Sagnac Effect and Rotating-Frame Synchronization

Split a beam of light on a spinning turntable, send the halves round the rim in opposite directions, and bring them back together. They arrive apart. The gap is

\Delta t = \frac{4\,\vec{\Omega}\cdot\vec{A}}{c^2}

and it is the most quietly load-bearing formula in navigation: what a ring-laser gyroscope measures, what an airliner's fibre-optic gyro measures, what forces GPS to do time transfer in a non-rotating frame, and what a cold-atom interferometer measures ten billion times better than light can. The formula is also the source of the effect's most repeated error, which the base commits twice: that $\Delta t$ does not depend on the signal's speed. It does — as $1/v^2$ . What is independent of speed is the phase, and the gap between those two statements is worth $c^2/v^2 \approx 10^{10}$ : the entire reason atom gyroscopes exist.

Key takeaways

Signature $(+,-,-,-)$ : $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}x^2$ , timelike positive. The base's rotating-frame metric is mostly-plus — re-derived here from Minkowski, not transcribed.
The effect. Counter-propagating signals round a closed loop on a platform rotating at $\vec\Omega$ arrive split by $\Delta t = 4\vec\Omega\cdot\vec A/c^2$ for light; exactly, on a rim: $\Delta t = 4\pi\Omega r^2/(c^2 - \Omega^2 r^2)$ .
The owned result — THE SPEED LEDGER. For a signal of speed $v$ , $\;\Delta t = 4\vec\Omega\cdot\vec A/v^2$ . It is not speed-independent. A $343$ m/s sound wave sees a delay $\mathbf{7.6\times10^{11}}$ times larger than light in the same loop. What is universal is the phase: $\;\Delta\phi = \dfrac{4}{\hbar}\left(\dfrac{p}{v}\right)\vec\Omega\cdot\vec A$ — one formula, both worlds, with $p/v = \hbar\omega/c^2$ for a photon and $p/v = m$ for a massive particle. Convention-free: $\;\Delta\phi_{\text{matter}}/\Delta\phi_{\text{light}} = \dfrac{mc^2}{\hbar\omega}$ , which for $^{87}$ Rb ( $mc^2 = 80.96$ GeV) against a $633$ nm photon ( $1.959$ eV) is $\mathbf{4.1\times10^{10}}$ .
The circulation theorem. $\;\Delta t = \dfrac{2}{c^2}\oint\vec u\cdot\mathrm{d}\vec\ell = \dfrac{4}{c^2}\vec\Omega\cdot\vec A$ , since $\nabla\times\vec u = 2\vec\Omega$ . Hence shape-independence and axis-independence: shifting the axis adds $-(\vec\Omega\times\vec d)\cdot\oint\mathrm{d}\vec\ell$ , and $\oint\mathrm{d}\vec\ell = \vec 0$ round any closed loop. A gyroscope need not know where it is pivoting.
First order, not second. Sagnac is $O(v/c)$ ; time dilation is $O(v^2/c^2)$ . At his 1913 $\beta = 6.9\times10^{-9}$ he beat any second-order effect in the same instrument by $2/\beta = \mathbf{2.9\times10^{8}}$ — why he saw a fringe in 1913 and Michelson and Morley saw nothing in 1887.
Three experiments, recomputed from geometry alone. Sagnac 1913 ( $860$ cm², $2$ rev/s, $436$ nm): his reported $0.07$ is the reversal shift — one sense gives $\mathbf{0.0331}$ , CW-vs-CCW doubles it to $\mathbf{0.066}$ . Michelson–Gale 1925 ( $41.77^\circ$ N, $570$ nm): $\mathbf{0.2363}$ vs the paper's computed $0.236$ , observed $0.230$ . Wettzell "G" ( $4\times4$ m, $632.8$ nm): beat $\mathbf{348.6}$ Hz.
Harress's paradox, resolved. Fill the loop with a co-rotating medium of index $n$ and $\Delta t$ does not change — Fresnel drag exactly repays what the slowing costs, and $n$ cancels identically. That is the true invariance the base over-generalised into a false one.
KEY IDEA: the delay is geometry, the phase is physics. $\Delta t$ counts a loop that fails to close and every theory predicts it alike; $\Delta\phi = (4/\hbar)(p/v)\,\vec\Omega\cdot\vec A$ is the invariant statement. The base's three defects — a mostly-plus metric, a signal-speed independence contradicting its own eq[5] by $c^2/v^2$ , and a matter-wave card printing $4\pi m/h$ where its own eq[5] demands $8\pi m/h$ — are each repaired in a warning card below.

