of $0.256$ ms. The star does not freeze: it goes out.

On the cycloid, the last stretch from $r_s = 2.95$ km to $r = 0$ takes the surface $6.6\ \mu$s of its own time — the event horizon is crossed without anything locally marking it.

The number, computed. $R_0^3 = (6.957\times10^8)^3 = 3.3673\times10^{26}\ \mathrm{m^3}$; $2GM_\odot = 2.6542\times10^{20}\ \mathrm{m^3s^{-2}}$; the ratio is $1.2687\times10^{6}\ \mathrm{s^2}$, whose root is $1126.4$ s, times $\pi/2$ gives $1769.2$ s $= 29.49$ min. The last stretch, from $R = r_s = 2953$ m to $0$, starts at $\cos\eta = 2r_s/R_0 - 1$, i.e. $\pi - \eta = 2\sqrt{r_s/R_0}$ to leading order, and the cycloid's $\tau(\pi) - \tau(\eta)$ is then $6.6\ \mu$s: the free-fall time from $r_s$, $\tfrac23 r_s^{3/2}/\sqrt{2GM}$. The Sun will not do this — it has pressure, and will end as a white dwarf — but the number is the right order for the core collapse that does produce holes, where the starting radius is a few thousand kilometres and the time is milliseconds.

Why the "frozen star" language is wrong, and was used for decades. The Soviet literature called these objects zamyorzshiye zvyozdy, frozen stars, because in Schwarzschild time the surface never crosses $r_s$. But no observer uses Schwarzschild time, and the observable quantity is the received flux, which decays exponentially with an e-folding time of order $r_s/c$ — here $98.5\ \mu$s, and the $3\sqrt3$ comes from the last photons having to climb out of the photon sphere at $3GM/c^2$ (unit 37). The star is gone, not frozen. Wheeler's "black hole", popularised in 1967, replaced the metaphor with a better one.

40 · collapse

Two Penrose diagrams, and which regions are real

Eternal Schwarzschild. Four regions. III is a second asymptotically flat universe; IV is a white hole.

Formed by collapse. The green line is the star's surface, the shaded wedge its interior. Regions III and IV are simply absent.

The extra regions of unit 36's maximal extension are not a prediction. They are what the eternal solution requires, and no hole is eternal: every real one has a past in which there was a star.

What "maximal extension" was and was not claiming. Unit 36 extended the Schwarzschild chart until every geodesic either continues forever or hits a curvature singularity. That is a mathematically canonical operation on a given metric, and it produced regions III and IV. But the metric it was applied to — Schwarzschild everywhere, for all time — is not the metric of anything that formed. Replace the region below the star's surface with the FLRW interior and the extension terminates there instead: $r = 0$ inside the star is a regular centre, a perfectly ordinary timelike worldline, until the surface passes it. So III and IV never appear, and no white hole and no second universe are predicted by any collapse.

Reading the right-hand diagram. Time runs upwards, light at 45°, and every point is a 2-sphere except on $r = 0$. The star's surface starts at $i^-$ (past timelike infinity: it has been sitting there forever in this idealisation), falls, crosses the horizon — the solid 45° line — and reaches the singularity. Region I is everything outside; region II is the hole. $\mathcal I^+$ is where light that escapes ends up; the horizon is exactly the boundary of the past of $\mathcal I^+$, which is the definition of an event horizon and is manifestly a global, teleological one: you cannot tell where it is without knowing the whole future.

The apparent horizon, which is local and different. The boundary of the trapped region on each slice is the apparent horizon, defined by $\theta_{\text{out}} = 0$ and computable from data on one slice. For Oppenheimer–Snyder it forms inside the star, before the surface crosses $r_s$, and it is a spacelike surface that coincides with the event horizon only after the collapse settles. Numerical relativity locates apparent horizons, because event horizons cannot be found until the simulation is over.

Kerr's diagram, for completeness. The maximal extension of Kerr is much worse: an infinite ladder of regions, a timelike ring singularity that can be avoided, a passage through $r = 0$ to $r \lt 0$, and closed timelike curves near the ring. Every one of those features lies beyond the Cauchy horizon $r_-$, and every one of them is believed to be destroyed by the mass-inflation instability of section 1. The physical Kerr hole is regions I and II and nothing more.

40 · singularities

Congruences, and what expansion means

null congruence and its expansion

Let $k$ be a null geodesic field, $k\cdot k = 0$, $k^\nu\nabla_\nu k^\mu = 0$, filling a region. Pick an auxiliary null $n$ with $k\cdot n = -1$; the screen is the spacelike 2-plane orthogonal to both, with projector $h^\mu{}_\nu$. Decompose

$$h^\alpha{}_\mu h^\beta{}_\nu\nabla_\beta k_\alpha = \underbrace{\tfrac12\theta\,h_{\mu\nu}}_{\text{trace}} + \underbrace{\sigma_{\mu\nu}}_{\text{trace-free sym}} + \underbrace{\omega_{\mu\nu}}_{\text{antisym}} .$$

$\theta$ is the expansion, $\sigma$ the shear, $\omega$ the twist; the $\tfrac12$ is $1/(D-2)$.

what $\theta$ measures

If $\mathcal A(\lambda)$ is the cross-sectional area of an infinitesimal bundle of the congruence, then

$$\theta = \frac{1}{\mathcal A}\frac{\mathrm d\mathcal A}{\mathrm d\lambda} .$$

Follow a thin bundle and watch its cross-section. $\theta$ is the fractional rate at which that area grows: positive for a spreading bundle, negative for a converging one, and $-\infty$ at a caustic, where the area closes to nothing. Schematic.

Reading the figure. Each line is one null geodesic of the congruence, drawn against its affine parameter $\lambda$ running left to right; the ellipses are the screen's cross-sections of the bundle at two values of $\lambda$. The top bundle is the "flat, outgoing" example of the next slide, the bottom one its "flat, ingoing" partner. Shear would squash the ellipses without changing their area, and twist would rotate them; neither changes $\mathcal A$, which is why $\theta$ alone carries the area.

Why the screen, and why $k$ is not enough. For a timelike congruence the orthogonal complement of the tangent is a 3-space and the decomposition is the familiar one (unit 22's rate-of-strain tensor, in fact, with $\theta$ the divergence). For a null congruence $k$ is orthogonal to itself, so the orthogonal complement contains $k$ and is three-dimensional and degenerate. Quotienting by $k$ — equivalently, projecting with the help of an auxiliary $n$ — gives a genuine 2-dimensional Euclidean space. The quantities $\theta$, $\sigma$, $\omega$ turn out not to depend on the choice of $n$, which is the technical fact that makes the construction worth doing.

