Part D · Minkowski · spine

The equivalence principle, and why flat spacetime cannot hold gravity

One experimental coincidence — inertial mass equals gravitational mass — predicts falling light and slow clocks before any geometry exists, and then proves that no field on Minkowski space can be gravity.

needs 27, 05, 16~2½ h slides + 5 h problemsproblems: ./mr problems 30

This is the last unit of Part D, and its job is to end the part. Units 26 to 29 built flat spacetime and put everything on it: particles, fields, conserved currents. This unit takes one experimental fact — that all bodies fall alike — pushes it as far as it will go, and then shows exactly where it breaks the stage it is standing on.

The argument has three movements. First, the equivalence principle is stated — three times, because there are three genuinely different claims that go by the name, distinguished by experiments that can falsify one and not another. Second, three consequences are derived from it with no differential geometry whatsoever: light climbing loses frequency, clocks low down run slow, light bends. Two of these are measured to parts in $10^5$ and one of them runs a satellite navigation system. Third, the honest reckoning: an accelerating frame in flat spacetime reproduces a gravitational field's clock rates exactly, and a rotating frame in flat spacetime already has non-Euclidean space — so non-constant metric components are not gravity. What is gravity is the tidal tensor of unit 05, and no coordinate change removes it. Therefore there is no global inertial frame, and therefore gravity is not a force field on Minkowski space.

How to read this deck

One photon runs through all five sections: a $14.4$ keV gamma ray climbing $22.5$ m of Harvard tower in 1960. Section 2 predicts its shift from the equivalence principle alone; section 3 recomputes it exactly, from a metric on flat spacetime; section 5 says which part of the answer flat spacetime cannot supply. Every number quoted was computed, and the working is in the notes.

Proofs are on the slides where they are short and in these notes under Full proof where they are not; imported results are marked cited. Conventions as in units 26 and 27: $x^0 = ct$, signature $(-,+,+,+)$, Greek indices $0$ to $3$, Latin $1$ to $3$, Einstein summation as in unit 16. $g = 9.80665\ \mathrm{m\,s^{-2}}$ throughout, and $\Phi$ is the Newtonian potential with $\vec g = -\nabla\Phi$, so $\Phi$ is negative near a mass.

30 · overview

One photon, five sections

the question

A photon climbs a 22.5 m tower. Does its frequency change, and by how much?

Nothing in units 26–29 answers this: on Minkowski space a photon's frequency is constant along its worldline between inertial observers at rest relative to one another. The answer comes from one experimental fact about falling bodies — all of them fall alike — and it is $\Delta\nu/\nu = -2.455\times10^{-15}$, measured in 1960 to ten per cent.

the route, in four steps

1. Swap the tower, standing in the Earth's field $g$, for a rocket of the same height $h$ accelerating at $g$ in empty space. The equivalence principle says no experiment inside can tell them apart.

2. Light from the floor needs $h/c$ to reach the ceiling; by then the ceiling has gained speed $gh/c$, away from the source.

3. A receiver receding at $v$ sees the frequency lowered by $v/c$ (unit 26's Doppler shift, to first order), so $\Delta\nu/\nu = -gh/c^2$.

4. $h = 22.5$ m: $-2.455\times10^{-15}$. Pound and Rebka measured $(2.57\pm0.26)\times10^{-15}$.

The plan. 1 $m_{\mathrm i} = m_{\mathrm g}$, how well it is known, three statements of it · 2 redshift, clock rates, bending, with no geometry · 3 Rindler: the same redshift exactly, on flat spacetime · 4 the rotating disc, whose space is not Euclidean · 5 tides: what no coordinate change removes.

Getting around. Words with a dotted underline are glossary terms: click one for its definition; G lists every term in the unit. A solid underline, like the tower, is a piece of the running example, shown on hover.

Where each step of the route is done. Step 1 is Einstein's elevator, and what it actually claims — and it needs more than the statement that stones fall alike, because the experiment inside the box uses light, not stones; Three principles, three different claims names the stronger principle (EEP) that licenses it. Steps 2 and 3 are The rocket argument, with the error in each first-order step bounded in its notes. Step 4 is Worked example 1 — Pound and Rebka's tower, which also explains why they measured twice that, a source-up minus source-down difference. Section 3 then gets the same number with no approximation at all, in The redshift again, exactly.

Why this example. It is the shortest route in physics from a principle to a number to a measurement. The principle is a statement about falling stones; the number involves no stones at all, only light and a clock; and the measurement was done in a stairwell with equipment that fits in a room. It also has the rare property that three quite different derivations — the accelerating rocket here, the Rindler metric in section 3, and unit 36's Schwarzschild solution — give the same leading answer, so it is a place to watch the theory become consistent with itself.

What section 5 is for. Sections 2 to 4 could leave the impression that gravity is a coordinate effect: an accelerating frame reproduces the clock rates, a rotating frame reproduces the non-Euclidean space. Section 5 kills that impression with the tidal tensor, which is the only object in the unit that no change of coordinates touches. The slogans "gravity is not a force" and "gravity is real" are both true, and the tidal tensor is precisely the difference between them.

What the problem sheet adds. The Rindler chart derived from scratch with its horizon; the counting argument that leaves 20 unremovable second derivatives; the rotating disc's light cylinder; the orbital radius at which the two GPS effects cancel; a perpetual motion machine that runs unless clocks slow down; and Deser's argument, which says what happens if you insist on gravity as a field on flat spacetime anyway.

section 1

Two masses, and three principles

A coincidence measured to one part in $10^{15}$, and the three different things it is taken to mean.

30 · masses

Two masses Newton did not have to make equal

In Newton's theory two constants attach to a body, and they appear in different laws.

the two masses

Inertial mass $m_{\mathrm i}$ is the constant in $\vec F = m_{\mathrm i}\ddot{\vec x}$: the resistance to being accelerated by any force.

Gravitational mass $m_{\mathrm g}$ is the charge in $\vec F = -m_{\mathrm g}\nabla\Phi$: the coupling to the gravitational field, exactly as $q$ couples to $\vec E$.

Nothing in the axioms relates them. If they differ, free fall reads $$\ddot{\vec x} = -\frac{m_{\mathrm g}}{m_{\mathrm i}}\,\nabla\Phi ,$$ and the trajectory depends on the body.

the Eötvös parameter

For two bodies $A$, $B$ in the same field, with free-fall accelerations $a_A$, $a_B$, the dimensionless $$\eta_{AB} := 2\,\frac{a_A - a_B}{a_A + a_B} = \frac{(m_{\mathrm g}/m_{\mathrm i})_A - (m_{\mathrm g}/m_{\mathrm i})_B}{\text{mean}} .$$ $\eta = 0$ for every pair is the claim under test.

The letter is the literature's. This $\eta$ is a number and has nothing to do with the Minkowski form $\eta_{\mu\nu}$, which returns in section 3.

what would falsify it

Any composition dependence. Bodies differ in nuclear binding energy per nucleon, in neutron-to-proton ratio, in electrostatic energy: if any of these carried its own gravitational charge at a different rate, $\eta\ne0$ for a beryllium–titanium pair even though both fall at $9.8\ \mathrm{m\,s^{-2}}$ to a part in $10^{4}$.

Why this is not a triviality. The electric analogue is instructive: the ratio $q/m_{\mathrm i}$ is emphatically not universal — it is $1.76\times10^{11}$ C kg$^{-1}$ for an electron and $9.58\times10^{7}$ for a proton, a factor of 1836 — and that is why no electrostatic analogue of the equivalence principle exists and why unit 28's field cannot be geometrised the way gravity will be. The gravitational case is special, and it is an experimental fact rather than a theorem.

Newton knew it mattered. Principia, Book III, Proposition VI: he swung pendulums of gold, silver, lead, glass, sand, salt, wood, water and wheat in pairs of equal-length boxes and found no difference "by an experiment that would have detected one part in a thousand". The pendulum period is $2\pi\sqrt{(m_{\mathrm i}/m_{\mathrm g})(\ell/g)}$, so period equality is exactly $\eta = 0$. Bessel repeated it in 1832 with better technique and reached $2\times10^{-5}$.

The normalisation in $\eta$. Dividing by the mean acceleration makes $\eta$ dimensionless and antisymmetric in $A \leftrightarrow B$, and makes it independent of the source: a fractional difference in $m_{\mathrm g}/m_{\mathrm i}$ is the same number whether the attractor is the Earth, the Sun or the galaxy, which is why bounds from very different experiments can be listed in one column.

30 · masses

How well it is known

experiment	date	method	$\|\eta\|$ below
Newton, Principia III.6	1687	paired pendulums	$10^{-3}$
Bessel	1832	paired pendulums	$2\times10^{-5}$
Eötvös, Pekár, Fekete	1922	torsion balance, Earth	$5\times10^{-9}$
Roll, Krotkov, Dicke	1964	torsion balance, Sun	$10^{-11}$
Braginsky, Panov	1972	torsion balance, Sun	$10^{-12}$
Eöt-Wash (Adelberger)	2008	rotating torsion balance	$2\times10^{-13}$
MICROSCOPE	2022	drag-free satellite	$1.5\times10^{-15}$

Twelve orders of magnitude in three centuries, and the answer has never moved off zero.

Each point is a bound: $|\eta|$ above it is ruled out. The shaded region is everything excluded by the best result to date — it has only ever grown downward, and $\eta = 0$ has stayed inside what is left.

Reading the plot. The vertical axis is logarithmic, so equal steps are factors of a thousand. Pendulums (Newton, Bessel) bought two orders of magnitude in a century and a half; the torsion balance bought eight more; the satellite two more again. Nothing on the plot is a measurement of a nonzero $\eta$: each point is the size below which the result was consistent with zero.

MICROSCOPE's number, exactly. Touboul et al., Phys. Rev. Lett. 129, 121102 (2022), final result for a titanium–platinum pair: $\eta(\mathrm{Ti},\mathrm{Pt}) = [-1.5 \pm 2.3\,(\text{stat}) \pm 1.5\,(\text{syst})]\times10^{-15}$, consistent with zero and bounding $|\eta|$ at the $1.5\times10^{-15}$ level. Two concentric cylindrical test masses of different metals were held on the same orbit by electrostatic forces in a drag-free satellite; the measured quantity is the difference in the forces needed. An orbit is a very long free fall, which is why space wins here.

Eötvös's own numbers. The 1922 paper (published posthumously; the work dates from 1889–1909) reports differences below $5\times10^{-9}$ for eight materials. It was accurate enough that a 1986 reanalysis claiming a composition-dependent "fifth force" in the residuals launched a decade of experiments — all of which found nothing, and in passing produced the modern generation of torsion balances.

30 · masses

Why a torsion balance works

why a torsion balance

Hang two different materials on a horizontal beam. A difference $\eta$ in $m_{\mathrm g}/m_{\mathrm i}$ gives a differential acceleration $\eta a_\perp$ along the horizontal component of the source's field — the centrifugal residue of the Earth's rotation, $a_\perp\approx1.69\ \mathrm{cm\,s^{-2}}$ at mid-latitude, or the Sun's field swinging through $360^\circ$ each day.

Each mass is pushed $\tfrac12\eta a_\perp$ off the mean, in opposite senses, so the torque is $\tau = 2\cdot m\,\tfrac12\eta a_\perp\,d = m\,\eta a_\perp\,d$, and it twists a fibre of stiffness $\kappa$ through $\theta = \tau/\kappa$. The signal is periodic at a known frequency, which is what buys the extra orders of magnitude.

P12 designs one: for $\eta = 10^{-13}$, two 10 g masses 2 cm from the axis on a fibre with an 800 s period twist by $0.69$ nrad.

Seen from above. The shared part of the pull, $m a_\perp$ on each end, has no torque about the fibre; only the $\pm\tfrac12 m\eta a_\perp$ differences turn the beam. The vertical part of the field is taken up by the fibre.

What to see in the figure. Both masses are pulled the same way, so a beam of identical materials feels no torque at all — the fibre just hangs slightly off vertical. A composition difference makes one pull a little harder than the mean and the other a little weaker; those two small excess forces point in opposite senses relative to the mean, at opposite ends of the beam, and so add as a torque.

Eöt-Wash's number. Schlamminger et al., Phys. Rev. Lett. 100, 041101 (2008): $\eta(\mathrm{Be},\mathrm{Ti}) = (0.3\pm1.8)\times10^{-13}$. The balance sits on a turntable rotating at about $2/3$ mHz, which moves the signal to a frequency where the fibre's thermal noise and the local gravity gradients are quiet — the same trick as a lock-in amplifier.

The torque estimate, worked. Two masses $m = 10$ g at $d = 2$ cm on opposite ends give a moment of inertia $I = 2md^2 = 8.0\times10^{-6}\ \mathrm{kg\,m^2}$; a period $T = 800$ s fixes $\kappa = I(2\pi/T)^2 = 4.93\times10^{-10}\ \mathrm{N\,m\,rad^{-1}}$. The differential acceleration between the two masses is $\eta a_\perp = 10^{-13}\times1.69\times10^{-2} = 1.69\times10^{-15}\ \mathrm{m\,s^{-2}}$, by the definition of $\eta$ (difference over mean), so each is $\tfrac12\eta a_\perp$ off the mean and $\tau = 2m\,(\tfrac12\eta a_\perp)\,d = 3.38\times10^{-19}$ N m, giving $\theta = \tau/\kappa = 6.85\times10^{-10}$ rad. Detecting a nanoradian over hours is ordinary optical-lever work; detecting it against drifts is the whole difficulty, and it is why the balance is rotated.

30 · masses

Einstein's elevator, and what it actually claims

Einstein, 1907, on the patent-office argument he later called "the happiest thought of my life": a man falling freely from a roof does not feel his own weight.

the thought experiment

A windowless box. Inside, release two balls of different composition.

Box at rest on the Earth: both fall at $g$.
Box in deep space with proper acceleration $g$: both appear to fall at $g$ — they are inertial and the floor rises.

Because $\eta = 0$, no mechanical experiment inside distinguishes the two. Einstein's step was to strike the word "mechanical".

Two balls of different composition in each box. On the left they fall at $g$; on the right they float, inertial, and the floor comes up to meet them at $g$. Seen from inside, the two pictures are the same.

The logical shape. The elevator is not an argument that gravity is acceleration. It is an argument that a certain class of experiments cannot tell them apart, and the interesting work is in specifying the class. Restrict it to mechanics and you have a statement equivalent to $\eta = 0$ and nothing more. Extend it to all local non-gravitational physics — optics, nuclear decay rates, atomic spectra — and you have a much stronger claim, which is what generates the redshift in section 2.

