The geometry of spacetime: special and general relativity

MotivationH

Light that cannot be outrun

James Clerk Maxwell's equations (1865) predict electromagnetic waves travelling at a speed $c$ fixed by two constants of electricity and magnetism — and that speed is the speed of light. But relative to what? A light wave seemed to need a medium, the "ether", and the Earth moves through it; yet Albert Michelson and Edward Morley (1887) found no trace of that motion. In 1905 Albert Einstein took two principles at face value: the laws of physics are the same for all inertial observers, and the speed of light is the same for all of them. The price is that time and space are not separate. In 1908 Hermann Minkowski recast the theory as geometry: "Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality."

Gravity did not fit. Newton's force acts instantaneously, and Newton's theory left a small anomaly: Mercury's perihelion advances by about $43''$ per century more than the pulls of the other planets explain (Le Verrier, 1859; Newcomb, 1882). Einstein's general relativity (1915) makes gravity the curvature of spacetime and explains the $43''$ exactly. Its predictions are now engineering: the clocks on GPS satellites run slow by about $7$ microseconds a day because of their speed and fast by about $45$ because they sit higher in the Earth's gravity; uncorrected, the net $38\,\mu\mathrm s$ a day would put positions off by kilometres within a day. Gravitational waves (detected in 2015) and the image of a black hole's shadow (2019) are geometric predictions too.

IntuitionH

Pythagoras with a minus sign

Measure time in years and distance in light-years, so that $c = 1$ . An event is a point $(t, x, y, z)$ of spacetime, and a particle's history is a curve, its worldline. Light from an event travels along a light cone at $45^\circ$ in a spacetime diagram; massive particles stay inside.

The basic quantity that all inertial observers agree on is not the time between two events, nor their distance, but the interval

\Delta s^2 = -\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2,

Pythagoras with a minus sign. When $\Delta s^2 \lt 0$ , the events can be joined by a particle, and $\Delta\tau = \sqrt{-\Delta s^2}$ is the time shown by a clock that travels between them uniformly — its proper time. A clock moving at speed $v$ over coordinate time $\Delta t$ covers $\Delta x = v\Delta t$ , so it records $\Delta\tau = \Delta t\sqrt{1 - v^2} = \Delta t/\gamma$ : moving clocks run slow.

Changes of inertial frame are Lorentz transformations, the linear maps preserving the interval. In one space dimension they are "hyperbolic rotations": they slide points along the hyperbolas $t^2 - x^2 = \text{const}$ just as rotations slide points along circles. The minus sign reverses the triangle inequality: a bent worldline between two events is shorter in proper time than the straight one. That is the twin paradox, and it is a theorem of geometry.

Gravity enters through Einstein's "happiest thought" (1907): a person falling freely does not feel their own weight. In a small freely falling laboratory the laws of special relativity hold; gravity is not a force acting in spacetime but the curvature of spacetime, and freely falling bodies follow its straightest lines, the geodesics. What cannot be transformed away is the relative acceleration of nearby falling bodies — tidal forces — and that is curvature. (The sphere is a good analogy for curvature, as long as one remembers that spacetime's geometry is Lorentzian, not Riemannian: on a sphere, initially parallel geodesics converge, as do two apples falling towards the Earth's centre.)

DefinitionsH

Definitions

Definition— Minkowski spacetime and causal character

Minkowski spacetime is $\R^4$ with the bilinear form $\eta(v, w) = -v^0 w^0 + v^1 w^1 + v^2 w^2 + v^3 w^3$ , i.e. $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$ (we take $c = 1$ ). A vector $v \ne 0$ is timelike if $\eta(v, v) \lt 0$ , null (lightlike) if $\eta(v, v) = 0$ , spacelike if $\eta(v, v) > 0$ ; it is causal if timelike or null, and future-pointing if moreover $v^0 > 0$ .

Definition— proper time

A timelike curve $\gamma : [a, b] \to \R^4$ has timelike tangent vectors everywhere. Its proper time is

\tau[\gamma] = \int_a^b \sqrt{-\eta(\gamma'(\lambda), \gamma'(\lambda))} \, d\lambda,

independent of the parametrisation. For a future-pointing causal vector $v = (t, \mathbf x)$ we write $\tau(v) = \sqrt{t^2 - |\mathbf x|^2}$ . Parametrised by proper time, a worldline has 4-velocity $u = d\gamma/d\tau$ with $\eta(u, u) = -1$ .

