Part F · Einstein · spine

Schwarzschild: the first exact solution, and what its two singularities are

Solve the vacuum field equations for a round, static star, once and slowly. The answer breaks down in two places: one is a badly chosen clock, and the other is the end of time.

needs 35, 13~3½ h slides + 7 h problemsproblems: ./mr problems 36

Unit 35 produced an equation. This unit produces the first thing that equation says, and it is worth pausing on how unusual that is: Einstein published the field equations in November 1915 believing an exact solution would be out of reach, and Karl Schwarzschild, serving on the Russian front, sent him one within weeks. Everything in unit 37 — Mercury's 43 arcseconds, the bending of starlight, the radar echo — is read off the metric derived here.

The unit has two halves of different character. The first half is a computation: impose symmetry, solve $R_{\mu\nu} = 0$, fix one constant. It is done in full, every Christoffel symbol and every Ricci component, because this is the one place in the course where the machinery of units 31–33 runs from start to finish on a problem whose answer matters, and because doing it once by hand is what makes every later citation of "the Schwarzschild metric" feel earned. The second half is interpretation: the metric has two places where its components misbehave, and deciding what each one is requires two different tools — a change of chart for one, an invariant for the other.

How to read this deck

One object runs through every section: a black hole of ten solar masses. Section 1 says what "static and spherically symmetric" means and derives the four-term ansatz. Section 2 solves the vacuum equations and fixes $r_s = 29.53$ km. Section 3 proves the hole is the only round vacuum geometry. Section 4 reads its conserved energy and angular momentum off two Killing fields and names the theorem that is doing it. Section 5 decides which of its two singularities is real. Section 6 asks what an infalling and a distant observer each report. Section 7 builds the whole spacetime, four regions of it, and draws it.

Conventions

Unit 35's, unchanged: signature $(-,+,+,+)$; the curvature sign $R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$ with $R_{\sigma\nu} = R^\mu{}_{\sigma\mu\nu}$; factors of $c$ kept everywhere. The time coordinate is $t$, in seconds, so $g_{tt}$ carries a $c^2$. A dot is $\mathrm d/\mathrm d\tau$ along a timelike curve, or $\mathrm d/\mathrm d\lambda$ for an affine parameter. $\mathrm d\Omega^2 = \mathrm d\theta^2 + \sin^2\theta\,\mathrm d\varphi^2$.

Constants used for every number: $c = 299\,792\,458$ m/s, $GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3s^{-2}}$, $GM_\oplus = 3.986004418\times10^{14}\ \mathrm{m^3s^{-2}}$.

36 · overview

One hole, derived, then tested for its two singularities

1–2 · derive it

Static and round gives a four-term ansatz; $R_{\mu\nu} = 0$ fixes it up to one constant; unit 35's Newtonian limit fixes the constant.

3 · it is the only one

Birkhoff: round and empty already forces static. A pulsating star is silent; a hollow shell is flat inside.

4 · its symmetries

Four Killing fields; two constants of motion $E$ and $L$, for free. This is unit 13's theorem.

5–7 · its singularities

At $r = r_s$ a chart fails and a new chart repairs it. At $r = 0$ an invariant diverges and nothing repairs it. Then the whole spacetime, four regions, drawn.

the running example

A black hole of ten solar masses. Every section returns to it with a number: its radius $29.53$ km, the curvature at its horizon, the tidal stretch across a body falling in, the $65.7\ \mu$s of proper time between horizon and centre, and the diagram of the whole thing.

getting around

Words with a dotted underline are glossary terms: click one for its definition. G lists every term, N shows the notes, where the longer proofs live. A solid underline, like the metric, shows a piece of the running example on hover.

Why a black hole, and why ten solar masses. The exterior Schwarzschild metric describes the outside of the Sun just as well, and unit 37 uses it there. But the Sun's surface sits at $235\,000$ Schwarzschild radii, so both of the metric's singular places are buried inside matter, where the vacuum equations do not hold. To see what the metric says about them, one needs an object with no surface outside $r_s$: a black hole. Ten solar masses is typical of those found in X-ray binaries and of the components of the first gravitational-wave sources; it is small enough that its horizon is violent, which makes the contrast with the supermassive hole of the Milky Way's centre, $4.3\times10^{6}M_\odot$, instructive in section 5.

Prerequisites, and where each is used. Unit 32's Christoffel formula and its Lagrangian shortcut (worked example 1); unit 33's Ricci and Riemann tensors and geodesic deviation (sections 2 and 5); unit 31's Killing fields and Frobenius theorem (sections 1 and 4); unit 35's Newtonian limit and Killing's equation (sections 2 and 4); unit 13's Noether theorem (section 4); unit 30's Rindler horizon, which is the model for the whole of section 7.

section 1

The ansatz

What "static" and "spherically symmetric" mean as statements about a metric, and why together they leave two functions of one variable.

36 · ansatz

Static and spherically symmetric, defined

definition · round

$(M, g)$ is spherically symmetric if $SO(3)$ acts on $M$ by isometries and every orbit is a round 2-sphere (or a point). The areal radius of a point is $$r(p) := \sqrt{\operatorname{Area}(\text{orbit of } p)/4\pi}.$$

definition · unchanging

$(M,g)$ is stationary if it has a Killing field $\xi$ (unit 31) that is timelike, and static if moreover $\xi$ is orthogonal to a family of hypersurfaces.

the physics

Outside a non-rotating star nothing depends on direction or on time. "Round" is the first statement, "static" the second. A rotating star is stationary but not static: the orthogonality fails, and that failure is frame dragging (unit 40).

what $r$ is not

$r$ is not a distance from a centre. It is defined by the area of a sphere, and the "centre" need not even be part of the spacetime. Radial distances will turn out longer than differences of $r$.

Why define $r$ by area. In flat space the sphere of radius $r$ has area $4\pi r^2$, and there are three candidate definitions of "radius" that agree: area, circumference over $2\pi$, and distance from the centre. In curved space they disagree, and only the first two are available without first knowing where the centre is. Areal radius is also a scalar function on spacetime, fixed by the geometry with no choice: that is what makes it a legitimate coordinate to solve for things in.

Static, unpacked. By Frobenius (unit 31), $\xi$ is orthogonal to a family of hypersurfaces exactly when the distribution $\xi^\perp$ is involutive. Take one such hypersurface $\Sigma$, put coordinates $x^i$ on it, and carry them along the flow of $\xi$; call the flow parameter $t$. Then $\xi = \partial_t$, and because the flow is by isometries, $\partial_tg_{\mu\nu} = 0$ (unit 31's Killing condition in adapted coordinates). Because $\partial_t$ is orthogonal to the slices, $g_{ti} = 0$. So a static metric has the form $-N(x)^2c^2\mathrm dt^2 + h_{ij}(x)\,\mathrm dx^i\mathrm dx^j$, with nothing depending on $t$ and no cross terms. Unit 30's Rindler metric is of exactly this form — the reminder that "static" describes a symmetry, not the presence of gravity.

Why the distinction matters physically. Time reversal $t\mapsto-t$ is an isometry of a static metric: a film of the geometry run backwards is the same film. A rotating star run backwards rotates the other way, which is a different geometry. That is the whole content of "stationary but not static".

36 · ansatz

Why the ansatz has four terms

theorem · the Schwarzschild ansatz

A static, spherically symmetric metric, in a region where $\xi$ is timelike and $\mathrm dr \ne 0$, has coordinates in which

$$\mathrm ds^2 = -A(r)\,c^2\mathrm dt^2 + B(r)\,\mathrm dr^2 + r^2\,\mathrm d\Omega^2$$

with $A, B \gt 0$: the ansatz. Proof on the next slide; it rests on one lemma.

lemma · rotation fixes nothing

Rotation by $90^\circ$ fixes no nonzero vector of $\mathbb R^2$, and the symmetric bilinear forms it preserves are exactly the multiples of the dot product.

Proof. $\rho(x,y) = (-y, x)$. If $b$ is $\rho$-invariant, $b(e_1,e_1) = b(e_2,e_2)$ and $b(e_1,e_2) = b(e_2,-e_1) = -b(e_1,e_2)$, so $b(e_1,e_2) = 0$. $\square$

The rotations about the axis through $p$ fix $p$ and the radial direction, and turn the sphere's tangent plane $S_p$. A cross term $\mathrm dr\,\mathrm d\theta$ would be a vector $n$ in $S_p$ left fixed by every turn; an angular metric other than $\lambda\,\mathrm d\Omega^2$ would be an ellipse left fixed. The lemma says neither exists.

Reading the figure. Fix a point $p$ and look at the rotations of space that leave it where it is: the rotations about the axis through $p$. Each one is an isometry, so its differential at $p$ is a linear isometry of the tangent space there. It maps the sphere through $p$ to itself, so it turns the tangent plane $S_p$ of that sphere — by any angle, in particular by $90^\circ$ — and it leaves alone the two directions out of the sphere, the radial direction $\nabla r$ and the static Killing field $\xi$. The metric at $p$ must look the same before and after. A term $g_{r\theta}\,\mathrm dr\,\mathrm d\theta + g_{r\varphi}\,\mathrm dr\,\mathrm d\varphi$ pairs $\partial_r$ with a vector $n$ in $S_p$ (the one with $g(\partial_r, s) = g(n, s)$ for $s\in S_p$), and invariance says $n$ equals its own rotation; the restriction of $g$ to $S_p$ is a symmetric form, drawn by its unit "circle", and invariance says the figure equals its own rotation. The lemma kills the first and forces the second to be round.

Why no $\mathrm dt\,\mathrm d\varphi$? That is the term a rotating star has, and the lemma is exactly what forbids it here: it would be a nonzero invariant vector on the sphere.

36 · ansatz

The ansatz, proved in three moves

proof, in three moves

1. Static: $g_{ti} = 0$, no $t$-dependence.
2. At a point $p$, the rotations fixing $p$ form an $SO(2)$ acting on the sphere's tangent plane and fixing the $(t, r)$ directions. By the lemma, cross terms $\mathrm dr\,\mathrm d\theta$, $\mathrm dr\,\mathrm d\varphi$ vanish and the angular part is $\lambda\,\mathrm d\Omega^2$; the definition of $r$ makes $\lambda = r^2$.
3. What remains, $-Ac^2\mathrm dt^2 + B\,\mathrm dr^2$, cannot depend on $t$ (step 1) or on the angles (rotations), so $A, B$ are functions of $r$. $\square$

Ten functions of four variables have become two functions of one — and for the hole those two will turn out to be fixed by a single length, $29.53$ km.

What the ansatz says, drawn (schematic): a two-dimensional $(t, r)$ spacetime with metric $-Ac^2\mathrm dt^2 + B\,\mathrm dr^2$, and at each of its points a round sphere of area $4\pi r^2$. The spheres depend on $r$ only, never on $t$.

Full proof. Step 2 needs care. Fix $p$ and let $H \cong SO(2)$ be its isotropy group, the rotations about the axis through $p$'s orbit point. Each $h\in H$ is an isometry fixing $p$, so its differential $\mathrm dh_p$ is a linear isometry of $T_pM$. $H$ maps each orbit to itself, so $\mathrm dh_p$ preserves the 2-plane $S_p$ tangent to the orbit and acts on it as a rotation, and preserves the orthogonal 2-plane $N_p = S_p^\perp$. On $N_p$ it acts trivially: $N_p$ is spanned by $\xi$ (which commutes with the rotations — built into "static and spherically symmetric", the two symmetries being taken together) and by $\nabla r$ ($r$ is rotation-invariant, so $\mathrm dh_p$ fixes its gradient). Now let $n \in N_p$. The linear form $s \mapsto g(n, s)$ on $S_p$ is $H$-invariant, so the vector representing it in $S_p$ is fixed by a $90^\circ$ rotation, and by the lemma it is zero: there are no cross terms between $N_p$ and $S_p$. The restriction of $g$ to $S_p$ is an $H$-invariant symmetric form, so by the lemma a multiple of the round metric, and the multiple is $r^2$ by the definition of areal radius. With $t$ the Killing parameter and $r$ the areal radius as coordinates on $N$ (legitimate where $\mathrm dr \ne 0$), $g_{tr} = 0$ because $\partial_t$ is orthogonal to the slices and $\partial_r$ is tangent to them. $\square$

Cited: that the orthogonal planes $N_p$ fit together into surfaces, so that $(t, r)$ are coordinates on them — Frobenius again, applied to the orthogonal distribution. A careful account is in R. M. Wald, General Relativity (1984), §6.1, and in Hawking and Ellis, The Large Scale Structure of Space-Time (1973), Appendix B.

The residual freedom. The theorem leaves exactly one: $t \mapsto kt$ for a constant $k \gt 0$, which multiplies $A$ by $k^{-2}$. Keep it in mind; it is spent in section 2.

The figure is the shape of the result: a metric of the form $h_{ab}(x)\,\mathrm dx^a\mathrm dx^b + r(x)^2\,\mathrm d\Omega^2$ on a product of a two-dimensional space with a sphere is called a warped product, and every spherically symmetric metric has this form (step 2 alone). "Static" then makes the two-dimensional part $-Ac^2\mathrm dt^2 + B\,\mathrm dr^2$ with coefficients independent of $t$. Birkhoff, in section 3, starts from the warped product without the static step.

section 2

Solving $R_{\mu\nu} = 0$, slowly

Nine Christoffel symbols, four Ricci components, two lines of algebra, one constant from Newton.

36 · solving · worked example

Worked example 1 — the nine Christoffel symbols

unit 32's shortcut: Euler–Lagrange for $\mathcal L = g_{\mu\nu}\dot x^\mu\dot x^\nu$

$\mathcal L = -Ac^2\dot t^2 + B\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\,\dot\varphi^2$, $\ ' = \mathrm d/\mathrm dr$.

$t$: $\frac{\mathrm d}{\mathrm d\lambda}(-2Ac^2\dot t) = 0$, i.e. $A\ddot t + A'\dot r\dot t = 0$: $$\ddot t + \frac{A'}{A}\dot r\dot t = 0 .$$

$r$: $\frac{\mathrm d}{\mathrm d\lambda}(2B\dot r) = 2B\ddot r + 2B'\dot r^2$ equals $\partial_r\mathcal L = -A'c^2\dot t^2 + B'\dot r^2 + 2r\dot\theta^2 + 2r\sin^2\theta\,\dot\varphi^2$. Divide by $2B$: $$\begin{aligned}\ddot r &+ \frac{c^2A'}{2B}\dot t^2 + \frac{B'}{2B}\dot r^2\\ &- \frac rB\dot\theta^2 - \frac{r\sin^2\theta}{B}\dot\varphi^2 = 0 .\end{aligned}$$

$\theta$, $\varphi$: the sphere's, with $r$ variable (unit 32).

Read off each $\Gamma^\mu_{\alpha\beta}$ as the coefficient of $\dot x^\alpha\dot x^\beta$ (halved when $\alpha\ne\beta$). Nine of forty survive:

$$\begin{aligned} \Gamma^t_{tr} &= \frac{A'}{2A}, & \Gamma^r_{tt} &= \frac{c^2A'}{2B},\\ \Gamma^r_{rr} &= \frac{B'}{2B}, & \Gamma^r_{\theta\theta} &= -\frac rB,\\ \Gamma^r_{\varphi\varphi} &= -\frac{r\sin^2\theta}{B}, & \Gamma^\theta_{r\theta} &= \Gamma^\varphi_{r\varphi} = \frac1r,\\ \Gamma^\theta_{\varphi\varphi} &= -\sin\theta\cos\theta, & \Gamma^\varphi_{\theta\varphi} &= \cot\theta . \end{aligned}$$

checks

$B = 1/A$ gives unit 32's worked example 5 ✓. $A = B = 1$: flat space in spherical coordinates ✓. $[\Gamma^r_{tt}\dot t^2] = [c^2/\text{length}]\,\mathrm s^2/[\lambda]^2$, a length per $[\lambda]^2$, like $\ddot r$ ✓.