The core insight

The Sagnac delay is not about light, nor about time dilation: it is the statement that a rotating frame has no consistent global "now". March Einstein's synchronisation convention from clock to clock around the rim and you return holding a clock that disagrees with itself by $2\vec\Omega\cdot\vec A/c^2$ — half the Sagnac delay, and exactly the one-way correction GPS applies. The loop fails to close. That failure is a circulation, $\oint\vec u\cdot\mathrm{d}\vec\ell$ , and by Stokes a flux of vorticity; rigid rotation has vorticity $2\vec\Omega$ everywhere, so the answer can only ever be the enclosed area times $\Omega$ — whatever shape the loop, wherever the axis, and (at leading order) whatever fills it. But not regardless of how fast the signal goes — that is where everybody trips.

Where this lives today

Ring-laser and fibre-optic gyroscopes. The circulation theorem is a specification document: shape-independence lets a fibre wind into an $N$ -turn coil enclosing $NA$ ; axis-independence lets it bolt anywhere in the airframe. A ring laser reads out a beat note, $\Delta f = 4\vec\Omega\cdot\vec A/\lambda L$ — the two cavity modes sit at different frequencies since they see different path lengths, and the difference is an audio tone. The largest on Earth is the $4\times4$ m "G" ring laser at the Geodetic Observatory Wettzell in Bavaria, set horizontally in bedrock so its area vector points along the local vertical. Its Sagnac beat, computed above from geometry alone, is $348.6$ Hz — a tone whose pitch is Earth's rotation rate, and G is steady enough to watch the length of day and Chandler wobble drift it.

Global timekeeping. A signal crossing the rotating Earth accumulates a Sagnac correction $2\vec\Omega\cdot\vec A_E/c^2$ , with $\vec A_E$ the area swept by the vector from Earth's axis to the signal — hence the $414.8$ ns above. That is why GPS does time transfer in the non-rotating Earth-Centred Inertial frame. The full relativistic clock budget — $-7.2$ , $+45.8$ , net $+38.6$ µs/day — belongs to `aq-sr-gps-relativistic-corrections`, this corpus's canonical owner. The Sagnac term is a separate correction, applied per signal path rather than per clock, and does not appear in that budget.

Matter-wave gyroscopes. Cold-atom Sagnac interferometers enclosing square centimetres compete with optical gyros enclosing square metres. What the $mc^2/\hbar\omega$ ledger buys is not raw sensitivity but stability — an atom's mass is a constant of nature and cannot drift, unlike a laser's wavelength. See Barrett et al., Comptes Rendus Physique 15 (2014) 875.

Origins · 2 minSagnac effect and how this was discovered

Sagnac effect

1911–1925 · Harress spins glass and blames the drag · Sagnac spins a turntable and claims the ether · Michelson and Gale spin the Earth and settle it