Definition— Lorentz and Poincaré groups

The Lorentz group is $O(1, 3) = \{\Lambda \in GL_4(\R) : \Lambda^{\mathsf T} \eta \Lambda = \eta\}$ , the linear maps preserving the interval; with translations it generates the Poincaré group of all isometries of Minkowski spacetime. The boost with rapidity $\varphi$ in the $x$ -direction acts on $(t, x)$ by

\begin{pmatrix} t' \\ x' \end{pmatrix} = \begin{pmatrix} \cosh\varphi & \sinh\varphi \\ \sinh\varphi & \cosh\varphi \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix},

with velocity $v = \tanh\varphi$ and Lorentz factor $\gamma = \cosh\varphi = 1/\sqrt{1 - v^2}$ .

Definition— Lorentzian manifold

A Lorentzian manifold $(M, g)$ is a smooth four-manifold with a smooth symmetric tensor $g$ of signature $(-, +, +, +)$ at every point. Tangent vectors are timelike, null or spacelike as above; a time orientation is a continuous choice of future light cone. $g$ has a unique torsion-free compatible (Levi-Civita) connection $\nabla$ , with Christoffel symbols $\Gamma^\mu_{\alpha\beta} = \frac12 g^{\mu\nu}(\partial_\alpha g_{\nu\beta} + \partial_\beta g_{\nu\alpha} - \partial_\nu g_{\alpha\beta})$ , and a geodesic is a curve with $\nabla_{\gamma'}\gamma' = 0$ :

\frac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda} = 0.

The Riemann tensor $R^\mu{}_{\nu\alpha\beta}$ measures the failure of covariant derivatives to commute; its contractions are the Ricci tensor $R_{\nu\beta} = R^\mu{}_{\nu\mu\beta}$ and the scalar curvature $R = g^{\nu\beta} R_{\nu\beta}$ , and the Einstein tensor is $G_{\mu\nu} = R_{\mu\nu} - \frac12 R \, g_{\mu\nu}$ .

Definition— Einstein's field equations

Matter is described by a symmetric stress–energy tensor $T_{\mu\nu}$ (for dust of density $\rho$ moving with 4-velocity $u$ , $T = \rho \, u \otimes u$ ). Einstein's field equations are

G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu},

with the cosmological constant $\Lambda$ . In vacuum with $\Lambda = 0$ they reduce to $R_{\mu\nu} = 0$ . Freely falling test particles follow timelike geodesics and light follows null geodesics.

Definition— Schwarzschild metric

In units $G = c = 1$ , with $M$ the mass (physically $GM/c^2$ , a length), the Schwarzschild metric on $r > 2M$ is

ds^2 = -\Big(1 - \frac{2M}{r}\Big) dt^2 + \frac{dr^2}{1 - 2M/r} + r^2 \big(d\theta^2 + \sin^2\theta \, d\phi^2\big).

The radius $r_s = 2M$ is the Schwarzschild radius: about $3$ km for the Sun and $9$ mm for the Earth.

Worked examplesH

Examples

Example— Muons reach the ground

Muons are created by cosmic rays about $15$ km up and live, at rest, about $2.2\,\mu$ s; even at nearly the speed of light they would cover only about $660$ m in that time. They reach sea level in large numbers (Rossi and Hall, 1941) because at $v \approx 0.995$ their Lorentz factor is about $10$ : in the Earth's frame their clocks run ten times slow. In the muon's frame the same fact appears as the contraction of the atmosphere's thickness by the factor $\gamma$ .

Example— The twins, in numbers

One twin stays on Earth; the other flies to a star $3$ light-years away at $v = \frac35$ and returns at the same speed. In Earth's frame the trip takes $5 + 5 = 10$ years. With $\gamma = 1/\sqrt{1 - 9/25} = \frac54$ , the traveller's clock records $10/\gamma = 8$ years. Directly: each leg is the vector $(5, 3)$ , of proper time $\sqrt{25 - 9} = 4$ , while the Earth twin's straight worldline $(10, 0)$ has proper time $10 > 4 + 4$ . The situation is not symmetric: only the traveller's worldline bends.

Example— Simultaneity and velocity addition

In the frame of an observer moving with velocity $v$ , events with $t' = 0$ are those with $t = vx$ : a tilted line in the original diagram. Events simultaneous for one observer are not simultaneous for another. Composing two boosts adds rapidities, so velocities combine as $\tanh(\varphi + \psi)$ :

u \oplus v = \frac{u + v}{1 + uv}, \qquad \text{e.g.} \quad \tfrac35 \oplus \tfrac35 = \tfrac{15}{17} \lt 1.

