Einstein Synchronization of Clocks

Two clocks sit a kilometre apart, both perfect. Do they read the same time? The question sounds trivial and is not, because there is no way to look at both at once: every comparison involves sending something, and everything you can send takes time to arrive. Newton did not have this problem — absolute time supplied a universal "now", and Absolute Space and Absolute Time shows what he built on it. Once you insist instead that light travels at $c$ for everybody, the universal now is gone (The Relativity of Simultaneity, 1859) and "the time over there" must be manufactured. Einstein manufactured it in §1 of the 1905 paper, with a rule so simple it looks like a triviality. Fire a light pulse from clock $A$ at $A$ -time $t_1$ , bounce it off $B$ , catch it back at $t_2$ — then declare that the reflection happened at $t_B = \tfrac{1}{2}(t_1 + t_2)$ . Nothing forces the even split, since no single-clock experiment can ever see the two legs separately (a point Poincaré had made seven years earlier), so Einstein called it what it is: a definition. But a definition still has to work. The rule is transitive: set $B$ from $A$ , $C$ from $B$ , and $C$ from $A$ directly, and the answers agree — a theorem, not an assumption, and what entitles you to speak of "the time in this frame" at all. Slow clock transport agrees with it too, by a route that sends no light. And it fails, measurably, the moment the frame stops being inertial: carry the rule once around the equator and it misses closure by $207$ nanoseconds — $62$ metres of GPS range error, corrected for in every receiver on the planet.

Key takeaways

Distant simultaneity is not discovered, it is defined. Einstein's rule: bounce light off the far clock and stamp the reflection at the midpoint of the round trip, $t_B = \tfrac12(t_1+t_2)$ — timed on one clock, so no prior synchronization is smuggled in.
Transitivity is a theorem. Run a pulse round a closed circuit $A \to B \to C \to A$ : $A$ 's own clock says it took (path length) $/c$ , and the chain of Einstein stamps sums to exactly the same number. The clocks close — which is what licenses the phrase "the time in this frame".
Slow clock transport agrees. Carry a clock from $A$ to $B$ at speed $u$ and it lags the light-synchronized value by $(L/u)\big(1-\sqrt{1-u^2/c^2}\big) = Lu/(2c^2) + \cdots$ , which vanishes as $u \to 0$ . A kilometre at walking pace costs $\mathbf{5.6\ fs}$ . Two different procedures, one answer.
And it fails on a rotating platform. Transitivity was a theorem in an inertial frame. Carry the rule around a closed loop on the spinning Earth and it does not close: the mismatch $\Delta t = 2\omega A/c^2$ is $\mathbf{207\ ns} = \mathbf{62\ m}$ of range around the equator.
Key idea: Einstein synchronization is a convention that behaves like a fact — consistent, transitive, and independently reproduced by slow clock transport — for exactly as long as the frame is inertial. The rotating Earth is the boundary, and the 207 ns by which the equator fails to close is that boundary made visible.

The core insight

"The same time over there" is not measured; it is stipulated, because only the round trip is observable on a single clock. Einstein's stipulation — split the round trip evenly — is transitive within one inertial frame, and slow clock transport independently reproduces it. Neither guarantee survives rotation: on the spinning Earth the rule fails to close a loop by $2\omega A/c^2$ .

Where this lives today

The rule is not a museum piece; it runs in every timing system spanning more than a few metres. National metrology laboratories compare their caesium and optical clocks by two-way satellite time transfer — Einstein's rule with a geostationary mirror, good to tens of picoseconds across an ocean. Where a one-way link is unavoidable, as in every GPS pseudo-range, the Sagnac term of equation (3) is applied explicitly. The failure mode is the instructive part. GPS does not try to be clever about synchronization on the rotating Earth; it refuses to work there. Every clock in the system, on the ground and in orbit, is referred to a coordinate time defined in a non-rotating Earth-centred frame, and every measurement is transformed into that frame before use — the convention preserved by moving to a frame in which it is consistent.