The word "local" is doing all the work, and section 5 is about what it costs. Inside a large enough box for long enough, the two situations are trivially distinguishable: on the Earth the field points at a centre, so two balls released side by side approach each other, and two released one above the other separate. In the accelerating box they do neither. That difference is the tidal tensor, it is not small in principle, and no restatement of the principle removes it.

Einstein's own account is in his 1922 Kyoto lecture and in the 1907 Jahrbuch review: "I was sitting in a chair in the patent office at Bern when all of a sudden a thought occurred to me: if a person falls freely he will not feel his own weight. I was startled. This simple thought made a deep impression on me. It impelled me toward a theory of gravitation." The gap between that thought and the field equations was eight years, and most of it was spent acquiring the mathematics of units 31 to 34.

30 · masses

Three principles, three different claims

WEP

Weak. The worldline of a freely falling test body, given its initial position and velocity, is independent of its mass and composition.

"Test body": small enough to ignore its own field, structureless, uncharged.

EEP

Einstein. WEP, plus: in any local freely falling frame the outcome of any non-gravitational experiment is independent of the frame's velocity (local Lorentz invariance) and of where and when it is performed (local position invariance).

SEP

Strong. EEP with "non-gravitational" deleted, and "test body" weakened to include bodies with appreciable gravitational self-energy — stars, planets, the Earth.

30 · masses

Nested, and strictly

Each ring holds the theories that satisfy a principle; each named theory sits where it is known to belong. The experiment written on a boundary is the one that tests whether nature is inside it.

these are not rephrasings of one another

WEP $\Leftarrow$ EEP $\Leftarrow$ SEP, and both arrows are strict: theories are known that satisfy WEP and violate EEP (add a scalar field coupling to electromagnetism), and theories that satisfy EEP and violate SEP — every metric theory with an extra field does, including Brans–Dicke. General relativity is the standard example of a theory satisfying all three.

What to see in the figure. The rings are sets of theories, not regions of space. A theory in the outer ring but not the middle one keeps free fall universal and still lets an atomic clock's rate depend on where it sits; one in the middle ring but not the innermost passes every non-gravitational laboratory test and still lets the Earth and the Moon fall differently. That is why the three boundaries need three different experiments.

Where the strictness comes from. EEP $\Rightarrow$ WEP is immediate. The converse fails because WEP constrains only the motion of neutral structureless bodies, and says nothing about, say, whether the fine-structure constant is the same in the basement as on the roof. A theory in which a scalar field $\varphi$ multiplies the electromagnetic Lagrangian and couples to matter only through that term leaves free fall universal while making atomic energy levels depend on position: WEP holds, local position invariance fails.

Schiff's conjecture (1960) says the gap is illusory: any complete, self-consistent theory that obeys WEP necessarily obeys EEP. It is a conjecture, not a theorem; the "TH$\epsilon\mu$" formalism of Lightman and Lee (1973) proves versions of it under stated assumptions about how matter is built. It is worth knowing about because it is the reason a torsion-balance result is quoted as evidence for the far stronger EEP.

Why SEP is genuinely separate. A body held together by gravity carries negative binding energy $U$, which by unit 27's $E = mc^2$ contributes $U/c^2$ to its inertial mass. SEP demands it contribute exactly the same amount to its gravitational mass. In Brans–Dicke theory it does not, because the effective gravitational constant depends on the ambient scalar field, which a self-gravitating body itself modifies. The fractional self-energy is $|U|/Mc^2 = 3GM/5Rc^2 = 4.17\times10^{-10}$ for a uniform sphere of the Earth's mass and radius — about $4.6\times10^{-10}$ with the real density profile — so the effect is tiny but the Earth is very well tracked.

30 · masses

An experiment for each

WEP · Eöt-Wash, MICROSCOPE

Do two compositions fall alike? $|\eta| \lt 1.5\times10^{-15}$.

Falsified by any composition-dependent differential acceleration.

EEP · clock comparison

Does a clock's rate depend only on its potential, not on what it is made of or when it is read? Gravity Probe A (1976) flew a hydrogen maser to 10 000 km: prediction confirmed to $7\times10^{-5}$.

Two Galileo satellites launched into the wrong, eccentric orbit in 2014 turned into a redshift experiment: local position invariance holds to $2.5\times10^{-5}$.

SEP · lunar laser ranging

Do the Earth and Moon, with different self-energies, fall alike toward the Sun? A violation polarises the lunar orbit toward the Sun — the Nordtvedt effect. Corner cubes left in 1969 give $\eta_{\mathrm N} = (4.4\pm4.5)\times10^{-4}$.

this is the checkpoint's second clause

A torsion balance cannot falsify EEP: it contains no clock. A maser cannot falsify SEP: a hydrogen atom has no gravitational self-energy worth the name. Only a body the size of the Earth tests SEP, and only because the Moon's distance is known to millimetres. And the running example is an EEP experiment: a tower full of gamma rays tests local position invariance, and bounds $\eta$ not at all.

Gravity Probe A (Vessot and Levine, Phys. Rev. Lett. 45, 2081 (1980)): a hydrogen maser on a Scout rocket to $10^4$ km altitude, compared against an identical ground maser with a two-way microwave link that cancels the first-order Doppler shift. The predicted fractional rate difference peaks near $4\times10^{-10}$ and the measurement agreed with it to $7\times10^{-5}$ — still, decades later, among the sharpest tests of local position invariance.

The Galileo satellites. GSAT-0201 and 0202 were stranded in eccentric orbits by an upper-stage failure in August 2014. The eccentricity makes each onboard clock's rate oscillate with a 13-hour period at an amplitude $2GM_\oplus e/ac^2 \approx 5\times10^{-11}$, worth a clock-bias swing of some $\pm0.4\ \mu$s per orbit — a redshift experiment nobody paid for. It is that modulation which tests local position invariance; the constant $4.6\times10^{-10}$ by which the satellite clock outruns a ground one carries no information, because it never changes. Delva et al. and Herrmann et al., Phys. Rev. Lett. 121, 231101 and 231102 (2018), used three years of it.

The Nordtvedt effect, mechanically. Nordtvedt (1968) noticed that if the Earth's gravitational binding energy fell at a slightly different rate from ordinary matter, the Earth and Moon would accelerate differently toward the Sun. The Moon's orbit would then be displaced along the Earth–Sun line by roughly $\eta_{\mathrm N}\times13$ m, oscillating at the synodic month. Lunar laser ranging measures the Earth–Moon distance to about a centimetre by timing a pulse off the Apollo and Lunokhod retroreflectors, so a 13 m signature at $\eta_{\mathrm N}\sim10^{-3}$ is visible. Williams, Turyshev and Boggs, Phys. Rev. Lett. 93, 261101 (2004): $\eta_{\mathrm N} = (4.4\pm4.5)\times10^{-4}$, consistent with zero. In the standard post-Newtonian parametrisation $\eta_{\mathrm N} = 4\beta - \gamma - 3$, which is zero in general relativity ($\beta = \gamma = 1$).

Falsifiability, concretely. Each row names what would kill it: a nonzero $\eta$ for some material pair; a clock rate that depends on the clock's construction and not only on where it sits; a lunar orbit polarised toward the Sun. None of the three has happened.

30 · masses

What the WEP says mathematically

free fall is a structure, not a force

WEP says: through each event $p$ and each admissible four-velocity $u$ at $p$ there passes exactly one free-fall worldline, and it does not depend on which body travels it.

So gravity determines a map $$(p, u) \;\longmapsto\; \text{a curve through } p \text{ with tangent } u ,$$ a family of preferred curves on spacetime — a path structure, the same kind of datum as the family of straight lines on an affine space.

In Newtonian language: the equation $\ddot{\vec x} = -\nabla\Phi$ has no $m$ in it. In unit 12's language, the Lagrangian $L = \tfrac12 m|\dot{\vec x}|^2 - m\Phi$ has an overall $m$ that the Euler–Lagrange equations divide out.

Free fall in a uniform field, drawn in spacetime (exaggerated). One curve leaves $p$ for each initial velocity; a beryllium ball and a titanium ball launched alike share one of them, whatever they are made of.

What to see in the figure. Time runs up, height across. Each curve is $x = vt - \tfrac12 gt^2$ for a different launch velocity $v$: a fan of parabolas through the one event $p$, one per direction. The two balls drawn on the highlighted curve are different materials; WEP is the statement that they never separate. Pick a different $p$ and a different fan comes out, which is the "family of curves, one through each point in each direction" of the box.

Why state it this way. The physics slogan "gravity is geometry" is worth very little until one says which geometric object. This slide says it: the datum the WEP hands you is a family of curves, one through each point in each direction, and the mathematical object whose job is to encode such a family is a connection. That is not a metaphor — unit 32 defines a connection and proves its geodesics have exactly this existence-and-uniqueness property, by Picard–Lindelöf from unit 01 applied to the geodesic equation.

A caution about "test body". The uniqueness claim is false for extended or self-gravitating bodies: a body with a quadrupole moment feels a torque and a composition-dependent force in an inhomogeneous field (unit 10), and one with appreciable self-energy is the subject of SEP. "Test body" is not a hedge, it is part of the statement.

30 · masses

A property of the arena

The same launch, $p$ and $u$ fixed, in a uniform electric field instead (exaggerated). Now $q/m$ picks the curve: bodies that start together part company.

why this is the whole programme

A field that acts on every body identically is indistinguishable from a property of the arena. Electromagnetism is not like this: unit 28's Lorentz force carries $q/m$, so charged and neutral bodies take different paths and the field is visibly a thing on spacetime.

A rule assigning a preferred family of curves to a manifold is an affine connection; its preferred curves are geodesics. Unit 32 defines both. Everything after this unit is the construction of that object.

The one available candidate is already in hand: unit 27's free particle extremises proper time. If the metric is allowed to vary from place to place, extremal proper time becomes a nontrivial equation — unit 32.

What to see in the figure. Compare it with the previous slide's fan. There, the curves differ because the launch velocities differ, and the body does not matter. Here the launch is the same for all of them, $x = vt + \tfrac12(q/m)Et^2$, and the curves differ only because the bodies do. A field of the first kind can be absorbed into the arena; one of the second kind cannot, because no single family of curves describes it.

The comparison with charge, sharpened. Unit 28's equation of motion is $m\,\mathrm{d}u^\mu/\mathrm{d}\tau = qF^\mu{}_\nu u^\nu$. Divide by $m$ and the right-hand side keeps a factor $q/m$: two bodies at the same event with the same velocity follow different worldlines. There is no way to read the electromagnetic field as a property of spacetime, and the modern statement of that — it is a connection on a $U(1)$ bundle over spacetime, not on spacetime's own tangent bundle — is beyond this course but is the same observation.

section 2

Three consequences, before any geometry

Light loses frequency climbing, clocks low down run slow, and light bends — all from a box and a Doppler shift.

30 · consequences

The rocket argument

A rocket of height $h$, proper acceleration $g$, far from everything. The floor emits a photon upward; work in the inertial frame momentarily comoving at that instant.

three lines

1. To first order the photon reaches the ceiling after time $t = h/c$.
2. By then the ceiling has speed $\Delta v = gt = gh/c$, away from the source.
3. Unit 26's Doppler factor for a receding receiver is $1 - \Delta v/c$ to first order.

gravitational redshift, from the equivalence principle $$\boxed{\ \frac{\Delta\nu}{\nu} \;=\; -\frac{gh}{c^2} \;=\; \frac{\Delta\Phi}{c^2}\ }$$

The rocket in the inertial frame of the floor at emission (exaggerated: $gh/c^2 = 0.3$ here). When the photon arrives, the ceiling's worldline has tilted: it is moving away. No gravity anywhere in the picture.

$\Delta\Phi = \Phi_{\text{emit}} - \Phi_{\text{receive}}$ in a general static field. Checks. $gh/c^2$ dimensionless ✓; $g\to0$, no shift ✓; receiver recedes, so redder ✓.

What to see in the figure. Time runs up and height across, in the inertial frame in which the floor is at rest at the instant of emission. Floor and ceiling are hyperbolae (unit 27's uniform acceleration), both vertical at the bottom of the picture: at emission nothing is moving. The photon goes at $45^\circ$. Where it meets the ceiling, the ceiling's tangent has tilted away from vertical — that tilt is the speed $\Delta v$ of step 2, and the receiver is running away from the light, which is step 3's Doppler shift. The figure is drawn with the exact hyperbolae of section 3, so the arrival is a little later than $h/c$ (by the fraction $gh/2c^2$) — the second-order effect the first note below bounds.

Full justification of the three lines. Each contains an approximation, and each error is second order in the small quantity $gh/c^2$, which is $2.5\times10^{-15}$ for the tower. (1) The floor is itself accelerating, so the true flight time differs from $h/c$ by a fraction of order $gh/c^2$; since the answer is already first order, the correction is second order. Section 3 does the calculation exactly and confirms it. (2) $\Delta v = gt$ uses Newtonian kinematics, valid because $\Delta v/c = gh/c^2 \ll 1$; unit 27's exact hyperbolic motion gives $\Delta v = c\tanh(g\tau/c)$, differing at third order. (3) The exact Doppler factor of unit 26 is $\sqrt{(1+\beta)/(1-\beta)}$, whose expansion is $1 + \beta + \beta^2/2 + \dots$; the $\beta^2$ term is again second order. For the tower every neglected term is below $10^{-29}$.

The step to the general static field. The equivalence principle licenses the replacement of "rocket accelerating at $g$" by "laboratory at rest in a field of strength $g$" only locally, so $gh$ must be read as the work per unit mass done against the field along the path, which is $-\Delta\Phi$ for a conservative field (unit 04). Hence $\Delta\nu/\nu = \Delta\Phi/c^2$ with $\Delta\Phi = \Phi_{\text{emit}} - \Phi_{\text{receive}}$: emitting low, where $\Phi$ is more negative, and receiving high gives $\Delta\Phi \lt 0$ and a redshift. That form is what section 2's clock slide and the GPS example use, and it is exactly what unit 36's Schwarzschild solution reproduces to first order.

A photon does not "lose energy climbing out of a potential well". That phrase gets the right number for the wrong reason: it assigns the photon a mass $h\nu/c^2$ and computes $mgh$. The derivation above uses no such thing — the photon's energy is whatever the receiver measures, and the entire content is that the receiver is not in the same inertial frame as the emitter was. Section 3 will say the same thing a third way, as a statement about clocks rather than about light.