Why the shortcut is legitimate. Unit 32 proved that the Euler–Lagrange equations of the energy functional $\int g_{\mu\nu}\dot x^\mu\dot x^\nu\,\mathrm d\lambda$ are the geodesic equations $\ddot x^\mu + \Gamma^\mu_{\alpha\beta}\dot x^\alpha\dot x^\beta = 0$, with the Levi-Civita $\Gamma$. Since $\Gamma^\mu_{\alpha\beta}$ is symmetric in its lower indices, the coefficient of $\dot x^\alpha\dot x^\beta$ with $\alpha \ne \beta$ in the $\mu$-th equation is $2\Gamma^\mu_{\alpha\beta}$ — hence the halving, which is where $\Gamma^t_{tr} = A'/2A$ comes from.

The angular equations, written out. $\theta$: $\frac{\mathrm d}{\mathrm d\lambda}(2r^2\dot\theta) = 2r^2\sin\theta\cos\theta\,\dot\varphi^2$, so $\ddot\theta + \frac2r\dot r\dot\theta - \sin\theta\cos\theta\,\dot\varphi^2 = 0$. $\varphi$: $\frac{\mathrm d}{\mathrm d\lambda}(2r^2\sin^2\theta\, \dot\varphi) = 0$, so $\ddot\varphi + \frac2r\dot r\dot\varphi + 2\cot\theta\,\dot\theta\dot\varphi = 0$. Neither contains $A$ or $B$, because $g_{\theta\theta}$ and $g_{\varphi\varphi}$ do not.

Two symbols checked against the Koszul formula (unit 32), $\Gamma^\mu_{\alpha\beta} = \tfrac12g^{\mu\nu}(\partial_\alpha g_{\nu\beta} + \partial_\beta g_{\nu\alpha} - \partial_\nu g_{\alpha\beta})$: $\Gamma^r_{tt} = \tfrac12 g^{rr}(-\partial_rg_{tt}) = \tfrac1{2B}\,\partial_r(Ac^2) = c^2A'/2B$ ✓; $\Gamma^r_{\theta\theta} = \tfrac12g^{rr}(-\partial_r r^2) = -r/B$ ✓. Problem P2 checks the rest and runs unit 32's metric-compatibility test.

Why thirty-one vanish. Every metric component is independent of $t$ and $\varphi$, and the metric is diagonal. For a diagonal metric $\Gamma^\mu_{\alpha\beta}$ with three distinct indices is zero, and every surviving symbol involves a derivative of some $g_{\mu\mu}$ by $r$ or $\theta$ — which is exactly the list.

36 · solving

The Ricci tensor, and the two tools that shorten it

unit 33's contraction, written out $$\begin{aligned}R_{\sigma\nu} = {}&\partial_\mu\Gamma^\mu_{\nu\sigma} - \partial_\nu\Gamma^\mu_{\mu\sigma}\\ &+ \Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\mu_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}\end{aligned}$$

tool 1 · the trace of $\Gamma$ $$\Gamma^\mu_{\mu\lambda} = \partial_\lambda\ln\sqrt{-g},$$ $$\sqrt{-g} = c\sqrt{AB}\,r^2\sin\theta .$$

So $\Gamma^\mu_{\mu r} = \frac{A'}{2A} + \frac{B'}{2B} + \frac2r$ and $\Gamma^\mu_{\mu\theta} = \cot\theta$.

tool 2 · symmetry kills the off-diagonal

$R_{\mu\nu}$ is built from $g$ alone, so every isometry preserves it. Under $t\mapsto -t$, $R_{tr}$, $R_{t\theta}$, $R_{t\varphi}$ change sign: they vanish. $R_{r\theta}$, $R_{r\varphi}$ would define an invariant vector on the sphere: zero by the lemma. $R_{\theta\varphi} = 0$ and $R_{\varphi\varphi} = \sin^2\theta\,R_{\theta\theta}$ by the lemma's second half.

Four components remain: $R_{tt}$, $R_{rr}$, $R_{\theta\theta}$, and $R_{\varphi\varphi}$, which comes free.

Full proof of tool 1. $\Gamma^\mu_{\mu\lambda} = \tfrac12g^{\mu\nu}(\partial_\mu g_{\nu\lambda} + \partial_\lambda g_{\nu\mu} - \partial_\nu g_{\mu\lambda})$. The first and third terms cancel after swapping the dummy indices $\mu\leftrightarrow\nu$ (using $g^{\mu\nu}$ symmetric). What remains is $\tfrac12g^{\mu\nu}\partial_\lambda g_{\mu\nu} = \tfrac12\operatorname{tr}(g^{-1}\partial_\lambda g) = \tfrac12\partial_\lambda\ln|\det g|$ by Jacobi's formula for the derivative of a determinant, which unit 35 used in varying $\sqrt{-g}$. Here $\det g = -Ac^2\cdot B\cdot r^2\cdot r^2\sin^2\theta$. $\square$

Full proof of tool 2. An isometry $\phi$ satisfies $\phi^*g = g$; the Levi-Civita connection is determined by $g$ (unit 32), so $\phi$ also preserves it, and therefore preserves every tensor built from it: $\phi^*\mathrm{Ric} = \mathrm{Ric}$. The map $t\mapsto -t$ is an isometry of the ansatz (it appears only as $\mathrm dt^2$), and it pulls $R_{tr}\,\mathrm dt\,\mathrm dr$ back to $-R_{tr}\,\mathrm dt\,\mathrm dr$, evaluated at the same point since nothing depends on $t$. So $R_{tr} = -R_{tr} = 0$; likewise $R_{t\theta}, R_{t\varphi}$. For $R_{r\theta}$, $R_{r\varphi}$: the rotations fixing a point act trivially on $\partial_r$, so $s\mapsto\mathrm{Ric}(\partial_r, s)$ is an invariant linear form on the sphere's tangent plane, zero by the lemma. The angular block of $\mathrm{Ric}$ is an invariant symmetric form on that plane, so a multiple of the round metric $\mathrm d\theta^2 + \sin^2\theta\,\mathrm d\varphi^2$, which says both $R_{\theta\varphi} = 0$ and $R_{\varphi\varphi} = \sin^2\theta\,R_{\theta\theta}$. $\square$

Why bother. The Ricci tensor has ten components, each a sum of up to $4 + 4 + 16 + 16$ terms. The two tools cut the work to three components, and the symmetry argument doubles as a check: if a direct computation of $R_{tr}$ gives something nonzero, there is an error upstream.

36 · solving · worked example

Worked example 2 — $R_{\theta\theta}$, every term

$R_{\theta\theta} = \partial_\mu\Gamma^\mu_{\theta\theta} - \partial_\theta\Gamma^\mu_{\mu\theta} + \Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{\theta\theta} - \Gamma^\mu_{\theta\lambda}\Gamma^\lambda_{\mu\theta}$

Term 1. Only $\Gamma^r_{\theta\theta} = -r/B$ is nonzero: $$\partial_r\Big(-\frac rB\Big) = -\frac1B + \frac{rB'}{B^2}.$$

Term 2. $\Gamma^\mu_{\mu\theta} = \cot\theta$ (tool 1): $$-\partial_\theta\cot\theta = \frac{1}{\sin^2\theta}.$$

Term 3. Only $\lambda = r$ survives: $$\Big(\frac{A'}{2A} + \frac{B'}{2B} + \frac2r\Big)\Big(-\frac rB\Big).$$

Term 4. Three pairs $(\mu,\lambda)$ survive: $(r,\theta)$, $(\theta,r)$, $(\varphi,\varphi)$: $$-\Big[2\cdot\Big(-\frac rB\Big)\frac1r + \cot^2\theta\Big] = \frac2B - \cot^2\theta .$$

Sum. The $\theta$-dependence cancels, $\csc^2\theta - \cot^2\theta = 1$, and the $2/B$ terms cancel: $$R_{\theta\theta} = 1 - \frac1B + \frac{r}{2B}\Big(\frac{B'}{B} - \frac{A'}{A}\Big).$$

checks

$A = B = 1$: $R_{\theta\theta} = 0$ ✓. Dimensionless, as $R_{\theta\theta}$ must be since $\theta$ is ✓. The angles dropped out, as tool 2 promised ✓.

Term 4 in detail. $\Gamma^\mu_{\theta\lambda}\Gamma^\lambda_{\mu\theta}$ summed over $\mu,\lambda$. For the product to be nonzero, $\Gamma^\mu_{\theta\lambda}$ must be on the list with one lower index $\theta$: the candidates are $\Gamma^r_{\theta\theta}$ ($\mu = r, \lambda = \theta$), $\Gamma^\theta_{\theta r}$ ($\mu = \theta,\lambda = r$), and $\Gamma^\varphi_{\theta\varphi}$ ($\mu = \lambda = \varphi$). The partners are $\Gamma^\theta_{r\theta} = 1/r$, $\Gamma^r_{\theta\theta} = -r/B$ and $\Gamma^\varphi_{\varphi\theta} = \cot\theta$. So the sum is $(-r/B)(1/r) + (1/r)(-r/B) + \cot^2\theta = -2/B + \cot^2\theta$, and with the leading minus sign, $2/B - \cot^2\theta$.

The sum in detail. $-\frac1B + \frac{rB'}{B^2} + \csc^2\theta - \frac{rA'}{2AB} - \frac{rB'}{2B^2} - \frac2B + \frac2B - \cot^2\theta = 1 - \frac1B + \frac{rB'}{2B^2} - \frac{rA'}{2AB}$, which is the boxed form.

The same computation for the other two is problem P3, and its answers are on the next slide. Doing one component in full on the slide and the other two on paper is the right division of labour: the first time, every term needs to be seen; the second and third time, the skill is in not losing a term, and that is learnt only by doing it.

36 · solving

All four Ricci components

the Ricci tensor of the ansatz $$\begin{aligned} R_{tt} &= c^2\Big[\frac{A''}{2B} - \frac{A'}{4B}\Big(\frac{A'}{A} + \frac{B'}{B}\Big) + \frac{A'}{rB}\Big],\\ R_{rr} &= -\frac{A''}{2A} + \frac{A'}{4A}\Big(\frac{A'}{A} + \frac{B'}{B}\Big) + \frac{B'}{rB},\\ R_{\theta\theta} &= 1 - \frac1B + \frac{r}{2B}\Big(\frac{B'}{B} - \frac{A'}{A}\Big),\qquad R_{\varphi\varphi} = \sin^2\theta\,R_{\theta\theta}. \end{aligned}$$

All other components vanish.

checks

$A = B = 1$: all zero ✓. Dimensions: $R_{tt}$ is $c^2/\text{length}^2$ since $g_{tt}$ carries $c^2$; $R_{rr}$ is $1/\text{length}^2$; $R_{\theta\theta}$ is a pure number ✓.

the combination to look for

$R_{tt}$ and $R_{rr}$ share the bracket with $A''$. Multiply $R_{tt}$ by $B/(c^2A)$ and add $R_{rr}$: the $A''$ terms and the brackets cancel.

Where these come from. $R_{tt}$: term 1 is $\partial_r\Gamma^r_{tt} = \partial_r(c^2A'/2B)$; term 2 vanishes ($\Gamma^\mu_{\mu t} = 0$); term 3 is $\Gamma^\mu_{\mu r}\Gamma^r_{tt}$; term 4 is $-2\Gamma^t_{tr}\Gamma^r_{tt}$. Adding: $c^2\big[\frac{A''}{2B} - \frac{A'B'}{2B^2} + \big(\frac{A'}{2A} + \frac{B'}{2B} + \frac2r\big)\frac{A'}{2B} - \frac{A'^2}{2AB}\big]$, which collects to the boxed form. $R_{rr}$ is problem P3, where every term is laid out.

An independent check of the whole table. Flat space, $A = B = 1$, is a weak test: it only checks that no term survives when everything is constant. A sharper check is a line times a round 3-sphere, $A = 1$, $B = (1 - r^2/a^2)^{-1}$, a static spacetime whose spatial slices are 3-spheres of radius $a$ (unit 39's closed universe, frozen). Then $R_{tt} = 0$, and unit 33's formula for a maximally symmetric 3-space, $R_{ij} = (2/a^2)h_{ij}$, predicts $R_{rr} = 2B/a^2$ and $R_{\theta\theta} = 2r^2/a^2$. From the table: $B'/B = 2r/(a^2 - r^2)$, so $R_{rr} = \frac{B'}{rB} = \frac{2}{a^2 - r^2} = \frac{2B}{a^2}$ ✓, and $R_{\theta\theta} = \frac{r^2}{a^2} + \frac r2\big(1 - \frac{r^2}{a^2}\big)\frac{2r}{a^2 - r^2} = \frac{r^2}{a^2} + \frac{r^2}{a^2} = \frac{2r^2}{a^2}$ ✓. A formula that passes a non-trivial curved case as well as the flat one is very probably right.

36 · solving

Solving: two lines and one constant

Line 1. The combination: $$\frac{B}{c^2A}R_{tt} + R_{rr} = \frac1r\Big(\frac{A'}{A} + \frac{B'}{B}\Big) = \frac{(AB)'}{rAB}.$$ Vacuum makes it zero, so $AB$ is constant. Asymptotic flatness — $A, B \to 1$ as $r\to\infty$ — makes the constant $1$: $$B = 1/A .$$

Line 2. Put $B = 1/A$, so $B'/B = -A'/A$, into $R_{\theta\theta}$: $$R_{\theta\theta} = 1 - A - rA' = 1 - (rA)' = 0 .$$ So $rA = r - r_s$ for a constant $r_s$: $$A = 1 - \frac{r_s}{r},\qquad B = \Big(1 - \frac{r_s}{r}\Big)^{-1}.$$

the third equation holds by itself

With $f = 1 - r_s/r$: $f' = r_s/r^2$, $f'' = -2r_s/r^3$, and $B = 1/A$ kills the bracket in $R_{tt}$: $$R_{tt} = c^2f\Big(\frac{f''}2 + \frac{f'}r\Big) = c^2f\Big(-\frac{r_s}{r^3} + \frac{r_s}{r^3}\Big) = 0\ ✓$$

why three equations had a solution

Three equations, two unknowns — overdetermined, yet consistent. That is unit 33's contracted Bianchi identity at work: $\nabla^\mu G_{\mu\nu} = 0$ holds for every metric and ties the components together.

Line 1, the algebra. $\frac{B}{c^2A}R_{tt} = \frac{A''}{2A} - \frac{A'}{4A}\big(\frac{A'}{A} + \frac{B'}{B}\big) + \frac{A'}{rA}$. Adding $R_{rr}$, the first two terms cancel against the first two of $R_{rr}$ exactly, leaving $\frac{A'}{rA} + \frac{B'}{rB}$. $\square$

Spending the residual freedom. Line 1 gives $AB = k$ for a constant $k \gt 0$. Section 1 left the freedom to rescale $t$ by a constant: replacing $t$ by $\tilde t = \sqrt{k}\,t$ turns $-Ac^2\mathrm dt^2$ into $-(A/k)c^2\mathrm d\tilde t^2$, so $\tilde A\tilde B = (A/k)B = 1$. Choosing $t$ to be the time of a clock at rest far away — which is what asymptotic flatness means — is exactly this choice. So "$AB = 1$" is not an extra assumption; it is a choice of units for $t$.