Example— Constant acceleration

The worldline $t(\tau) = \frac1a \sinh(a\tau)$ , $x(\tau) = \frac1a \cosh(a\tau)$ has 4-velocity $(\cosh a\tau, \sinh a\tau)$ , of length $-1$ , and constant proper acceleration $a$ — what an accelerometer on board reads. It is a hyperbola asymptotic to the light cone $t = x$ : light emitted from events with $t > x$ never catches up with the traveller, who has a horizon (the Rindler horizon), a first model of a black-hole horizon.

ProofH

Special relativity: the Lorentz group and the twin paradox

Theorem— Boosts are Lorentz transformations, and rapidities add

Let $\Lambda(\varphi)$ be the boost matrix and $\eta = \operatorname{diag}(-1, 1)$ . Then $\Lambda(\varphi)^{\mathsf T} \eta \Lambda(\varphi) = \eta$ , $\det \Lambda(\varphi) = 1$ and $\Lambda(\varphi)\Lambda(\psi) = \Lambda(\varphi + \psi)$ . So the boosts form a one-parameter group isomorphic to $(\R, +)$ , and velocities satisfy $|u \oplus v| \lt 1$ whenever $|u|, |v| \lt 1$ .

Proof

Write $c = \cosh\varphi$ , $s = \sinh\varphi$ . Then $\Lambda^{\mathsf T}\eta\Lambda = \begin{pmatrix} -c^2 + s^2 & -cs + sc \\ -sc + cs & -s^2 + c^2 \end{pmatrix} = \eta$ , since $c^2 - s^2 = 1$ , and $\det\Lambda = c^2 - s^2 = 1$ . The product formula is the addition theorem $\cosh(\varphi + \psi) = \cosh\varphi\cosh\psi + \sinh\varphi\sinh\psi$ , $\sinh(\varphi + \psi) = \sinh\varphi\cosh\psi + \cosh\varphi\sinh\psi$ . For velocities: $1 + uv > 0$ , and $(1 + uv) \mp (u + v) = (1 \mp u)(1 \mp v) > 0$ , so $|u + v| \lt 1 + uv$ .

∎

Theorem— The four components of the Lorentz group

Every $\Lambda \in O(1, 3)$ satisfies $\det\Lambda = \pm 1$ and $|\Lambda^0{}_0| \ge 1$ . Hence $O(1, 3)$ has at least four connected components, distinguished by the signs of $\det\Lambda$ and of $\Lambda^0{}_0$ ; in fact it has exactly four, and the identity component $SO^+(1, 3)$ consists of products $R_1 B R_2$ of spatial rotations and a boost.

Proof

Taking determinants in $\Lambda^{\mathsf T}\eta\Lambda = \eta$ gives $(\det\Lambda)^2 = 1$ . Applying $\Lambda$ to $e_0 = (1, 0, 0, 0)$ : $\Lambda e_0$ is again a unit timelike vector, so $-(\Lambda^0{}_0)^2 + \sum_{i=1}^3 (\Lambda^i{}_0)^2 = -1$ and $(\Lambda^0{}_0)^2 \ge 1$ . The functions $\det\Lambda$ and $\operatorname{sign}\Lambda^0{}_0$ are continuous on $O(1, 3)$ with values in $\{\pm 1\}$ , so they are constant on connected components, and all four combinations occur ( $I$ , the space reflection $\operatorname{diag}(1, -1, 1, 1)$ , time reversal $\operatorname{diag}(-1, 1, 1, 1)$ and $-I$ ). For the decomposition and connectedness of $SO^+(1, 3)$ , write $\Lambda e_0 = (\cosh\varphi, \sinh\varphi \, \mathbf n)$ , rotate $\mathbf n$ to the $x$ -axis, undo the boost, and observe that what is left fixes $e_0$ and is therefore a rotation (Naber, The Geometry of Minkowski Spacetime, Ch. 1); since rotations and boosts are connected to the identity, so is $SO^+(1, 3)$ .

∎

The orthochronous Lorentz transformations ( $\Lambda^0{}_0 \ge 1$ ) are exactly those that preserve the future light cone: they never reverse the order of causally related events. The interval and the speed of light are the invariants; everything else is "relative".