The physical picture

You want to set a clock on a mountaintop by a clock in the valley. You cannot see both at once, and whatever you send arrives late. So you send the fastest thing there is — a light pulse, up and back. Your own clock, one clock, no controversy, says the excursion took $t_2 - t_1$ . Now the only question that matters: when, during that interval, did the pulse touch the mountaintop? Nature will not tell you. Any experiment that could catch the pulse arriving up there needs a clock up there, already set — which is the thing you are trying to do. The two legs can never be timed separately; all you ever get is their sum. So you decide. Einstein's decision was half and half. It has no more empirical warrant than any other split — a point The Conventionality of Simultaneity (1861) takes apart — and yet it is not arbitrary the way a coin flip is. It is the only split that makes light isotropic; it composes consistently around closed loops; and a clock walked slowly up the mountain agrees with it. Three good reasons to like a convention — and not one of them makes it a fact.

Figure

Einstein's rule, drawn twice. Left: a pulse leaves A at t₁, reflects at B, returns at t₂, and the reflection is STAMPED at the midpoint ½(t₁+t₂) — only A's single clock is ever read. Right: the same rule as a flash at the midpoint M, both clocks starting when the light arrives after equal distances at c.

The two pictures are the same rule: in one you stamp a time onto the far event, in the other you start the far clock. Either way the content is identical — a one-way trip from $X$ to $Y$ is declared to cost $d_{XY}/c$ of coordinate time — and the whole article is about what happens when you demand that of every pair of clocks at once.

Figure

PUNCHLINE — the rule is consistent exactly as long as the frame is inertial. Left: chain it round a 300–400–500 m triangle at rest and the stamps sum to 4.003 μs, exactly what A's own clock reads for the closed-path pulse. GAP = 0. Right: chain it eastward round the equator and it misses closure by 2ωA/c² = 207 ns — 62 m of range.

The left panel is a theorem, the right its boundary — and they are the same computation run twice. Both ask: does the closed-path light time equal the sum of the one-way stamps? In an inertial frame it does, identically. On a spinning platform an eastward pulse must chase a receding finish line, so the real loop time is longer than the stamps admit, and the shortfall is $207$ nanoseconds.

The core equations

The rule is a stipulation, so it needs no derivation — but it does need an input, and the input is the one thing a single clock can honestly measure. Clock $A$ emits at $t_1$ , the pulse bounces off $B$ and returns at $A$ 's reading $t_2$ . Only $A$ is read, so no synchronization is presupposed and $t_2 - t_1 = 2L/c$ is a genuine measurement. The stipulation cuts that interval in half.

t_B \equiv \tfrac{1}{2}\left(t_1 + t_2\right), \qquad t_2 - t_1 = \frac{2L}{c} \qquad (1)

where: $t_1$ , $t_2$ — $A$ 's readings at emission and return; $t_B$ — the reading assigned to $B$ at the reflection; $L$ — the rest-frame distance $A$ to $B$ ; $c$ — the two-way speed of light (The Constancy of the Speed of Light). Physical reading: the second equation is data; the first is a decision. Combine them and $B$ is stamped at $t_1 + L/c$ . That combination is worth isolating, because every later argument uses it and nothing else. Write $\sigma_{XY}$ for the coordinate time the rule assigns to a one-way trip from $X$ to $Y$ .

\sigma_{XY} \equiv (\text{Y's stamp on arrival}) - (\text{X's reading at departure}) = \frac{d_{XY}}{c} \qquad (2)

where: $\sigma_{XY}$ — the coordinate time the rule assigns to a one-way trip $X \to Y$ ; $d_{XY}$ — the ruler distance. Physical reading: Einstein synchronization is the demand that every one-way trip cost $d/c$ . Whether that demand can be met for all pairs at once is a genuine question with a genuine answer — and the answer differs between an inertial frame and a merry-go-round. Now the boundary. On a platform rotating at $\omega$ the clocks are not inertial: a point at radius $r$ sweeps past the underlying inertial frame at $\mathbf{v} = \boldsymbol{\omega}\times\mathbf{r}$ . Chain the rule round a closed loop and each step under-counts the true transit by $v\,\mathrm{d}l/c^2$ ; summing, and applying Stokes' theorem, gives the non-closure.