30 · consequences · example

Worked example 1 — Pound and Rebka's tower

the running example, evaluated

1. $h = 22.5$ m (Jefferson Laboratory, Harvard), $g = 9.80665\ \mathrm{m\,s^{-2}}$, $c^2 = 8.987552\times10^{16}\ \mathrm{m^2\,s^{-2}}$.

2. $\dfrac{gh}{c^2} = \dfrac{9.80665\times22.5}{8.987552\times10^{16}} = \mathbf{2.455\times10^{-15}}$.

3. Pound and Rebka measured the difference between source-up and source-down, which is $2gh/c^2 = 4.910\times10^{-15}$.

4. Their result: $(5.13\pm0.51)\times10^{-15}$, i.e. $(2.57\pm0.26)\times10^{-15}$ per one-way trip.

5. Ratio measured/predicted $= 1.04\pm0.10$. Inside the error bar.

how anyone measures $10^{-15}$

The Mössbauer effect: a $^{57}$Fe nucleus in a crystal emits a $14.4$ keV gamma ray with the recoil taken by the whole crystal, so the line is absurdly narrow — $\Gamma/E = 3\times10^{-13}$. A shift of $10^{-15}$ is a detectable fraction of that width.

The shift was cancelled by moving the source on a loudspeaker cone: the speed needed is $v = c\,(2gh/c^2) = 1.47\ \mu\mathrm{m\,s^{-1}}$.

check

$g$ from $GM_\oplus/R_\oplus^2 = 9.7983\ \mathrm{m\,s^{-2}}$ instead of standard gravity gives $2.453\times10^{-15}$ — a $0.09\%$ change, a hundred times inside the experiment's error bar ✓.

The arithmetic. $9.80665\times22.5 = 220.6496\ \mathrm{m^2\,s^{-2}}$; divided by $c^2 = 8.987551787\times10^{16}$ that is $2.455058\times10^{-15}$. Doubling: $4.910116\times10^{-15}$. The published ratio of observed to predicted is $1.05\pm0.10$ in Pound and Rebka's own normalisation; using the number above, $5.13/4.910 = 1.045$.

Why they measured a difference. Absolute frequency comparison at $10^{-15}$ was impossible; a difference was not. Any systematic that does not reverse when the source and absorber are exchanged — temperature-dependent second-order Doppler shift from lattice vibrations, chief among them — cancels from the difference. The temperature effect is the serious one: it is $-2.4\times10^{-15}$ per kelvin, larger than the signal, so the two ends of the tower had to be matched to better than $0.1$ K.

The reference. R. V. Pound and G. A. Rebka Jr., "Apparent weight of photons", Phys. Rev. Lett. 4, 337 (1960). Pound and Snider repeated it in 1965 with better temperature control and reached a ratio of $0.9990\pm0.0076$ — a $0.8\%$ test, and still the best terrestrial one of its kind until optical clocks arrived.

Two modern versions. NIST compared two optical aluminium-ion clocks separated by $33$ cm in 2010, where $gh/c^2 = 3.6\times10^{-17}$. In 2022 a JILA strontium lattice clock resolved the redshift within a single millimetre-tall atom cloud, $gh/c^2 = 1.09\times10^{-19}$. The tower has shrunk by four orders of magnitude, and the principle under test has not changed.

The loudspeaker speed. A first-order Doppler shift $v/c$ cancels a fractional shift $2gh/c^2$ when $v = 4.910\times10^{-15}\times2.998\times10^8\ \mathrm{m\,s^{-1}} = 1.47\times10^{-6}\ \mathrm{m\,s^{-1}}$ — one and a half microns per second, about a millimetre in ten minutes. That number is a good answer to "is this experiment hard?"

30 · consequences · example

Worked example 2 — the clocks that run the satnav

the GPS budget

A clock moving with speed $v$ at potential $\Phi$ runs, relative to coordinate time in the Earth-centred inertial frame, at $$\frac{\mathrm d\tau}{\mathrm dt} = 1 + \frac{\Phi}{c^2} - \frac{v^2}{2c^2} + O(c^{-4}).$$ First term: section 2's redshift. Second: unit 26's time dilation, expanded.

Potential. $a = 26\,561$ km, $R_\oplus = 6378.137$ km, $GM_\oplus = 3.986004\times10^{14}\ \mathrm{m^3\,s^{-2}}$: $$\frac{GM_\oplus}{c^2}\Big(\frac1{R_\oplus} - \frac1a\Big) = 5.2837\times10^{-10} \;\Rightarrow\; +45.65\ \mu\mathrm{s/day}.$$

Speed. $v = \sqrt{GM_\oplus/a} = 3874\ \mathrm{m\,s^{-1}}$, so $v^2/2c^2 = 8.349\times10^{-11} \Rightarrow -7.21\ \mu\mathrm{s/day}$.

Net. $+38.44\ \mu$s/day; $+38.54$ with the ground clock's own $465\ \mathrm{m\,s^{-1}}$ of rotation restored.

what it costs to ignore

Position is inferred from time-of-flight, so a clock error $\delta t$ is a range error $c\,\delta t$: $$c \times 38.5\ \mu\mathrm s = \mathbf{11.5\ km\ per\ day}.$$

The satellite oscillators are therefore set low before launch, by a fractional $4.4647\times10^{-10}$, so that they tick at $10.23$ MHz once in orbit.

Where $\mathrm d\tau/\mathrm dt$ comes from. Two ingredients already proved. Unit 26: for a clock moving at speed $v$ in an inertial frame, $\mathrm d\tau/\mathrm dt = \sqrt{1 - v^2/c^2} = 1 - v^2/2c^2 + O(v^4/c^4)$. This unit: a static clock at potential $\Phi$ ticks slow by the factor $1 + \Phi/c^2$ relative to one at $\Phi = 0$, which is the boxed redshift read as a statement about clocks. To first order in both small quantities the two corrections simply add, which is all that is claimed; the cross terms are $O(c^{-4})$, around $10^{-20}$ here, three orders below what the system needs.

The arithmetic, in full. $1/R_\oplus = 1.567855\times10^{-7}$, $1/a = 3.764919\times10^{-8}$, difference $1.191363\times10^{-7}\ \mathrm{m^{-1}}$. Times $GM_\oplus$: $4.74858\times10^{7}\ \mathrm{m^2\,s^{-2}}$. Divided by $c^2$: $5.28373\times10^{-10}$. Times $86400$ s: $4.5651\times10^{-5}$ s, i.e. $45.65\ \mu$s. Orbital speed: $GM_\oplus/a = 1.50070\times10^{7}\ \mathrm{m^2\,s^{-2}}$, so $v = 3873.9\ \mathrm{m\,s^{-1}}$ and $v^2/2c^2 = 8.3488\times10^{-11}$, which is $7.213\ \mu$s/day. Net $38.438\ \mu$s/day. The ground clock at the equator moves at $\omega_\oplus R_\oplus = 465.1\ \mathrm{m\,s^{-1}}$, worth $1.203\times10^{-12}$ or $0.104\ \mu$s/day, which restores to $38.542$. Published figures range from $38.4$ to $38.6\ \mu$s/day depending on whether the ground clock is put on the equator or on the geoid, and on whether the Earth's quadrupole is included; the honest statement is "about $38.5$".

The pre-launch offset. The specified fractional rate offset for the GPS Block satellites is $-4.4647\times10^{-10}$, so that the nominal $10.23$ MHz clock is built to run at $10.229\,999\,995\,43$ MHz on the ground. Multiply $4.4647\times10^{-10}$ by $86400$ s: $38.57\ \mu$s/day — the design figure, and the one this slide's arithmetic should reproduce, as it does.

The 11.5 km. $c\times38.5\ \mu\mathrm s = 2.998\times10^8\times3.85\times10^{-5} = 1.154\times10^4$ m. This is the figure usually quoted as "GPS would fail within minutes", which is a little unfair to it: a constant rate offset common to all satellites is partly absorbed into the receiver's own clock solution. What is not absorbed is the eccentricity correction, which differs between satellites and is applied explicitly in every receiver as $-2\,\vec r\cdot\vec v/c^2$, worth up to $45$ ns, or $14$ m. Relativity is in the receiver firmware, not merely in the design margin.

30 · consequences · example

One term alone

both terms, always

The two effects have opposite signs and differ by a factor of six. Keep only the potential term and the prediction is $+45.65$ — $7.2\ \mu$s/day fast. Keep only the speed term and it is $-7.21$, off by $45.7\ \mu$s/day: worse than dropping both, which is off by $38.4$.

It is the cleanest working demonstration that special and general relativity are one subject.

Four predictions for the daily gain of an orbiting clock over a ground clock, against the right answer (dashed). Section 2's redshift alone overshoots by $7.2\ \mu$s; unit 26's time dilation alone has the wrong sign.

What to see in the figure. Each bar is what a navigator would build into the satellite clocks if they believed only part of relativity. The dashed line is the potential term plus the speed term, $+38.44\ \mu$s/day for an equatorial ground clock taken at rest, $+38.54$ with its own rotation restored — the two are indistinguishable at this scale. The bracket on each bar is its error. Neither half on its own is close: the bar nearest the truth still leaves $7.2\ \mu$s/day, which is $2.2$ km of range a day.

Why the speed term cannot be dropped. It is tempting to think the redshift is the "general" correction and the speed the "special" one, so that a theory of gravity owes only the first. The satellite does not know the difference: the clock-rate formula above is one expansion of one quantity, $\mathrm d\tau/\mathrm dt$, and for a circular orbit its two terms are $(GM_\oplus/c^2)(1/R_\oplus - 1/a)$ and $GM_\oplus/2ac^2$: the same order in $1/c^2$, the same order in the orbit's size, and for GPS within a factor of six of each other.

30 · consequences

Clocks, and the first non-constant metric

Read the redshift as a statement about clocks. A signal emitted at rate $\nu_1$ from potential $\Phi_1$ is received at rate $\nu_2$ at $\Phi_2$ with $\nu_2/\nu_1 = 1 + (\Phi_1-\Phi_2)/c^2$. Nothing accumulates in between — the field is static — so this is a permanent ratio of clock rates, not a transient.

the uniform-field metric

Let $t$ label the ticks of a reference clock at $\Phi = 0$. Then a static clock at $\Phi$ has $\mathrm d\tau = (1 + \Phi/c^2)\,\mathrm dt$, so $$-c^2\mathrm d\tau^2 = -\Big(1 + \frac{2\Phi}{c^2}\Big)c^2\,\mathrm dt^2 + \dots$$ to first order in $\Phi/c^2$: the $00$ component of a metric that varies from place to place.

Notation: from here on $g_{\alpha\beta}$, always with two indices, is a metric component; a bare $g$ is still the acceleration, $9.80665\ \mathrm{m\,s^{-2}}$ for the Earth.

what is and is not established

Established: $g_{00} = -(1 + 2\Phi/c^2)$ to first order. Not established: the spatial part. The dots hide $g_{ij}$, and the equivalence principle alone cannot fix it — the redshift is a statement about clocks only.

That gap has a famous cost: the next slide's light deflection comes out exactly half the true value.

Checks. $\Phi/c^2$ is dimensionless ✓. $\Phi\to0$: $\eta_{\mu\nu}$ back, and unit 26's Minkowski space ✓. Newtonian limit: a slow particle extremising $\int\mathrm d\tau$ in this metric obeys $\ddot{\vec x} = -\nabla\Phi$ — unit 35 does it properly.

Why "permanent ratio" needs saying. A common objection is that the redshift must be a one-off effect of the emission event, not an ongoing difference in rates. Here is the argument that it is ongoing. The field is static, so the geometry the signal traverses is the same for every wavecrest. Send $N$ crests over an interval $\Delta\tau_1$ by the emitter's clock; since nothing in the intervening region changes with time, exactly $N$ crests arrive, over an interval $\Delta\tau_2$ by the receiver's clock. Hence $\nu_1\Delta\tau_1 = \nu_2\Delta\tau_2$ and the ratio of measured frequencies is the ratio of clock rates, permanently. Section 3 makes this argument exact, where "the field is static" becomes "the metric components do not depend on the time coordinate".

Full derivation of the metric line. By definition of proper time (unit 26), $-c^2\mathrm d\tau^2 = \mathrm ds^2$ along the clock's worldline, and the clock is static, so only the $\mathrm dt^2$ term contributes: $\mathrm ds^2 = g_{00}c^2\mathrm dt^2$. With $\mathrm d\tau = (1+\Phi/c^2)\mathrm dt$ we get $g_{00} = -(1+\Phi/c^2)^2 = -(1 + 2\Phi/c^2) + O(\Phi^2/c^4)$. For the Earth's surface $\Phi/c^2 = -6.95\times10^{-10}$, so the dropped term is $5\times10^{-19}$.

The honest status of this metric. It is a first-order expression, obtained from one principle, with its spatial part unknown. It is not a solution of anything — there is no field equation yet. What it does establish, and this is the point of the slide, is that the equivalence principle by itself already forces the metric components to be functions of position. Whether that is gravity is the question of sections 3 and 5, and the answer will be: not on its own.

Where this reappears. Unit 35 takes $g_{00} = -(1+2\Phi/c^2)$ as the Newtonian limit that fixes the constant in the field equations, and unit 36 finds the exact Schwarzschild $g_{00} = -(1-2GM/rc^2)$, whose expansion is this line with $\Phi = -GM/r$.

30 · consequences

Light bends — and by half the right amount

Back in the rocket. A photon crosses the cabin horizontally, width $L$, in time $L/c$; in that time the cabin has risen by $\tfrac12 g(L/c)^2$. So the photon's path, drawn in the cabin, is bent downward, and the deflection angle over the crossing is $gL/c^2$. In a field that varies along the ray, the same argument applied to each short stretch $\mathrm d\ell$ gives $\mathrm d\theta = (g_\perp/c^2)\,\mathrm d\ell$, with $g_\perp$ the field's component across the ray.

The same crossing seen twice, at four equal instants (exaggerated). Left, in the inertial frame: the photon goes straight and the cabin rises under it. Right, in the cabin: the cabin is still, so the photon's path is the parabola $y = -gx^2/2c^2$, leaving at the angle $gL/c^2$.

What to see in the figure. The four dots are the photon at times $0$, $L/3c$, $2L/3c$ and $L/c$. In the inertial frame they lie on a straight horizontal line, and the cabin's floor climbs $\tfrac12 g t^2$ under them. Subtract the cabin's height at each instant and the dots fall by $\tfrac12 g t^2$ with $t = x/c$: a parabola, the trajectory of a ball thrown horizontally at speed $c$. That is the equivalence principle at its most literal — light falls like everything else, because in the cabin "falling" is only the floor coming up.