Full proof that $\theta = \dot{\mathcal A}/\mathcal A$. Let $\eta_1,\eta_2$ be two deviation vectors in the screen, Lie-dragged along $k$, so $\mathcal L_k\eta_i = 0$ and hence $k^\nu\nabla_\nu\eta_i^\mu = \eta_i^\nu\nabla_\nu k^\mu$ (unit 31's $\mathcal L_XY = [X,Y]$). The area they span is $\mathcal A = \sqrt{\det(h(\eta_i,\eta_j))}$, and Jacobi's formula (unit 22, unit 35) gives $\dot{\mathcal A}/\mathcal A = \tfrac12\mathrm{tr}(h^{-1}\dot h)$, which on substituting the deviation equation is the trace of the projected $\nabla k$ — that is, $\theta$. The $\tfrac12\theta h_{\mu\nu}$ normalisation is chosen so this comes out with no factor.

40 · singularities

Expansion in three examples, and the sign that matters

three examples

Flat, outgoing from a point. A sphere of radius $\lambda$: $\mathcal A\propto\lambda^2$, so $\theta = 2/\lambda \gt 0$.
Flat, ingoing toward a centre a distance $\lambda$ away: $\theta = -2/\lambda \lt 0$.
Schwarzschild, outgoing from radius $r$. $\theta \propto (1 - r_s/r)$: positive outside the horizon, zero on it, negative within.

checks

With $\lambda$ of dimension length, $k$ is dimensionless and $\theta$ is $\mathrm m^{-1}$ ✓. At $r\to\infty$ the Schwarzschild value $\to 2/r$, recovering flat space ✓.

Light leaving a sphere of radius $r$, outwards and inwards. Far out the outgoing value is the flat $2/r$; it falls to zero on the horizon and is negative within, where both fronts shrink. Both expansions negative at once is the whole content of "trapped" (Trapped surfaces, and Penrose's theorem).

Reading the figure. The curves are $\theta_{\text{out}} = (2/r)(1 - r_s/r)$ and $\theta_{\text{in}} = -2/r$, in units of $1/r_s$, for the radial null geodesics normal to a round sphere, with the affine parameter normalised as in the trapped-surface notes. Only the signs are invariant — rescaling $\lambda$ rescales $\theta$ by a positive factor — and the signs are what the singularity theorems use: positive-and-negative outside $r_s$, zero-and-negative on it, negative-and-negative inside.

The twist and Frobenius. $\omega = 0$ if and only if the congruence is hypersurface orthogonal — that is unit 31's Frobenius theorem, applied to the distribution orthogonal to $k$. The null geodesics normal to a closed 2-surface are automatically twist-free, since they are orthogonal to the family of surfaces obtained by dragging it. That is why the focusing theorem's hypothesis $\omega = 0$ is free in every application here.

40 · singularities

Raychaudhuri, and focusing

Raychaudhuri's equation (null form; cited, Wald §9.2) $$\frac{\mathrm d\theta}{\mathrm d\lambda} = -\tfrac12\theta^2 - \sigma^2 + \omega^2 - R_{\mu\nu}k^\mu k^\nu ,$$

writing $\sigma^2 = \sigma_{\mu\nu}\sigma^{\mu\nu}$ and $\omega^2 = \omega_{\mu\nu}\omega^{\mu\nu}$.

Term by term. $-\tfrac12\theta^2$: converging bundles converge faster. $-\sigma^2$: shear focuses. $+\omega^2$: rotation defocuses, the only term with the wrong sign. $-R_{\mu\nu}k^\mu k^\nu$: gravity focuses, if matter obliges.

where the physics enters

Einstein's equation turns the last term into matter, the trace term dropping out on null $k$:

$$R_{\mu\nu}k^\mu k^\nu = \tfrac{8\pi G}{c^4}\,T_{\mu\nu}k^\mu k^\nu ,$$

so "$\ge 0$" is the null energy condition.

focusing theorem

If $\omega = 0$ and $R_{\mu\nu}k^\mu k^\nu\ge0$, and $\theta(0) = \theta_0 \lt 0$ at some point, then $\theta\to-\infty$ within affine parameter

$$\lambda \le \frac{2}{|\theta_0|} .$$

The proof is one inequality integrated once; the next slide does it and plots it.

Why the rotation term escapes. $+\omega^2$ is why a rotating dust cloud need not collapse to a point in Newtonian gravity, and it was the standing hope through the 1950s — Lifshitz and Khalatnikov argued that realistic collapse would be generic, rotating, and singularity-free. Penrose's theorem killed that hope by using null congruences orthogonal to a closed surface, for which Frobenius forces $\omega = 0$. Choosing the right congruence is the entire trick.

Raychaudhuri's own paper. Amal Kumar Raychaudhuri published the timelike version in 1955, working alone in Calcutta and motivated by exactly the question of whether the Big Bang singularity was an artefact of symmetry. Landau found it independently. It is the one equation in this section that is pure geometry — it is an identity about congruences, true on any pseudo-Riemannian manifold, with no field equation in it until you substitute for $R_{\mu\nu}$.

40 · singularities

Focusing, proved and plotted

proof, three lines

The hypotheses give $\theta' \le -\tfrac12\theta^2$. Put $u = 1/\theta$: then $u' = -\theta'/\theta^2 \ge \tfrac12$, so $u(\lambda) \ge u(0) + \lambda/2$. But $u(0) = 1/\theta_0 \lt 0$, so $u$ reaches $0$ — that is, $|\theta|\to\infty$ — by $\lambda = -2/\theta_0$. $\square$

checks

$\theta$ is $\mathrm m^{-1}$, $\theta'$ and $R_{\mu\nu}k^\mu k^\nu$ both $\mathrm m^{-2}$ ✓. Flat space, ingoing from a sphere of radius $r$: $\theta_0 = -2/r$, and the bound gives $\lambda\le r$ — exactly the distance to the focus ✓.

Start converging, $\theta_0 \lt 0$. With nothing but $-\tfrac12\theta^2$ acting, $\theta = \theta_0/(1 + \theta_0\lambda/2)$ reaches $-\infty$ exactly at $\lambda = 2/|\theta_0|$. Every other term the hypotheses allow only adds focusing, so the caustic can come earlier, never later.

Reading the figure. Both axes are scaled by $|\theta_0|$, so the solid curve is the same for every starting value: $\theta/|\theta_0| = -1/(1 - \lambda|\theta_0|/2)$. The dashed curve integrates $\theta' = -\tfrac12\theta^2 - s$ with a constant $s = \tfrac12\theta_0^2$ standing in for $\sigma^2 + R_{\mu\nu}k^\mu k^\nu$; it diverges at $\lambda|\theta_0| = \pi/2 \approx 1.57$, solvable in closed form with an arctangent. Any non-negative $s$, constant or not, keeps the dashed curve below the solid one — that comparison is the proof.