\Delta t = \frac{1}{c^{2}}\oint \mathbf{v}\cdot \mathrm{d}\boldsymbol{\ell} = \frac{2\omega A}{c^{2}} \qquad (3)

where: $A$ — the area the loop encloses, projected perpendicular to $\boldsymbol{\omega}$ ; $\omega$ — the platform's angular speed. Physical reading: the failure depends on enclosed area, not path length — an out-and-back path (zero area) closes perfectly even on the spinning Earth, while the equator does not. This is the synchronization failure; the interferometer built to see it is Sagnac Effect and Rotating-Frame Synchronization (S7). Finally, the independent check. Set $B$ not by light but by carrying a clock there, slowly. Granting the rate factor $\sqrt{1-u^2/c^2}$ that The Light Clock and Time Dilation (S1) derives, a clock walked from $A$ to $B$ at speed $u$ takes coordinate time $L/u$ and accumulates only $(L/u)\sqrt{1-u^2/c^2}$ of its own.

\Delta_{\text{transport}} = \frac{L}{u}\left(1 - \sqrt{1 - \frac{u^{2}}{c^{2}}}\right) = \frac{Lu}{2c^{2}} + \frac{Lu^{3}}{8c^{4}} + \cdots \qquad (4)

where: $\Delta_{\text{transport}}$ — how far the transported clock lags the Einstein-synchronized value on arrival; $u$ — the transport speed. Physical reading: the deficit is proportional to $u$ , so it vanishes in the limit of infinitely slow transport. Walk, do not run, and you reproduce Einstein's convention exactly — by a route that never sends a light signal.

Where the results come from

Derivation ATransitivity in an inertial frame, and its failure on a rotating one

The rule tells you how to set clock $B$ from clock $A$ . It says nothing about whether the answers agree when you set $B$ from $A$ , $C$ from $B$ , and $C$ from $A$ independently. If they disagreed, "the time in this frame" would be meaningless and Frames, Observers, and Networks of Clocks would have nothing to stand on. That they agree is a theorem — and the lines that prove it show exactly where it dies.

Step 1. Restate the rule as a claim about one-way transit: since $A$ measures $t_2 - t_1 = 2d_{AB}/c$ on its own clock, the midpoint stamp sets $B$ 's reading to $t_1 + d_{AB}/c$ , so the rule asserts $\sigma_{XY} = d_{XY}/c$ for every ordered pair. Why: consistency now means — can all these one-way assignments hold at once?

Step 2. Test the circuit $A \to B \to C \to A$ with a single pulse and mirrors at $B$ and $C$ . Clock $A$ times the whole excursion by itself, no convention involved, and gets $t_{\text{loop}} = (d_{AB} + d_{BC} + d_{CA})/c$ . Why: in an inertial frame the closed-path speed of light is $c$ , isotropically — the empirical content of the postulate, and what Michelson and Morley failed to disturb.

Step 3. Sum the rule's stamps around the same circuit: $\sigma_{AB} + \sigma_{BC} + \sigma_{CA} = (d_{AB} + d_{BC} + d_{CA})/c$ — identical to the measured $t_{\text{loop}}$ . The pulse's return to $A$ is stamped, by the chain of synchronized clocks, at exactly the reading $A$ 's own clock shows: the offsets composed around the loop sum to zero. Why: transitivity is not an extra postulate — it is the statement that the closed-path light time equals the perimeter over $c$ , which in an inertial frame is true.

Step 4. Now spin the platform and watch Step 2 die. Take a circular loop of radius $R$ rotating at $\omega$ and work in the underlying inertial frame. An eastward pulse must cover the circumference plus the distance the finish line runs away: $c\,t_+ = 2\pi R + \omega R\,t_+$ , so $t_+ = 2\pi R/(c - \omega R)$ — strictly longer than $2\pi R/c$ . But the rule still stamps the loop at $2\pi R/c$ . Why: the platform's clocks are not inertial. Each slides sideways through the frame in which light is isotropic, so "half the round trip" is no longer where the pulse actually was.

Step 5. Compute the shortfall. $\Delta t = \dfrac{2\pi R}{c - \omega R} - \dfrac{2\pi R}{c} \approx \dfrac{2\pi R \cdot \omega R}{c^{2}} = \dfrac{2\omega\,(\pi R^{2})}{c^{2}} = \dfrac{2\omega A}{c^{2}}$ — equation (3); for a general loop the same sum is $c^{-2}\oint \mathbf{v}\cdot\mathrm{d}\boldsymbol{\ell} = 2\omega A$ by Stokes. Around the equator ( $\omega_\oplus = 7.292\times10^{-5}\ \mathrm{rad/s}$ , $A = 1.278\times10^{14}\ \mathrm{m^2}$ ) this is $\mathbf{207\ ns}$ . Why: the answer depends on enclosed area, so a there-and-back path closes exactly even here. Ashby states the same result: a clock Einstein-synchronized eastward all the way round the equator ends up lagging the one it started from by $207.4\ \mathrm{ns}$ . *Verified in verify_aq-sr-einstein-synchronization.py: the loop identity $\sum \sigma = t_{\text{loop}}$ proved symbolically for a general triangle; the 300–400–500 m circuit gives $4.003\ \mathrm{\mu s}$ both ways, gap $0$ ; the exact $2\pi R/(c-\omega R) - 2\pi R/c$ agrees with $2\omega A/c^2$ to six figures; and $2\omega_\oplus \pi R_\oplus^2/c^2 = 207.4\ \mathrm{ns}$ , $\times c = 62.2\ \mathrm{m}$ .*