The angle. The slope of $y = -gx^2/2c^2$ is $-gx/c^2$, so after a width $L$ the ray has turned through $gL/c^2$. For a cabin $3$ m wide on the Earth that is $9.81\times3/8.99\times10^{16} = 3.3\times10^{-16}$ rad, and the droop at the far wall is $gL^2/2c^2 = 4.9\times10^{-16}$ m, a few millionths of an atomic radius — which is why nobody has seen it in a lift.

30 · consequences

Deflection by the Sun, added up along the ray

deflection by the Sun, from the equivalence principle alone

Adding up $\mathrm d\theta = (g_\perp/c^2)\,\mathrm d\ell$ along a ray grazing a mass $M$ at impact parameter $b$ gives $$\Delta\theta_{\mathrm{EP}} = \frac{2GM}{c^2 b}.$$ For the Sun: $2GM_\odot/c^2 = 2953.25$ m, $R_\odot = 6.957\times10^8$ m, so $\Delta\theta = 4.245\times10^{-6}$ rad $= \mathbf{0.876''}$.

The integrand is largest at closest approach and falls off as $|x|^{-3}$ either side, so almost all of the bend happens within a few $b$ of the Sun: the ray is straight, then turns, then is straight again.

Top: a ray passing the Sun at impact parameter $b$ (not to scale), with the field's component across it. Bottom: that component along the ray, in units of $GM/b^2$, against $x/b$. The area under the whole curve, tails included, is $2$: that is the whole calculation.

What to see in the figure. The arrows on the ray are $g_\perp$ at a few points: longest at closest approach, shrinking either side, and always pointing across the ray towards the line through the Sun's centre. The curve underneath is $g_\perp b^2/GM = (1 + x^2/b^2)^{-3/2}$, whose integral over all $x/b$ is exactly $2$ — so $\int g_\perp\,\mathrm dx = 2GM/b$ and $\Delta\theta = 2GM/c^2 b$. Half the area lies within $|x| \lt b/\sqrt3$ of closest approach.

Full derivation of the integral. Take the ray along the $x$-axis at height $b$ above a mass at the origin, and treat the deflection as small (it is $10^{-6}$ rad). The component of $\vec g$ perpendicular to the ray at position $x$ is $g_\perp = (GM/r^2)(b/r) = GMb/(b^2+x^2)^{3/2}$. In the equivalence-principle picture the photon acquires transverse velocity $\mathrm dv_\perp = g_\perp\,\mathrm dt = g_\perp\,\mathrm dx/c$, so the total deflection is $\Delta\theta = (1/c^2)\int_{-\infty}^{\infty} GMb\,(b^2+x^2)^{-3/2}\,\mathrm dx$. Substituting $x = b\tan\psi$ turns the integral into $(GM/bc^2)\int_{-\pi/2}^{\pi/2}\cos\psi\,\mathrm d\psi = 2GM/bc^2$. Dimensions: $GM/c^2$ is a length, divided by $b$ gives a pure number, correctly read as radians ✓. Limit $b\to\infty$: no deflection ✓.

The arithmetic. $2GM_\odot/c^2 = 2\times1.32712440\times10^{20}/8.987552\times10^{16} = 2953.25$ m. Divided by $R_\odot = 6.957\times10^8$ m: $4.24500\times10^{-6}$ rad. One radian is $206\,264.806''$, so that is $0.87560''$; the full general-relativistic value is twice it, $1.75119''$.

30 · consequences

The factor of two, and 1919

the factor of two

Einstein published $0.83''$ on this argument in 1911 (solar data of the day) and $1.75''$ in 1915. The true value is $$\Delta\theta = \frac{4GM_\odot}{c^2R_\odot} = \mathbf{1.751''} .$$ The missing half comes from $g_{ij}$ — the spatial part of the metric that the clocks slide could not determine.

Eddington's 1919 eclipse expedition measured $1.98\pm0.12''$ (Sobral, four-inch) and $1.61\pm0.30''$ (Príncipe).

why this is good news

The equivalence principle is exactly strong enough to predict the effects that involve only clocks, and exactly too weak for the ones that involve space. That is a sharp, falsifiable boundary, and 1919 is where it was drawn.

Deflection at the Sun's limb: the two predictions (lines) against the 1919 plates and modern radio interferometry (points, one-sigma bars). Both sets of plates that were kept disfavour the equivalence-principle value.

What to see in the figure. The two horizontal lines are the predictions, $0.876''$ from this section's argument and $1.751''$ from general relativity. Sobral's four-inch result sits $2\sigma$ above the second and nine $\sigma$ above the first; Príncipe's sits within $0.5\sigma$ of the second and $2.5\sigma$ above the first. The hollow point is the discarded astrographic result, drawn without a bar because none was published for it. The last point, radio interferometry, has a bar too small to draw.

Where the other half really comes from. The full calculation (unit 37) integrates a null geodesic in the Schwarzschild metric. Half the deflection comes from $g_{00}$ — the effect this slide computes, which can be phrased as light slowing in a position-dependent effective refractive index — and half from $g_{rr}$, the fact that radial distances near a mass exceed what the circumference suggests. The equivalence principle is local and cannot see the latter, because over a small enough patch the spatial geometry is flat to the accuracy that matters.

1919, honestly. The Sobral astrographic plates showed about $0.93''$ and were discarded on grounds of focus drift; the surviving Sobral four-inch plates gave $1.98\pm0.12''$ and the Príncipe plates $1.61\pm0.30''$. The discarding was contentious for decades; the modern verdict is that it was defensible on the stated technical grounds, and in any case irrelevant now — radio interferometry confirms the general-relativistic value to about $10^{-4}$, most sharply by very-long-baseline interferometry on quasars occulted by the Sun.

section 3

Rindler: a varying metric on a flat spacetime

Take unit 27's hyperbola seriously as a laboratory floor, and see how much of gravity comes for free.

30 · rindler

What a metric in arbitrary coordinates is

the definition

$(V,\eta)$ is unit 26's vector space with its symmetric bilinear form. Let $x^\mu \mapsto y^\alpha(x)$ be a smooth invertible change of coordinates on a region. At each point the coordinate basis is $$e_{(\alpha)} := \frac{\partial x^\mu}{\partial y^\alpha}\,\hat e_\mu ,$$ the columns of unit 03's Jacobian — a basis of $V$ that varies from point to point.

The metric components are the matrix of $\eta$ in that basis: $$g_{\alpha\beta}(y) := \eta\big(e_{(\alpha)}, e_{(\beta)}\big) = \frac{\partial x^\mu}{\partial y^\alpha}\frac{\partial x^\nu}{\partial y^\beta}\,\eta_{\mu\nu}.$$

this is unit 16, unchanged

$\eta$ is a $(0,2)$-tensor: a bilinear map $V\times V\to\mathbb R$. Unit 16 proved that the matrix of such a map transforms as $g' = S^{\!\top}\!gS$ under a change of basis. The only new thing here is that $S$ is a function of position.

So $g_{\alpha\beta}(y)$ is not a new object. It is the same $\eta$, written in a basis that moves.

Why the coordinate basis is the right one. A curve $y^\alpha(\lambda)$ has velocity $\mathrm dx^\mu/\mathrm d\lambda = (\partial x^\mu/\partial y^\alpha)(\mathrm dy^\alpha/\mathrm d\lambda)$ by unit 03's chain rule, so the components of the velocity in the coordinate basis are exactly $\mathrm dy^\alpha/\mathrm d\lambda$. That is the whole reason for this choice, and it is why the line element $\mathrm ds^2 = g_{\alpha\beta}\,\mathrm dy^\alpha\mathrm dy^\beta$ computes the squared length of an infinitesimal displacement in the new coordinates. Unit 31 makes the same construction on a manifold, where there is no $x^\mu$ to refer back to.

Nondegeneracy and signature survive. $S$ is invertible at each point (it is a Jacobian of a diffeomorphism), so $g = S^\top\eta S$ is nondegenerate, and by Sylvester's law of inertia its signature is $(-,+,+,+)$ at every point. That is worth noting: whatever the components look like, there is always a basis at each point in which the metric is $\mathrm{diag}(-1,1,1,1)$. Section 5 asks how much more than that can be arranged, and the answer to that question is the whole of general relativity.

30 · rindler

Polar coordinates: a basis that moves

the trap this whole section is about

Non-constant $g_{\alpha\beta}$ therefore proves nothing about the spacetime. Polar coordinates on the Euclidean plane give $\mathrm dl^2 = \mathrm dr^2 + r^2\mathrm d\phi^2$, and the plane is flat.

The coordinate basis of polar coordinates is $e_{(r)} = (\cos\phi, \sin\phi)$ and $e_{(\phi)} = (-r\sin\phi,\, r\cos\phi)$ — the columns of the Jacobian of $(r,\phi)\mapsto(r\cos\phi, r\sin\phi)$. So $$g_{rr} = |e_{(r)}|^2 = 1,\qquad g_{r\phi} = 0,\qquad g_{\phi\phi} = |e_{(\phi)}|^2 = r^2 .$$

The component $r^2$ is the length of a basis vector, squared. The plane did nothing; the basis grew.

The coordinate basis at three points of the Euclidean plane. $e_{(r)}$ always has length $1$; $e_{(\phi)}$ has length $r$, so it grows as the point moves out. That growth, not any property of the plane, is the varying $g_{\phi\phi}$.

What to see in the figure. At each marked point the two arrows are the coordinate basis: $e_{(r)}$ along the ray, $e_{(\phi)}$ along the circle. They are always at right angles, which is $g_{r\phi} = 0$. The radial arrow is the same length everywhere; the angular one is $r$ times a unit vector, because increasing $\phi$ by $\mathrm d\phi$ moves a point a distance $r\,\mathrm d\phi$. Nothing in the picture is curved except the coordinate lines themselves.

The polar-coordinate warning, quantitatively. In $\mathrm dl^2 = \mathrm dr^2 + r^2\mathrm d\phi^2$ the component $g_{\phi\phi} = r^2$ grows without bound, vanishes at $r = 0$, and has a nonzero derivative everywhere — and yet a circle of radius $r$ has circumference exactly $2\pi r$ and the angles of every triangle sum to $\pi$. Any argument of the form "the metric components vary, therefore spacetime is curved" is wrong, and section 3 exists to make that unforgettable before unit 33 defines curvature properly.

30 · rindler

The Rindler chart

Unit 27's Born-rigid family: one hyperbola per label $X \gt 0$, of constant proper acceleration $c^2/X$, all sharing the same lines of simultaneity through the origin. Take them as a static laboratory.

Rindler coordinates

With a reference acceleration $g$ and $X_0 := c^2/g$, $$ct = X\sinh\frac{gT}{c},\qquad x = X\cosh\frac{gT}{c},$$ for $X \gt 0$, $T\in\mathbb R$. $X$ labels the observer; $T$ is the proper time of the one at $X_0$.

the Rindler metric $$\mathrm ds^2 = -\Big(\frac{gX}{c}\Big)^{\!2}\mathrm dT^2 + \mathrm dX^2 + \mathrm dy^2 + \mathrm dz^2 .$$

Checks. $(gX/c)\,\mathrm dT$ is in metres ✓. At $X = X_0$ the coefficient is $c^2$, so $T$ is proper time there ✓. Small $gT/c$ near $X_0$: Minkowski back ✓.

two lines

Write $\theta := gT/c$. Then $$\mathrm d(ct) = \sinh\theta\,\mathrm dX + X\cosh\theta\,\tfrac gc\mathrm dT,$$ $$\mathrm dx = \cosh\theta\,\mathrm dX + X\sinh\theta\,\tfrac gc\mathrm dT.$$ Form $-\mathrm d(ct)^2 + \mathrm dx^2$. The cross terms cancel exactly ($-2\sinh\cosh + 2\cosh\sinh$), $\mathrm dX^2$ keeps $\cosh^2-\sinh^2 = 1$, and $\mathrm dT^2$ keeps $(gX/c)^2(\sinh^2-\cosh^2) = -(gX/c)^2$. $\square$

The coefficient $gX/c$ is $\sqrt{-g_{TT}}$, and it fixes the lapse function — the rate at which a static observer's proper time runs per unit of coordinate time, $\mathrm d\tau/\mathrm dT = gX/c^2 = X/X_0$. It depends on position — and the spacetime is exactly Minkowski.

Where the chart comes from, rather than being guessed. Unit 27 showed that a worldline of constant proper acceleration $\alpha$ is $x^2 - c^2t^2 = c^4/\alpha^2$, parametrised by rapidity $\theta = \alpha\tau/c$. Put $X := c^2/\alpha$, so that $X$ is the hyperbola's distance from the origin, and note that unit 27's Born-rigidity proof showed every member of the family is at the same rapidity on the same line through the origin. That makes $(\theta, X)$ a coordinate system. Scaling $\theta$ to a time by $T := c\theta/g$ makes $T$ the proper time of the member at $X_0 = c^2/g$, since that member's proper time is $c\theta/\alpha = c\theta/g$. The only choice in the whole construction is which member gets to keep the clock.

The proper time of a general static observer. Setting $\mathrm dX = 0$ in the metric gives $c^2\mathrm d\tau^2 = (gX/c)^2\mathrm dT^2$, so $\mathrm d\tau = (gX/c^2)\,\mathrm dT = (X/X_0)\,\mathrm dT$. Observers higher up (larger $X$) age faster, in exact proportion to $X$. That is the next slide but one.

The proper acceleration, recovered from the metric. Unit 27 gives it directly as $c^2/X$. It is worth seeing that this is what the $gX/c$ lapse encodes: the "gravitational field strength" a Rindler observer measures is $c^2/X$, which blows up as $X\to0$ and falls to zero as $X\to\infty$. A uniform gravitational field — the same $g$ at every height — is therefore not what a uniformly accelerating family provides, and unit 27's Born-rigid rod said the same thing from the other side: the back must accelerate harder than the front.

Naming. Wolfgang Rindler named and popularised the chart in the 1960s; the coordinates themselves go back to Kottler (1914) and Møller (1943), and the underlying hyperbolic motion to Born (1909). "Rindler coordinates", "Kottler–Møller coordinates" and "the uniformly accelerated frame" all name this chart, sometimes with $X$ replaced by $\xi = (c^2/g)\ln(gX/c^2)$, which makes the metric $\mathrm ds^2 = e^{2g\xi/c^2}(-c^2\mathrm dT^2 + \mathrm d\xi^2)$ — conformally flat and convenient, but with a less direct physical reading of the coordinate.