Why the last check is exact and not merely consistent. In flat space $\sigma = \omega = 0$ and $R = 0$, so the inequality is an equality: $\theta' = -\tfrac12\theta^2$ integrates to $\theta(\lambda) = -2/(r-\lambda)$, which blows up at $\lambda = r$. The bound $2/|\theta_0|$ is attained exactly when nothing but the $-\tfrac12\theta^2$ term acts, and every additional non-negative term makes focusing happen sooner. That is the whole logic of the singularity theorems: they never need to compute anything, only to bound.

What "$\theta\to-\infty$" is and is not. It is a caustic — a conjugate point, where neighbouring geodesics of the congruence cross and the cross-sectional area goes to zero. Nothing physical goes wrong there: the crossing point of a bundle of light rays at a focus is a perfectly ordinary event, and in flat space it is the centre of the sphere. The equation says the congruence fails to be a congruence past that point, not that spacetime ends. Turning caustics into incompleteness needs a global argument, which is Trapped surfaces, and Penrose's theorem.

40 · singularities

The four energy conditions

Each is a statement about $T_{\mu\nu}$ alone, with no field equation in it. For unit 35's perfect fluid, $T_{\mu\nu} = (\rho + p/c^2)u_\mu u_\nu + p\,g_{\mu\nu}$, they reduce to the right-hand column.

condition	statement	perfect fluid
null (NEC)	$T_{\mu\nu}k^\mu k^\nu\ge0$, $k$ null	$\rho c^2 + p \ge 0$
weak (WEC)	$T_{\mu\nu}u^\mu u^\nu\ge0$, $u$ timelike	$\rho\ge0$ and $\rho c^2+p\ge0$
strong (SEC)	$\big(T_{\mu\nu}-\tfrac12Tg_{\mu\nu}\big)u^\mu u^\nu\ge0$	$\rho c^2+p\ge0$ and $\rho c^2+3p\ge0$
dominant (DEC)	WEC, and $-T^\mu{}_\nu u^\nu$ causal	$\rho c^2 \ge \|p\|$

WEC: every observer measures non-negative energy density. SEC: gravity attracts — by Einstein's equation, $R_{\mu\nu}u^\mu u^\nu\ge0$. DEC: energy does not flow faster than light.

SEC, derived

With $u\cdot u = -c^2$: $T_{\mu\nu}u^\mu u^\nu = \rho c^4$ and $T = -\rho c^2+3p$, so the combination is $\rho c^4 + \tfrac12(-\rho c^2+3p)c^2$, that is

$$\tfrac{c^2}2\big(\rho c^2+3p\big).$$

checks

Each entry is an energy density ✓. Dust ($p = 0$, $\rho \gt 0$) satisfies all four ✓. Radiation ($p = \rho c^2/3$): NEC, WEC, SEC ✓, and DEC gives $\rho c^2\ge\rho c^2/3$ ✓.

SEC $\not\Rightarrow$ WEC and WEC $\not\Rightarrow$ SEC

They are not ordered. DEC $\Rightarrow$ WEC $\Rightarrow$ NEC is the only chain. SEC implies NEC but says nothing about $\rho\ge0$.

NEC, derived. For null $k$, $g_{\mu\nu}k^\mu k^\nu = 0$, so $T_{\mu\nu}k^\mu k^\nu = (\rho + p/c^2)(u\cdot k)^2$, and $(u\cdot k)^2 \gt 0$ for any timelike $u$ and non-zero null $k$. Hence NEC $\iff \rho + p/c^2 \ge 0 \iff \rho c^2 + p \ge 0$.

WEC, derived. Take $u_{\text{obs}}$ timelike, boosted relative to the fluid with $u\cdot u_{\text{obs}} = -\gamma c^2$. Then $T_{\mu\nu}u^\mu_{\text{obs}}u^\nu_{\text{obs}} = (\rho+p/c^2)\gamma^2c^4 - pc^2 = c^4[\gamma^2\rho + (\gamma^2-1)p/c^2]$. As $\gamma\to1$ this is $\rho c^4$, requiring $\rho \ge 0$; as $\gamma\to\infty$ it is dominated by $\gamma^2(\rho c^4 + pc^2)$, requiring $\rho c^2 + p \ge 0$. Both together are also sufficient.

Why DEC has the odd second clause. $-T^\mu{}_\nu u^\nu$ is the energy–momentum flux measured by the observer $u$; requiring it to be causal (timelike or null, future-directed) says the energy does not stream past the observer faster than light. For a perfect fluid the flux in the rest frame is $(\rho c^2, 0,0,0)$ and in a boosted frame it acquires momentum $\propto(\rho c^2+p)v$; the causality requirement works out to $|p| \le \rho c^2$. DEC is the condition Hawking's area theorem and the positive-energy theorem use, and it is the strongest of the four for ordinary matter.

Their status, honestly. None of these is derived from anything. They are hypotheses imposed on $T_{\mu\nu}$ because all known classical matter satisfies them and because without some such hypothesis Einstein's equation has no content — any metric whatever solves it for a suitably chosen $T$. Which condition a theorem uses is therefore the most important thing about it, and Who violates what is why.

40 · singularities

Who violates what

matter	NEC	WEC	SEC	DEC
dust, $p = 0$	✓	✓	✓	✓
radiation, $p = \rho c^2/3$	✓	✓	✓	✓
stiff fluid, $p = \rho c^2$	✓	✓	✓	✓ (marginal)
$\Lambda \gt 0$: $p = -\rho c^2$	✓ (marg.)	✓	✗	✓ (marg.)
slow-rolling scalar (inflation)	✓	✓	✗	✓
Casimir vacuum	✗	✗	✗	✗

$\Lambda$, in one line

$\rho_\Lambda c^2 = \Lambda c^4/8\pi G$, $p_\Lambda = -\rho_\Lambda c^2$: so $\rho c^2 + p = 0$ ✓ but $\rho c^2 + 3p = -2\rho_\Lambda c^2 \lt 0$ ✗. $\square$

Each perfect fluid is a ray $p = w\rho c^2$ from the origin, and each condition is a half-plane or a wedge. Ordinary matter sits inside all of them. $\Lambda$ ($w = -1$) lies on the NEC's edge and below the SEC's; slow-roll inflation lies between the two lines — the band where the SEC fails and the NEC holds.