Go deeper · Derivation B · 2 minSlow clock transport reproduces the same synchronization

Derivation BSlow clock transport reproduces the same synchronization

Einstein's rule uses light — a loaded choice, since light is the very thing whose one-way speed is in question, so a sceptic may fairly call the convention rigged. The alternative is brutally simple: set two clocks side by side, then carry one to $B$ . No signals, no isotropy assumption, no light. The remarkable fact, noticed by Eddington in 1924, is that this completely different procedure gives the same answer — provided you walk slowly enough, and you can say exactly how slowly.

Step 1. Grant the rate factor. A clock moving at speed $u$ ticks at $\sqrt{1-u^2/c^2}$ times the coordinate rate — a result this article does not derive and does not own; it belongs to The Light Clock and Time Dilation (S1). Why: we are testing the transport procedure, not re-deriving dilation. Borrowing the factor by name is legitimate; borrowing the derivation would not be.

Step 2. Do the bookkeeping. The journey of length $L$ at speed $u$ takes coordinate time $L/u$ , during which the travelling clock accumulates $(L/u)\sqrt{1-u^2/c^2}$ of its own. On arrival it reads less than the Einstein-synchronized clock at $B$ by $\Delta = (L/u)\big(1 - \sqrt{1-u^2/c^2}\big)$ . Why: the transported clock moved, so it is the one that fell behind. The question is by how much — and whether the deficit can be made to vanish.

Step 3. Expand. With $x = u^2/c^2$ , $1 - \sqrt{1-x} = x/2 + x^2/8 + \cdots$ , so $\Delta = \dfrac{Lu}{2c^{2}} + \dfrac{Lu^{3}}{8c^{4}} + \cdots$ : the leading term is linear in $u$ and goes to zero with it. Slow transport and Einstein synchronization coincide exactly as $u \to 0$ . Why: the $1/u$ from the longer journey and the $u^2$ from the rate factor fight, and the $u^2$ wins. Go slowly enough and the disagreement is as small as you please, at no cost but patience.

Step 4. Put numbers on the patience. Carrying a clock $L = 1\ \mathrm{km}$ at $u = 1\ \mathrm{m/s}$ — an unhurried walk, a journey of $1000\ \mathrm{s}$ — costs $\Delta = Lu/(2c^2) = \mathbf{5.6\ fs}$ . Slow to $1\ \mathrm{cm/s}$ and the trip takes $28$ hours but the deficit falls to $5.6\times10^{-17}\ \mathrm{s}$ . Why: $\Delta \propto u$ , so a hundredfold reduction in speed buys a hundredfold reduction in error at a hundredfold cost in time. No obstruction of principle — only of temper.

Step 5. Beware the arithmetic. Evaluating $\Delta = (L/u)\big(1-\sqrt{1-u^2/c^2}\big)$ in double precision returns exactly zero: at $u = 1\ \mathrm{m/s}$ , $u^2/c^2 \approx 1.1\times10^{-17}$ falls below the machine epsilon of $1$ , so the square root rounds to $1.0$ and the bracket annihilates. Why: catastrophic cancellation — a $5.6\ \mathrm{fs}$ effect silently reported as nothing. Use the series $Lu/(2c^2)$ , or arbitrary precision. *Verified in verify_aq-sr-einstein-synchronization.py: the series $1-\sqrt{1-x} = x/2 + x^2/8 + O(x^3)$ by sympy; $\Delta \to 0$ as $u \to 0$ symbolically; the exact expression evaluated at 50-digit precision agrees with $Lu/(2c^2)$ to a relative $3\times10^{-18}$ at $u = 1\ \mathrm{m/s}$ , giving $5.563\ \mathrm{fs}$ and $5.563\times10^{-17}\ \mathrm{s}$ ; and the float64 expression is confirmed to return exactly $0.0$ .*