Theorem— Reversed Cauchy–Schwarz and reverse triangle inequality

Let $v = (t_1, \mathbf x)$ and $w = (t_2, \mathbf y)$ be future-pointing causal vectors, where $\mathbf x, \mathbf y$ lie in a real inner product space of any dimension and $|\mathbf x| \le t_1$ , $|\mathbf y| \le t_2$ . Then

\tau(v)\,\tau(w) \le t_1 t_2 - \langle \mathbf x, \mathbf y \rangle = -\eta(v, w) \qquad\text{and}\qquad \tau(v) + \tau(w) \le \tau(v + w).

Proof

By Cauchy–Schwarz, $\langle \mathbf x, \mathbf y \rangle \le |\mathbf x||\mathbf y|$ , so it suffices to show $\sqrt{(t_1^2 - |\mathbf x|^2)(t_2^2 - |\mathbf y|^2)} \le t_1 t_2 - |\mathbf x||\mathbf y|$ . The right side is non-negative because $t_1 \ge |\mathbf x| \ge 0$ and $t_2 \ge |\mathbf y| \ge 0$ , and

(t_1 t_2 - |\mathbf x||\mathbf y|)^2 - (t_1^2 - |\mathbf x|^2)(t_2^2 - |\mathbf y|^2) = (t_1|\mathbf y| - t_2|\mathbf x|)^2 \ge 0.

\tau(v + w)^2 - \big(\tau(v) + \tau(w)\big)^2 = 2\big(t_1 t_2 - \langle \mathbf x, \mathbf y \rangle\big) - 2\,\tau(v)\tau(w) \ge 0

by the first inequality.

∎

Theorem— Straight worldlines are the longest

Let $q - p$ be future timelike. Every future-directed timelike curve from $p$ to $q$ has proper time at most $\tau(q - p)$ , with equality only for the straight segment. On the other hand there are causal curves from $p$ to $q$ with arbitrarily small proper time.

Proof

For a broken line $p = p_0, p_1, \dots, p_n = q$ with future timelike steps, applying the reverse triangle inequality $n - 1$ times gives $\sum_k \tau(p_{k+1} - p_k) \le \tau(q - p)$ . A smooth timelike curve's proper time is the limit of the proper times of inscribed broken lines as the subdivision is refined (by the definition of the integral, since the integrand is continuous). Hence $\tau[\gamma] \le \tau(q - p)$ . If $\gamma$ is not straight, some inscribed triangle $p, r, q$ has $r$ off the segment, and equality in the reverse Cauchy–Schwarz inequality fails strictly; refinement only decreases the sums, so the inequality is strict. For the last claim, zigzag between $p$ and $q$ along steps that are nearly null: each step has tiny proper time, and the total tends to $0$ .

∎

So proper time has no minimum but a maximum, achieved by inertial motion: the opposite of the Euclidean situation, and the resolution of the twin paradox. The traveller's worldline is not straight; there is no symmetry between the twins, and no appeal to gravity or acceleration "forces" is needed — only geometry.

ProofH

General relativity: curvature and the field equations

Geodesics extremise proper time. In local coordinates, consider curves $x^\mu(\lambda)$ and the energy functional $E[x] = \frac12 \int g_{\mu\nu}(x) \dot x^\mu \dot x^\nu \, d\lambda$ . Its Euler–Lagrange equations are

\frac{d}{d\lambda}\big(g_{\mu\nu}\dot x^\nu\big) - \frac12 \partial_\mu g_{\alpha\beta} \dot x^\alpha \dot x^\beta = 0 \iff \ddot x^\mu + \Gamma^\mu_{\alpha\beta}\dot x^\alpha \dot x^\beta = 0,

the geodesic equation (expand the first term and raise the index with $g^{\mu\nu}$ ). Along a solution $g(\dot x, \dot x)$ is constant, and the critical points of $E$ among affinely parametrised curves are the critical points of proper time. Timelike geodesics locally maximise proper time: within a convex normal neighbourhood the geodesic between two timelike-related points is the longest causal curve, the Lorentzian analogue of the theorem above (O'Neill, Ch. 10; Hawking–Ellis, §4.5). Beyond a conjugate point they stop maximising.