Reading the figure. The table's "perfect fluid" column, drawn. NEC is $\rho c^2 + p \ge 0$: above the solid line $p = -\rho c^2$. SEC adds $\rho c^2 + 3p \ge 0$: above the dashed line $p = -\rho c^2/3$. WEC adds $\rho \ge 0$: right of the vertical axis. DEC is $|p| \le \rho c^2$: the shaded wedge. All the conditions are unchanged when $\rho$ and $p$ are scaled together, which is why each fluid is a ray and only its equation-of-state parameter $w$ matters. The inflation dot is drawn at $w = -0.8$; slow roll needs $V \gt \dot\phi^2$, which is $w \lt -1/3$, and $w \ge -1$ always, since $\rho c^2 + p = \dot\phi^2$.

The inflation entry, derived. A homogeneous scalar field has $\rho c^2 = \tfrac12\dot\phi^2 + V(\phi)$ and $p = \tfrac12\dot\phi^2 - V(\phi)$. Then $\rho c^2 + p = \dot\phi^2 \ge 0$ (NEC holds, always) and $\rho c^2 + 3p = 2\dot\phi^2 - 2V$, which is negative whenever $V \gt \dot\phi^2$ — the slow-roll condition. So inflation is, by construction, a SEC-violating epoch, and it has to be: unit 39 showed that $\ddot a \gt 0$ and $\rho c^2 + 3p \ge 0$ are incompatible.

40 · singularities

The violations are measured, not hypothetical

this is not hypothetical

The accelerating expansion measured in 1998 is a measured SEC violation: $\ddot a \gt 0$ requires $\rho c^2 + 3p \lt 0$, unit 39's acceleration equation read backwards. And $68.5\%$ of the universe is made of it.

quantum fields break all of them

Casimir's plates have negative energy density between them, and so do squeezed states of light. All four pointwise conditions fail; what survives is averaged: $\int T_{\mu\nu}k^\mu k^\nu\,\mathrm d\lambda\ge0$ along a complete achronal null geodesic.

Which condition a theorem uses decides whether it survives contact with the real universe.

The Casimir entry. Between two conducting plates a distance $d$ apart the renormalised vacuum energy density is $-\pi^2\hbar c/720d^4$, negative, and measured to a few per cent (Lamoreaux 1997, Mohideen–Roy 1998). The energy is negative relative to the unbounded vacuum, which is what the energy conditions compare against, so the violation is real and not a bookkeeping choice. This is the standing reason nobody believes the pointwise conditions are fundamental.

What is at stake. If NEC can be violated over macroscopic regions, traversable wormholes and warp drives become solutions of Einstein's equation — Morris and Thorne pointed this out in 1988 and it is why quantum energy inequalities became a research field. The known quantum energy inequalities (Ford–Roman) bound how negative an energy density can be and for how long: the larger the magnitude, the shorter the interval, with the product fixed by $\hbar$. Every bound proved so far is tight enough to forbid a macroscopic traversable wormhole, and none is a proof that all of them are. The subject remains open.

40 · singularities

Trapped surfaces, and Penrose's theorem

trapped surface

A compact spacelike 2-surface without boundary, both of whose orthogonal null congruences have $\theta \lt 0$. If $\theta_{\text{out}} = 0$, it is marginally trapped.

None exists in flat space: a round sphere has $\theta_{\text{out}} = +2/r$. In Schwarzschild every sphere with $r \lt r_s$ is trapped — the outward front shrinks, not merely slows.

Penrose singularity theorem, 1965 (cited)

Let $(\mathcal M,g)$ be globally hyperbolic: it has a Cauchy surface, a spacelike slice that every inextendible causal curve crosses exactly once (unit 35, Data on a slice). Let that surface be non-compact, let $R_{\mu\nu}k^\mu k^\nu\ge0$ for all null $k$ (the NEC), and let $\mathcal M$ contain a closed trapped surface. Then $\mathcal M$ is future null geodesically incomplete.

Flash a sphere and look a moment later. Normally the outgoing front encloses more area and the ingoing less. On a trapped surface both enclose less: even the light aimed outwards converges. Schematic.

Reading the figure. The solid circle is the 2-surface $S$; the dashed circles are where the two families of light rays orthogonal to it have reached a short affine parameter later, so their areas compare $\mathcal A$ before and after. Left: $\theta_{\text{out}} \gt 0 \gt \theta_{\text{in}}$, a round sphere in flat space or outside $r_s$. Right: $\theta_{\text{out}} \lt 0$ and $\theta_{\text{in}} \lt 0$, a sphere of areal radius $r \lt r_s$ in Schwarzschild. What the picture cannot show is that the right-hand surface is perfectly smooth and locally unremarkable — nothing at $S$ signals the trap except the two signs.

The two words, more carefully. "Inextendible" means the curve runs as far as the spacetime lets it, in both directions; "exactly once" rules out curves that miss the slice or cross it twice. Minkowski space with $t = 0$ is globally hyperbolic. Anti-de Sitter space is not (light reaches its boundary in finite time), nor is any spacetime with closed timelike curves. Unit 35 met the property as the conclusion of the Choquet-Bruhat–Geroch theorem: the maximal development of good data is globally hyperbolic.

The Schwarzschild computation, since it is the example. The outgoing radial null geodesics from a sphere of areal radius $r$ have $\theta_{\text{out}} \propto (1-r_s/r)/r$ and the ingoing ones $\theta_{\text{in}} \propto -1/r$, up to a positive normalisation depending on the choice of affine parameter. So outside the horizon one is positive and one negative — an ordinary sphere — on the horizon $\theta_{\text{out}} = 0$, marginally trapped, and inside both are negative. The trapped region is exactly the black-hole interior, which is not a coincidence but is also not a theorem in general: the boundary of the trapped region is the apparent horizon, and it coincides with the event horizon only in stationary spacetimes.

40 · singularities

What Penrose's theorem says, and what it does not

read the conclusion exactly

Some null geodesic cannot be extended past a finite affine parameter. It does not say a curvature invariant diverges, nor where, nor that the singularity is hidden.

why it was a shock

Before 1965 the consensus was that Schwarzschild's singularity came from exact spherical symmetry, which no star has. Penrose's proof uses no symmetry — only a trapped surface, an open condition that survives perturbation. Nobel Prize, 2020.

In one sentence: focusing makes $\partial J^+(S)$ compact; a non-compact Cauchy surface makes it non-compact. Notes for the full argument.

Where each hypothesis does its work. Trapped surface: supplies $\theta_0 \lt 0$ for both congruences, so the focusing theorem applies to $\partial J^+(S)$. NEC: makes the $-R_{\mu\nu}k^\mu k^\nu$ term in Raychaudhuri non-positive, which is what lets focusing run. Global hyperbolicity: makes $\partial J^+(S)$ a well-behaved 3-manifold generated by null geodesics leaving $S$ orthogonally. Non-compact Cauchy surface: $\partial J^+(S)$ would project homeomorphically into it, and a compact set cannot map onto a non-compact one — the topological contradiction. Take away any one and the theorem is false: de Sitter space with a compact Cauchy surface has trapped surfaces and no singularity.