Limiting cases

Case	Result	Meaning
Galilean limit, $c \to \infty$	$t_2 - t_1 \to 0$ , $t_B \to t_1$	Instantaneous signalling makes synchronization trivial and simultaneity absolute — Newton's universal now (Absolute Space and Absolute Time). The convention problem exists because $c$ is finite.
Non-rotating frame, $\omega \to 0$	$\Delta t = 2\omega A/c^2 \to 0$	The loop closes; transitivity is restored. The theorem holds in exactly the frames it claims to.
Zero enclosed area, $A \to 0$	$\Delta t \to 0$	An out-and-back path closes exactly, even on the spinning Earth: it is area, not distance, that breaks the rule.
Slow transport, $u \to 0$	$\Delta = Lu/(2c^2) \to 0$	A completely independent procedure reproduces $\varepsilon = \tfrac12$ . Convention or not, it is a well-supported one.
Non-standard split, $\varepsilon \neq \tfrac12$	one-way speeds change; two-way stays $c$	No experiment can tell the difference — the subject of The Conventionality of Simultaneity (1861).

Common trapNobody has ever measured the one-way speed of light — and no experiment using synchronized clocks can, because the synchronization already assumed the answer

This is the trap the whole subject is built around. To time a one-way trip you need clocks at both ends that already agree; to make them agree you must decide how long light takes to go one way; and that is the very number you set out to measure. The circle cannot be broken by better clocks or longer baselines. What is measured — to parts in $10^{18}$ — is the two-way speed: emit and return, one clock, no convention. Every quoted "measurement of $c$ " is two-way, and since 1983 not even that, because the metre is defined from $c$ . The isotropy of the one-way speed is a stipulation nothing contradicts, not a discovery — and slow clock transport, the strongest thing you can say in its favour, is not a proof either.

Worked examples

1 · Why GPS cannot Einstein-synchronize around the Earth

Chain Einstein's rule eastward around the equator, clock to neighbouring clock, all the way round. By how much does the last clock fail to agree with the first — and what does that cost a navigation system?

Step 1. Get the geometry. $R_\oplus = 6.378\times10^{6}\ \mathrm{m}$ , so the enclosed area is $A = \pi R_\oplus^{2} = 1.278\times10^{14}\ \mathrm{m^{2}}$ . The ground itself is moving: $\omega_\oplus R_\oplus = 465\ \mathrm{m/s}$ , a $\beta$ of $1.55\times10^{-6}$ .

Step 2. Apply equation (3): $\Delta t = \dfrac{2\omega_\oplus A}{c^{2}} = \dfrac{2(7.292\times10^{-5})(1.278\times10^{14})}{(2.998\times10^{8})^{2}} = \mathbf{207\ ns}$ . (With the precise WGS-84 constants this is $207.4\ \mathrm{ns}$ , the figure Ashby quotes: a clock synchronized eastward all the way round the equator ends up lagging the one it started from by that amount.)

Step 3. Convert to what a navigator cares about. A timing error $\Delta t$ becomes a range error $c\,\Delta t = (2.998\times10^{8})(2.07\times10^{-7}) = \mathbf{62\ m}$ . That is not a subtlety; that is the difference between the right street and the wrong one.

Step 4. See what the fix must be. No clever synchronization repairs this: the failure is a theorem about rotating frames. So GPS abandons the attempt, declares one non-rotating Earth-centred coordinate time to be the truth, and applies an explicit Sagnac correction to every one-way transfer. The convention is kept; the rotating frame is given up. (The timing budget is GPS and Relativistic Timing Corrections (S7).)

2 · Walking a clock across the lab

Instead of light, carry a clock. How slowly must you walk before the transported clock agrees with the Einstein-synchronized one — and is the residual anywhere near measurable?

Step 1. Take $L = 1\ \mathrm{km}$ and a walking pace $u = 1\ \mathrm{m/s}$ ; the journey lasts $1000\ \mathrm{s}$ . Equation (4) gives $\Delta = \dfrac{Lu}{2c^{2}} = \dfrac{(1000)(1)}{2(8.988\times10^{16})} = \mathbf{5.6\ fs}$ — the transported clock arrives $5.6$ femtoseconds behind its light-synchronized twin.