Line 2, the integration. $(rA)' = 1$ integrates to $rA = r + C$; writing $C = -r_s$ is only naming the constant, whose sign is not yet known. It is fixed on the next slide, and turns out positive for positive mass.

Why $R_{tt} = 0$ comes free, precisely. For any metric, $\nabla^\mu G_{\mu r} = 0$. For the ansatz, written out with the Christoffel symbols above, this identity expresses $G_{\theta\theta}$ in terms of $G_{tt}$, $G_{rr}$ and $\partial_rG_{rr}$. So once $G_{tt}$ and $G_{rr}$ vanish identically in $r$, the angular equation is automatic. The direct check on the slide is independent of this argument, which is why both are given.

What was not assumed. Nothing about the source: the solution is valid in any region of vacuum outside a round static body, whatever the body is made of. That observation is the germ of section 3.

36 · solving

The constant, from the Newtonian limit

Far away the field is weak, and unit 35's limit (with unit 32's identification) says $g_{tt} \to -(1 + 2\Phi/c^2)c^2$ with $\Phi = -GM/r$. Compare $g_{tt} = -(1 - r_s/r)c^2$:

the Schwarzschild metric $$\mathrm ds^2 = -\Big(1 - \frac{r_s}{r}\Big)c^2\mathrm dt^2 + \frac{\mathrm dr^2}{1 - r_s/r} + r^2\,\mathrm d\Omega^2,$$

with the Schwarzschild radius $\ r_s = \dfrac{2GM}{c^2}$.

checks

Dimensions: $GM/c^2$ is $\mathrm{m^3s^{-2}}/\mathrm{m^2s^{-2}} = $ m ✓. $M\to0$: Minkowski in spherical coordinates ✓. $r\gg r_s$: unit 32's $\Gamma^r_{tt} = GMf/r^2 \to GM/r^2$, Newton ✓.

object	$r_s$	$r_s/R$
Earth	$8.870$ mm	$1.39\times10^{-9}$
Sun	$2.953$ km	$4.245\times10^{-6}$
the hole	$29.53$ km	—

only outside the matter

The Sun's $r_s$ lies deep inside the Sun, where $T_{\mu\nu}\ne0$ and this metric does not apply. The Sun has no horizon. The hole has no surface: its $r_s$ is in vacuum.

The matching, in full. Unit 32 showed that for a slow particle in a weak static field the geodesic equation reduces to $\ddot x^i = -\partial_i\Phi$ when $g_{00} = -(1 + 2\Phi/c^2)$ with $x^0 = ct$, i.e. $g_{tt} = -(1 + 2\Phi/c^2)c^2$; unit 35 showed the field equations then give $\nabla^2\Phi = 4\pi G\rho$, so outside a mass $M$ the potential is $-GM/r$. Matching $-(1 - r_s/r) = -(1 - 2GM/(rc^2))$ term by term gives $r_s = 2GM/c^2$. The weak-field identification is made at large $r$, where $r$, the areal radius, and the Newtonian radial coordinate agree to the order that matters.

What $M$ means. $M$ is whatever mass a distant astronomer infers from Kepler's third law for wide orbits — unit 08's $GM = 4\pi^2a^3/T^2$. For a star it is not the sum of the rest masses of its particles: it includes thermal energy and the (negative) gravitational binding energy, all divided by $c^2$. General relativity answers "how much does the star weigh" with this single number, and $GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3s^{-2}}$ is known to eleven figures even though $G$ and $M_\odot$ separately are known to five.

The numbers. $2GM_\oplus/c^2 = 2\times3.986004418\times10^{14}/8.98755179\times10^{16} = 8.870\times10^{-3}$ m. $2GM_\odot/c^2 = 2953.25$ m. For ten solar masses, $29\,532.5$ m. Ratios: $8.870\times10^{-3}/6.378137\times10^{6} = 1.39\times10^{-9}$, and $2953.25/6.957\times10^{8} = 4.245\times10^{-6}$ — twice unit 35's $|\Phi|/c^2$ at the same surfaces, as it must be, since $r_s/R = 2GM/(Rc^2) = 2|\Phi|/c^2$.

History. Schwarzschild's letter reached Einstein in December 1915; the paper appeared in January 1916. His radial coordinate was not the areal one; the form above is due to Droste and to Hilbert, both in 1916. Schwarzschild died in May 1916 of an illness contracted at the front.

section 3

Birkhoff: round and empty forces static

The static assumption was never needed. What that buys: silence, and a flat cavity.

36 · Birkhoff

Birkhoff's theorem

A spherically symmetric solution of $R_{\mu\nu} = 0$, in a region where $\nabla r$ is spacelike, is isometric to part of the Schwarzschild metric for some $r_s$. In particular it is static: it has a fourth Killing field it was never asked to have.

Drop "static" from the ansatz. Spherical symmetry alone allows (cited: section 1's argument without step 1)

$$\mathrm ds^2 = -A(t,r)c^2\mathrm dt^2 + B(t,r)\,\mathrm dr^2 + r^2\,\mathrm d\Omega^2 .$$

proof

1. The one new component is $$R_{tr} = \frac{\partial_tB}{rB}.$$ Vacuum: $\partial_t B = 0$, so $B = B(r)$.
2. $R_{\theta\theta}$ contains no $t$-derivatives, so $R_{\theta\theta} = 0$ reads as before: $\frac{\partial_rA}{A} = \frac{B'}{B} + \frac{2(B - 1)}{r}$, a function of $r$ alone. So $A = e^{h(t)}a(r)$.
3. New time $\tilde t = \int e^{h(t)/2}\,\mathrm dt$: then $-Ac^2\mathrm dt^2 = -a(r)c^2\mathrm d\tilde t^2$. The metric is now the static ansatz, and section 2 finishes it. $\square$

Why the time-dependent ansatz is general. Section 1's argument without step 1 still gives a metric of the form $h_{ab}\,\mathrm dx^a\mathrm dx^b + r^2\mathrm d\Omega^2$, with $h$ a Lorentzian metric on the two-dimensional orbit space. Where $\nabla r$ is spacelike, $r$ can be one coordinate. For the other, take a function $t$ whose level curves are the integral curves of $\nabla r$ — in two dimensions any nonvanishing vector field has such a function locally, since Frobenius is automatic for a line field. Then $\nabla t \perp \nabla r$, so $h_{tr} = 0$. Cited: the full construction is in Hawking and Ellis, Appendix B, and in Wald, §6.1.

Full proof of step 1. With $A$, $B$ depending on $t$ the Christoffel list gains $\Gamma^t_{tt} = \dot A/2A$, $\Gamma^t_{rr} = \dot B/(2Ac^2)$, $\Gamma^r_{tr} = \dot B/2B$ (dots are $\partial_t$ here). In $R_{tr} = \partial_\mu\Gamma^\mu_{rt} - \partial_r\Gamma^\mu_{\mu t} + \Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{rt} - \Gamma^\mu_{r\lambda}\Gamma^\lambda_{\mu t}$, the first two terms are $\partial_t(A'/2A) + \partial_r(\dot B/2B) - \partial_r(\dot A/2A + \dot B/2B)$, which is zero by equality of mixed partials of $\ln A$. In the last two, every product cancels against a partner except the one in which $\Gamma^\mu_{\mu r}$ contributes its $2/r$ (from $\Gamma^\theta_{\theta r} + \Gamma^\varphi_{\varphi r}$), times $\Gamma^r_{rt} = \dot B/2B$. The net is $\dot B/(rB)$. Problem P4 lays out every term. $\square$

Step 2 in detail. With $B = B(r)$, $R_{\theta\theta} = 1 - \frac1B + \frac r{2B}\big(\frac{B'}B - \frac{\partial_rA}{A}\big)$ exactly as before — the symbols entering $R_{\theta\theta}$ are $\Gamma^r_{\theta\theta}$, $\Gamma^\mu_{\mu r}$ and the angular ones, and none is differentiated in $t$. Setting it to zero and solving for $\partial_rA/A$ gives a function of $r$; integrating in $r$ at fixed $t$ gives $\ln A = F(r) + h(t)$.

After step 3, $A$ and $B$ are functions of $r$ alone, so every $t$-derivative in every component is zero and section 2's computation applies verbatim: $A = 1 - r_s/r$, $B = 1/A$.

History and the name. Jørg Jebsen proved it in 1921 and George Birkhoff independently in 1923; the theorem bears the second name. In the region where $\nabla r$ is timelike — inside the horizon — the same argument with the roles of $t$ and $r$ exchanged gives the Schwarzschild interior, which is not static there; "static" in the theorem is a statement about the exterior.

36 · Birkhoff

Two consequences: silence, and a flat cavity

no monopole radiation

A star that pulsates, collapses or explodes, keeping round, has exactly the Schwarzschild metric outside it at every moment, with the same $M$. Nothing propagates outwards: a spherically symmetric gravitational wave does not exist.

The same is true of charge: Gauss's law (units 23, 24) fixes the field outside a round charge distribution, so a pulsating charged sphere does not radiate either. Gravitational waves start at the quadrupole (unit 38).

The hole was once a star that collapsed inside $29.53$ km. At every moment of that collapse the geometry outside the star's surface was Schwarzschild with the same $r_s$: a distant planet's orbit never flinched.

A pulsating star in a spacetime diagram (schematic). Birkhoff applies to the whole vacuum region to the right of the moving surface: it is one Schwarzschild geometry, with one $r_s$, however the surface moves. The planet's worldline cannot tell.

The pulsating star, precisely. Birkhoff applies in the vacuum region $r \gt R(t)$ outside the star, however $R(t)$ moves. It says the metric there is Schwarzschild for some $r_s$; could $r_s$ change with time? No: the constant $r_s$ is a single number for the whole connected vacuum region, and far away it is fixed by Kepler's law for a distant satellite, which cannot know the star is pulsating, because no signal carrying that news exists. Energy conservation is the physical reason: radiation would carry away mass, and there is no mass change available to carry.

The analogy with electromagnetism, and where it stops. For charge, the absence of monopole radiation follows from charge conservation plus Gauss's law, and the first radiating multipole is the dipole. For gravity the dipole is also silent: the time derivative of the mass dipole moment is the total momentum, which is conserved. So gravitational radiation starts at the quadrupole, which is why it is so weak and why it took until 2015 to see it. Unit 38 derives this.

Reading the figure. Time runs up, radius across; each point stands for a whole sphere. The shaded region is the star, where $T_{\mu\nu} \ne 0$ and the vacuum equations say nothing. The dashed line is $r_s$, inside the matter for a star that is not a black hole: there is no horizon anywhere in this picture. Everything to the right of the star's surface is the region Birkhoff's theorem is about, and it is static, though nothing in the setup said so.

36 · Birkhoff

Inside a hollow shell: flat, but slow

the relativistic shell theorem

Inside an empty spherical cavity — a hollow shell, round, of any mass — spacetime is flat.

Proof. Birkhoff gives Schwarzschild with some $r_s$ for $0 \lt r \lt r_{\text{shell}}$. The cavity's centre is a regular point of spacetime. But the curvature scalar $K = R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}$ of that metric — a number at each point, the same in every chart — is $12r_s^2/r^6$ (computed in section 5), which diverges at $r = 0$ unless $r_s = 0$. $\square$

flat is not the same as unaffected

A clock in the cavity runs slow relative to one at infinity by a constant factor. Newton's shell theorem (unit 05) predicted no force; general relativity adds a redshift.

The rate $\mathrm d\tau/\mathrm dt = \sqrt{-g_{tt}}/c$ of a static clock, for a thin shell at $R = 2r_s$. Inside it is flat — no slope, so no force — but sits at $\sqrt{1 - r_s/R} = 0.707$: a redshift without gravity.

Reading the figure. A static clock's rate relative to coordinate time is $\sqrt{-g_{tt}}/c$, which is $\sqrt{1 - r_s/r}$ outside the shell and the constant $\sqrt k$ inside. Continuity of $g_{tt}$ across the shell (next paragraph) joins the two at $r = R$. A force would show as a slope — unit 32's weak-field reading, $g_{tt} \approx -(1 + 2\Phi/c^2)c^2$, makes the slope of the curve the gravitational field — and inside there is none. What is left is the level: light climbing out of the cavity to infinity is redshifted by $1 + z = 1/\sqrt{k} = 1.414$ for this shell.

The cavity, precisely. Inside the shell, Birkhoff gives $A = k(1 - r_s/r)$, $B = (1 - r_s/r)^{-1}$, where now the constant $k$ cannot be set to $1$: $t$ was already fixed by the clocks at infinity, and matching across the shell determines $k$ (cited: continuity of the metric across a thin shell, Israel's junction conditions, 1966). Regularity at $r = 0$ forces $r_s = 0$: with $r_s \ne 0$ the curvature invariant diverges at $r\to 0$, but $r = 0$ is the centre of an empty cavity, which a timelike geodesic passes through in finite proper time. So inside, $\mathrm ds^2 = -k\,c^2\mathrm dt^2 + \mathrm dr^2 + r^2\mathrm d\Omega^2$ — Minkowski, with clocks running at the rate $\sqrt{k}$ relative to infinity. For a thin shell of mass $M$ and radius $R$, continuity of $g_{tt}$ across it gives $k = 1 - r_s/R$ with $r_s = 2GM/c^2$: exactly the rate of a static clock just outside the shell.

Newton's version. Unit 05 proved by Gauss's theorem that a uniform spherical shell exerts no force inside. The relativistic statement is stronger — flatness, meaning no tidal force either, and no need for uniformity beyond spherical symmetry — and says nothing about shells that are not exactly round, where Newton's theorem also fails.

section 4

Killing fields, and constants for free

Four symmetries, two conserved quantities, one theorem from unit 13 in geometric clothing.

36 · Killing

The four Killing fields of Schwarzschild

Unit 31: $K$ is a Killing vector field if $\mathcal L_Kg = 0$, in coordinates $$K^\lambda\partial_\lambda g_{\mu\nu} + g_{\lambda\nu}\partial_\mu K^\lambda + g_{\mu\lambda}\partial_\nu K^\lambda = 0 .$$

$\partial_t$ is Killing

$K = \partial_t$ has constant components $(1,0,0,0)$, so the last two terms vanish, and the first is $\partial_tg_{\mu\nu} = 0$ ✓.

the rotations are Killing

$\partial_\varphi$ likewise. The other two, $R_x = -\sin\varphi\,\partial_\theta - \cot\theta\cos\varphi\,\partial_\varphi$ and $R_y$, are unit 31's rotation fields of the sphere; they do not touch $t, r$, and $g$ restricted to each sphere is $r^2$ times the round metric, so unit 31's check applies unchanged.

the whole symmetry algebra

For $r_s \ne 0$ the Killing fields are exactly $\mathbb R\,\partial_t \oplus \mathfrak{so}(3)$: four, against Minkowski's ten. Cited: Hawking and Ellis, §5.5.

$\partial_t$ is not always time

$g(\partial_t,\partial_t) = -(1 - r_s/r)c^2$. Timelike for $r \gt r_s$, null at $r_s$, spacelike inside. The field still exists inside, and still generates an isometry, but there it translates along a direction of space.

Unit 31's check, recalled for $\partial_\varphi$. Components $(0,0,0,1)$, constant, so $\mathcal L_{\partial_\varphi}g = \partial_\varphi g_{\mu\nu} = 0$ since no component depends on $\varphi$. For $R_x$, the components depend on $\theta$ and $\varphi$ and the check is a genuine computation; unit 31 did it on the unit sphere, and since $R_x$ has no $t$ or $r$ components and $g_{\theta\theta}, g_{\varphi\varphi}$ are $r^2$ times the unit sphere's, every term of the Killing equation is $r^2$ times the corresponding unit-sphere term, or zero. Problem P5 does it explicitly.