How much matter it takes. Schoen and Yau (1983) proved that a trapped surface forms whenever enough mass is concentrated in a small enough region, with an explicit constant — so the hypothesis is not merely plausible but checkable from initial data. Combined with the theorem, that is the modern statement of gravitational collapse: concentrate mass past a threshold and a singularity is unavoidable, whatever the matter is doing, provided it obeys the NEC.

Christodoulou's refinement. In 2009 Christodoulou proved trapped-surface formation from vacuum initial data — a strong enough pulse of pure gravitational radiation focuses itself into a hole, no matter at all. That closes the last gap: general relativity produces singularities from its own field, without borrowing an energy condition from matter that might violate it.

40 · singularities

Hawking's theorem, and where $\Lambda$ hits it

Hawking's cosmological theorem, 1967 (cited)

Suppose $R_{\mu\nu}u^\mu u^\nu\ge0$ for every timelike $u$ (the SEC, via Einstein); no closed timelike curves; and a compact spacelike hypersurface with $\theta \le -C \lt 0$ everywhere. Then every past-directed timelike geodesic is incomplete: the universe has a beginning.

the hypothesis $\Lambda$ breaks

$\Lambda \gt 0$ violates the SEC and satisfies the NEC. So:

Penrose's theorem is untouched — black holes in our accelerating universe still contain singularities.
Hawking's is not applicable — and the failure is not a technicality, because de Sitter space is geodesically complete. The conclusion is genuinely false under the violated hypothesis.

check the two conditions on $\Lambda$

NEC: $\rho_\Lambda c^2 + p_\Lambda = 0 \ge 0$ ✓, marginal — so Raychaudhuri's gravity term vanishes for null rays and focusing still runs on $-\tfrac12\theta^2$ alone.
SEC: $\rho_\Lambda c^2 + 3p_\Lambda = -2\rho_\Lambda c^2 \lt 0$ ✗ — and the geodesic congruence of de Sitter space defocuses, forever.

The cleanest illustration in the course of why a theorem's hypotheses are its content. "The universe began" follows from general relativity plus the SEC — and the SEC is measured to be false today.

what survives

The Borde–Guth–Vilenkin theorem (2003) replaces the SEC by a kinematic condition — positive average expansion along a past-directed causal geodesic — and still concludes past-incompleteness. No energy condition at all.

Why an inflating universe evades Hawking and not BGV. Hawking's argument focuses timelike geodesics using SEC; with $\Lambda \gt 0$ they defocus and no conjugate point forms. BGV instead argues that a past-directed geodesic observer in an expanding region sees the comoving matter blueshift without bound, and the affine parameter to the point where that blueshift diverges is finite. It is a much weaker statement — it says the inflating description breaks down, not that curvature diverges — and it holds in the multiverse scenarios inflation produces. It is the honest modern successor to the phrase "the universe had a beginning".

The de Sitter counterexample, made concrete. De Sitter space is the maximally symmetric vacuum solution with $\Lambda \gt 0$: $\mathrm ds^2 = -c^2\mathrm dt^2 + e^{2Ht}\mathrm d\vec x^2$ with $H = c\sqrt{\Lambda/3}$ in flat slicing, or a closed slicing that contracts and re-expands. In the global (closed) form it is geodesically complete, has compact Cauchy surfaces, and has no singularity anywhere. It satisfies every hypothesis of Hawking's theorem except the strong energy condition, and it is not singular. That single example is why the SEC clause cannot be dropped, and why the theorem's statement must carry it explicitly.

A caution about "the SEC is violated, so cosmology is fine". The SEC is violated today and during inflation. It is comfortably satisfied during the radiation and matter eras, which is where the classical Big Bang singularity would be. So the Hawking theorem remains a good argument about the hot early universe and a bad one about the universe as a whole. Distinguishing the two is exactly the kind of care that stating the hypotheses out loud enforces.

section 7

Four laws, one temperature, and the edge of the course

Three theorems about horizons that look exactly like thermodynamics — and then the one result that turns the resemblance into an identity, which this course cannot derive.

40 · the four laws

The four laws of black-hole mechanics

	black holes	thermodynamics	status here
0	$\kappa$ is constant over the horizon of a stationary hole	$T$ is constant in equilibrium	cited (Bardeen–Carter–Hawking 1973, with DEC; Rácz–Wald 1996 for bifurcate horizons)
1	$\delta(Mc^2) = \dfrac{\kappa c^2}{8\pi G}\,\delta A + \Omega_{\mathrm H}\delta J + \Phi_{\mathrm H}\delta Q$	$\mathrm dE = T\,\mathrm dS + \text{work}$	proved — it is the Christodoulou formula, differentiated
2	$\delta A \ge 0$ in any classical process	$\delta S \ge 0$	cited (Hawking 1971) — proof sketched opposite, needs the NEC
3	$\kappa = 0$ cannot be reached in finitely many steps	$T = 0$ is unreachable	cited (Israel 1986)

the first law, checked on the hole

Differentiating $A(M,J)$ numerically at $M = 10M_\odot$, $a_* = 0.9$ and forming $\dfrac{\kappa c^2}{8\pi G}\dfrac{\partial A}{\partial M}\Big/c^2$ returns $1.0000$; and $\dfrac{\kappa c^2}{8\pi G}\dfrac{\partial A}{\partial J} + \Omega_{\mathrm H}$ returns $0$ to nine figures ✓.

dimensions

$\kappa c^2 A/G$ is $\mathrm{(m\,s^{-2})(m^2s^{-2})(m^2)/(m^3kg^{-1}s^{-2})} = \mathrm{kg\,m^2s^{-2}} = \mathrm J$ ✓; $\Omega_{\mathrm H}\delta J$ is J ✓.

Where the analogy stood in 1973. Bardeen, Carter and Hawking wrote the four laws down and explicitly said the analogy was formal: a black hole has zero temperature classically, since it absorbs everything and emits nothing, so identifying $\kappa$ with a temperature was in their words "a mathematical analogy rather than a physical identity". The paper's abstract says so. Their intention was partly to rebut Bekenstein, who had proposed the year before that $A$ really is an entropy. Bekenstein was right and they knew within eighteen months.