Step 2. Go slower. At $u = 1\ \mathrm{cm/s}$ the walk becomes a $28$ -hour crawl and the deficit drops proportionally to $\Delta = \mathbf{5.6\times10^{-17}\ s}$ . In the limit the two procedures agree exactly — Eddington's point.

Step 3. Ask whether $5.6\ \mathrm{fs}$ is real. A transportable optical clock good to a fractional frequency of $10^{-18}$ accumulates about $10^{-18} \times 1000\ \mathrm{s} = 1\ \mathrm{fs}$ of its own error over the journey. The transport deficit is several times that — no longer a theorist's fiction, but a quantity at the edge of measurement.

Step 4. Do not compute it naively. In double precision at $u = 1\ \mathrm{m/s}$ , $(L/u)\big(1-\sqrt{1-u^2/c^2}\big)$ returns exactly $0.0$ : $u^2/c^2 = 1.1\times10^{-17}$ vanishes against $1$ inside the root. Use $Lu/(2c^2)$ . A real $5.6\ \mathrm{fs}$ effect, silently reported as nothing, by code that looks perfectly correct.

Physics of the portrait

The portrait here is a procedure. Einstein wrote it down in §1 of the 1905 paper, before anything else — but its intellectual owner is arguably Poincaré, and its modern home is not a textbook but a satellite constellation.

Origins · 1 minClock Synchronization and how this was discovered

Clock Synchronization

1898–present · Poincaré's convention · Einstein's §1 · Eddington's walk · the Sagnac term in every receiver

The idea is a rule, and it has three fathers and one enemy. Henri Poincaré (1854–1912) stated the problem and half the answer in La Mesure du Temps (1898): we have no direct intuition of distant simultaneity, distant clocks can only be set by signals, and doing so requires stipulating that light takes equal time each way. He called it a convention and meant it. Albert Einstein (1879–1955) opened "Zur Elektrodynamik bewegter Körper" (1905) with the same stipulation, written as an equation, and then treated it as constitutive: there is no truer time underneath for it to approximate. Arthur Eddington (1882–1944), in the 1924 second edition of The Mathematical Theory of Relativity, showed that a clock carried slowly between the points reproduces the same setting — an independent witness using no light at all. The enemy is rotation: Sagnac's rotating interferometer (1913) showed that counter-propagating beams do not agree on a spinning platform, and the same fact makes Einstein's rule fail to close a loop on Earth by $207\ \mathrm{ns}$ around the equator. The convention Poincaré called free and Einstein made constitutive has a hard boundary — and every GPS receiver carries the correction in firmware.

History

1898

— Poincaré, La Mesure du Temps: distant clocks can only be set by light signals, on the convention that light takes equal time each way.

1905

— Einstein, "Zur Elektrodynamik bewegter Körper", §1: the rule is written down as a definition, and the whole theory is built on it.

1913

— Sagnac's rotating interferometer: counter-propagating light beams do not return together on a spinning platform.

1924

— Eddington, The Mathematical Theory of Relativity, 2nd ed.: slow clock transport gives the same synchronization as light signals.

1971

— Hafele and Keating fly caesium clocks east and west around the world; the rotating-Earth terms are part of the analysis.

1978

— the first GPS satellite is launched; the operational system carries the Sagnac correction from the outset.

Going further

Texts. Einstein, Annalen der Physik 17 (1905), §1 · Taylor and Wheeler, Spacetime Physics, 2nd ed., Ch. 1 · Rindler, Introduction to Special Relativity, 2nd ed., §2 · Eddington, The Mathematical Theory of Relativity, 2nd ed. (1924) · Ashby, "Relativity in the Global Positioning System", Living Reviews in Relativity 6 (2003).