30 · rindler

The wedge, and the horizon

what the chart covers

$(T,X)\mapsto(ct,x)$ with $X\gt0$ is a bijection onto the Rindler wedge $\;x \gt |ct|$, and onto nothing else.

proof

$x^2 - c^2t^2 = X^2 \gt 0$ and $x = X\cosh \gt 0$, so the image lies in the wedge. Conversely, given $x\gt|ct|$, set $X := \sqrt{x^2-c^2t^2} \gt 0$ and $\theta := \operatorname{artanh}(ct/x)$, which is defined because $|ct/x| \lt 1$; these invert the map, and both are unique. $\square$

At $X\to0$ the lapse $gX/c\to0$ and the proper acceleration $c^2/X\to\infty$. The boundary $x = |ct|$ is the Rindler horizon of unit 27.

Hyperbolae are the static observers, $X = $ const; rays from the origin are $T = $ const. The chart fills the right wedge and stops at the null lines $x = |ct|$, the horizon.

What to see in the figure. The thick hyperbola is the reference observer $X = X_0 = c^2/g$; the others are $X = X_0/2$, $3X_0/2$ and $2X_0$, each a worldline of constant proper acceleration $c^2/X$, so the inner ones bend harder. The rays are $gT/c = 0, \pm\tfrac12, \pm1$: each is one instant of the whole family's common time, crossing every hyperbola at right angles in the Minkowski sense. All of it is squeezed between the two dashed null lines, and nothing in the chart reaches the other three wedges.

The horizon is a coordinate artefact, and the spacetime is not. At $X = 0$ the metric component $g_{TT}$ vanishes, so the coordinates degenerate: this is a coordinate singularity. Nothing whatever happens there physically. Every event on the null line $x = ct$ is an ordinary event of Minkowski space; an inertial observer crosses it without noticing, and the spacetime is flat on both sides. The chart fails, not the world. That distinction is the entire content of unit 36's discussion of $r = 2GM/c^2$ in the Schwarzschild solution, and meeting it here first — where the spacetime is known to be flat, so there is nothing to argue about — is the reason this section exists.

Three quarters of Minkowski space are missing. The chart covers only the right wedge. The left wedge $x \lt -|ct|$ carries a mirror-image family; the future and past wedges $|ct| \gt |x|$ carry no static worldlines at all, because there the hyperbolae $x^2 - c^2t^2 = $ const are spacelike curves. A Rindler observer's experience is genuinely confined to the right wedge: events in the future wedge can never send it a signal, which is unit 27's last-signal calculation.

What "horizon" means here. It is the boundary of the region from which a signal can ever reach the family — a causal boundary, defined by the observers, not by the spacetime. A different family of observers in the same Minkowski space has a different horizon, or none. This is the precise sense in which horizons are observer-dependent, and the sense in which a black hole horizon is not: there the boundary is the same for every observer who stays outside for ever, which unit 40 makes exact.

Unruh, in one sentence, and then dropped. A quantum field in the Minkowski vacuum, restricted to the Rindler wedge, is a thermal state at temperature $T_{\mathrm U} = \hbar g/2\pi ck_{\mathrm B}$ — for $g = 9.8\ \mathrm{m\,s^{-2}}$, about $4\times10^{-20}$ K. This course does no quantum field theory and proves nothing about it; the fact is mentioned because it is the reason the Rindler wedge is famous outside relativity, and because it shows the wedge is not merely a pedagogical device.

30 · rindler

The redshift again, exactly

static redshift in a static metric

If $g_{\alpha\beta}$ does not depend on the time coordinate $T$, then for two static observers at $X_1$, $X_2$, $$\frac{\nu_2}{\nu_1} = \frac{\sqrt{-g_{TT}(X_1)}}{\sqrt{-g_{TT}(X_2)}} .$$ For the Rindler metric $\sqrt{-g_{TT}} = gX/c$, so $\;\nu_2/\nu_1 = X_1/X_2$, exactly.

proof

A light signal from $X_1$ to $X_2$ is a curve $T(X)$ with $\mathrm ds^2 = 0$. Because the metric has no $T$-dependence, a signal emitted at $T_1+\Delta T$ is the $T$-translate of one emitted at $T_1$, so it arrives at $T_2+\Delta T$: the coordinate interval is preserved. Proper times are $\Delta\tau_i = \sqrt{-g_{TT}(X_i)}\,\Delta T/c$, and frequencies are their reciprocals. $\square$

Light signals between two static observers, drawn in the Rindler chart. Every signal is the same curve slid up in $T$, so equal gaps in $T$ leave and arrive. The clocks' ticks (equal proper time) are wider apart in $T$ low down.

What to see in the figure. The vertical lines are two static observers, at $X_1 = X_0/2$ and $X_2 = 3X_0/2$ — an absurdly tall tower, chosen so that the effect can be seen. Each curve is a light signal, $T(X) = (c/g)\ln(X/X_1) + $ const: all are translates of one another, so signals leaving $\Delta T$ apart arrive $\Delta T$ apart. The tick marks on each worldline are equal intervals of that observer's own proper time, $\Delta T = (X_0/X)\,\Delta\tau$: twice as far apart in $T$ at the bottom, two-thirds as far at the top. Count ticks between two arriving signals and the receiver has more of them than the sender did: it sees the signal's frequency lowered by $X_1/X_2 = 1/3$.

Full proof of the translation step. Null curves in the $(T,X)$ plane satisfy $(gX/c)^2\mathrm dT^2 = \mathrm dX^2$, i.e. $\mathrm dT/\mathrm dX = \pm c/(gX)$, an ordinary differential equation whose right-hand side is independent of $T$ — an autonomous equation in unit 02's language. Hence its solutions form a one-parameter family closed under $T$-translation: $T(X) = (c/g)\ln(X/X_1) + \text{const}$. Two signals from $X_1$ separated by $\Delta T$ therefore arrive at $X_2$ separated by exactly $\Delta T$. Note the flight time $(c/g)\ln(X_2/X_1)$ diverges as $X_1\to0$, which is the coordinate face of the horizon.

Why this is a theorem about staticity, not about gravity. Nothing in the proof used the Rindler form: any metric whose components are independent of a time coordinate, and for which the observers sit at fixed spatial coordinates, has this redshift. Unit 36 applies the identical argument to Schwarzschild and gets $\nu_2/\nu_1 = \sqrt{(1-r_s/r_1)/(1-r_s/r_2)}$. The mathematics is one line in both cases; the difference between the two situations is nowhere in this calculation, and that is the lesson.

30 · rindler

The tower, exactly

the running example, exactly

Floor at $X_1$, ceiling at $X_2 = X_1 + h$. Then $$\frac{\nu_2}{\nu_1} = \frac{X_1}{X_1+h} = \frac1{1 + g_1h/c^2},$$ with $g_1 = c^2/X_1$ the floor's proper acceleration. Expanding: $\Delta\nu/\nu = -g_1h/c^2 + O((g_1h/c^2)^2)$.

Section 2's three-line argument was right to first order and no further, and now we know exactly what it dropped: for the tower, $(gh/c^2)^2 = 6.0\times10^{-30}$.

Check. $h\to0$: ratio $1$ ✓. $X_1\to\infty$ at fixed $h$: $g_1\to0$, no shift ✓. $X_1\to0$: the floor approaches the horizon and $\nu_2/\nu_1\to0$ — infinite redshift ✓.

The received frequency ratio against $g_1h/c^2$: exact (solid) and section 2's first-order answer (dashed). They agree where the tower lives, at the far left; the first-order line reaches zero at $g_1h/c^2 = 1$ and goes negative, while the exact ratio only tends to zero, as the floor nears the horizon.

What to see in the figure. The horizontal axis is $u = g_1h/c^2 = h/X_1$, the height of the tower measured in units of the floor's distance from the horizon. For the Harvard tower $u = 2.5\times10^{-15}$, which is indistinguishable from the left edge; the gap between the curves there is $u^2 = 6.0\times10^{-30}$. To the right the first-order formula fails visibly, and at $u = 1$ it predicts that the signal arrives with no frequency at all. The exact ratio $1/(1+u)$ is never zero: the infinite redshift is reached only as the floor approaches the horizon, $u\to\infty$.

Which acceleration appears. Writing the same result with $g_2 = c^2/X_2$, the ceiling's proper acceleration, gives $\Delta\nu/\nu_1 = -g_2h/c^2$ exactly — with no correction at all. Both statements are exact; they differ because $g$ is not the same at the two ends. For the tower $g_1 - g_2 = g\,(gh/c^2) = 2.4\times10^{-14}\ \mathrm{m\,s^{-2}}$, so the distinction is academic here and will not be academic near a black hole.

An aside worth having. The clock-rate ratio $X_2/X_1$ is unbounded: a Rindler observer at $X_1 = X_0/100$ ages a hundred times slower than the one at $X_0$, in flat spacetime, with no gravitating matter anywhere. Any intuition of the form "large time dilation implies strong gravity" dies here.

30 · rindler · example

Worked example 3 — the tower as a Rindler wedge

the Harvard tower, re-derived

1. Model the laboratory floor as the Rindler observer with proper acceleration $g$: $X_1 = c^2/g = 9.1648\times10^{15}$ m.

2. In light years, $X_1 = 9.1648\times10^{15}/9.4607\times10^{15} = 0.9687$ ly.

3. Ceiling: $X_2 = X_1 + 22.5$ m. Exact ratio $$\frac{\nu_1}{\nu_2} = \frac{X_2}{X_1} = 1 + \frac{gh}{c^2} = 1 + 2.4551\times10^{-15}.$$

4. So section 2's answer is exact here, read the right way: $\nu_1/\nu_2 = 1 + g_1h/c^2$ with the floor's proper acceleration, or $\Delta\nu/\nu = -g_2h/c^2$ with the ceiling's. Only $\Delta\nu/\nu = -g_1h/c^2$ stops at first order — P4 sorts the three out.

5. The horizon lies at $X = 0$: proper distance $0.9687$ light years below the floor.

what the model gets wrong, and by how much

The Earth's field is not the Rindler field. Rindler has $g(X) = c^2/X$, so over $22.5$ m it varies by a fractional $2.5\times10^{-15}$; the Earth's $g = GM/r^2$ varies by $2h/R_\oplus = 7.1\times10^{-6}$ — nine orders of magnitude more.

That difference is the tidal tensor, and it is the subject of section 5. The redshift does not notice it, because the redshift only sees $\int g\,\mathrm d\ell$.

checks

$c^2/g$: $\mathrm{(m^2s^{-2})/(m\,s^{-2})} = \mathrm m$ ✓. $X_2/X_1 - 1 = h/X_1 = hg/c^2$ ✓ — the same number as section 2, not merely close to it.

The arithmetic. $c^2/g = 8.987552\times10^{16}/9.80665 = 9.164752\times10^{15}$ m. One Julian light year is $9.460730\times10^{15}$ m, so $X_1 = 0.968715$ ly — unit 27's number, unchanged, because it is the same hyperbola. Then $h/X_1 = 22.5/9.164752\times10^{15} = 2.455058\times10^{-15}$, identical to $gh/c^2$ as it must be. Do not try to compute $X_2/X_1 - 1$ by forming $X_2$ and $X_1$ in double precision and subtracting: the cancellation destroys every significant figure and returns $2.44\times10^{-15}$, which is wrong in the third digit. Rearrange first.

How good is the Rindler model of a laboratory? Two separate questions. (i) Does it get the redshift right? Yes, to the accuracy of any experiment ever done in a tower, because the redshift depends only on the potential difference. (ii) Does it get the field right? No: a Rindler field falls off as $1/X$ from a horizon $10^{16}$ m away, so over any laboratory it is uniform to one part in $10^{15}$, whereas the Earth's field varies by one part in $10^{5.1}$ over the same tower. An experiment sensitive to the variation — a gradiometer — distinguishes the elevator from the Earth immediately, which is exactly what the "local" in the equivalence principle is conceding.

The horizon under the floor. It is worth pausing on how strange this is and then on why it is not strange at all. An observer who has accelerated at $1g$ forever has a causal horizon $0.97$ ly behind them, from beyond which no signal will ever reach them. Nothing is there; Minkowski space is homogeneous. The horizon is a fact about the observer's motion, not about a place. A laboratory on the Earth has no such horizon, because it has not been accelerating for ever and will not continue to — which is one more difference the local equivalence principle is happy to ignore and section 5 is not.

30 · rindler

The inoculation

the belief this section exists to destroy

"The metric components depend on position, therefore spacetime is curved, therefore there is gravity."

exhibit A

Polar coordinates on the plane: $g_{\phi\phi} = r^2$. Flat.

exhibit B

Rindler: $g_{TT} = -(gX/c)^2$, a lapse varying with height, an infinite redshift surface, a horizon, and observers who feel weight. Minkowski space.

exhibit C

Rotating coordinates (section 4): non-Euclidean spatial geometry. Minkowski space.

what does distinguish them

In every example above there exists a global coordinate change bringing $g_{\alpha\beta}$ to $\eta_{\mu\nu}$ everywhere at once. Section 5 shows that near the Earth there is not — and identifies exactly what obstructs it.

Why this slide is a slide. Every general-relativity course loses some readers to this confusion, and the confusion is not stupid: the metric really does become position-dependent when gravity is present, and the position dependence really does produce redshift, horizons and apparent forces. The error is the converse. Rindler is the cleanest available counterexample because it has every symptom of a gravitational field — weight, a redshift, a horizon, an infinite-redshift surface — in a spacetime that is flat by construction, not by a computation one might have got wrong.

The right question, stated now and answered in unit 33. Given $g_{\alpha\beta}(y)$ on a region, does there exist a coordinate change making it $\eta_{\mu\nu}$ everywhere? This is a question about the integrability of a system of partial differential equations, and it has a clean answer: yes if and only if a certain tensor built from second derivatives of $g$ vanishes identically. That tensor is the Riemann tensor. Everything between here and unit 33 is the construction of it, and this unit's last section is the physical argument for why it has to be nonzero near the Earth.

A useful habit. When a metric is handed to you, the first question is never "what do the components look like?" but "what is invariant?" — the length of a curve, the proper time along a worldline, the angle between two vectors at a point, and, from unit 33 on, the curvature scalars. The components are a choice of description; the invariants are the spacetime.

section 4

The rotating disc

A frame in flat spacetime whose space is already not Euclidean — the observation Einstein said pushed him to Riemann.