Why "exactly four" is cited. Proving there are no others requires showing that the Killing equation's solution space, which unit 31 noted is determined by the value and antisymmetrised derivative of $K$ at one point, is four-dimensional here. The standard route uses the fact that isometries preserve curvature, and the Riemann tensor of Schwarzschild picks out the $(t,r)$ plane at every point (section 5 shows its structure); an isometry must therefore preserve the foliation by spheres and the function $r$. That argument is sound but long, and nothing in this course depends on it.

The warning box, and what it will cost. Inside the horizon the conserved quantity that $\partial_t$ produces is still conserved, but it is no longer an energy in any observer's sense: it is a momentum along a spatial direction, and its sign is not fixed. Unit 40 exploits the analogous fact for a rotating hole — where $\partial_t$ becomes spacelike outside the horizon, in the ergoregion — to extract energy from a black hole: the Penrose process.

36 · Killing

A Killing field conserves $g(K,\dot x)$

theorem

If $K$ is Killing and $x(\lambda)$ is an affinely parametrised geodesic, then $g(K, \dot x)$ is constant along it.

proof — one line

$$\frac{\mathrm d}{\mathrm d\lambda}\big(K_\mu\dot x^\mu\big) = \dot x^\nu\nabla_\nu\big(K_\mu\dot x^\mu\big) = \underbrace{(\nabla_\nu K_\mu)\,\dot x^\nu\dot x^\mu}_{\text{antisymmetric}\ \times\ \text{symmetric}\ =\ 0} + K_\mu\underbrace{\dot x^\nu\nabla_\nu\dot x^\mu}_{\text{geodesic}\ =\ 0} = 0 .\ \square$$

the ingredient from unit 35

$\mathcal L_Kg = 0$ is equivalent to Killing's equation $\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0$: replace the partials by covariant derivatives (the $\Gamma$ terms cancel, $\Gamma$ being symmetric) and use $\nabla g = 0$.

checks

Minkowski with $K = \partial_x$: $g(\partial_x,\dot x) = \dot x^1$, and free particles have constant velocity ✓. Nothing was assumed about signature or dimension ✓.

Full proof, with every step justified. Along the curve, $\frac{\mathrm d}{\mathrm d\lambda}$ of a scalar is $\dot x^\nu\partial_\nu$ of it, which equals $\dot x^\nu\nabla_\nu$ of it since the covariant derivative of a scalar is its partial derivative. The Leibniz rule for $\nabla$ (unit 32) splits $\nabla_\nu(K_\mu\dot x^\mu)$ into $(\nabla_\nu K_\mu)\dot x^\mu + K_\mu\nabla_\nu\dot x^\mu$; strictly $\dot x$ is only defined along the curve, but $\dot x^\nu\nabla_\nu\dot x^\mu$ is the covariant derivative along the curve, which is all that appears. The geodesic equation says it is zero. For the first term: by Killing's equation $\nabla_\nu K_\mu = -\nabla_\mu K_\nu$, so $T_{\nu\mu} := \nabla_\nu K_\mu$ is antisymmetric, and $T_{\nu\mu}\dot x^\nu\dot x^\mu = T_{\mu\nu}\dot x^\mu\dot x^\nu = -T_{\nu\mu}\dot x^\nu\dot x^\mu$ (rename, then use antisymmetry), so it is zero. $\square$

Killing's equation from $\mathcal L_Kg = 0$, as unit 35 showed: in $K^\lambda\partial_\lambda g_{\mu\nu} + g_{\lambda\nu}\partial_\mu K^\lambda + g_{\mu\lambda}\partial_\nu K^\lambda$, replace every $\partial$ by $\nabla$. The correction terms this introduces are $-K^\lambda(\Gamma^\sigma_{\lambda\mu}g_{\sigma\nu} + \Gamma^\sigma_{\lambda\nu}g_{\mu\sigma})$ from the first and $+g_{\lambda\nu}\Gamma^\lambda_{\mu\sigma}K^\sigma + g_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}K^\sigma$ from the other two, which cancel by the symmetry of $\Gamma$. Then $\nabla g = 0$ removes the first term and the remaining two are $\nabla_\mu K_\nu + \nabla_\nu K_\mu$.

For null geodesics too. The proof never used $g(\dot x,\dot x) = -c^2$, so it holds for light — which is how unit 37 gets the light-bending orbit equation.

36 · Killing

This is Noether's theorem

	unit 13	here
configuration space	$Q$, coordinates $q^i$	spacetime, coordinates $x^\mu$
Lagrangian	$L(q,\dot q)$	$\mathcal L = g_{\mu\nu}(x)\,\dot x^\mu\dot x^\nu$
symmetry	one-parameter group $\phi_s$ leaving $L$ invariant	the flow of $K$, leaving $g$ invariant
generator	$\delta q^i = \frac{\mathrm d}{\mathrm ds}\phi_s(q)^i\big\|_0$	$K^\mu$
charge	$\dfrac{\partial L}{\partial\dot q^i}\,\delta q^i$	$\dfrac{\partial\mathcal L}{\partial\dot x^\mu}K^\mu = 2\,g(K,\dot x)$

the claim, said out loud

The theorem on the previous slide is unit 13's Noether theorem, applied to the geodesic Lagrangian and the flow of a Killing field. The Killing condition is exactly "the Lagrangian is invariant".

why invariant

The flow $\phi_s$ acts on velocities by its differential; then $\mathcal L(\phi_s x, \mathrm d\phi_s\dot x) = (\phi_s^*g)(\dot x,\dot x) = g(\dot x,\dot x)$, since $\phi_s^*g = g$ (unit 31). $\square$

Full proof that the two charges agree. $\partial\mathcal L/\partial\dot x^\mu = 2g_{\mu\nu}\dot x^\nu$, so the Noether charge is $2g_{\mu\nu}\dot x^\nu K^\mu = 2g(K,\dot x)$. Unit 13 proved it constant along solutions of the Euler–Lagrange equations whenever $L$ is invariant under the group; unit 32 proved those solutions are the affinely parametrised geodesics. So the previous slide's theorem is a corollary of unit 13's, and the previous slide's proof is an independent second proof of it, written in the language of connections. Two proofs of one fact, from two directions, is the reason to give both.

The cyclic-coordinate special case. When $K = \partial_t$ or $\partial_\varphi$, the flow is a translation of one coordinate, and "invariant" means $\mathcal L$ does not contain that coordinate — unit 12's cyclic coordinate. The Noether charge is then just the conjugate momentum $\partial\mathcal L/\partial\dot t$ or $\partial\mathcal L/\partial\dot\varphi$. The rotations $R_x$, $R_y$ are Killing but not coordinate translations in these coordinates; for them the geometric statement is genuinely more convenient than hunting for coordinates in which they become cyclic.

Why this is worth saying. Unit 13's theorem was proved for point mechanics in $\mathbb R^n$. It has now reappeared, unchanged, as a statement about geodesics in a curved spacetime. Energy is conserved because the geometry does not change in time; angular momentum is conserved because it does not change under rotation. Nothing about gravity was needed except that free fall is geodesic.

36 · Killing · worked example

Worked example 3 — $E$, $L$, and the radial equation

timelike geodesics of the metric, $f = 1 - r_s/r$, $\tau$ proper time

Step 1 — the two constants. From $\partial_t$ and $\partial_\varphi$:

$$E := -g(\partial_t,\dot x) = f\,c^2\,\dot t,$$ $$L := g(\partial_\varphi,\dot x) = r^2\sin^2\theta\,\dot\varphi .$$

The specific energy and specific angular momentum: per unit rest mass.

Step 2 — the plane. By rotation invariance, start the orbit on $\theta = \pi/2$ with $\dot\theta = 0$; the $\theta$ equation keeps it there (uniqueness, unit 01). So $L = r^2\dot\varphi$.

Step 3 — the normalisation. $g(\dot x,\dot x) = -c^2$: $$-fc^2\dot t^2 + \frac{\dot r^2}{f} + r^2\dot\varphi^2 = -c^2 .$$

Step 4 — substitute $\dot t = E/(fc^2)$, $\dot\varphi = L/r^2$, and multiply by $f$:

$$\dot r^2 = \frac{E^2}{c^2} - \Big(1 - \frac{r_s}{r}\Big)\Big(c^2 + \frac{L^2}{r^2}\Big).$$

Step 5 — radial, from rest at infinity. $L = 0$, and $\dot r\to0$ as $r\to\infty$ gives $E = c^2$:

$$\dot r^2 = \frac{c^2r_s}{r} = \frac{2GM}{r}.$$

Newton's escape speed, exactly — in proper time and areal radius. At the hole's horizon, $|\dot r| = c$; at $10r_s$, $c/\sqrt{10} = 9.48\times10^{7}$ m/s.

checks

Every term is $\mathrm{m^2s^{-2}}$ ✓. $r_s\to0$: $\dot r^2 = E^2/c^2 - c^2 - L^2/r^2$, a free particle ✓.

Step 1, the sign. $g(\partial_t,\dot x) = g_{tt}\dot t = -fc^2\dot t$. The minus in the definition of $E$ makes it positive for a future-directed particle outside the horizon, where $\dot t \gt 0$ and $f \gt 0$. Dimensions: $E$ is $\mathrm{m^2s^{-2}}$, an energy per unit mass; $L$ is $\mathrm{m^2s^{-1}}$, unit 08's specific angular momentum.

Step 1, the physical reading. For a slow particle far away, $\dot t \approx 1 + v^2/2c^2 + r_s/2r$ (unit 27's time dilation and unit 30's gravitational one), so $E \approx (1 - r_s/r)c^2(1 + v^2/2c^2 + r_s/2r) = c^2 + \tfrac12v^2 - GM/r + O(c^{-2})$: rest energy plus Newton's energy, per unit mass. Unit 37 relies on this reading.

Step 2, in full. The $\theta$ geodesic equation (worked example 1) is $\ddot\theta + \frac2r\dot r\dot\theta - \sin\theta\cos\theta\,\dot\varphi^2 = 0$. With initial data $\theta = \pi/2$, $\dot\theta = 0$, the function $\theta\equiv\pi/2$ solves it (since $\cos(\pi/2) = 0$); by Picard–Lindelöf (unit 01) it is the solution. That we may choose such initial data is the rotation invariance: an isometry maps geodesics to geodesics (unit 32), and some rotation carries any initial position and velocity into the equatorial plane.

Step 4, the algebra. $-fc^2\cdot\frac{E^2}{f^2c^4} + \frac{\dot r^2}f + \frac{L^2}{r^2} = -c^2$; multiply by $f$: $-\frac{E^2}{c^2} + \dot r^2 + \frac{fL^2}{r^2} = -fc^2$, which rearranges to the boxed equation. It is exact: no approximation has been made. Unit 37 reads the whole of Mercury off it.

Step 5, and why it is remarkable. Newton's energy equation for radial fall from rest at infinity is $\tfrac12v^2 = GM/r$. The relativistic equation is identical in form, but $\dot r$ is $\mathrm dr/\mathrm d\tau$ with $r$ the areal radius and $\tau$ the falling body's own clock. The coincidence is special to radial fall with $E = c^2$ in these variables; with $L \ne 0$ an extra term $+r_sL^2/r^3$ appears on the right, which is the whole of unit 37.

section 5

Two singularities, one of them real

A chart to dismiss one, an invariant to convict the other. Neither tool can do the other's job.

36 · singularities

Where the components misbehave

In the metric, two places:

$r = r_s$: $g_{tt}\to0$ and $g_{rr}\to\infty$.
$r = 0$: $g_{tt}\to\infty$ and $g_{rr}\to0$.

Neither fact, by itself, says anything about the geometry. Components are a chart's description, and charts fail for their own reasons.

two failures seen before

The plane in polar coordinates, $\mathrm dr^2 + r^2\mathrm d\phi^2$: $\det g = r^2 \to 0$ at the origin (unit 05). Rindler's metric on flat spacetime: $g_{tt}\to0$ at the Rindler horizon (unit 30). Both are flat.

The two components that misbehave, against $r$. At $r_s$ one passes through zero and the other through infinity, and both change sign: $t$ and $r$ swap roles, timelike and spacelike, inside. At $r = 0$ (the thick line) the roles stay swapped and the components run off again.

Reading the figure. $-g_{tt}/c^2 = f = 1 - r_s/r$ and $g_{rr} = 1/f$. For $r \gt r_s$ both are positive: $\partial_t$ timelike, $\partial_r$ spacelike, as in Minkowski space, to which both curves tend at large $r$. Between $0$ and $r_s$ both are negative, so $\partial_t$ is spacelike and $\partial_r$ timelike — the coordinate called $r$ is a time there, which the horizon theorem later in this section turns into the statement that $r$ must decrease. The picture shows where the chart is in trouble; it cannot show whether the geometry is, which is the point of the next slide.

The polar example, precisely. In $\mathrm dr^2 + r^2\mathrm d\phi^2$ the metric is degenerate at $r = 0$, yet the plane is perfectly smooth there, as $x = r\cos\phi$, $y = r\sin\phi$ shows. The failure is that $\phi$ is not a coordinate at the origin. At $r = r_s$ the failure will turn out to be that $t$ is not a coordinate there.

36 · singularities

Two kinds of singularity, and how each is proved

two kinds

A coordinate singularity is a place where a chart fails but the geometry extends smoothly: some other chart covers it.

A curvature singularity is an end of spacetime reached by a geodesic in finite affine parameter, along which some curvature scalar is unbounded.

the strategy, and its asymmetry

To dismiss a singularity, exhibit a chart in which the metric is smooth and non-degenerate there. To convict one, exhibit an invariant that diverges. A finite invariant proves nothing; a failed search for a chart proves nothing.

A curvature scalar is a function built from $g$ and its derivatives whose value at a point is the same in every chart — the Kretschmann slide computes the one that matters here. For $r_s$, the chart is worked example 4.

Why a finite invariant proves nothing. A spacetime can have every polynomial curvature scalar finite, even zero, and still be singular: exact plane gravitational waves have all such scalars zero everywhere, and there are geodesically incomplete flat spacetimes (Minkowski with a point removed, or a cone-shaped identification). So "$K$ is finite at $r_s$" is suggestive, not a proof. The proof that $r_s$ is harmless has to be the chart.

Why a divergent invariant does prove something. A curvature scalar is a function on spacetime whose value at a point does not depend on any chart. If the metric extended as a $C^2$ tensor to a manifold containing the endpoint of a curve along which the scalar is unbounded, then the scalar, built from $g$ and its first two derivatives, would be continuous at that endpoint and hence bounded near it — contradiction. So no extension, in any coordinates whatever, makes it regular.

The definition of a singularity is subtle, and the one on the slide is the working definition physicists use. The mathematically careful notion is geodesic incompleteness: a geodesic that cannot be extended to all values of its affine parameter, in a spacetime that cannot be extended. Unit 40 states the Penrose singularity theorem in those terms. For Schwarzschild the two notions agree at $r = 0$.