The first law, derived. In geometrised units $M^2 = M_{\mathrm{irr}}^2 + J^2/4M_{\mathrm{irr}}^2$ with $A = 16\pi M_{\mathrm{irr}}^2$. Differentiate at fixed $Q$: $$2M\,\delta M = \Big(2M_{\mathrm{irr}} - \frac{J^2}{2M_{\mathrm{irr}}^3}\Big)\delta M_{\mathrm{irr}} + \frac{J}{2M_{\mathrm{irr}}^2}\delta J,$$ and substituting $\delta M_{\mathrm{irr}} = \delta A/(32\pi M_{\mathrm{irr}})$ together with $M_{\mathrm{irr}}^2 = Mr_+/2$ and $\Omega_{\mathrm H} = a/2Mr_+$ reduces it, after some algebra, to $\delta M = (\kappa/8\pi)\delta A + \Omega_{\mathrm H}\delta J$ with $\kappa = (r_+-r_-)/2(r_+^2+a^2)$. The slide's numerical check is the honest way to confirm that algebra, and it is the check the reader should run.

The electrostatic term. $\Phi_{\mathrm H} = Qr_+/(4\pi\varepsilon_0(r_+^2+a^2))$ is the electrostatic potential of the horizon, and it plays the role of a chemical potential conjugate to $Q$ — the same structure as $\Omega_{\mathrm H}$ conjugate to $J$. It is included for completeness; nothing astrophysical has it.

The third law's two versions. The "unattainability" version (Israel 1986) is a theorem: no process taking finite advanced time and obeying the weak energy condition can turn a non-extremal hole into an extremal one. The stronger "Nernst" version — that $S\to0$ as $T\to0$ — is false for black holes: an extremal Kerr hole has $\kappa = 0$ and $A = 8\pi G^2M^2/c^4 \ne 0$, so its entropy is large at zero temperature. That failure is a real one, and it is one of the clues that the entropy counts something structural rather than thermal.

40 · the four laws

Why the area cannot decrease

Hawking's area theorem (1971)

In a globally hyperbolic spacetime obeying the NEC, with no naked singularities to the future, the area of the event horizon is non-decreasing along any spacelike foliation.

the argument, in four steps

1. The horizon is a null hypersurface generated by null geodesics with no future endpoint (they cannot leave it).
2. Suppose $\theta \lt 0$ at some point of a generator. By the focusing theorem — NEC supplies the hypothesis — a caustic forms within affine parameter $2/|\theta|$.
3. A generator that reaches a caustic leaves the horizon and enters its interior, so it has a future endpoint. Contradiction with 1.
4. Hence $\theta \ge 0$ everywhere on the horizon, and $\theta = \dot{\mathcal A}/\mathcal A$ gives $\delta A\ge0$. $\square$

the merger check, a real measurement

GW150914 merged holes of $36$ and $29$ solar masses into one of $62$ with $a_* = 0.67$: the mass fell by $3\,M_\odot$, radiated as waves. The area did not. In units of $16\pi(GM_\odot/c^2)^2$ the initial areas sum to $36^2+29^2 = 2137$ and the final one is $62^2\times\tfrac12(1+\sqrt{1-0.67^2}) = 3348$ — up by $57\%$ while the mass fell by $4.6\%$ ✓. P12 does this in full.

what breaks it

The NEC. Hawking radiation violates it — the horizon-crossing flux has negative energy — and an evaporating hole's area does decrease. The area theorem is a classical theorem, and the second law it stands for has to be generalised — Bekenstein and Hawking, next.

Step 3, made precise. A point on a null generator of $\partial J^+$ beyond a caustic is connected to the surface by a timelike curve (the crossing of two null geodesics can be smoothed into a timelike one), so it lies in the interior of $J^+$, not on its boundary. The horizon being the boundary of the past of $\mathcal I^+$, its generators cannot have future endpoints — once on the horizon, always on the horizon. Those two facts contradict each other unless no caustic forms, hence $\theta\ge0$. The argument is Penrose's from the singularity theorem, run backwards: there the caustic produced the contradiction that proved incompleteness; here it produces the contradiction that proves monotonicity.

The GW150914 numbers, honestly. The masses are $36^{+5}_{-4}$ and $29^{+4}_{-4}$ solar masses with a final mass of $62\pm4$ and $3.0\pm0.5$ radiated (Abbott et al., PRL 116, 061102, 2016). The slide's inequality uses $A \ge 16\pi G^2M_{\mathrm{irr}}^2/c^4$ with $M_{\mathrm{irr}} \le M$, so writing $M^2$ for the initial holes overestimates their area and the conclusion is conservative — the true increase is larger. The LIGO collaboration published a direct test of the area law from the ringdown in 2021 and found it satisfied at $95\%$ confidence.

Why "no naked singularities" is in the hypotheses. The proof needs the horizon's generators to be well-defined and the spacetime to the future to be predictable; a naked singularity can emit anything, including generators that end. The area theorem therefore rests on cosmic censorship, which is a conjecture. This is the standard state of affairs in the subject: the physically important results are theorems modulo a conjecture nobody can prove.

40 · the edge

Bekenstein and Hawking: the analogy becomes an identity

Bekenstein, 1972. Drop a box of entropy $S$ into a hole. The outside world's entropy falls and nothing outside compensates: the second law fails, unless the hole has an entropy of its own. The only candidate that never decreases is $A$. Dimensional analysis leaves one free number:

$$S_{\mathrm{BH}} = \alpha\,\frac{k_{\mathrm B}c^3A}{G\hbar} .$$

Hawking, 1974. Computed the spectrum a collapsing body radiates, using quantum field theory on the collapse background. It is thermal, at

Hawking temperature and Bekenstein–Hawking entropy (cited) $$T_{\mathrm H} = \frac{\hbar\kappa}{2\pi c\,k_{\mathrm B}},\qquad S_{\mathrm{BH}} = \frac{k_{\mathrm B}c^3A}{4G\hbar} .$$

the two results determine each other

Put $T_{\mathrm H}$ and $S_{\mathrm{BH}}$ into $\mathrm dE = T\,\mathrm dS$:

$$\frac{\hbar\kappa}{2\pi ck_{\mathrm B}}\cdot\frac{k_{\mathrm B}c^3}{4G\hbar}\,\delta A = \frac{\kappa c^2}{8\pi G}\,\delta A ,$$

which is exactly the first law's first term. So $\alpha = 1/4$ is forced once $T_{\mathrm H}$ is known, and $\hbar$ cancels out of the combination — the classical law survives untouched.

dimensions

$\hbar\kappa/ck_{\mathrm B}$: $\mathrm{(J\,s)(m\,s^{-2})/(m\,s^{-1}\cdot J\,K^{-1})} = \mathrm K$ ✓. $c^3A/G\hbar$: $\mathrm{(m^3s^{-3})(m^2)/(m^3kg^{-1}s^{-2}\cdot J\,s)}$, dimensionless ✓.