30 · disc

Rotating coordinates, and their metric

Minkowski space in cylindrical coordinates: $$\mathrm ds^2 = -c^2\mathrm dt^2 + \mathrm d\rho^2 + \rho^2\mathrm d\phi^2 + \mathrm dz^2 .$$ Pass to a frame rotating at angular velocity $\omega$ by $\phi = \phi' + \omega t$, so $\mathrm d\phi = \mathrm d\phi' + \omega\,\mathrm dt$.

the rotating metric $$\begin{aligned}\mathrm ds^2 = \;&-\Big(1 - \frac{\omega^2\rho^2}{c^2}\Big)c^2\mathrm dt^2 + 2\omega\rho^2\,\mathrm d\phi'\,\mathrm dt\\ &+ \mathrm d\rho^2 + \rho^2\mathrm d\phi'^2 + \mathrm dz^2 .\end{aligned}$$

Two new features at once: the lapse $\sqrt{1-\omega^2\rho^2/c^2}$ again depends on position, and there is now a cross term $g_{0\phi'}\ne0$ — the metric is not diagonal.

The cylinder of radius $\rho$, cut and unrolled: arc length $\rho\phi$ across, $ct$ up. A co-rotating observer is a tilted line; the new time axis tilts with it, but the slices $t = $ const do not. That mismatch is $g_{0\phi'}$.

Checks. $\omega\rho/c$ dimensionless ✓; $\omega\to0$ gives cylindrical Minkowski back ✓; $g_{\phi'\phi'} = \rho^2$ unchanged, because the substitution does not touch $\mathrm d\phi'^2$'s coefficient ✓.

What to see in the figure. Cut the cylinder $\rho = $ const along a line and lay it flat: arc length $\rho\phi$ across, $ct$ up, and the metric on it is plain $-c^2\mathrm dt^2 + \rho^2\mathrm d\phi^2$, so light still runs at $45^\circ$. An inertial observer at fixed $\phi$ is a vertical line. A co-rotating one, at fixed $\phi'$, advances $\rho\omega t$ in time $t$: a straight line of slope $\omega\rho/c$ from the vertical, drawn here at $\omega\rho/c = 0.4$. Its own simultaneity would be the line tilted by the same angle the other way, as in unit 26's boost picture — but the rotating chart keeps the old slices $t = $ const, which are horizontal. The time axis and the space axis of the chart are therefore not orthogonal, and $g_{0\phi'} = \eta(e_{(0)}, e_{(\phi')})$ measures by how much.

The substitution, in full. $\rho^2\mathrm d\phi^2 = \rho^2(\mathrm d\phi' + \omega\mathrm dt)^2 = \rho^2\mathrm d\phi'^2 + 2\omega\rho^2\mathrm d\phi'\mathrm dt + \omega^2\rho^2\mathrm dt^2$. Adding $-c^2\mathrm dt^2$ collects the $\mathrm dt^2$ coefficient as $-(c^2 - \omega^2\rho^2) = -c^2(1-\omega^2\rho^2/c^2)$. Nothing else changes. Writing $x^0 = ct$, the components are $g_{00} = -(1-\omega^2\rho^2/c^2)$, $g_{0\phi'} = \omega\rho^2/c$, $g_{\phi'\phi'} = \rho^2$, $g_{\rho\rho} = g_{zz} = 1$.

What the cross term is. $g_{0i}\ne0$ says the coordinate time axis is not $\eta$-orthogonal to the coordinate spatial axes — that is, "the set of events with the same $t$" is not the set of events the rotating observers call simultaneous. It is the metric face of unit 26's relativity of simultaneity, and it is why the next slide has to define what "distance" even means before it can compute a circumference. In general relativity the same component appears around a rotating mass (unit 40's Kerr solution) and is called frame dragging; it is not the same phenomenon, but it is the same slot in the metric.

30 · disc

The light cylinder

the light cylinder

$g_{tt}$ vanishes at $\rho = c/\omega$ and is positive beyond. There, and outside, no observer can co-rotate: the curve $\rho,\phi',z$ constant is null, then spacelike.

For a millisecond pulsar at $716$ Hz the light cylinder is at $66.6$ km — a few stellar radii out, and where its magnetosphere is believed to end.

The co-rotating observer at radius $\rho$ moves, in the inertial frame, at $\omega\rho$. The lapse $\sqrt{1-\omega^2\rho^2/c^2}$ is unit 26's $1/\gamma$ at that speed, and it reaches zero exactly where the speed reaches $c$.

Co-rotating observers seen from above, each with its speed $\omega\rho$ in the inertial frame (arrows to scale with one another). The speed grows linearly outward and reaches $c$ on the dashed circle $\rho = c/\omega$; beyond it, no observer can keep up.

What to see in the figure. The arrows are drawn at radii $\tfrac14$, $\tfrac12$ and $\tfrac34$ of $c/\omega$, with lengths in proportion to $\omega\rho$; the arrow on the dashed circle would have length $c$. A rigid rotation is a statement about angles, and the speed it demands at radius $\rho$ is $\omega\rho$ whatever the material — so for any $\omega$ there is a radius at which co-rotation would need light speed. That is why "the light cylinder" belongs to the rotating chart, not to any object.

The light cylinder is a real object. For PSR J1748−2446ad, the fastest known pulsar at $716$ Hz, $\omega = 2\pi\times716 = 4499\ \mathrm{s^{-1}}$ and $c/\omega = 2.998\times10^8/4499 = 6.66\times10^4$ m $ = 66.6$ km, against a stellar radius of roughly $12$ km. A co-rotating magnetosphere cannot extend past it, which is why the light cylinder appears in every pulsar emission model. Note the honesty required: the real star's spacetime is not flat, so the number is an order-of-magnitude statement about a flat-space idealisation, not a prediction.

Beyond the light cylinder, nothing goes wrong. Just as with the Rindler horizon, the chart degenerates and the spacetime does not. Events at $\rho \gt c/\omega$ are ordinary events; an inertial observer sails through. Only the family of co-rotating observers ceases to exist, because it would have to move faster than light.

30 · disc

What "the distance between two rotating observers" means

radar distance

An observer at fixed spatial coordinates bounces a light signal off a neighbour at $y^i + \mathrm dy^i$ and measures the round-trip proper time $\Delta\tau$. Define $\mathrm dl := c\,\Delta\tau/2$.

the spatial metric $$\gamma_{ij} = g_{ij} - \frac{g_{0i}g_{0j}}{g_{00}} ,\qquad \mathrm dl^2 = \gamma_{ij}\,\mathrm dy^i\mathrm dy^j .$$

A symmetric bilinear form on each observer's orthogonal complement of their four-velocity — positive definite, by the signature.

When $g_{0i} = 0$ this is just $g_{ij}$, and no one notices the definition is needed. When $g_{0i}\ne0$ — a rotating frame — it is not, and the correction is the whole effect.

Radar between two neighbouring riders on the rim, in the rotating chart (exaggerated: $\omega\rho/c = 0.4$). Both are vertical lines. Light takes longer going with the rotation than against it, but only the round trip $\Delta x^0$ enters $\mathrm dl$.

What to see in the figure. Coordinate time $x^0$ up, arc length $\rho\phi'$ along the rim across. In this chart the two riders sit still, so their worldlines are vertical. Solving $\mathrm ds^2 = 0$ with the rotating metric at rim speed $\beta = \omega\rho/c$ gives light slopes $\mathrm dx^0/\mathrm dy = 1/(1-\beta)$ one way and $1/(1+\beta)$ the other: the out and back legs are no longer mirror images, which is the cross term at work. Their sum is $2\,\mathrm dy/(1-\beta^2)$, the rider's clock reads $\sqrt{1-\beta^2}$ of that, and $\mathrm dl = \mathrm dy/\sqrt{1-\beta^2}$ — the next slide but one, in miniature.

Why a definition is needed at all. "Distance" in a spacetime is not given. A spacetime has a metric on four-vectors; a spatial distance is a number attached to a pair of worldlines, and it needs a rule. Radar is the honest rule: it uses only the observer's own clock and the invariant $\mathrm ds^2 = 0$ for light, and it is what any real measurement of a distance actually does. Unit 26 already used radar to define simultaneity, so this is the same idea one step further.

30 · disc

Radar distance, from one quadratic

proof

Set $\mathrm ds^2 = 0$ and solve for $\mathrm dx^0$: a quadratic $$g_{00}(\mathrm dx^0)^2 + 2g_{0i}\mathrm dx^0\mathrm dy^i$$ $$+\ g_{ij}\mathrm dy^i\mathrm dy^j = 0,$$ whose two roots are the outward and return legs, with difference $\Delta x^0 = 2\sqrt D/(-g_{00})$ for $$D := (g_{0i}\mathrm dy^i)^2 - g_{00}\,g_{ij}\mathrm dy^i\mathrm dy^j .$$ The waiting observer's proper time is $\Delta\tau = \sqrt{-g_{00}}\,\Delta x^0/c$, so $\mathrm dl = c\Delta\tau/2 = \sqrt{D}/\sqrt{-g_{00}}$, and $\mathrm dl^2 = D/(-g_{00})$ is the boxed form. $\square$

The rim unrolled, its two ends the same place. The riders' local "now" (dashed), carried once round, returns to the same rider later than it left.

The catch. Each rider's spatial metric is fine, but the slices do not join up: carried round the rim, the riders' simultaneity returns late by $\Delta x^0 = 2\pi\omega\rho^2c/(c^2-\omega^2\rho^2)$ — the Sagnac effect.

What to see in the figure. The strip is the cylinder $\rho = $ const cut open, in the inertial frame, with arc length across and $ct$ up; its left and right edges are the same line. The tilted lines are one rider and the copy of the same rider one circumference on. The dashed line is what that rider calls simultaneous, tilted by the rim speed exactly as a boosted observer's simultaneity is in unit 26. Followed all the way round it meets the copy of the rider at a later time, drawn here for $\omega\rho/c = 0.4$. Neighbours can agree on "now"; the whole ring cannot.

The geometric statement behind the formula. Let $u$ be the four-velocity of the static observer, so $u \propto e_{(0)}$. The subspace $u^\perp = \{w : \eta(u,w) = 0\}$ is spacelike and three-dimensional, and $\gamma$ is exactly the restriction of the metric to it, written in terms of the coordinate basis. Indeed, the projector onto $u^\perp$ is $P = \mathbb 1 + u\otimes u^\flat/c^2$ (unit 16's index-raising), and applying it to the coordinate vectors $e_{(i)}$ and pairing gives $\eta(Pe_{(i)}, Pe_{(j)}) = g_{ij} - g_{0i}g_{0j}/g_{00}$. So the radar definition and the orthogonal-projection definition agree, which is the reassurance one wants.

Positive definiteness. $-g_{00} \gt 0$ for a static observer (the worldline is timelike), and $\gamma$ is the metric restricted to a spacelike subspace, so $\gamma_{ij}v^iv^j \gt 0$ for $v\ne0$. In the rotating frame that requires $\rho \lt c/\omega$, which is the light cylinder again: outside it there is no static observer and hence no spatial metric to speak of.

The catch, stated and set aside. The three-dimensional "space" so defined is a metric on the quotient of the wedge by the observers' worldlines, and in a rotating frame that quotient is not a hypersurface of the spacetime: the orthogonal complements do not fit together into a family of slices, because a loop around the axis does not close up in time. (Concretely: the neighbour a static observer calls simultaneous lies at $\mathrm dx^0 = -(g_{0i}/g_{00})\,\mathrm dy^i$, and carrying that once around the rim leaves $\Delta x^0 = 2\pi\omega\rho^2c/(c^2-\omega^2\rho^2)$ rather than zero, so co- and counter-propagating signals differ by the Sagnac time $4\pi\omega\rho^2/(c^2 - \omega^2\rho^2)$ — a real, measured effect, used in ring-laser gyroscopes and corrected for in GPS.) So "the geometry of the rotating disc" means the geometry of the observers' measurements, which is exactly what is wanted here, and not the intrinsic geometry of a slice, which does not exist. Frobenius' integrability condition, in unit 34's language, is what fails.

30 · disc

Circumference over diameter is $\pi\gamma$

Apply $\gamma_{ij} = g_{ij} - g_{0i}g_{0j}/g_{00}$ to the rotating metric. Radially, $g_{0\rho} = 0$, so $\gamma_{\rho\rho} = 1$. Around the rim, $$\gamma_{\phi'\phi'} = \rho^2 - \frac{(\omega\rho^2/c)^2}{-(1-\omega^2\rho^2/c^2)} = \frac{\rho^2}{1 - \omega^2\rho^2/c^2} .$$

the Ehrenfest paradox, resolved as geometry

Radius measured by rotating rulers: $R = \rho$. Circumference measured by rotating rulers: $$C = \int_0^{2\pi}\frac{\rho\,\mathrm d\phi'}{\sqrt{1-\omega^2\rho^2/c^2}} = 2\pi\gamma\rho,\qquad \gamma = \big(1-\tfrac{\omega^2\rho^2}{c^2}\big)^{-1/2}.$$ So $C/2R = \pi\gamma \gt \pi$: the rotating observers' space is not Euclidean.

The bare $\gamma$ is unit 26's Lorentz factor at the rim speed $\omega\rho$; the spatial metric $\gamma_{ij}$ always carries indices. They meet here: $\gamma_{\phi'\phi'} = \gamma^2\rho^2$.

The lab's view of the riders' rulers (exaggerated: rim speed $0.6c$, $\gamma = 1.25$). Rulers along the rim move along their length and are contracted; rulers along a radius move sideways and are not. More rim rulers fit: $20$ where $16$ fitted at rest.

What to see in the figure. Both discs have the same radius in the lab. The riders lay rulers of one fixed rest length end to end. On the disc at rest, $16$ fit round the rim and $16/2\pi = 2.55$ along a radius. On the spinning disc each rim ruler moves along its own length at $0.6c$ and is shortened to $0.8$ of it, so $16/0.8 = 20$ fit; the radial rulers move across their length and keep it. The riders, counting rulers, find $C/R$ larger by exactly $\gamma$ — the radar result, seen from outside.

Reconciling this with unit 27. Unit 27 presented Ehrenfest's paradox as a contradiction: Born rigidity demands the rim's proper length be unchanged, so the lab circumference must be both $2\pi R$ and $2\pi R/\gamma$, and the Herglotz–Noether theorem says the disc cannot be spun up Born-rigidly at all. This slide is the other half. Take the disc as already rotating, ask what its own observers measure, and there is no paradox and no contradiction: the lab measures $2\pi R$, the riders measure $2\pi\gamma R$, and the two numbers differ because the riders' "circumference" is a sum of proper lengths along a curve that the lab does not regard as simultaneous. The contradiction in unit 27 was with rigidity, not with geometry.