36 · singularities

The Riemann tensor, in the static frame

Orthonormal frame: $e_{\hat t} = \frac{\partial_t}{c\sqrt f}$, $e_{\hat r} = \sqrt f\,\partial_r$, $e_{\hat\theta} = \frac{\partial_\theta}r$, $e_{\hat\varphi} = \frac{\partial_\varphi}{r\sin\theta}$. The six sectional components, up to the symmetries:

$$\begin{aligned} R_{\hat t\hat r\hat t\hat r} &= -\frac{r_s}{r^3}, & R_{\hat\theta\hat\varphi\hat\theta\hat\varphi} &= \frac{r_s}{r^3},\\ R_{\hat t\hat\theta\hat t\hat\theta} &= \frac{r_s}{2r^3}, & R_{\hat r\hat\theta\hat r\hat\theta} &= -\frac{r_s}{2r^3}, \end{aligned}$$

and $\hat\varphi$ like $\hat\theta$. All others vanish. Not flat: vacuum, yet curved — unit 35's point that the Weyl tensor survives $R_{\mu\nu} = 0$.

read through geodesic deviation (unit 33)

For a freely falling body momentarily at rest, $u = c\,e_{\hat t}$, and $\ddot\xi^{\hat\imath} = -c^2R^{\hat\imath}{}_{\hat t\hat\jmath\hat t}\,\xi^{\hat\jmath}$:

$$\ddot\xi^{\hat r} = +\frac{2GM}{r^3}\xi^{\hat r},\qquad \ddot\xi^{\hat\theta} = -\frac{GM}{r^3}\xi^{\hat\theta}.$$

Stretched radially, squeezed sideways: unit 05's Newtonian tidal tensor, exactly. The next slide draws it.

Full derivation, for the general $f$. For $\mathrm ds^2 = -fc^2\mathrm dt^2 + f^{-1}\mathrm dr^2 + r^2\mathrm d\Omega^2$ the orthonormal components are $R_{\hat t\hat r\hat t\hat r} = \tfrac12f''$, $R_{\hat t\hat\theta\hat t\hat\theta} = \frac{f'}{2r}$, $R_{\hat r\hat\theta\hat r\hat\theta} = -\frac{f'}{2r}$, $R_{\hat\theta\hat\varphi\hat\theta\hat\varphi} = \frac{1 - f}{r^2}$. Two of them from the definition, using worked example 1's symbols with $A = f$, $B = 1/f$: $R^\theta{}_{t\theta t} = \partial_\theta\Gamma^\theta_{tt} - \partial_t\Gamma^\theta_{\theta t} + \Gamma^\theta_{\theta\lambda}\Gamma^\lambda_{tt} - \Gamma^\theta_{t\lambda}\Gamma^\lambda_{\theta t} = \Gamma^\theta_{\theta r}\Gamma^r_{tt} = \frac1r\cdot\frac{c^2ff'}2$. Lower with $g_{\theta\theta} = r^2$ and divide by $g_{\theta\theta}|g_{tt}| = r^2fc^2$: $R_{\hat\theta\hat t\hat\theta\hat t} = \frac{f'}{2r}$, which by pair symmetry is $R_{\hat t\hat\theta\hat t\hat\theta}$. Similarly $R^\theta{}_{\varphi\theta\varphi} = \partial_\theta\Gamma^\theta_{\varphi\varphi} + \Gamma^\theta_{\theta r}\Gamma^r_{\varphi\varphi} - \Gamma^\theta_{\varphi\varphi}\Gamma^\varphi_{\theta\varphi} = -\cos2\theta - f\sin^2\theta + \cos^2\theta = (1 - f)\sin^2\theta$; lowering and dividing by $r^4\sin^2\theta$ gives $(1-f)/r^2$. With $f = 1 - r_s/r$: $f' = r_s/r^2$, $f'' = -2r_s/r^3$, $1 - f = r_s/r$, giving the slide.

The deviation equation. Unit 33: $\frac{\mathrm D^2\xi^\rho}{\mathrm d\tau^2} = -R^\rho{}_{\sigma\mu\nu}u^\sigma\xi^\mu u^\nu$. With $u = c\,e_{\hat t}$ and $\xi = \xi^{\hat r}e_{\hat r}$, $\ddot\xi^{\hat r} = -c^2R^{\hat r}{}_{\hat t\hat r\hat t}\xi^{\hat r} = -c^2R_{\hat r\hat t\hat r\hat t} \xi^{\hat r}$ (raising a spatial index in an orthonormal frame changes nothing) $= -c^2(-r_s/r^3)\xi^{\hat r} = (2GM/r^3)\xi^{\hat r}$, using $c^2r_s = 2GM$. The transverse one is $-c^2(r_s/2r^3) = -GM/r^3$. Unit 05 found $-\partial_i\partial_j\Phi$ for $\Phi = -GM/r$ has eigenvalues $2GM/r^3$ radially and $-GM/r^3$ twice. The agreement is exact, not approximate — though "$r$" and "proper time" are now the relativistic quantities.

36 · singularities

Tides, drawn: a ring becomes an ellipse

Free particles on a small ring round a falling body, all at rest relative to it. With $\omega_{\rm tide}^2 := GM/r^3$ (a rate, not section 7's conformal factor), the deviation equations of the previous slide are

$$\ddot\xi^{\hat r} = 2\omega_{\rm tide}^2\,\xi^{\hat r},\qquad \ddot\xi^{\hat\theta} = -\omega_{\rm tide}^2\,\xi^{\hat\theta},$$

solved by $\xi^{\hat r}\propto\cosh(\sqrt2\,\omega_{\rm tide}\tau)$ and $\xi^{\hat\theta}\propto\cos\omega_{\rm tide}\tau$: the ring grows along the line to the hole and shrinks across it.

checks

Trace of the tidal matrix $2 - 1 - 1 = 0$: that is $R_{\hat t\hat t} = 0$, vacuum ✓ — a small ball is deformed without, at first, changing its volume. Dimensions $1/\text{length}^2$ ✓. $r_s\to0$: flat ✓.

The ring in the $(\hat r, \hat\theta)$ plane, at $\omega_{\rm tide}\tau = 0.6$ (computed from the two solutions, short enough that $r$ barely changes). Arrows: the tidal acceleration at each particle's starting point, twice as strong outwards along $\hat r$ as inwards across it.

Reading the figure. The arrows are the vector field $\xi\mapsto(2\omega_{\rm tide}^2\xi^{\hat r}, -\omega_{\rm tide}^2\xi^{\hat\theta})$ evaluated on the starting circle — the quadrupole pattern every tide has, unit 05's Earth–Moon tide in particular. Integrating for a time $\tau$ with zero initial relative velocity gives the ellipse with semi-axes $\cosh(\sqrt2\cdot0.6) = 1.38$ and $\cos 0.6 = 0.83$ times the starting radius. The third direction, $\hat\varphi$, is squeezed like $\hat\theta$, so in three dimensions a ball becomes a cigar pointing at the hole. Near $r = 0$ the rate $\omega_{\rm tide}$ grows without bound — the physical face of $K\to\infty$ in the next slide.

Why the frame is the static one, and a caution. The static observers ($r$ fixed) are not in free fall; the deviation equation is for a geodesic momentarily at rest with respect to them, whose four-velocity at that instant is $c\,e_{\hat t}$. The frame exists only for $r \gt r_s$. For a body crossing the horizon one needs a frame boosted radially, and problem P8 shows that these particular components do not change under that boost.

36 · singularities

The Kretschmann scalar

the Kretschmann scalar $$K := R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta} = \frac{12r_s^2}{r^6} = \frac{48G^2M^2}{c^4r^6}.$$

proof

In an orthonormal frame, raising an index multiplies by $\pm1$ and each $\hat t$ appears an even number of times, so $K$ is a sum of squares. Each of the six components appears four times ($abab$, $baba$, $abba$, $baab$): $$K = 4\Big[2\Big(\frac{r_s}{r^3}\Big)^2 + 4\Big(\frac{r_s}{2r^3}\Big)^2\Big] = \frac{12r_s^2}{r^6}.\ \square$$

checks

$[K] = \text{length}^{-4}$ ✓. $M\to0$: $K = 0$ ✓. $r\to\infty$: $K\to0$, flat far away ✓.

the verdict

At $r_s$: $K = 12/r_s^4$, finite. No verdict yet — that needs a chart.

At $r = 0$: $K \to \infty$, and a falling body reaches $r = 0$ in finite proper time (section 6). No chart can repair it. Real.

where	$K$ ($\mathrm m^{-4}$)
Earth's surface	$1.40\times10^{-44}$
horizon, the hole	$1.58\times10^{-17}$
horizon, $4.3\times10^6M_\odot$	$4.61\times10^{-40}$

The sum, carefully. In the orthonormal frame, $R^{\alpha\beta\gamma\delta} = \eta^{\alpha\alpha'} \cdots R_{\alpha'\beta'\gamma'\delta'}$, and each $\hat t$ index raised contributes a factor $-1$. A nonzero component here contains $\hat t$ either zero or two times, so every term $R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}$ (no sum) is $+(R_{\alpha\beta\gamma\delta})^2$. The nonzero components are of the form $R_{abab}$ with $a \ne b$, and for each unordered pair $\{a,b\}$ the four index orderings $abab, baba, abba, baab$ carry the values $x, x, -x, -x$, all with square $x^2$. Six pairs: two with $|x| = r_s/r^3$ ($\hat t\hat r$, $\hat\theta\hat\varphi$) and four with $|x| = r_s/2r^3$. So $K = 4(2 + 4\cdot\tfrac14)r_s^2/r^6 = 12r_s^2/r^6$. With $r_s = 2GM/c^2$, $12\cdot4G^2M^2/ (c^4r^6) = 48G^2M^2/(c^4r^6)$.

Why not the Ricci scalar? $R = 0$ everywhere, since the metric is a vacuum solution, so the simplest invariant is useless here. The next simplest quadratic, $R_{\mu\nu}R^{\mu\nu}$, also vanishes. The Kretschmann scalar is the first one that sees vacuum curvature; in vacuum it equals the square of the Weyl tensor, $C_{\alpha\beta\gamma\delta}C^{\alpha\beta\gamma\delta}$. Kretschmann is the same Erich Kretschmann who, in 1917, made the objection to general covariance that unit 31 discussed.

The numbers. Earth: $r_s = 8.870\times10^{-3}$ m, $R_\oplus = 6.378137\times10^6$ m, $12r_s^2/R^6 = 1.40\times10^{-44}\ \mathrm{m^{-4}}$. The hole: $12/(29\,532.5)^4 = 1.58\times10^{-17}$. The Milky Way's central hole, $M = 4.3\times10^{6}M_\odot$ from the orbits of the stars around it: $r_s = 1.270\times10^{10}$ m, $12/r_s^4 = 4.61\times10^{-40}$. Because $K$ at the horizon scales as $M^{-4}$, the horizon of a large enough hole is less curved than the surface of the Earth: problem P9 finds the crossover at about $5.8\times10^{7}M_\odot$.

The logic, once more. The table says the horizon is not special in its curvature. It does not say the horizon is a regular part of spacetime; that is the next slide's job.

36 · singularities · worked example

Worked example 4 — ingoing Eddington–Finkelstein coordinates

remove the horizon singularity by an explicit change of chart

Step 1 — follow light. Radial null curves have $\mathrm ds^2 = 0$, $\mathrm d\Omega = 0$: $c\,\mathrm dt = \pm\mathrm dr/f$. Define the tortoise coordinate by $\mathrm dr_*/\mathrm dr = 1/f$: $$r_* = r + r_s\ln\Big|\frac r{r_s} - 1\Big| .$$ Check: $\frac{\mathrm dr_*}{\mathrm dr} = 1 + \frac{r_s}{r - r_s} = \frac{r}{r - r_s} = \frac1f$ ✓. Ingoing light: $ct + r_* = $ const.

Step 2 — label by incoming light. $v := ct + r_*$. Then $c\,\mathrm dt = \mathrm dv - \mathrm dr/f$.

Step 3 — substitute. $$-f(\mathrm dv - \tfrac{\mathrm dr}f)^2 + \tfrac{\mathrm dr^2}f = -f\,\mathrm dv^2 + 2\,\mathrm dv\,\mathrm dr .$$ So, in ingoing Eddington–Finkelstein coordinates $(v, r, \theta, \varphi)$: $$\mathrm ds^2 = -\Big(1 - \frac{r_s}r\Big)\mathrm dv^2 + 2\,\mathrm dv\,\mathrm dr + r^2\,\mathrm d\Omega^2 .$$

Step 4 — inspect. Every component is smooth for $r \gt 0$. And $\det g = -r^4\sin^2\theta \ne 0$, including at $r = r_s$. The chart runs straight through; the region $0 \lt r \lt r_s$ is glued on smoothly.

checks

$r_s\to0$: $v = ct + r$ and $-\mathrm dv^2 + 2\mathrm dv\,\mathrm dr = -c^2\mathrm dt^2 + \mathrm dr^2$ ✓. $v$ is a length ✓.

Step 3 in full. $-fc^2\mathrm dt^2 = -f(\mathrm dv - \mathrm dr/f)^2 = -f\,\mathrm dv^2 + 2\,\mathrm dv\,\mathrm dr - \mathrm dr^2/f$. Adding $+\mathrm dr^2/f$ cancels the last term. The angular part is untouched.

Step 4 in full. The $(v, r)$ block is $\begin{pmatrix}-f & 1\\ 1 & 0\end{pmatrix}$, with determinant $-1$ for every $r$, including where $f = 0$. Times $r^2\cdot r^2\sin^2\theta$ from the angles. A metric with smooth components and nonzero determinant is a smooth Lorentzian metric; there is nothing else to check.

What exactly has been shown. The Schwarzschild chart covers $r \gt r_s$. The map $(t, r) \mapsto (v, r)$ is a diffeomorphism from that region onto $\{r \gt r_s\}$ in the new chart, and it is an isometry by construction. But the new metric also makes sense for $0 \lt r \le r_s$. So the exterior is isometric to part of a larger spacetime, in which $r = r_s$ is an ordinary null hypersurface. That is what "the singularity at $r_s$ is a coordinate artefact" means, stated as a theorem. The inside part, $0 \lt r \lt r_s$, is isometric to Schwarzschild's chart for $r \lt r_s$, by the same map with $r_* = r + r_s\ln(1 - r/r_s)$ there.

Where it came from. Eddington wrote this form in 1924 without drawing the conclusion; Finkelstein rediscovered it in 1958 and did. Lemaître had shown in 1933 that $r_s$ was not a real singularity, using freely falling coordinates. For a quarter of a century the result was available and not believed.

Why "tortoise". $r_*\to-\infty$ as $r\to r_s^+$: in terms of $r_*$ the horizon is infinitely far away, like the finish line in Zeno's race between Achilles and the tortoise.

36 · singularities

The horizon is a one-way null surface

theorem

(i) $r = r_s$ is a null hypersurface.
(ii) For $r \lt r_s$, $r$ strictly decreases along every future-directed causal curve.

proof

(i) Normal $\mathrm dr$: in EF, $g^{rr} = f = 0$ there, so the normal is null.

(ii) Time-orient by $-\partial_r$, which is null ($g_{rr} = 0$) and nowhere zero. Future causal $X$ then has $X^v \ge 0$. From $g(X,X)\le0$: $$2X^vX^r \le fX^{v\,2} \lt 0\quad(f \lt 0,\ X^v \gt 0),$$ so $X^r \lt 0$; if $X^v = 0$, $X \propto -\partial_r$ and again $X^r \lt 0$. $\square$

Light cones in the $(r, v - r)$ plane, where ingoing light runs at $45^\circ$. Outgoing light tips over: outward outside $r_s$, vertical on it, inward inside. Every future inside points toward $r = 0$.

(i) in full. A hypersurface $\{\phi = \text{const}\}$ is null when its normal covector $\mathrm d\phi$ is null, $g^{\mu\nu}\partial_\mu\phi\,\partial_\nu\phi = 0$. The inverse of the $(v,r)$ block $\begin{pmatrix}-f & 1\\1 & 0\end{pmatrix}$ is $\begin{pmatrix}0 & 1\\ 1 & f\end{pmatrix}$, so $g^{rr} = f$, which vanishes at $r_s$. A null hypersurface contains its own normal direction: the vector $g^{\mu\nu}\partial_\nu r = (1, f, 0, 0)$ is, at $r_s$, $\partial_v$, which is tangent to the surface. So the horizon is generated by light rays that neither fall in nor escape: outgoing light exactly at $r_s$ stays at $r_s$.