Theorem— The Newtonian limit

Let $g = -(1 + 2\Phi)\,dt^2 + (1 - 2\Phi)\,(dx^2 + dy^2 + dz^2)$ with $\Phi$ a time-independent function with $|\Phi| \ll 1$ , and consider a slowly moving particle ( $|dx^i/dt| \ll 1$ ). To first order in $\Phi$ and in the velocity, the geodesic equation is Newton's law $\dfrac{d^2 x^i}{dt^2} = -\partial_i \Phi$ . For dust at rest with density $\rho$ , the $00$ component of the field equations is, to the same order, Poisson's equation $\Delta \Phi = 4\pi\rho$ .

Proof

For slow motion $dt/d\tau \approx 1$ and the terms with $\dot x^i \dot x^j$ or $\dot x^i \dot t$ are of higher order, so $\ddot x^i \approx -\Gamma^i_{00}$ . With $g_{00} = -(1 + 2\Phi)$ time-independent, $\Gamma^i_{00} = -\frac12 g^{ii}\partial_i g_{00} = \partial_i\Phi + O(\Phi^2)$ . Hence $d^2x^i/dt^2 = -\partial_i\Phi$ . For the field equations, take the trace of $R_{\mu\nu} - \frac12 R g_{\mu\nu} = 8\pi T_{\mu\nu}$ to get $R = -8\pi T$ and rewrite them as $R_{\mu\nu} = 8\pi\big(T_{\mu\nu} - \frac12 T g_{\mu\nu}\big)$ . For dust at rest, $T_{00} = \rho$ and $T = g^{00}T_{00} \approx -\rho$ , so the right side of the $00$ equation is $8\pi(\rho - \frac12\rho) = 4\pi\rho$ . On the left, to first order in a static field, $R_{00} = \partial_\mu\Gamma^\mu_{00} - \partial_0\Gamma^\mu_{0\mu} + O(\Gamma^2) = \partial_i\partial_i\Phi$ .

∎

This is how the constant $8\pi G/c^4$ is fixed: Einstein's theory must reproduce Newton's where Newton is accurate. The field equations are also forced by consistency. The contracted Bianchi identity $\nabla^\mu G_{\mu\nu} = 0$ holds for every metric, so the equations imply $\nabla^\mu T_{\mu\nu} = 0$ : energy and momentum are conserved, and matter moves consistently with the geometry it creates. The vacuum equations are the Euler–Lagrange equations of the Einstein–Hilbert action $\int R \sqrt{-\det g} \, d^4x$ (Hilbert, 1915).

Theorem— Birkhoff's theorem (Jebsen 1921, Birkhoff 1923)

Every spherically symmetric solution of the vacuum equations $R_{\mu\nu} = 0$ is locally isometric to a piece of the Schwarzschild solution (or of Minkowski space). In particular it is static.

We do not prove Birkhoff's theorem here (Hawking–Ellis, Appendix B; Carroll, §5.2). That the Schwarzschild metric is Ricci-flat is a direct, if lengthy, computation of its Christoffel symbols and curvature; the third experiment below does it symbolically. A consequence: a spherical star that pulsates radially emits no gravitational waves, and the field outside it is the same as that of a static star of the same mass.

$r = 2M$ is not a singularity of spacetime. The metric component $g_{rr} = (1 - 2M/r)^{-1}$ blows up at $r = 2M$ , but the curvature does not: the Kretschmann scalar $R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta} = 48M^2/r^6$ is finite there and blows up only at $r = 0$ . The trouble is with the coordinates. With the ingoing Eddington–Finkelstein coordinate $v = t + r + 2M\log|r/2M - 1|$ the metric becomes

ds^2 = -\Big(1 - \frac{2M}{r}\Big)dv^2 + 2\,dv\,dr + r^2 \, d\Omega^2,

which is smooth for all $r > 0$ . At $r \lt 2M$ the vector $\partial_r$ is timelike and every future-directed causal curve has decreasing $r$ : nothing, not even light, gets out. The sphere $r = 2M$ is the event horizon of a black hole; $r = 0$ is a genuine curvature singularity.

ProofH

Light, planets and black holes: Schwarzschild geodesics

Conserved quantities. The Schwarzschild metric does not depend on $t$ or $\phi$ , so along a geodesic with affine parameter $\lambda$ (proper time for massive particles) the quantities

E = \Big(1 - \frac{2M}{r}\Big)\frac{dt}{d\lambda}, \qquad L = r^2 \frac{d\phi}{d\lambda}

are conserved (the $t$ - and $\phi$ -components of the Euler–Lagrange equations above). By spherical symmetry we may take the orbit in the plane $\theta = \pi/2$ . Normalising $g(\dot x, \dot x) = -\kappa$ , with $\kappa = 1$ for particles and $\kappa = 0$ for light, gives

\Big(\frac{dr}{d\lambda}\Big)^2 = E^2 - \Big(1 - \frac{2M}{r}\Big)\Big(\kappa + \frac{L^2}{r^2}\Big).