$\hbar$ appears in each of $T$ and $S$ and cancels in their product. That is the signature of a classical relation with a quantum origin.

Bekenstein's argument, in more detail. Lower a box of energy $E$ and entropy $S$ slowly to just outside the horizon and release it. By the rope argument of section 2 almost all of $E$ can be extracted as work, so the hole's mass barely rises and $\delta A$ is nearly zero; yet $S$ has left the universe. If $S_{\mathrm{BH}} = \alpha k_{\mathrm B}c^3A/G\hbar$, the requirement that $S_{\text{outside}} + S_{\mathrm{BH}}$ never decrease — the generalised second law — bounds $\alpha$ from below. Bekenstein got $\alpha \sim \ln2/8\pi$ from an information-theoretic argument about one bit; Hawking's calculation replaced the estimate with $1/4$.

The Euclidean argument, as an indication of how it goes. Continue $t \to -i\tau$ in the near-horizon Schwarzschild metric; the $(\tau, r)$ part becomes $\mathrm d\rho^2 + \rho^2(\kappa\,\mathrm d\tau/c)^2$ in a suitable radial variable $\rho$, which is flat polar coordinates — and is smooth at $\rho = 0$ only if the angle $\kappa\tau/c$ has period $2\pi$, i.e. $\tau$ has period $2\pi c/\kappa$. In quantum statistical mechanics, a field theory periodic in imaginary time with period $\beta\hbar$ is a thermal state at $T = 1/k_{\mathrm B}\beta$. Matching gives $T = \hbar\kappa/2\pi ck_{\mathrm B}$ immediately. It is a beautiful argument, it is three lines, and it is not available here: every step after "in quantum statistical mechanics" needs machinery this course has not built.

Three independent derivations, all agreeing. (i) Hawking's original Bogoliubov computation: the in-vacuum of the collapsing star is not the out-vacuum, and the mismatch is a Planck spectrum. (ii) The Euclidean periodicity argument above (Gibbons–Hawking 1977). (iii) The Unruh effect plus the equivalence principle: a uniformly accelerated detector in Minkowski space sees a thermal bath at $T = \hbar\alpha/2\pi ck_{\mathrm B}$ (Unruh 1976), and unit 30's Rindler observer is the near-horizon static observer. The third is the one this course comes closest to touching, since unit 30 built Rindler coordinates in full — and it still needs a quantum field to put in them.

40 · the edge

Worked example 5 — $T$, $S$ and the lifetime of the running hole

ten solar masses, $a_* = 0.9$: $\kappa = 9.238\times10^{11}\ \mathrm{m\,s^{-2}}$, $r_+ = 21.203$ km

Step 1 — the temperature.

$$T_{\mathrm H} = \frac{\hbar\kappa}{2\pi ck_{\mathrm B}} = \frac{9.743\times10^{-23}}{2.601\times10^{-14}} = 3.75\ \mathrm{nK}.$$

A Schwarzschild hole of the same mass would be at $6.17$ nK: spinning it up has cooled it.

Step 2 — the area. $A = 4\pi(r_+^2+a^2) = 7.869\times10^{9}\ \mathrm{m^2}$, that is $7869\ \mathrm{km^2}$.

Step 3 — the entropy.

$$\frac{S_{\mathrm{BH}}}{k_{\mathrm B}} = \frac{c^3A}{4G\hbar} = 7.53\times10^{78} .$$

Step 4 — compare. The Sun's thermodynamic entropy is about $10^{34}$ J/K, i.e. $7\times10^{56}k_{\mathrm B}$: the hole holds $10^{22}$ times more, on $10^{58}$ baryons' worth of matter.

Step 5 — the lifetime. Radiating as a black body at $T_{\mathrm H}$, a Schwarzschild hole evaporates in $t = 5120\pi G^2M^3/\hbar c^4$ (cited):

$$t = 6.6\times10^{77}\ \mathrm s = 2.1\times10^{70}\ \text{years} .$$

checks

$T\propto1/M$ and $S\propto M^2$, so $t\propto M^3$: $2\times10^{70}$ yr for ten solar masses against the standard $2\times10^{67}$ for one ✓. The hole is far colder than the $2.725$ K microwave background, so it is currently absorbing: net loss begins once the background has cooled by $7.3\times10^{8}$, in $3.6\times10^{11}$ years ✓.

Step 1, the arithmetic. Numerator $1.0546\times10^{-34}\times9.238\times10^{11} = 9.743\times10^{-23}$. Denominator $2\pi\times2.9979\times10^{8}\times1.3806\times10^{-23} = 2.601\times10^{-14}$. Ratio $3.746\times10^{-9}$ K. For the Schwarzschild case put $\kappa = c^4/4GM = 1.5216\times10^{12}$ and the same denominator gives $6.170\times10^{-9}$ K, the familiar $T = 6.17\ \mathrm{nK}\times(10M_\odot/M)$.

Step 3, the arithmetic. $c^3 = 2.6944\times10^{25}$; $c^3A = 2.1203\times10^{35}$; $4G\hbar = 4\times6.6743\times10^{-11}\times1.0546\times10^{-34} = 2.8155\times10^{-44}$; the quotient is $7.531\times10^{78}$.

Step 5, the crossover. $2.7255/3.746\times10^{-9} = 7.28\times10^{8}$, and in a $\Lambda$-dominated future the scale factor grows as $e^{H_\Lambda t}$ with $H_\Lambda = H_0\sqrt{\Omega_\Lambda} = 1.808\times10^{-18}\ \mathrm{s^{-1}}$ for unit 39's Planck parameters, so $t = \ln(7.28\times10^{8})/H_\Lambda = 1.13\times10^{19}$ s $= 3.6\times10^{11}$ years. Only then does the hole begin to lose mass, and the $2\times10^{70}$ years of evaporation follow after that — a delay that is, on this scale, instantaneous.

Why the $T\propto1/M$ sign is the strange one. The hole's heat capacity is negative: absorbing energy makes it colder. That is the same instability as a self-gravitating gas sphere (unit 39's Jeans problem in disguise), and it is why a hole in equilibrium with a heat bath is unstable and why the microcanonical rather than canonical ensemble is the right one. It is also why evaporation runs away at the end: the last gram goes in under a nanosecond.

40 · the edge

Where this course stops, precisely

everything in this unit up to the four laws

Killing horizons, $\kappa$, $\Omega_{\mathrm H}$, the ergosphere, the Penrose ceiling, Carter's constant, Raychaudhuri, the energy conditions, the singularity theorems, the area theorem: all Lorentzian differential geometry plus Einstein's equation, all available with units 31–35 in hand.