(ii), the time orientation. Outside, is $-\partial_r$ (at fixed $v$) future-directed in the usual sense? $\partial_t$ at fixed $r$ is $c\,\partial_v$, and $g(c\partial_v, -\partial_r) = -c\,g_{vr} = -c \lt 0$; two causal vectors with negative inner product lie in the same half of the cone. So yes: the orientation by $-\partial_r$ agrees with the one given by $\partial_t$ outside, and extends it through the horizon. For causal $X$ in the same half-cone, $g(X, -\partial_r) = -X^v \le 0$, i.e. $X^v \ge 0$.

(ii), the inequality. $g(X, X) = -fX^{v\,2} + 2X^vX^r + r^2|X_\Omega|^2 \le 0$, where $|X_\Omega|^2 \ge 0$ is the angular part. So $2X^vX^r \le fX^{v\,2} - r^2|X_\Omega|^2 \le fX^{v\,2}$, and inside, $f \lt 0$. If $X^v \gt 0$ then $X^r \le fX^v/2 \lt 0$. If $X^v = 0$ then $g(X,X) = r^2|X_\Omega|^2 \le 0$ forces $X_\Omega = 0$, so $X = X^r\partial_r$, and future-directedness makes it a positive multiple of $-\partial_r$. In both cases $X^r \lt 0$.

The figure. In EF coordinates, with $\tilde v := v - r$ as vertical axis, ingoing light ($v$ const) has $\mathrm d\tilde v/\mathrm dr = -1$, and outgoing light ($\mathrm dr/\mathrm dv = f/2$) has direction $(\mathrm dr, \mathrm d\tilde v) \propto (1 - r_s/r,\ 1 + r_s/r)$. That vector points right for $r \gt r_s$, straight up at $r_s$, and left for $r \lt r_s$ — the figure is drawn from exactly this formula.

So what "black" means. Nothing inside can reach $r \gt r_s$: a causal curve starting at $r \lt r_s$ has $r$ decreasing for ever. This is a theorem about the future light cones, not about forces; no rocket helps, because a rocket's worldline is a causal curve.

36 · singularities

Nothing happens at the horizon, locally

the event horizon

The boundary of the set of events from which some future causal curve escapes to arbitrarily large $r$. Here, $r = r_s$.

a global object

Whether an event lies inside depends on the entire future. No measurement made near one event can reveal a horizon there: by unit 32's local flatness theorem, a small enough freely falling laboratory at $r_s$ is Minkowski space to first order.

What a small laboratory can feel is tides. Radial stretch across a body of length $\ell = 2$ m at the horizon, $\Delta a = 2GM\ell/r_s^3$:

hole	$\Delta a$ ($\mathrm{m\,s^{-2}}$)	in $g$
$10\,M_\odot$	$2.06\times10^{8}$	$2.1\times10^{7}$
$4.3\times10^6M_\odot$	$1.11\times10^{-3}$	$1.1\times10^{-4}$

Since $r_s\propto M$, $\Delta a \propto M^{-2}$: small holes kill before the horizon, large ones do not notice it. Problem P8 finds the threshold.

The definition, precisely. Section 7 gives the clean version: the event horizon is the boundary of the causal past of future null infinity $\mathscr I^+$. The version on the slide says the same thing for this spacetime without needing the compactification. That it is $r = r_s$: every causal curve from $r \lt r_s$ has $r$ decreasing (previous slide), so none escapes; from $r \gt r_s$, the outgoing radial light ray has $\mathrm dr/\mathrm dv = f/2 \gt 0$ with $f$ increasing, so it escapes. The boundary between the two sets is $r = r_s$.

Teleology. Imagine a shell of matter, far outside, that will fall onto the hole a million years from now and double its mass. Then the true event horizon today is already outside $r_s$ of the present hole, growing in anticipation of matter that has not yet arrived. Nothing local marks it. This is why numerical relativity tracks apparent horizons (unit 40) instead: those are defined on a single slice.

The numbers. $\Delta a = 2GM\ell/r_s^3 = c^6\ell/(4G^2M^2)$. For $10M_\odot$: $2\times1.32712 \times10^{21}\times2/(29\,532.5)^3 = 2.06\times10^{8}\ \mathrm{m\,s^{-2}}$, which divided by $g = 9.80665$ is $2.1\times10^{7}$. For $4.3\times10^{6}M_\odot$: $r_s = 1.2699\times10^{10}$ m and $\Delta a = 1.11\times10^{-3}\ \mathrm{m\,s^{-2}}$ — a ten-thousandth of $g$, and only about 180 times the Earth's own tide across a person standing on it, $6.1\times10^{-6}\ \mathrm{m\,s^{-2}}$. Someone falling into the Milky Way's central hole would cross its horizon without noticing, and would then have at most about a minute (problem P13).

Why these numbers are legitimate at the horizon. The table uses the static frame's tidal component, and there is no static observer at $r_s$. Problem P8 proves that for a radially falling body the radial and transverse tidal components are the same as the static ones at the same $r$ — they are invariant under radial boosts — so the formula holds for the only observers who are actually there.

section 6

Two observers, two stories

A body falls in. Its own clock reads a finite time; a distant clock reads infinity. Both are right, about different things.

36 · observers

Static observers: their rulers and their clocks

static observer

A worldline of fixed $r, \theta, \varphi$ — tangent to $\partial_t$, so it exists only where $\partial_t$ is timelike: $r \gt r_s$.

Clock. $\mathrm d\tau = \sqrt f\,\mathrm dt$: unit 30's gravitational time dilation, now exact. At $r = 2r_s$, $\sqrt{1/2} = 0.7071$.

Ruler. The proper radial distance between $r_1$ and $r_2$ at fixed $t$: $$\ell = \int_{r_1}^{r_2}\frac{\mathrm dr}{\sqrt{1 - r_s/r}} \gt r_2 - r_1 .$$

holding still costs more and more

A static observer's proper acceleration is $\dfrac{GM}{r^2\sqrt{1 - r_s/r}}$, which diverges at $r_s$. At $2r_s$ from the hole: $5.4\times10^{11}\ \mathrm{m\,s^{-2}}$.

The clock and the acceleration both fail at $r_s$, where no static observer can exist. The ruler does not: the next slide measures it, and draws it.

The acceleration, in full. A static observer has $u = \dot t\,\partial_t$ with $\dot t = 1/\sqrt f$ (normalisation). Its acceleration $a^\mu = \dot u^\mu + \Gamma^\mu_{\alpha\beta}u^\alpha u^\beta$ has only an $r$-component, $a^r = \Gamma^r_{tt}\dot t^2 = \frac{c^2ff'}{2}\cdot\frac1f = \frac{c^2f'}{2} = \frac{GM}{r^2}$. Its magnitude is $\sqrt{g_{rr}}\,a^r = \frac{GM}{r^2\sqrt f}$. At large $r$ it is Newton's $g$; at $r_s$ it diverges. At $r = 2r_s$ for the hole: $1.32712\times10^{21}/((59\,065)^2\times0.70711) = 5.38\times10^{11}\ \mathrm{m\,s^{-2}}$.

Unit 30 revisited. A Rindler observer at proper distance $\ell$ from the Rindler horizon has proper acceleration $c^2/\ell$, also divergent at the horizon. Near $r_s$, write $r = r_s + \varepsilon$. The proper distance to the horizon is $\ell \approx 2\sqrt{r_s\varepsilon}$ (the integrand is $\approx\sqrt{r_s/\varepsilon}$), and the acceleration is $GM/(r^2\sqrt f) \approx \frac{c^2r_s/2}{r_s^2\sqrt{\varepsilon/r_s}} = \frac{c^2}{2\sqrt{r_s\varepsilon}} = \frac{c^2}{\ell}$. The same law, to leading order: near the horizon, static observers are Rindler observers. Section 7 shows why — Schwarzschild time translation is a boost there.

36 · observers

The ruler, drawn: Flamm's paraboloid

for the hole

With $x = r/r_s$, the antiderivative is $r_s\big[\sqrt{x(x-1)} + \ln(\sqrt x + \sqrt{x - 1})\big]$. From $r_s$ to $2r_s$: $$\ell = r_s\big[\sqrt2 + \ln(1 + \sqrt2)\big] = 2.296\,r_s = 67.79\ \text{km},$$ against $\Delta r = 29.53$ km. Finite, though $g_{rr}\to\infty$.

The picture. Take the slice $t = $ const, $\theta = \pi/2$: metric $\mathrm dr^2/f + r^2\mathrm d\varphi^2$. Find a surface of revolution $z(r)$ in flat space with the same metric, $\mathrm dz^2 + \mathrm dr^2 + r^2\mathrm d\varphi^2$. Matching, $z'^2 = 1/f - 1 = r_s/(r - r_s)$: $$z = 2\sqrt{r_s(r - r_s)} .$$

The equatorial slice outside the hole, drawn to scale as a surface in flat space: each ring is a circle of circumference $2\pi r$, and radial distance is length along the curve. From $r_s$ to $2r_s$ the curve (blue) is $2.296\,r_s$ long while $r$ grows by $r_s$ (red). The throat is vertical, which is $g_{rr}\to\infty$.

The integral. $\frac{\mathrm d}{\mathrm dx}\big[\sqrt{x(x-1)} + \ln(\sqrt x + \sqrt{x-1})\big] = \frac{2x - 1}{2\sqrt{x(x-1)}} + \frac{1}{2\sqrt{x(x-1)}} = \frac{x}{\sqrt{x(x-1)}} = \sqrt{\frac{x}{x - 1}} = \frac{1}{\sqrt{1 - 1/x}}$ ✓, where the derivative of the logarithm is $\frac{1/(2\sqrt x) + 1/(2\sqrt{x-1})}{\sqrt x + \sqrt{x - 1}} = \frac{1}{2\sqrt x\sqrt{x-1}}$. At $x = 1$ the bracket is $0$; at $x = 2$ it is $\sqrt2 + \ln(1 + \sqrt2) = 1.41421 + 0.88137 = 2.29559$. Times $29.5325$ km, $67.79$ km. The integrand blows up at $x = 1$ like $(x-1)^{-1/2}$, which is integrable: the horizon is a finite proper distance from any exterior point. Far away the excess grows only logarithmically: $\ell - \Delta r \approx \tfrac12r_s\ln(r_2/r_1)$ for $r_1, r_2 \gg r_s$.

The embedding, in full. On the surface $z = z(r)$ in $\mathbb R^3$ with cylindrical coordinates $(r, \varphi, z)$, the flat metric $\mathrm dr^2 + r^2\mathrm d\varphi^2 + \mathrm dz^2$ restricts to $(1 + z'^2)\,\mathrm dr^2 + r^2\mathrm d\varphi^2$. Setting $1 + z'^2 = 1/(1 - r_s/r) = r/(r - r_s)$ gives $z' = \sqrt{r_s/(r - r_s)}$, and $\int\mathrm dr\,\sqrt{r_s}\,(r - r_s)^{-1/2} = 2\sqrt{r_s(r - r_s)}$. Squaring, $z^2 = 4r_s(r - r_s)$: a parabola in the $(r, z)$ plane, so the surface is a paraboloid of revolution — Ludwig Flamm, 1916, the first picture anyone drew of the Schwarzschild geometry. Length along the profile is $\int\sqrt{1 + z'^2}\,\mathrm dr = \int\mathrm dr/\sqrt f$, which is $\ell$, so the figure is the integral on the slide made visible.

What the picture is not. The third dimension $z$ has no physical meaning; it is there only so that a curved 2-surface can be seen with Euclidean eyes. Nor does the picture show time: it is one slice. And it stops at $r_s$ only because this slice of the Schwarzschild chart does — section 7 continues it through the throat.

36 · observers

The infaller's clock: finite

From worked example 3, radial fall from rest at infinity: $\dot r = -c\sqrt{r_s/r}$. Separate and integrate:

$$\tau(r_1\to r_2) = \frac{2r_s}{3c}\Big[\Big(\frac{r_1}{r_s}\Big)^{3/2} - \Big(\frac{r_2}{r_s}\Big)^{3/2}\Big].$$

Horizon to centre:

proper time left after crossing $$\tau = \frac{2r_s}{3c} = \frac{4GM}{3c^3}.$$

The hole: $65.7\ \mu$s. A $4.3\times10^6M_\odot$ hole: $28.2$ s.

why the crossing itself is legitimate

The equation was derived in the Schwarzschild chart, which misses $r_s$. In EF, $\partial_t = c\,\partial_v$, so $E = -c\,g(\partial_v,\dot x)$ is still conserved and the same radial equation holds for all $r \gt 0$. And $$\frac{\mathrm dv}{\mathrm d\tau} = c\dot t + \frac{\dot r}{f} = \frac{c}{1 + \sqrt{r_s/r}},$$ finite and positive everywhere: $v$ passes smoothly through its horizon value.

checks

$[r_s/c] = $ s ✓. $M\to0$: zero time, nothing to fall into ✓.

The integral. $\mathrm d\tau = -\mathrm dr\,\sqrt{r}/(c\sqrt{r_s})$, and $\int r^{1/2}\mathrm dr = \tfrac23r^{3/2}$. So $\tau(r_1\to r_2) = \frac{2}{3c\sqrt{r_s}}(r_1^{3/2} - r_2^{3/2})$, which is the slide's form. From $r_s$ to $0$: $\frac{2}{3c\sqrt{r_s}}r_s^{3/2} = \frac{2r_s}{3c}$.

The radial equation in EF. In EF, $E = -g(c\partial_v, \dot x) = c(f\dot v - \dot r)$, and the normalisation is $\dot v(2\dot r - f\dot v) = -c^2$ for radial motion. Since $f\dot v = E/c + \dot r$, this is $\dot v(\dot r - E/c) = -c^2$; multiply by $f$ and use $f\dot v = E/c + \dot r$ again: $\dot r^2 - E^2/c^2 = -fc^2$. The same equation, derived without ever dividing by $f$, so valid at and inside $r_s$.

The $\mathrm dv/\mathrm d\tau$ computation. $v = ct + r_*$, so $\dot v = c\dot t + \dot r/f$. With $E = c^2$, $c\dot t = c/f$; with $\dot r = -c\sqrt{r_s/r}$, $\dot v = \frac cf\big(1 - \sqrt{r_s/r}\big) = \frac{c(1 - \sqrt{r_s/r})}{(1 - \sqrt{r_s/r})(1 + \sqrt{r_s/r})} = \frac{c}{1 + \sqrt{r_s/r}}$. The factor $1 - \sqrt{r_s/r}$ that vanishes at the horizon cancels exactly: the Schwarzschild $\dot t$ blows up there and $\dot r/f$ blows up with the opposite sign, and the two infinities are the chart's, not the body's.

The numbers. $2\times29\,532.5/(3\times299\,792\,458) = 6.567\times10^{-5}$ s. For $4.3\times10^{6} M_\odot$: $2\times1.2699\times10^{10}/(3c) = 28.24$ s. A body released from rest just outside the horizon crosses it with almost no inward speed and has more: in the limit, $\pi r_s/(2c) = 154.7\ \mu$s for the hole, and problem P13 proves that this is the maximum over all possible trajectories inside. Firing a rocket to hold off the fall, contrary to intuition, shortens the time left.

What the infaller feels. Nothing special on crossing. Tides grow like $r^{-3}$ and become unbounded as $r\to0$, which is the physical content of the curvature singularity: the body is pulled apart radially and crushed sideways within the last fraction of the $65.7\ \mu$s.