For particles, the radial motion is that of a unit-mass particle with energy $E^2$ in the effective potential $V(r) = (1 - 2M/r)(1 + L^2/r^2)$ .

Theorem— The orbit equation

With $u = 1/r$ as a function of $\phi$ , geodesics satisfy

\frac{d^2u}{d\phi^2} + u = \frac{\kappa M}{L^2} + 3Mu^2.

For $\kappa = 1$ the first term on the right is Newton's; the term $3Mu^2$ is the relativistic correction.

Proof

Since $dr/d\lambda = (dr/d\phi)(L/r^2) = -L \, du/d\phi$ , the energy equation becomes $L^2 (u')^2 = E^2 - (1 - 2Mu)(\kappa + L^2u^2)$ . Differentiating in $\phi$ : $2L^2 u'u'' = 2M\kappa u' + 6ML^2u^2u' - 2L^2 u u'$ . Divide by $2L^2u'$ (on stretches where $u' \ne 0$ ; circular orbits are treated separately).

∎

Theorem— Light bending (to first order in M/b)

A light ray with impact parameter $b \gg M$ is deflected by the angle $\delta = 4M/b$ , i.e. $4GM/(c^2 b)$ in physical units: $1.75''$ for starlight grazing the Sun, twice the value $2GM/(c^2b)$ obtained by treating light as a Newtonian particle.

Proof

With $\kappa = 0$ , $u'' + u = 3Mu^2$ . Without the mass, $u_0 = \frac{\sin\phi}{b}$ is a straight line with impact parameter $b$ . Write $u = u_0 + u_1$ with $u_1 = O(M/b^2)$ and linearise: $u_1'' + u_1 = \frac{3M}{b^2}\sin^2\phi = \frac{3M}{2b^2}(1 - \cos 2\phi)$ . A particular solution is $u_1 = \frac{3M}{2b^2}\big(1 + \frac13\cos 2\phi\big)$ , so

u = \frac{\sin\phi}{b} + \frac{M}{2b^2}(3 + \cos 2\phi).

The ray comes from and goes to infinity where $u = 0$ . Near $\phi = 0$ , setting $\phi = -\varepsilon$ with $\varepsilon$ small gives $-\varepsilon/b + 4M/(2b^2) = 0$ , so $\varepsilon = 2M/b$ ; by symmetry the same happens near $\phi = \pi$ . The asymptotic directions differ from a straight line by $2\varepsilon = 4M/b$ .

∎

The Newtonian value is half as large because it captures only the bending of the time part of the metric; the spatial curvature contributes the other half. Measured during the 1919 eclipse by expeditions to Príncipe and to Sobral in Brazil, and since by radio interferometry to better than one part in $10^4$ , the full value agrees with relativity.

Theorem— Perihelion precession (to first order in M/p)

A bound orbit with semi-latus rectum $p = L^2/M$ and eccentricity $e$ precesses by $\Delta\phi = 6\pi M/p = \dfrac{6\pi GM}{c^2 a(1 - e^2)}$ per revolution, where $a$ is the semi-major axis.

Proof

With $\kappa = 1$ , $u'' + u = \frac{1}{p} + 3Mu^2$ . The Newtonian solution is the ellipse $u_0 = \frac{1 + e\cos\phi}{p}$ . Linearising, $u_1'' + u_1 = 3Mu_0^2 = \frac{3M}{p^2}\big(1 + 2e\cos\phi + e^2\cos^2\phi\big)$ . The constant and $\cos 2\phi$ parts produce small bounded corrections to the shape. The term $\frac{6Me}{p^2}\cos\phi$ resonates with the free oscillation and produces the secular solution $\frac{3Me}{p^2}\,\phi\sin\phi$ . Keeping it,

u \approx \frac1p\Big(1 + e\cos\phi + \frac{3M}{p}e\,\phi\sin\phi\Big) \approx \frac1p\Big(1 + e\cos\big((1 - \tfrac{3M}{p})\phi\big)\Big),

using $\cos(\phi - \varepsilon\phi) \approx \cos\phi + \varepsilon\phi\sin\phi$ . Perihelia occur where the argument is a multiple of $2\pi$ , i.e. at $\phi = 2\pi n/(1 - 3M/p) \approx 2\pi n (1 + 3M/p)$ : each revolution the perihelion advances by $6\pi M/p$ . (The approximation is valid for $\phi \ll p/M$ ; the method of multiple scales, or Lindstedt–Poincaré, makes it uniform.)