Provenance. Anchors, all recomputed in verify_aq-sr-einstein-synchronization.py: $c = 2.998\times10^{8}\ \mathrm{m/s}$ , $\omega_\oplus = 7.292\times10^{-5}\ \mathrm{rad/s}$ , $R_\oplus = 6.378\times10^{6}\ \mathrm{m}$ , giving $A = \pi R_\oplus^2 = 1.278\times10^{14}\ \mathrm{m^2}$ , $\Delta t = 2\omega A/c^2 = 207.4\ \mathrm{ns}$ , $c\,\Delta t = 62.2\ \mathrm{m}$ , equatorial ground speed $465\ \mathrm{m/s}$ ; the 300–400–500 m circuit closing at $4.003\ \mathrm{\mu s}$ with gap exactly $0$ ; and slow transport at $L = 1\ \mathrm{km}$ giving $5.563\ \mathrm{fs}$ at $1\ \mathrm{m/s}$ and $5.563\times10^{-17}\ \mathrm{s}$ at $1\ \mathrm{cm/s}$ . Scope note: the rate factor $\sqrt{1-u^2/c^2}$ is granted by forward reference to The Light Clock and Time Dilation (S1), not derived here; the interferometric physics belongs to Sagnac Effect and Rotating-Frame Synchronization (S7) and the timing budget to GPS and Relativistic Timing Corrections (S7). What this article owns is the protocol, the transitivity theorem, the slow-transport equivalence, and the closure failure.

Builds on. Inertial Frames of Reference · The Constancy of the Speed of Light · The Relativity of Simultaneity · Absolute Space and Absolute Time. Leads to. The Conventionality of Simultaneity · Frames, Observers, and Networks of Clocks.

If it is only a convention, why does it matter which one I pick?

It does not matter physically — that is the point, and The Conventionality of Simultaneity (1861) makes it precisely: every measurable outcome is identical under any split of the round trip. What the choice buys is convenience. $\varepsilon = \tfrac12$ is the only split that makes the one-way speed of light the same in all directions, and so the only one in which the laws take their familiar isotropic form. Choose otherwise and nothing breaks — but your equations sprout cross-terms.

Isn't slow clock transport a way to sneak around the convention and measure the one-way speed?

It is the strongest argument for $\varepsilon = \tfrac12$ , and not a proof. The transported clock does reproduce Einstein's setting as $u \to 0$ , by a route that sends no light — genuinely striking. But the sceptic's reply is sharp: the rate factor $\sqrt{1-u^2/c^2}$ was itself worked out in coordinates that already assumed standard synchrony. Adopt a different $\varepsilon$ and the transport law changes to match, restoring the agreement. Well-motivated, not forced.

Why does the rotating-Earth failure depend on area rather than on how far you walk?

Because the accumulated error is $c^{-2}\oint \mathbf{v}\cdot\mathrm{d}\boldsymbol{\ell}$ , a circulation, and Stokes' theorem turns a circulation of $\boldsymbol{\omega}\times\mathbf{r}$ into $2\boldsymbol{\omega}\cdot\mathbf{A}$ . Physically: going east you chase a receding target and lose time; going west you meet an approaching one and gain it. Retrace your steps and the two cancel exactly — which is why an out-and-back path closes perfectly even on the spinning Earth. Only a path that encircles the axis fails to cancel.

Does the 207 ns mean the clocks on the equator are actually wrong?

No — it means there is no consistent "equatorial Einstein time" to be right or wrong about. Each pair of neighbouring clocks is synchronized impeccably; it is the chain that fails to close, which is a statement about the rotating frame, not about any clock in it. The practical response is GPS's: stop trying to define a global time by chaining a local rule around a rotating body, adopt a coordinate time in a non-rotating frame, and correct each one-way transfer explicitly. The clocks are fine. The frame is not inertial.

My code says the slow-transport deficit is exactly zero. Is the effect real?

Your code has a bug, and it is the classic one. At $u = 1\ \mathrm{m/s}$ , $u^2/c^2 \approx 1.1\times10^{-17}$ is smaller than the double-precision epsilon of $1$ , so $\sqrt{1-u^2/c^2}$ rounds to exactly $1.0$ , the bracket becomes exactly $0$ , and the expression vanishes. Catastrophic cancellation, not physics. Use the series $Lu/(2c^2)$ , or arbitrary precision, and the $5.6\ \mathrm{fs}$ reappears. This article's verifier computes it both ways and asserts the naive one is broken.

Two frames each synchronize perfectly. Why do they still disagree with each other?

Because each uses its own light signals and its own rulers, and both are entitled to. Every clock is consistent with every other clock in the same frame — this article's transitivity theorem — and no further fact makes one frame's slicing the correct one. The disagreement between frames is the relativity of simultaneity, established in The Relativity of Simultaneity (1859) and taken up in coordinate form by Einstein Clock Synchronization (S1).