36 · observers

The distant clock: infinite, and why both are right

Along the same fall, $\dot t = E/(fc^2) = 1/f$, so $$\frac{\mathrm dt}{\mathrm dr} = -\frac{1}{c\,f\sqrt{r_s/r}} \approx -\frac{r_s}{c}\,\frac{1}{r - r_s}\quad(r\to r_s),$$ and $t \approx -\frac{r_s}{c}\ln\!\big(\frac r{r_s} - 1\big) \to +\infty$.

the reconciliation

$t$ is the reading of static observers' synchronised clocks. The crossing event lies on $r = r_s$, where the Schwarzschild chart does not reach: it has no $t$-value. "$t\to\infty$" says the chart runs out, not that the body stops.

In EF the crossing has a finite $v$. What the distant observer sees is decided by the outgoing light rays, and the last ray to escape leaves just before the crossing.

One fall from rest at infinity into the hole, timed from $r = 6r_s$ by two clocks. By its own clock (red) the body crosses $r_s$ at $900\ \mu$s and reaches $r = 0$ at $965\ \mu$s. Against $t$ (blue) it creeps towards $r_s$ for ever, losing $98.5\ \mu$s per factor $e$ in $r - r_s$.

The logarithm. Near $r_s$, $f = (r - r_s)/r \approx (r - r_s)/r_s$ and $\sqrt{r_s/r}\approx1$, so $\mathrm dt/\mathrm dr \approx -\frac{r_s}{c(r - r_s)}$, which integrates to $t = -\frac{r_s}c\ln(r - r_s) +$ const. Every factor of $e$ closer to the horizon costs another $r_s/c = 98.5\ \mu$s of $t$.

The two curves, exactly. With $x = r/r_s$ and times in units of $r_s/c$: the proper time is the previous slide's formula, $\tau = \tfrac23(6^{3/2} - x^{3/2})$, which reaches $x = 1$ at $\tfrac23(6^{3/2} - 1) = 9.13$ ($900\ \mu$s) and $x = 0$ at $\tfrac23 6^{3/2} = 9.80$ ($965\ \mu$s); the difference is the $65.7\ \mu$s of the boxed result. For $t$, $\mathrm dt/\mathrm dx = -x^{3/2}/(x - 1)$; with $s = \sqrt x$ the integrand becomes $2s^4/(s^2 - 1) = 2\big(s^2 + 1 + \tfrac1{s^2 - 1}\big)$, so $t = F(6) - F(x)$ with $F(x) = \tfrac23x^{3/2} + 2\sqrt x + \ln\frac{\sqrt x - 1}{\sqrt x + 1}$. The logarithm is the divergence. Far from the hole the two curves agree, as they must: $\dot t = 1/f \to 1$.

No hand-waving about "infinite time dilation". The two clocks are never compared at one event, so there is no paradox to resolve by appeal to relativity of simultaneity. What is being compared is an event (the crossing) with a coordinate function ($t$) that is undefined there. Asked properly — at what $t_{\text{obs}}$ does the distant observer receive the signal the body emits at proper time $\tau$? — the answer is a finite, increasing function of $\tau$ for every $\tau$ before the crossing, tending to infinity as $\tau$ approaches the crossing. Both statements on the slide follow; the next slide draws that function's mechanism.

36 · observers

What the distant observer receives

what actually arrives far away

Light the body emits outward reaches a distant observer ever later and ever redder: $$1 + z \propto e^{ct/2r_s}.$$ e-folding time $2r_s/c$: $197\ \mu$s for the hole. The body does not freeze on the horizon; it fades out within a millisecond.

the frozen-star picture

"The distant observer sees the body hover at the horizon for ever" is wrong twice: she sees nothing after a few e-foldings, and what she would see is a statement about light, not about the body.

The fall in EF coordinates, from $3.5r_s$, flashing every $0.5\,r_s/c$ of its own time (vertical scale compressed). Outgoing light (blue) from the early flashes reaches the observer (green dots) at growing intervals; from near $r_s$ it clings to the horizon before escaping; from inside it falls to $r = 0$.

Reading the figure. Axes as on the horizon slide: $r$ across, $v - r$ up, so ingoing light would be a straight line (here at less than $45^\circ$, since the vertical is compressed to about a third of the horizontal scale). The red curve is the infaller, from $\mathrm dv/\mathrm dr = -\big(\sqrt{r_s/r}\,(1 + \sqrt{r_s/r})\big)^{-1}$, the ratio of the previous slides' $\mathrm dv/\mathrm d\tau$ and $\mathrm dr/\mathrm d\tau$. Each blue ray is outgoing light, $u = v - 2r_*$ constant, i.e. $v - r = $ const $+ \,r + 2r_s\ln|r/r_s - 1|$. The flashes are equally spaced in the infaller's time, yet the arrivals at $r = 3.9r_s$ are spaced $1.10, 1.16, 1.25, 1.37, 1.57, 1.98\ r_s/c$ apart and growing (at fixed $r$, arrival time is $v$ up to a constant). The last flash before the crossing, emitted at $1.19r_s$, leaves the top of the frame and arrives $3.46\,r_s/c$ after the one before it; flashes emitted closer to $r_s$ arrive later without bound, and the flash emitted inside never arrives.

The redshift, in full. An outgoing radial ray has $u := ct - r_* =$ const and reaches a distant observer at $r_{\text{obs}}$ at time $ct_{\text{obs}} = u + r_{*\text{obs}}$. So the rate at which the observer receives the infaller's signals is governed by $\mathrm du/\mathrm d\tau$ along the infaller. Near the horizon the infaller has nearly constant $v = v_H$, and $u = v - 2r_*$, with $r_* \approx r_s + r_s\ln(r/r_s - 1)$. So $r - r_s \approx r_s\exp\big((v_H - 2r_s - u)/2r_s\big)$, i.e. $r - r_s \propto e^{-u/2r_s}$. Near the horizon $\dot r \approx -c$, so $\mathrm d\tau \approx -\mathrm dr/c$, and $\frac{\mathrm d\tau}{\mathrm du} = \frac{r - r_s}{2r_sc}\propto e^{-u/2r_s}$. A signal emitted at fixed proper frequency therefore arrives with period stretched by $\mathrm du/\mathrm d\tau \propto e^{u/2r_s} \propto e^{ct_{\text{obs}}/2r_s}$. That is $1 + z$. The quantity $c^2/(2r_s) = c^4/4GM$ is the surface gravity of the horizon, unit 40's $\kappa$; the e-folding time is $c/\kappa$.

The number. $2r_s/c = 2\times29\,532.5/c = 1.970\times10^{-4}$ s. A millisecond is five e-foldings, so the redshift factor grows by about $e^5 \approx 150$ in that time, and the received power, which also suffers from the photons arriving less often, by roughly the square of that.

section 7

The whole spacetime

Compactify Minkowski space first, to learn the technique. Then build Kruskal's chart step by step, find four regions, and draw them.

36 · extension

Conformal compactification, on Minkowski first

conformal compactification

An embedding of $(M, g)$ into a manifold with boundary $(\tilde M, \tilde g)$, with $\tilde g = \omega^2g$ on $M$, the factor $\omega \gt 0$ on $M$ and $\omega = 0$ on the boundary. Infinity becomes a boundary at finite distance. (Most texts call the factor $\Omega$; here that letter is taken by the sphere's $\mathrm d\Omega^2$.)

lemma · cones survive

$\tilde g$ and $g$ have the same null, timelike and spacelike vectors. Proof: $\tilde g(X,X) = \omega^2g(X,X)$ and $\omega^2 \gt 0$. $\square$

Radial Minkowski: $\mathrm ds^2 = -\mathrm du\,\mathrm dv + r^2\mathrm d\Omega^2$, $u = ct - r$, $v = ct + r$ (unit 26). Squash each null coordinate into a finite interval, with $\ell$ any length: $$p = \arctan\frac v\ell,\quad q = \arctan\frac u\ell,\quad T = p + q,\ R = p - q.$$ Then

$$\begin{aligned}\mathrm ds^2 = \frac{\ell^2}{4\cos^2\!p\,\cos^2\!q}\Big(&-\mathrm dT^2 + \mathrm dR^2\\ &+ \sin^2\!R\,\mathrm d\Omega^2\Big)\end{aligned}$$

on $0 \le R \lt \pi$, $|T| + R \lt \pi$. The bracket is smooth up to the boundary; $\omega = 2\cos p\cos q/\ell$.

Full derivation. $\mathrm dv = \ell\sec^2p\,\mathrm dp$, $\mathrm du = \ell\sec^2q\,\mathrm dq$, so $-\mathrm du\,\mathrm dv = -\frac{\ell^2\,\mathrm dp\,\mathrm dq}{\cos^2p\cos^2q}$, and $-4\,\mathrm dp\, \mathrm dq = -(\mathrm dT^2 - \mathrm dR^2)$ since $\mathrm dT^2 - \mathrm dR^2 = (\mathrm dp + \mathrm dq)^2 - (\mathrm dp - \mathrm dq)^2 = 4\,\mathrm dp\,\mathrm dq$. For the angular part, $r = (v - u)/2 = \tfrac\ell2(\tan p - \tan q) = \frac{\ell\sin(p - q)}{2\cos p\cos q}$, so $r^2 = \frac{\ell^2\sin^2R}{4\cos^2p \cos^2q}$. Together this is the slide. The ranges: $u, v\in\mathbb R$ with $v \ge u$ ($r \ge 0$) become $p, q\in(-\pi/2, \pi/2)$ with $p \ge q$, i.e. $R \ge 0$ and $|T| + R \lt \pi$.

What the bracket is. $-\mathrm dT^2 + \mathrm dR^2 + \sin^2R\,\mathrm d\Omega^2$ is the Einstein static universe, $\mathbb R\times S^3$ with the round unit 3-sphere. So all of Minkowski space is conformal to a bounded, lens-shaped piece of it. Penrose introduced the construction in 1963–64; it is the standard way to talk about "infinity" precisely.

Why a conformal factor is the right thing to throw away. It changes lengths and times but, by the lemma, not which curves are causal. Causal structure — who can signal whom — is exactly what one wants to see at a glance, and it survives. Lengths, which do not survive, are what the original metric is still there for.

$\ell$ is arbitrary: changing it changes the picture's interior by a diffeomorphism but not its boundary structure. The arctangent is chosen because it maps $\mathbb R$ onto a bounded interval monotonically; any such function would do.

36 · extension

The Penrose diagram of Minkowski space

A Penrose diagram draws the $(R, T)$ half-plane of a compactification, one point per round sphere, light at $45^\circ$.

timelike infinity $i^\pm$: where every timelike geodesic ends and starts;
null infinity $\mathscr I^\pm$: where light rays end and start;
spatial infinity $i^0$: where spacelike slices end.

The blue line: a light ray comes in from $\mathscr I^-$, passes through $r = 0$, and leaves for $\mathscr I^+$.

Thin: $r = $ const, from $i^-$ to $i^+$. Dashed: $t = $ const, from $r = 0$ to $i^0$.

All of Minkowski space, radial directions only. Every timelike geodesic runs from $i^-$ to $i^+$.

Locating the boundary. $i^+$ is $p = q = \pi/2$, i.e. $T = \pi$, $R = 0$: reached as $u, v\to+\infty$ together, which is what a timelike geodesic does as $t\to\infty$. $i^0$ is $p = \pi/2$, $q = -\pi/2$: $T = 0$, $R = \pi$, reached as $r\to\infty$ at fixed $t$. $\mathscr I^+$ is $p = \pi/2$ with $q$ finite: $v\to\infty$ at fixed $u$, which is where an outgoing light ray goes. It is the segment $T + R = \pi$. Similarly $\mathscr I^-$ is $T - R = -\pi$.

The light ray. Ingoing light has $v$ constant, so $p$ constant, which is a line of slope $-1$ in the $(R, T)$ plane; it reaches $R = 0$ ($r = 0$), where it becomes outgoing, $u$ constant, slope $+1$. The drawn ray has $p = -0.37$ coming in and $q = -0.37$ going out: at the centre $u = v$.

Why $\mathscr I$ is pronounced "scri". Script I; the pronunciation is Penrose's. The notion matters because gravitational radiation is defined there: the energy a system radiates is the energy that crosses $\mathscr I^+$.

How to read any Penrose diagram. Light moves at $45^\circ$, so the causal future of a point is the wedge above it bounded by $45^\circ$ lines. Each interior point is a whole round sphere; the left edge, where the spheres have shrunk to points, is the centre $r = 0$.

36 · extension · worked example

Worked example 5 — Kruskal–Szekeres, step by step

regularise both horizons at once

Step 1 — tortoise (worked example 4): $r_* = r + r_s\ln(r/r_s - 1)$ on $r \gt r_s$.

Step 2 — both null coordinates. $u = ct - r_*$, $v = ct + r_*$. Since $c\,\mathrm dt = \tfrac12(\mathrm du + \mathrm dv)$ and $\mathrm dr_* = \tfrac12(\mathrm dv - \mathrm du) = \mathrm dr/f$: $$\mathrm ds^2 = -f\,\mathrm du\,\mathrm dv + r^2\mathrm d\Omega^2 .$$ Still singular: $f\to0$ as $u\to\infty$ or $v\to-\infty$.

Step 3 — exponentiate. $U = -e^{-u/2r_s}$, $V = e^{v/2r_s}$. Then $\mathrm dU\,\mathrm dV = -\frac{UV}{4r_s^2}\,\mathrm du\,\mathrm dv$ and $$UV = -e^{r_*/r_s} = \Big(1 - \frac r{r_s}\Big)e^{r/r_s}.$$

Step 4 — substitute. $f\,\mathrm du\,\mathrm dv = \frac{r - r_s}{r}\cdot\frac{4r_s^2\,\mathrm dU\,\mathrm dV}{(r/r_s - 1)e^{r/r_s}}$, so

$$\mathrm ds^2 = -\frac{4r_s^3}{r}e^{-r/r_s}\,\mathrm dU\,\mathrm dV + r^2\,\mathrm d\Omega^2 .$$

The factor $r - r_s$ has cancelled. These are Kruskal–Szekeres coordinates.

Step 5 — extend. $x\mapsto(1 - x)e^x$ has derivative $-xe^x \lt 0$ on $x \gt 0$: a bijection $(0,\infty)\to(-\infty, 1)$. So $r$ is a smooth positive function of $UV$ on all of $\{UV \lt 1\}$, and the metric is smooth and non-degenerate there. That is the maximal extension.

checks

$U, V$ dimensionless, $[r_s^3/r] = $ length$^2$ ✓. $U = 0$ or $V = 0$ $\Leftrightarrow$ $r = r_s$; $UV\to1$ $\Leftrightarrow$ $r\to0$ ✓.

Step 3 in full. $\mathrm dU = \frac1{2r_s}e^{-u/2r_s}\mathrm du = -\frac{U}{2r_s}\mathrm du$ and $\mathrm dV = \frac{V}{2r_s}\mathrm dv$, so $\mathrm dU\,\mathrm dV = -\frac{UV}{4r_s^2}\mathrm du\,\mathrm dv$. And $UV = -e^{(v - u)/2r_s} = -e^{r_*/r_s} = -e^{r/r_s}(r/r_s - 1)$, using $e^{\ln(r/r_s - 1)} = r/r_s - 1$.

Step 4 in full. From step 3, $\mathrm du\,\mathrm dv = -\frac{4r_s^2}{UV}\mathrm dU\,\mathrm dV = \frac{4r_s^2}{(r/r_s - 1)e^{r/r_s}}\mathrm dU\,\mathrm dV$. Multiply by $f = (r - r_s)/r = \frac{r_s}{r}(r/r_s - 1)$: the factor $(r/r_s - 1)$ cancels and leaves $\frac{4r_s^3}{r}e^{-r/r_s}\,\mathrm dU\,\mathrm dV$.