∎

For Mercury ( $a = 5.79 \times 10^{10}$ m, $e = 0.2056$ , about $415$ revolutions per century) this is $43''$ per century, the anomaly that Newtonian theory could not explain. Einstein computed it on 18 November 1915, a week before writing down the final field equations.

Theorem— Circular orbits and the innermost stable circular orbit

For $M > 0$ , circular orbits of radius $r > 0$ are critical points of $V$ , i.e. roots of $Mr^2 - L^2 r + 3ML^2 = 0$ . They exist only if $L^2 \ge 12M^2$ ; for $L^2 > 12M^2$ there are two, $r_\pm = \frac{L^2 \pm L\sqrt{L^2 - 12M^2}}{2M}$ (with $L > 0$ ), the outer stable (a minimum of $V$ ) and the inner unstable (a maximum). As $L^2 \downarrow 12M^2$ they merge at $r = 6M$ , the innermost stable circular orbit. For light, the effective potential $(1 - 2M/r)/r^2$ has a single critical point, the unstable photon sphere $r = 3M$ .

Proof

Differentiating, $V'(r) = \frac{2M}{r^2}\big(1 + \frac{L^2}{r^2}\big) - \big(1 - \frac{2M}{r}\big)\frac{2L^2}{r^3} = \frac{2(Mr^2 - L^2 r + 3ML^2)}{r^4}$ . A real root requires the discriminant $L^4 - 12M^2L^2$ to be non-negative; equivalently, completing the square, $(2Mr - L^2)^2 = L^2(L^2 - 12M^2)$ , and $L \ne 0$ since otherwise $Mr^2 = 0$ . The quadratic $Mr^2 - L^2r + 3ML^2$ is negative between its roots and positive outside, so $V$ increases beyond $r_+$ , decreases between $r_-$ and $r_+$ : $r_+$ is a minimum, $r_-$ a maximum. At $L^2 = 12M^2$ the double root is $r = L^2/(2M) = 6M$ . For light, $\frac{d}{dr}\frac{1 - 2M/r}{r^2} = \frac{6M - 2r}{r^4}$ vanishes only at $r = 3M$ , a maximum.

∎

Gas in an accretion disc spirals inwards through stable circular orbits until it reaches $r = 6M$ and then plunges; the ISCO sets how much energy a disc around a non-rotating black hole can radiate. Light rays with impact parameter below $3\sqrt3\,M$ fall in, which is why a black hole casts a shadow about $2.6$ times larger than its horizon — the dark disc imaged by the Event Horizon Telescope.

Beyond Schwarzschild. Rotating black holes are described by Kerr's metric (1963). The homogeneous, isotropic cosmologies are the Friedmann–Lemaître–Robertson–Walker metrics $-dt^2 + a(t)^2 \big(\frac{dr^2}{1 - kr^2} + r^2 d\Omega^2\big)$ , for which the field equations reduce to Friedmann's equation $\big(\frac{\dot a}{a}\big)^2 = \frac{8\pi\rho}{3} - \frac{k}{a^2} + \frac\Lambda3$ (Friedmann, 1922; Lemaître, 1927): the universe expands. Linearising the vacuum equations around Minkowski space gives a wave equation for the metric perturbation: gravitational waves travel at the speed of light.

Formal mathematicsF

Spacetime geometry in Lean

Mathlib has no Minkowski space or Lorentzian manifolds yet (the separate Physlib project builds them on top of Mathlib), so we formalise with Mathlib's real numbers, matrices, inner product spaces and derivatives, in units $c = G = 1$ . The metric and boosts in one space dimension, and the facts that boosts are Lorentz transformations and compose by adding rapidities:

etaLean 4Formally verified

/-- The Minkowski metric in one time and one space dimension, signature `(−, +)`:
`η((t, x), (t, x)) = −t² + x²`. -/
def eta : Matrix (Fin 2) (Fin 2) ℝ := !![-1, 0; 0, 1]