Einstein's own use of it. In the 1912 Prague period and again in the 1916 review, Einstein argued: by the equivalence principle a rotating frame is as good as a frame with a (centrifugal) gravitational field; the rotating frame's geometry is non-Euclidean; therefore gravity requires non-Euclidean geometry. The argument is heuristic — the rotating frame is flat spacetime, as this slide insists — but the conclusion is right, and the disc is what sent him to Grossmann and to Riemann's geometry. Stachel's essay "Einstein and the rigidly rotating disk" (1980) is the careful history.

Why "resolved as geometry" and not "resolved". The word paradox survives in the literature because several inequivalent questions share the name: what happens when a disc is spun up (a materials question, answered by Herglotz–Noether and elasticity); what riders measure (this slide); and what an inertial observer sees (a Lorentz-contracted rim moving past, no paradox at all). Keeping the three apart is most of the work.

30 · disc

How big the excess is

how big

Rim speed $0.1c$: $\gamma = 1.005038$, so $C/2R = 1.005\pi$ — a half-per-cent excess.

Rim speed $0.18c$ (a $12$ km sphere at $716$ Hz): $\gamma = 1.01660$.

A laboratory ultracentrifuge rim at $1000\ \mathrm{m\,s^{-1}}$: $\gamma - 1 = 5.6\times10^{-12}$.

and the spacetime is still flat

Every event here is an event of Minkowski space. The non-Euclidean geometry belongs to a family of observers, not to the spacetime — the lesson of the tower as a Rindler wedge, in the spatial slot instead of the temporal one.

Checks. $\omega\to0$: $\gamma\to1$, $C = 2\pi R$ ✓. $\rho\to c/\omega$: $C\to\infty$ ✓, consistent with the light cylinder.

The excess $C/2\pi R - 1 = \gamma - 1$ against rim speed, both axes logarithmic. It grows as $\beta^2/2$ — a straight line of slope two — until near $c$, where it runs away. The three examples on the left are marked.

What to see in the figure. On logarithmic axes the small-speed law $\gamma - 1 \approx \beta^2/2$ is a straight line of slope $2$: a hundred times the rim speed buys ten thousand times the excess. The centrifuge sits nine and a half orders of magnitude below the pulsar. Only in the last few per cent of $c$ does the curve leave the line and climb without bound, which is the light cylinder seen from inside the disc.

The arithmetic. $\gamma(0.1) = (1-0.01)^{-1/2} = 1.0050378$; $\gamma(0.1801) = (1-0.032427)^{-1/2} = 1.0166046$. For a centrifuge rim at $10^3\ \mathrm{m\,s^{-1}}$, $\beta^2 = 1.113\times10^{-11}$ and $\gamma - 1 \approx \beta^2/2 = 5.56\times10^{-12}$. Rim speeds of about $1\ \mathrm{km\,s^{-1}}$ are what the best laboratory rotors reach before the material fails, so the effect is unmeasurable mechanically — and yet it is not zero, and the Mössbauer rotor experiments of the 1960s measured the associated time dilation directly.

section 5

What no coordinate change can remove

The tidal tensor of unit 05, met again as the first curvature invariant — and the end of Part D.

30 · tides

How much of a metric a freely falling frame can flatten

Fix an event $p$. Expand a coordinate change about it: $$y^\mu = A^\mu{}_\nu x^\nu + \tfrac12 B^\mu{}_{\nu\rho}x^\nu x^\rho + \tfrac16 C^\mu{}_{\nu\rho\sigma}x^\nu x^\rho x^\sigma + \dots$$ and count what each order of coefficients can buy, order by order, against the Taylor coefficients of $g_{\mu\nu}$ at $p$.

At each order, what the coordinate change can adjust (solid) against what it must adjust (hollow). First order has $6$ to spare, second order matches exactly, third order is $20$ short.

the count, read

The $6$ spare at first order are the Lorentz group: unit 26's $3$ boosts and $3$ rotations.

The exact match at second order is the freely falling frame: every first derivative of the metric can be set to zero at a point, which is the equivalence principle written in components.

The $20$ left at third order are unremovable. Unit 33 shows they are exactly the components of the Riemann tensor in four dimensions, $n^2(n^2-1)/12 = 20$.

Cited. That the linear systems at each order are actually solvable — not merely of the right size — is proved in unit 32, by constructing normal coordinates. The count is what is proved here.

What to see in the figure. Each pair of bars is one order of the Taylor expansion at $p$. The hollow bar is the number of independent Taylor coefficients of the metric at that order — $g$, $\partial g$, $\partial\partial g$ — and the solid bar is the number of free coefficients in the coordinate change one order up — $A$, $B$, $C$. Where solid exceeds hollow there is freedom left over, where it falls short something survives: $6$, $0$, $-20$.

Full derivation of the counts. $g_{\mu\nu}$ is symmetric, so it has $4\cdot5/2 = 10$ independent components at $p$. Its first derivatives $\partial_\alpha g_{\mu\nu}$ carry one free index times $10$: $40$. Its second derivatives $\partial_\alpha\partial_\beta g_{\mu\nu}$ are symmetric in $\alpha\beta$ by Clairaut (unit 03), so $10$ index pairs times $10$: $100$. On the other side, $A^\mu{}_\nu$ has $16$ components; $B^\mu{}_{\nu\rho}$ is symmetric in $\nu\rho$, so $4\times10 = 40$; $C^\mu{}_{\nu\rho\sigma}$ is totally symmetric in its three lower indices, so $4\times\binom{4+2}{3} = 4\times20 = 80$. Subtracting, $100 - 80 = 20$.

What the count does and does not prove. It proves that at most $80$ of the $100$ second derivatives can be removed, so at least $20$ combinations survive every coordinate change — that direction is rigorous, because a surjective linear map from an $80$-dimensional space cannot reach a $100$-dimensional one. It does not by itself prove that exactly $20$ survive, nor that the first derivatives can all be killed; those require showing the relevant linear maps have full rank, which unit 32 does by exhibiting the coordinates (Riemann normal coordinates, built from geodesics) explicitly. The count is nonetheless the right thing to see first, because it says in advance how many numbers a theory of gravity has to be about.

The count in other dimensions. Repeat it in $n$ dimensions: $g$ has $n(n+1)/2$ components, its second derivatives $n^2(n+1)^2/4$, and $C$ has $n\binom{n+2}{3}$ free coefficients. The difference is $n^2(n^2-1)/12$ — which is $0$ for $n=1$, $1$ for $n=2$, $6$ for $n=3$ and $20$ for $n=4$. A one-dimensional metric is always flattenable, which is why a curve has no intrinsic curvature; a surface has exactly one invariant, which is Gauss's Theorema Egregium; and four dimensions has twenty, of which unit 35's field equations constrain ten and the other ten are free to propagate as unit 38's gravitational waves.

The historical name. Riemann posed and answered exactly this question — when can a metric be brought to constant form? — in his 1854 habilitation lecture, sixty years before it had a use. Einstein reached it through the disc and the elevator, and needed Grossmann to tell him it had already been answered.

30 · tides

The Newtonian version, proved in three lines

what a freely falling frame removes, and what it does not

Let $\vec x_0(t)$ be any twice-differentiable curve, and change to the frame $\vec x' = \vec x - \vec x_0(t)$. In it a free particle obeys $$\ddot{\vec x}' = -\nabla\Phi(\vec x_0+\vec x') - \ddot{\vec x}_0 .$$ Choosing $\ddot{\vec x}_0 = -\nabla\Phi(\vec x_0)$ kills the field at $\vec x' = 0$, for all $t$.

but the second derivatives are untouched

$\partial'_i\partial'_j$ of the new potential $\Phi(\vec x_0(t)+\vec x') + \ddot{\vec x}_0\cdot\vec x'$ is $$\partial_i\partial_j\Phi + \partial'_i\partial'_j\big(\ddot{\vec x}_0\cdot\vec x'\big) = \partial_i\partial_j\Phi ,$$ because the added term is linear in $\vec x'$. $\square$

The Earth's potential along a vertical line (solid), and the same potential seen from the falling frame (dashed): a straight line has been added. It flattens the slope at the chosen point, and cannot touch the bend.

What to see in the figure. The solid curve is $\Phi = -GM/r$ along a radial line, drawn over a stretch comparable to the Earth's radius so that its bend is visible. The falling frame adds $\ddot{\vec x}_0\cdot\vec x'$, a straight line through the chosen point chosen to cancel the slope there: the dashed curve is flat at $x' = 0$, so a particle there feels no field. But adding a straight line never changes a second derivative, and away from the chosen point the dashed curve still bends. The slope is the field, and is frame-dependent; the bend is the tidal tensor, and is not.

The frame change, in full. Newton's law in the original inertial frame is $\ddot{\vec x} = -\nabla\Phi(\vec x)$. Substituting $\vec x = \vec x_0(t) + \vec x'$ gives $\ddot{\vec x}_0 + \ddot{\vec x}' = -\nabla\Phi(\vec x_0 + \vec x')$, hence the boxed equation. This is unit 03's fictitious-force calculation for a linearly accelerating frame, with the acceleration chosen to cancel the field along one chosen worldline — one worldline, not a region, which is the whole limitation. Expanding the right-hand side to first order in $\vec x'$ recovers $\ddot x'_i = -T_{ij}x'_j$, unit 05's equation.

Why this is the theorem the counting argument is the relativistic version of. The Newtonian statement is: the first derivatives of $\Phi$ at a point are pure gauge, the second derivatives are not. The relativistic statement is: the first derivatives of $g_{\mu\nu}$ at a point are pure gauge, and $20$ combinations of the second derivatives are not. They are the same statement, and unit 33 makes the correspondence exact by proving that the Newtonian limit of the geodesic deviation equation is unit 05's tidal deviation equation with $R_{i0j0} \to \partial_i\partial_j\Phi/c^2$.

30 · tides

What the falling observer still measures

A ring of free test particles released at rest around a falling observer, and the same particles a moment later (exaggerated). Along the line to the Earth's centre the ring stretches; across it, it squeezes — at half the rate, eigenvalues $(-2, 1, 1)$.

so the tidal tensor is the invariant

Unit 05's $T_{ij} = \partial_i\partial_j\Phi$ survives every change to a freely falling frame, and unit 05's deviation equation $\ddot\xi_i = -T_{ij}\xi_j$ is what the falling observer still measures.

Both slogans are true. "Gravity is not a force": $\nabla\Phi$ is frame-dependent and can always be made zero at a point. "Gravity is real": $T_{ij}$ cannot, and in vacuum $\operatorname{tr}T = 0$ forces a stretch for every squeeze.

Checks. $[T] = \mathrm s^{-2}$ ✓ — the square of a frequency. Uniform field $\Phi = gz$: $T = 0$, and indeed a uniform field is entirely removable, which is the elevator ✓.

What to see in the figure. Each dot obeys unit 05's $\ddot\xi_i = -T_{ij}\xi_j$ with $T = (GM/r^3)\,\mathrm{diag}(-2, 1, 1)$, the first axis pointing at the Earth's centre. Released from rest on a circle of radius $\xi_0$, a dot at angle $\psi$ moves to $\big(\xi_0\cos\psi\cosh(\sqrt{2k}\,t),\ \xi_0\sin\psi\cos(\sqrt k\,t)\big)$ with $k = GM/r^3$; the figure takes $\sqrt{2k}\,t = 0.7$, which on the Earth's surface is about seven minutes. The circle becomes an oval, longer along the radius. In three dimensions the squeeze acts on both transverse axes, and to leading order the volume is unchanged: that is $\operatorname{tr}T = 0$. No choice of frame makes the oval a circle again, because the frame change only adds a term linear in position, which moves every dot alike.

The one-sentence version of this unit. A gravitational field is not $\nabla\Phi$, which any observer can abolish by jumping; a gravitational field is $\partial_i\partial_j\Phi$, which nobody can. Every remaining unit of this course is about the object that plays the role of $\partial_i\partial_j\Phi$ when the metric, rather than a potential, is the field.

30 · tides · example

Worked example 4 — how local is "local"?

a cubic metre, for one second, on the Earth

1. $T_{ij} = (GM_\oplus/r^3)(\delta_{ij} - 3n_in_j)$ from unit 05, with $\hat n$ the unit vector from the Earth's centre: eigenvalue $-2GM_\oplus/r^3$ along $\hat n$, and $+GM_\oplus/r^3$ twice across it.

2. $GM_\oplus/R_\oplus^3 = 3.986004\times10^{14}/(6.378137\times10^6)^3 = 1.5362\times10^{-6}\ \mathrm{s^{-2}}$.

3. Two balls $1$ m apart vertically: relative acceleration $2\times1.5362\times10^{-6} = 3.07\times10^{-6}\ \mathrm{m\,s^{-2}}$, separating.

4. After $1$ s the extra separation is $\tfrac12(3.07\times10^{-6})(1)^2 = \mathbf{1.54\ \mu m}$.

5. In the same second both have fallen $4.90$ m. The tidal residue is $3.1\times10^{-7}$ of it.

Two pairs of balls dropped near the Earth (not to scale). Each ball falls towards the centre: the lower of the vertical pair is nearer, so falls faster; the side-by-side pair fall along converging radii.

Check. Horizontally the two balls converge at $1.54\times10^{-6}\ \mathrm{m\,s^{-2}}$ — half the vertical rate, opposite sign, exactly the eigenvalue ratio $(-2,1,1)$ ✓.

What to see in the figure. Every arrow points at the Earth's centre and has length $GM/r^2$ at its own ball. The vertical pair's arrows are parallel but unequal — the lower one longer by $2(GM/r^3)\times1$ m — so the pair separates. The horizontal pair's arrows are equal but not parallel: each is tilted towards the other by the angle $1\ \mathrm m/r$, so the pair closes at $GM/r^2\times(1\ \mathrm m/r) = GM/r^3$ per metre. Those are the two eigenvalues, read off the geometry.