$T_{\mathrm H}$ and $S_{\mathrm{BH}}$ are not

They need a quantum field on a curved background; a vacuum, which in a non-stationary spacetime is not unique; a Bogoliubov transformation between in- and out-modes; and a renormalised $\langle T_{\mu\nu}\rangle$. That is a second course, and this one has deliberately not taught any of it.

three roads out, named

Quantum field theory in curved spacetime. Unruh's effect, then Hawking's. Birrell & Davies; Wald, QFT in Curved Spacetime and Black Hole Thermodynamics.
Numerical relativity. The two-body problem, not integrable and with no closed form; it produces unit 38's templates. Baumgarte & Shapiro.
Quantum gravity. What $A/4$ counts. String theory answers for extremal charged holes, loop quantum gravity with a tunable parameter, nobody for Kerr.

the open problem this leaves

Hawking radiation is thermal, so it depends only on $(M,J,Q)$: a hole formed from a book and one formed from a bomb of the same mass evaporate identically, and the difference is gone — which unitary quantum mechanics forbids. Fifty years on, the information paradox is unresolved.

Why a vacuum is the hard part. In Minkowski space the vacuum is picked out by Poincaré invariance: it is the state annihilated by the positive-frequency annihilation operators, and "positive frequency" means with respect to the global timelike Killing field. A general curved spacetime has no timelike Killing field, so there is no preferred splitting into positive and negative frequency, so there is no preferred vacuum and no observer-independent notion of a particle. Hawking's calculation works precisely because a collapse spacetime is stationary before and after — there are two natural vacua, and they disagree. The disagreement is the radiation. Everything in that paragraph requires a quantum field, and none of it can be faked classically.

What the paradox is not. It is not a claim that information loss is impossible; Hawking himself argued for thirty years that it happens and that quantum mechanics must give. It is not resolved by "the information is in the correlations", since Hawking's leading-order calculation shows the radiation is exactly thermal to all orders in the semiclassical expansion. The live proposal — the Page curve computed from replica wormholes and "islands", 2019–2020 — reproduces unitarity from a gravitational path integral, and whether it is a derivation or a consistency check is itself disputed.

An honest note on what a course can do. It would have been possible to state Hawking's formula on slide two and treat this unit as an exposition of results. The reason not to is the contract this course has kept for forty units: every claim proved or explicitly labelled with its source. Honouring it at the end means saying plainly that the last two formulas are borrowed, and naming what would have to be learnt to stop borrowing them. That is the difference between knowing physics and having read about it.

40 · closing

The running example, closed

what it is

A Kerr hole, $M = 10M_\odot$, $a_* = 0.9$. $r_g = 14.77$ km, $a = 13.29$ km, $J = 7.92\times10^{43}\ \mathrm{kg\,m^2s^{-1}}$.

Horizons at $21.20$ and $8.33$ km; an ergosphere reaching $29.53$ km at the equator and pinching to the horizon at the poles.

what it does

Turns rigidly at $1013$ Hz. Drags every zero-angular-momentum infaller round with it. Forbids any observer inside its ergosurface from standing still.

Holds $15.27\%$ of its rest energy — $2.73\times10^{47}$ J — in a form that can be taken out, against a ceiling of $29.3\%$ for a hole spun to extremality.

what it hides

A fourth constant of motion, $Q$, from a Killing tensor with no isometry behind it. Its geodesics are integrable in unit 17's exact sense.

A temperature of $3.75$ nK and an entropy of $7.5\times10^{78}k_{\mathrm B}$, neither of which this course can derive — and a singularity that Penrose's theorem guarantees, from the NEC alone.

One object, seven sections, and one honest boundary. Unit 36 solved a metric; unit 37 measured three numbers with it; this unit set it spinning and found that the classical theory, pushed to its limit, hands you a thermodynamic system and no way to explain it.

40 · checkpoint

Checkpoint

you can now do this, or you cannot

You show that inside the Kerr ergosphere no timelike worldline can hold $r$, $\theta$ and $\varphi$ constant — from $g_{tt}$, with the surface $r_{\mathrm E}(\theta)$ produced; you compute the Penrose process ceiling, $1 - 1/\sqrt2 = 29.3\%$, from the irreducible mass and the horizon inequality; and you state the Penrose singularity theorem with its energy condition named explicitly, and name a matter content that violates that condition.

Failing it is unambiguous. If your ergosphere argument cites the horizon rather than $g_{tt} \gt 0$, it is the wrong surface. If your $29.3\%$ arrives without $\delta M_{\mathrm{irr}}\ge0$, you have a number and not a bound. And if you answer the third clause with "$\Lambda$ violates the null energy condition", you have the wrong theorem and the wrong condition: Penrose's theorem uses the NEC, which $\Lambda$ satisfies marginally; it is Hawking's that uses the SEC, and that is the one $\Lambda \gt 0$ breaks.

Twelve problems, about six hours: ./mr problems 40. P2 is the ergosphere and the static limit; P5 separates the Hamilton–Jacobi equation for Carter's constant; P6 is the four-constant Liouville–Arnold reading; P8 is the energy-condition audit and the $\Lambda$ hypothesis; P11 is the Penrose ceiling end to end. When the checkpoint is true: ./mr done 40.

and that is the course

Forty-one units from $F = ma$ as an ODE to here.

The three clauses are separable and should be tested separately. Clause one is worked example 1 with P2: write $g_{tt}$, set it to zero, solve the quadratic, and say in one sentence why a static worldline needs $g_{tt} \lt 0$ — then add the repair, that stationary worldlines with $\Omega_- \lt \Omega \lt \Omega_+$ do exist there, so the ergosphere is not a trap.

Clause two is worked example 3 with P11. The chain is: $\xi_{\mathrm H}$ future-directed null on the horizon $\Rightarrow$ $\delta M c^2\ge\Omega_{\mathrm H}\delta J$ $\Rightarrow$ $\delta M_{\mathrm{irr}}\ge0$ $\Rightarrow$ the final mass is at least $M_{\mathrm{irr}}$ $\Rightarrow$ $\eta \le 1 - \sqrt{r_+/2r_g}$, maximised at extremality. Reciting $29.3\%$ without that chain is not passing.

Clause three is the singularity section with P8 and P9. The commonest failures are naming the wrong theorem, naming the wrong condition, and saying "a singularity forms" when the theorem says "geodesic incompleteness". A correct answer names the trapped surface, the non-compact Cauchy surface, the NEC, and null geodesic incompleteness — and then, for the violator, gives $\Lambda$ against Hawking's SEC, or the Casimir vacuum against all four.