Step 5 in full. $h(x) = (1 - x)e^x$ is smooth with $h'(x) = -xe^x$, negative for $x \gt 0$, and $h(0^+) = 1$, $h(x)\to-\infty$ as $x\to\infty$. So $h$ is a smooth bijection of $(0,\infty)$ onto $(-\infty,1)$ with nonvanishing derivative, and by the inverse function theorem (unit 04) its inverse is smooth. Then $r = r_sh^{-1}(UV)$ is smooth on $\{(U,V): UV \lt 1\}$, positive, and the coefficient $\frac{4r_s^3}{r}e^{-r/r_s}$ is smooth and positive. The $(U,V)$ block has zero diagonal and off-diagonal entries $-\frac{2r_s^3}{r}e^{-r/r_s} \ne 0$, so nonzero determinant. $\square$ The original exterior is the quadrant $U \lt 0 \lt V$; the rest is new.

Why "maximal". Every geodesic in the Kruskal spacetime either runs to infinite affine parameter or ends at $UV = 1$, where $K\to\infty$. So no further extension is possible. Cited: the geodesic completeness analysis is in Hawking and Ellis, §5.5.

The time-and-space version. $T = (V + U)/2$, $X = (V - U)/2$ give $-\mathrm dU\,\mathrm dV = -\mathrm dT^2 + \mathrm dX^2$: light cones at $45^\circ$ in the $(X, T)$ plane, which is what the next slide draws. Kruskal and Szekeres found the chart independently in 1960.

36 · extension

The Kruskal diagram

$T = \tfrac12(V + U)$, $X = \tfrac12(V - U)$: light at $45^\circ$.

Horizons: the diagonals $U = 0$, $V = 0$.
$r = $ const: hyperbolae $X^2 - T^2 = (r/r_s - 1)e^{r/r_s}$.
$t = $ const: straight lines, $T/X = \tanh(ct/2r_s)$.
Singularity: $T^2 - X^2 = 1$, top and bottom.

time translation is a boost

$t\mapsto t + s$ shifts $u, v$ by $cs$: $U\mapsto e^{-cs/2r_s}U$, $V\mapsto e^{cs/2r_s}V$. So $$\partial_t = \frac{c}{2r_s}\big(V\partial_V - U\partial_U\big),$$ a Lorentz boost in the $(X,T)$ plane — unit 30's Rindler picture, near the crossing point.

Shaded: region II. Dashed: $r = 1.2r_s, 1.5r_s$ outside and $0.5r_s$ inside. Thin lines through the centre: $t = 0, \pm1, \pm2$ in units of $r_s/c$. Thick curve: a body falling from region I through the horizon to the singularity.

The curves, derived. $UV = T^2 - X^2$, so $r = $ const is $T^2 - X^2 = (1 - r/r_s)e^{r/r_s}$: a hyperbola, opening sideways for $r \gt r_s$ and up/down for $r \lt r_s$. $r = 0$ is $T^2 - X^2 = 1$. In region I, $-V/U = e^{(u + v)/2r_s} = e^{ct/r_s}$; then $\frac TX = \frac{V + U}{V - U} = \frac{e^{ct/r_s} - 1}{e^{ct/r_s} + 1} = \tanh\frac{ct}{2r_s}$. As $t\to\pm\infty$, $T/X\to\pm1$: every $t = $ const line of region I passes through the origin, and $t = +\infty$ is the future horizon. This is the reconciliation slide as a picture — the infaller crosses the line $t = +\infty$ at a perfectly ordinary point.

The boost. $\frac{\mathrm d}{\mathrm ds}$ of $(e^{-cs/2r_s}U, e^{cs/2r_s}V)$ at $s = 0$ is $\frac{c}{2r_s}(-U, V)$. In $(X,T)$: $V\partial_V - U\partial_U = X\partial_T + T\partial_X$, the generator of boosts of the flat $(X,T)$ plane. Unit 30 showed Rindler time translation on flat spacetime is exactly such a boost with horizon at $X = \pm T$. The Schwarzschild horizon is locally a Rindler horizon, and the constant in front, $c/2r_s$, is the surface gravity divided by $c$ — the same constant that set the redshift e-folding time.

The body. Drawn schematically — its shape is not computed, but its causal character is right: it is everywhere steeper than $45^\circ$, it crosses $U = 0$ at a finite point, and it ends on the singularity, which it cannot avoid, because every future-directed timelike direction in region II points toward it.

36 · extension

Four regions, and which of them are real

I ($U \lt 0 \lt V$): the exterior, our universe, $r \gt r_s$.
II ($U, V \gt 0$): the black hole region. Every future causal curve reaches $r = 0$, within proper time $\pi r_s/2c$ at most.
III ($V \lt 0 \lt U$): a second exterior, isometric to I, asymptotically flat.
IV ($U, V \lt 0$): the white hole region, II run backwards: everything in it must leave.

I and III never meet

No causal curve joins them: from I, $V \gt 0$ and $V$ never decreases along a future curve; III has $V \lt 0$. The slice $T = 0$ joins them through a throat of radius $r_s$ — $29.53$ km for the hole — the Einstein–Rosen bridge, which no observer can cross.

The slice $T = 0$, drawn like Flamm's paraboloid: region I's exterior above, its mirror image region III below, joined at the throat $r = r_s$ (red). Profile to scale, rings in perspective. Later slices $T \gt 0$ pinch the throat off before anything can pass.

"$V$ never decreases". In Kruskal coordinates future-directed causal vectors have $\dot U \ge 0$ and $\dot V \ge 0$: the metric is $-F\,\mathrm dU\,\mathrm dV$ in that block with $F \gt 0$, so a causal vector has $\dot U\dot V \ge 0$, and the future orientation — agreeing with $\partial_t$ in region I, where $\partial_t \propto V\partial_V - U\partial_U$ has both components positive — selects $\dot U, \dot V \ge 0$. So a causal curve starting in I, with $V \gt 0$, keeps $V \gt 0$ and can never reach III, where $V \lt 0$. Symmetrically $U$ never decreases, so nothing in I or II reaches IV, and nothing in II reaches I (II has $U \gt 0$, I has $U \lt 0$).

The throat. On $T = 0$, $U = -V$ and $UV = -X^2 \le 0$, so $r \ge r_s$ with equality at $X = 0$. The slice is two copies of the exterior $\{r \gt r_s\}$ glued along the sphere $r = r_s$: a wormhole. Einstein and Rosen described it in 1935. It pinches off: the slices $T = $ const $\gt 0$ have throat radius shrinking to $0$ at $T = 1$, faster than light could cross. Fuller and Wheeler made the non-traversability explicit in 1962; the argument on the slide is the short form of it.

Region II's time limit. Inside, $r$ is a time coordinate, and the maximum proper time from horizon to singularity along any timelike curve is $\pi r_s/2c$ — for the hole, $154.7\ \mu$s. That is problem P13, with a proof that fits in four lines once one notices $c\,\mathrm d\tau \le |\mathrm dr|/\sqrt{r_s/r - 1}$.

The figure. On $T = 0$ the coordinate $X$ runs from $-\infty$ (far out in III) through $0$ (the throat) to $+\infty$ (far out in I), and $r$ goes down from infinity to $r_s$ and back up. The induced metric on each half is the exterior's $\mathrm dr^2/f + r^2\mathrm d\Omega^2$, so each half is the Flamm paraboloid of section 6, $z = \pm2\sqrt{r_s(r - r_s)}$, and the two glue smoothly at the vertical throat. The vertical scale is compressed a little to fit; the shape, not the numbers, is the point.

36 · extension

A hole formed by collapse: only I and II

an astrophysical hole has only I and II

A hole formed by collapse is vacuum only outside the star. Inside the star's surface, Birkhoff does not apply and the metric is something else. The surface, a timelike curve from region I into region II, cuts the diagram: everything to its left — III, IV and part of I and II — is replaced by the star's interior.

what survives the collapse

The exterior, the horizon, the black-hole region and its singularity. That is the part the checkpoint is about, and the part unit 37 observes.

A collapsing star, as a Penrose diagram (schematic; the conventions are the next slide's). The shaded star ends on the singularity. The event horizon starts at the centre before the surface reaches it, and is the 45° line to $i^+$.

Reading the figure. The left edge is $r = 0$, but for most of its length it is the centre of the star, an ordinary point of a smooth spacetime, like the left edge of the Minkowski diagram. The star's surface (the thick curve) is a timelike worldline of spheres, steeper than $45^\circ$ everywhere. Outside it, by Birkhoff, the geometry is a piece of Kruskal — part of region I and part of region II, nothing more. The event horizon (blue, dashed) is the boundary of what can still signal $\mathscr I^+$; it is a light ray moving outwards from the centre, which reaches the surface just as the surface crosses $r_s$ and from then on sits at $r = r_s$. Its starting point at the centre, before the star has collapsed, is the teleology of the horizon slide's notes made visible.

Where the interior comes from. Cited: Oppenheimer and Snyder (1939) solved the collapse of a homogeneous ball of pressureless dust exactly, gluing a piece of a closed Friedmann–Lemaître model (unit 39) inside to Schwarzschild outside; the figure is the shape of their solution. Nothing in it depends on the details of the star.

Why the white hole is not believed in. A white hole's horizon is the past boundary of region IV; it would have to exist from the beginning, and it is unstable to small amounts of infalling matter. Nothing in the collapse picture produces one.

36 · extension

The Penrose diagram of Schwarzschild

Compactify Kruskal as Minkowski was: $\tilde U = \arctan U$, $\tilde V = \arctan V$. The singularity $UV = 1$ becomes $$\tan\tilde U\tan\tilde V = 1 \iff \tilde U + \tilde V = \pm\tfrac\pi2,$$ a horizontal line: the singularity is a moment of time, not a place.

event horizon, precisely

The boundary of the causal past of region I's $\mathscr I^+$: the line $U = 0$.

The maximally extended Schwarzschild spacetime. Each exterior has its own $i^0$ and $\mathscr I^\pm$; the singularities are spacelike lines.

The singularity as a line. $\tan\tilde U\tan\tilde V = 1$ means $\cos\tilde U\cos\tilde V - \sin\tilde U \sin\tilde V = 0$, i.e. $\cos(\tilde U + \tilde V) = 0$. With $\tilde U, \tilde V \in(-\pi/2,\pi/2)$ and both of the same sign, $\tilde U + \tilde V = \pm\pi/2$. In the diagram's coordinates $\tilde T = \tilde U + \tilde V$, $\tilde X = \tilde V - \tilde U$, those are the lines $\tilde T = \pm\pi/2$.

Why "a moment of time". In region II every future-directed causal curve ends on the upper line, and the line is spacelike. Avoiding it is like avoiding next Tuesday. That is the same content as the horizon theorem ($r$ decreases along every future causal curve inside), now read off a picture.

The conformal factor. $-\mathrm dU\,\mathrm dV = -\sec^2\tilde U\sec^2\tilde V\,\mathrm d\tilde U\, \mathrm d\tilde V$, so the Kruskal metric is $\frac{4r_s^3}{r}e^{-r/r_s}\sec^2\tilde U\sec^2\tilde V\, (-\mathrm d\tilde U\,\mathrm d\tilde V) + r^2\mathrm d\Omega^2$. The prefactor diverges at the boundary $|\tilde U|$ or $|\tilde V| = \pi/2$ — which is infinity — and multiplying by $\omega^2 = \cos^2\tilde U \cos^2\tilde V$ tames it. Region I near its boundary looks exactly like Minkowski's diagram near $\mathscr I^\pm$ and $i^0$: asymptotic flatness, drawn.

The event horizon from the definition. The causal past of $\mathscr I^+$ of region I is the set of points from which a future causal curve reaches it; by the $U$-monotonicity of the previous slide's notes, these are exactly the points with $U \lt 0$. Its boundary is $U = 0$: the diagonal from the centre of the diagram to $i^+$, which bounds region I, and its continuation between regions II and III.

36 · summary

The running example, closed

	$r = r_s$ ($29.53$ km for the hole)	$r = 0$
in Schwarzschild's chart	$g_{tt}\to0$, $g_{rr}\to\infty$	$g_{tt}\to\infty$, $g_{rr}\to0$
Kretschmann $K = 12r_s^2/r^6$	$12/r_s^4 = 1.58\times10^{-17}\ \mathrm m^{-4}$: finite	$\to\infty$
an explicit chart	EF, Kruskal: smooth, $\det g\ne0$	none can exist
reached by an infaller in	finite $\tau$; $t = \infty$	$65.7\ \mu$s after the horizon
what it is	a null hypersurface, one-way: the event horizon	a spacelike end of time
verdict	coordinate singularity	curvature singularity

what the unit derived, and from what

Round + static + vacuum gave the metric, Newton gave its constant, Birkhoff removed "static", Noether gave $E$ and $L$, an invariant convicted $r = 0$, a chart acquitted $r_s$, and Kruskal's chart showed everything else the equations allow.

Where this goes. Unit 37 takes worked example 3's radial equation, keeps the $L$ term, and reads off Mercury's precession, the bending of light and the Shapiro delay. Unit 40 replaces "static" by "stationary" and finds Kerr, where $\partial_t$ becomes spacelike outside the horizon and the singularity is a ring; it also states the singularity theorems that say the curvature singularity found here is not an artefact of perfect symmetry.

What this unit did not do. It did not model collapse: the Oppenheimer–Snyder solution (1939) glues a collapsing ball of dust to the Schwarzschild exterior and is the honest version of the warning box on the collapse slide. It did not treat the interior of a static star (the Tolman–Oppenheimer–Volkoff equation), where $T_{\mu\nu}\ne0$ and the same ansatz gives a different, sourced solution. Both are reachable with exactly the tools used here.

36 · checkpoint

Checkpoint

you can now do this, or you cannot

You derive the Schwarzschild metric from the ansatz, with the Christoffel symbols and Ricci components produced and the constant fixed by the Newtonian limit; you obtain the geodesic constants of motion from its Killing vectors and name the unit-13 theorem they are an instance of; and you demonstrate, by an explicit coordinate transformation plus an invariant scalar, that exactly one of its two apparent singularities is real.

Failing it is unambiguous. If $R_{\theta\theta}$ does not come out as $1 - (rA)'$ once $AB = 1$, redo worked example 2. If you cannot say which theorem makes $fc^2\dot t$ constant, you have $E$ as a formula, not as a consequence. If your argument about $r_s$ uses $K$ rather than a chart, it proves nothing.

Thirteen problems, about seven hours: ./mr problems 36. P3 finishes the Ricci tensor; P4 proves Birkhoff; P6 is how long it takes to fall; P8 is tidal death at the horizon; P12 checks Kruskal. When the checkpoint is true: ./mr done 36.

The three clauses are separable and should be tested separately. Clause one is worked examples 1 and 2 with problems P2 and P3: reproduce the four Ricci components from the nine symbols, without looking, and solve. Clause two is worked example 3 with P5: produce $E$ and $L$, prove they are conserved by the Killing argument, and say "Noether's theorem, unit 13, applied to the geodesic Lagrangian". Clause three is worked example 4 and the Kretschmann slide with P9, P10 and P12.

The most common way to think clause three is passed when it is not: saying "$K$ is finite at $r_s$, so the horizon is not singular". That is the wrong tool for that job. The acquittal of $r_s$ is the explicit EF metric with its nonzero determinant; the conviction of $r = 0$ is the divergence of $K$ along a curve that reaches it in finite proper time. Each half needs its own tool, and the checkpoint asks for both.