The arithmetic. $R_\oplus^3 = (6.378137\times10^6)^3 = 2.5946\times10^{20}\ \mathrm m^3$; $GM_\oplus/R_\oplus^3 = 3.986004\times10^{14}/2.5946\times10^{20} = 1.53623\times10^{-6}\ \mathrm{s^{-2}}$. Vertical separation of $1$ m: relative acceleration $2\times1.53623\times10^{-6} = 3.07246\times10^{-6}\ \mathrm{m\,s^{-2}}$, so after $1$ s the extra separation is $1.5362\times10^{-6}$ m. Free fall in $1$ s is $g/2 = 4.9033$ m, and $1.5362\times10^{-6}/4.9033 = 3.133\times10^{-7}$. For a $10^{-18}$ m resolution, $t = \sqrt{2\times10^{-18}/3.07246\times10^{-6}} = 8.07\times10^{-7}$ s. And over the tower, $2(GM_\oplus/R_\oplus^3)h = 2\times1.53623\times10^{-6}\times22.5 = 6.9130\times10^{-5}\ \mathrm{m\,s^{-2}}$; divided by $g$ that is $7.049\times10^{-6}$, which is worked example 3's $2h/R_\oplus = 7.06\times10^{-6}$ again, as it has to be — the two differ only because $g \ne GM_\oplus/R_\oplus^2$.

Where the sign comes from. Unit 05's eigenvalue $-2GM/r^3$ lies along $\hat r$, so $\ddot\xi_r = +2(GM/r^3)\xi_r$ — vertical separations grow — while the two transverse eigenvalues $+GM/r^3$ give $\ddot\xi_\perp = -(GM/r^3)\xi_\perp$ — horizontal separations shrink. Physically: the lower ball is closer to the centre and falls faster; the two side-by-side balls fall along converging radii. The trace is zero, as it must be in vacuum.

30 · tides · example

Local means: below your resolution

what "local" means, quantitatively

The equivalence principle holds in a box of size $L$ for a time $\tau$ to whatever accuracy $(GM/r^3)L\tau^2$ — the vertical pair's extra separation — is below your instrument's resolution. It is a statement with an error bar, not an exact symmetry.

Two balls $1$ m apart, separations resolved to $10^{-18}$ m: the elevator is unmasked in $0.8\ \mu$s.

and this is why it is a principle about points

No box is small enough to make $T_{ij}$ vanish, only small enough to hide it: shrinking $L$ shrinks the effect linearly, never to zero. Over the tower's $22.5$ m it is $6.91\times10^{-5}\ \mathrm{m\,s^{-2}}$, or $7.05\times10^{-6}$ of $g$ — what no Rindler wedge has.

The extra separation of two balls dropped $1$ m apart vertically, against time, both axes logarithmic. Each horizontal line is a resolution; where the curve crosses it, that instrument unmasks the elevator.

What to see in the figure. The tidal separation is $\delta = (GM_\oplus/R_\oplus^3)Lt^2$, a straight line of slope $2$ on these axes, so every factor of ten in time is a factor of a hundred in signal. A resolution of $10^{-18}$ m catches it at $0.81\ \mu$s; an interatomic $10^{-10}$ m, at $8$ ms; a micron, at $0.81$ s. There is no time short enough for the effect to be zero — only short enough for it to be smaller than what you can see.

The $10^{-18}$ m is not a rhetorical number. LIGO measures a change in a $4$ km arm of about $10^{-18}$ m, so differential-displacement sensitivity at that level exists. The comparison should not be pushed too far — LIGO measures at hundreds of hertz, not over microseconds, and free-falls its mirrors only along one axis — but the point stands: the equivalence principle's "locality" is a statement about the ratio of a tidal effect to an experimental resolution, and modern resolutions are extraordinary.

The same calculation in orbit. This is why a "zero-gravity" laboratory on the International Space Station is advertised as a microgravity laboratory: at $r = 6.79\times10^6$ m, $GM/r^3 = 1.27\times10^{-6}\ \mathrm{s^{-2}}$, so an experiment $10$ m from the centre of mass feels a residual $2.5\times10^{-5}\ \mathrm{m\,s^{-2}}$, about $10^{-6}g$. Tides set the floor, and no engineering removes them.

30 · tides

There is no global inertial frame

the claim that ends Part D

Suppose gravity were a force field on Minkowski space. Then the inertial frames of unit 26 would exist globally, and in one of them free particles with no force on them move on straight lines.

But the WEP says gravity acts on every body identically, so "no force on them" cannot be arranged, and the frames that make free fall look force-free are the freely falling ones — which differ from place to place, by exactly $T_{ij}\ne0$.

Newtonian version, complete

Suppose a single change $\vec x' = \vec x - \vec x_0(t)$ made every free-fall worldline straight in a region. Then $\nabla\Phi(\vec x) = -\ddot{\vec x}_0(t)$ throughout it, a function of $t$ alone, so $\partial_i\partial_j\Phi \equiv 0$ there. Contrapositive: where $T\ne0$, no such frame exists. $\square$

Small freely falling boxes around the Earth, each with its own acceleration (arrows, to scale). Each box is an inertial frame; one frame for the whole ring would need every arrow equal, and no two of them are.

What to see in the figure. Each box falls with the field at its own centre, so inside it the field vanishes to first order. To merge two boxes into one frame, the frame would have to accelerate with both arrows at once. The boxes on opposite sides of the Earth accelerate in opposite directions; boxes at different heights by different amounts. That a single $\ddot{\vec x}_0(t)$ cannot match a field that changes from place to place is the proof on the slide, and the arrows are $-\nabla\Phi$ evaluated at a few points.

Reading the proof carefully. The Newtonian argument is complete and short, and the relativistic argument is its shadow. The premise "a single frame makes every free-fall worldline straight" is exactly what a global inertial frame would provide; the conclusion $\partial_i\partial_j\Phi \equiv 0$ is exactly the statement that there are no tides. Since there are tides, there is no such frame. What makes it feel like a large conclusion is that the same reasoning applies to any proposal of the form "gravity is a force on a fixed background": the background's straight lines are supposed to be observable, and the equivalence principle is the statement that they are not.

30 · tides

What survives, and what does not

what survives, and what does not

Survives: inertial frames at a point, to first order — the counting argument's first two orders. That is all the equivalence principle ever claimed.

Does not survive: one flat metric covering a region containing matter. So there is no arena on which to write "gravity is a field", because the arena was supposed to be the thing the field lives on.

what this does not yet prove

It does not prove that spacetime is curved — "curvature" has no definition until unit 33. It proves that flat spacetime with a global inertial frame is inconsistent with $T_{ij}\ne0$, and so that the next unit must build a spacetime without one. That is unit 31.

Read the two boxes as one sentence. Inertial frames survive at a point and to first order, which is all the elevator ever used; a single flat metric over a region does not, wherever there are tides. The first half is why special relativity remains true in every small laboratory; the second is why it is not the theory of the solar system.

The honest gap. One could keep a flat background and simply declare it unobservable, tracking free fall by a separate structure. Newton–Cartan theory does exactly that for Newtonian gravity: it keeps flat space and absolute time but replaces the force with a connection whose curvature is the tidal tensor. It is a real and consistent theory, and it shows that the step being taken here is forced by the equivalence principle plus relativity, not by the equivalence principle alone. The next slide closes the remaining escape route.

Why "Part D ends here". Part D's thesis was that special relativity is the geometry of a quadratic form on a vector space, and that everything from dynamics to electromagnetism to field theory is that geometry worked out. This slide says the vector space has to go: the affine structure of unit 26, which let any two events be subtracted and any frame be extended globally, is what tides forbid. Part E replaces it with a manifold, on which subtraction is local only, and the metric is a field.

30 · tides

Why not just put a field on flat spacetime anyway?

The natural objection: electromagnetism is a field on Minkowski space, so write gravity as one — a symmetric tensor field $h_{\mu\nu}$ sourced by mass-energy. Here is why the programme eats itself.

the Newtonian seed, complete

The field has energy. From unit 05's $\nabla^2\Phi = 4\pi G\rho$ and the divergence theorem, $$U = \tfrac12\!\int\!\rho\Phi\,\mathrm dV = -\frac1{8\pi G}\!\int\!|\nabla\Phi|^2\,\mathrm dV ,$$ an energy density $-|\nabla\Phi|^2/8\pi G$. By $E = mc^2$ that energy has mass; by the equivalence principle in its strong form — the only one that speaks about gravitational self-energy at all, as P7 makes precise — it gravitates. So $$\nabla^2\Phi = 4\pi G\Big(\rho - \frac{|\nabla\Phi|^2}{8\pi Gc^2}\Big) ,$$ and the corrected $\Phi$ changes the correction, for ever. A linear theory of gravity is not self-consistent.

Deser's argument — cited

The same for a massless spin-2 field $h_{\mu\nu}$ on Minkowski, coupled to the total stress-energy including its own: the iteration closes, and its sum is the Einstein–Hilbert action of unit 35 for $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} .$$

S. Deser, Gen. Rel. Grav. 1, 9 (1970). Needs units 29 and 35; P7 does the Newtonian half in full.

the punchline

In the resulting theory nothing measures $\eta$: every rod, clock and ray responds to $g$. You have not described gravity on Minkowski space — you have rediscovered general relativity under a misleading name.

Full derivation of the field energy. The potential energy of a mass distribution assembled from infinity is $U = \tfrac12\int\rho\Phi\,\mathrm dV$ with the sign convention $\nabla^2\Phi = 4\pi G\rho$ and $\Phi\to0$ at infinity; it is negative because $\Phi$ is. Substituting $\rho = \nabla^2\Phi/4\pi G$ and integrating by parts with the divergence theorem (unit 05), $\int\Phi\nabla^2\Phi\,\mathrm dV = \oint\Phi\nabla\Phi\cdot\mathrm d\vec S - \int|\nabla\Phi|^2\mathrm dV$, and the surface term vanishes because $\Phi\sim1/r$ and $|\nabla\Phi|\sim1/r^2$ while the area grows as $r^2$. Hence $U = -\frac1{8\pi G}\int|\nabla\Phi|^2\mathrm dV$, which is the only consistent way to read the field as the carrier of the energy. For a uniform sphere it evaluates to $-3GM^2/5R$, and the fractional self-energy $|U|/Mc^2 = 3GM/5Rc^2$ is $4.17\times10^{-10}$ for the Earth and $1.27\times10^{-6}$ for the Sun.

Why the sign of the correction matters. The correction term is negative: gravitational field energy is negative, so it reduces the effective source. That is the Newtonian ancestor of the statement that gravitational binding energy reduces a star's gravitational mass, which is what the Nordtvedt effect tests.

What "closes in one step" means, honestly. Deser's construction adds to the linear action a term coupling $h_{\mu\nu}$ to the stress-energy of the linear field, and then shows that with a judicious choice of variables (the densitised inverse metric and an independent connection, in the first-order Palatini form) the corrected action is already fully consistent: no further iteration is needed. Other treatments, iterating in $h$ with the metric held second-class, find an infinite series which sums to the same thing. Both routes end at the Einstein–Hilbert action. This course proves neither: it lacks unit 29's stress-energy tensor and unit 35's variational derivation, which is why the result is marked cited rather than sketched.

The philosophical residue. The spin-2 route is not a failure; it is a second derivation of general relativity, and it is the one that makes the graviton's spin and masslessness transparent. What it does not do is save flat spacetime. Feynman's Lectures on Gravitation (1962–63, published 1995) sets the whole programme out, starting from a quantum field theorist's prejudices and ending, lecture by lecture, in Einstein's theory.

30 · closing

The running example, closed

the photon in the tower, three times

Section 2, from the principle alone: a rocket and a Doppler shift give $\Delta\nu/\nu = -gh/c^2 = -2.455\times10^{-15}$.

Section 3, from a metric on flat spacetime: $\nu_1/\nu_2 = X_2/X_1$, exactly $1 + gh/c^2$ — the same number, with the error term now known to be $6\times10^{-30}$.

Unit 36, from the Schwarzschild solution: $\sqrt{(1-r_s/r_1)/(1-r_s/r_2)}$, whose expansion is the same again.

Measured: $(2.57\pm0.26)\times10^{-15}$. And $38.5\ \mu$s/day keeps a satellite fleet honest.

and what it did not prove

The redshift is a clock effect, and flat spacetime can produce it — the Rindler wedge does, with a horizon thrown in. The redshift is therefore evidence for the equivalence principle and not evidence for curvature.

What is evidence for curvature is $T_{ij}\ne0$: the $20$ numbers the counting argument could not remove.

What comes next.
31 builds a spacetime with no global chart: manifolds, tangent spaces, tensor fields.
32 gives it a metric and a connection, and free fall becomes a geodesic.
33 defines curvature and proves the $20$ are the Riemann tensor.
37 finishes the light-bending factor of two, and gets $43''$ per century.

The one idea to keep. Separate the two questions that "is there gravity here?" runs together. Do I feel weight, do clocks disagree, is there a horizon? — all of these are questions about a family of observers, and all of them have affirmative answers in flat spacetime. Can a single coordinate system make the metric constant over a region? — that is a question about the spacetime, it is answered by second derivatives of the metric, and near matter the answer is no. Every confusion in beginning general relativity is a confusion of the first sort of question for the second.

Reading. Rindler, Relativity: Special, General and Cosmological (2nd ed., 2006), chapter 1 for the equivalence principles and §3.7–3.8 for the wedge — all readable now. Misner, Thorne and Wheeler, Gravitation, §1.3 and chapter 6, for the elevator and the accelerated observer; chapter 6 uses the connection, so read it after unit 32. Will, Theory and Experiment in Gravitational Physics (2nd ed., 2018), chapter 2, is the authoritative account of WEP, EEP and SEP and their experimental status, and needs nothing from later units. For the history, Stachel's "Einstein and the rigidly rotating disk" and Pais, Subtle is the Lord, chapters 9 and 11–12.

An honest warning about the next four units. Units 31 to 34 are mathematics, and the physics goes quiet. This unit is what they are for: everything built there is built to give $T_{ij}$ a home. It is worth writing the last two slides' argument on a card and keeping it visible while doing them.

30 · checkpoint

Checkpoint

you can now

Derive the gravitational redshift from the equivalence principle with no geometry and match Pound–Rebka within its error bar; state the weak, Einstein and strong equivalence principles with an experiment that distinguishes them; and exhibit a non-constant metric on flat spacetime — thereby proving that non-constant metric components alone are not gravity.

it is false if

your redshift derivation quotes $mgh$ for a photon; or you cannot name an experiment that bounds SEP but not WEP, or WEP but not EEP; or, handed $\mathrm ds^2 = -(gX/c)^2\mathrm dT^2 + \mathrm dX^2$, you call it a gravitational field.

to check it

Close the deck. On paper: the rocket argument to $2.455\times10^{-15}$, compared with $(2.57\pm0.26)\times10^{-15}$; the three principles with Eöt-Wash, Gravity Probe A and lunar laser ranging against them; and the Rindler metric derived from $ct = X\sinh(gT/c)$, $x = X\cosh(gT/c)$. Then ./mr problems 30.