Franz Harress never knew what he had built. His Jena dissertation — Die Geschwindigkeit des Lichtes in bewegten Körpern, submitted 25 March 1912 on experiments run from 1909 — sent light round a ring of rotating glass prisms and found a shift, read as Fresnel drag in the moving glass: precisely the reading under which the effect becomes invisible, since drag cancels the index. He died in 1915, aged 30. Georges Sagnac, in 1913, did the clean experiment: an $860$ cm² loop on a turntable at two revolutions a second, photographed with the disc spun each way. He announced the shift as evidence of "l'existence de l'éther lumineux" — wrong about what he found, right about having found it. The effect shows rotation is absolute, which relativity had never denied. Albert Michelson and Henry Gale finished the story in 1925 with a Sagnac interferometer whose turntable was the planet: a rectangle of evacuated pipe, $2010$ by $1113$ feet, at Clearing, Illinois. Predicted $0.236 \pm 0.002$ fringes, measured $0.230 \pm 0.005$ . Michelson, who spent a lifetime failing to detect the Earth's motion through the ether, ended it detecting the Earth's rotation with the same instrument.

History

1912

— Franz Harress submits his Jena dissertation on 25 March, reporting a fringe displacement in rotating glass prisms, interpreted entirely as Fresnel drag. But the drag in a co-rotating medium cancels the refractive index exactly, so his own explanation is the one thing his data could never distinguish — it was a Sagnac interferometer, and nobody, including its builder, knew it.

1913

— Georges Sagnac publishes in the Comptes Rendus a fringe displacement of 0.07 on a turntable enclosing 860 cm² at 2 rev/s in indigo light, attributed to a stationary luminiferous ether. Recomputing from his own numbers, the one-sense shift at 436 nm is 0.033 and the CW-vs-CCW reversal shift is twice that, 0.066 — so his 0.07 is the reversal figure; read as one-sense it would demand 206 nm, deep ultraviolet, impossible on a 1913 bench.

1925

— Michelson and Gale, with Pearson, run it at planetary scale: a closed rectangle of evacuated pipe, 2010 by 1113 feet, at Clearing, Illinois. They predict 0.236 and observe 0.230; recomputing from the area, latitude and 570 nm gives 0.2363 — the first optical measurement of Earth's rotation.

1963

— Macek and Davis operate the first ring-laser gyroscope: the delay stops being a photographed fringe and becomes a counted beat frequency.

1976

— Fibre-optic gyroscopes exploit shape-independence, enclosing the area N times over with a coil: cheap, solid-state, eventually everywhere.

2003

— The Wettzell G ring laser reaches full operation, resolving Earth's rotation well enough to watch the length of day and Chandler wobble drift its 348.6 Hz beat note.

The physical picture

There are two honest routes to $\Delta t$ , and each catches the other's blind spot. The first is the chase argument, which Sagnac himself would have recognised: work in the inertial lab frame, where light goes at $c$ both ways and the loop is just moving mirrors. Split the beam on the rim — while the halves are in flight the splitter moves, so the co-rotating half chases a receding target and the counter-rotating half meets one approaching. The co-rotating beam has further to go and arrives late — no relativity used. The second is the metric route — write spacetime in rotating coordinates, read off the light cone, integrate. It makes shape- and axis-independence obvious in a way the chase argument never does, and connects the effect to its real parent: the failure of simultaneity to close. We do both, metric first.

The core equations


Rotating-frame metric, our lock $(+,-,-,-)$	$\mathrm{d}s^2 = \left(1 - \dfrac{\Omega^2r^2}{c^2}\right)c^2\mathrm{d}t^2 - 2\Omega r^2\,\mathrm{d}\phi\,\mathrm{d}t - \mathrm{d}r^2 - r^2\mathrm{d}\phi^2 - \mathrm{d}z^2$
Sagnac delay, light	$\Delta t = \dfrac{4\,\vec\Omega\cdot\vec A}{c^2}$
Sagnac delay, signal of speed $v$	$\Delta t = \dfrac{4\,\vec\Omega\cdot\vec A}{v^2}$
Exact, circular rim of radius $r$	$\Delta t = \dfrac{4\pi\Omega r^2}{c^2 - \Omega^2 r^2}$
The circulation theorem	$\Delta t = \dfrac{2}{c^2}\oint \vec u\cdot\mathrm{d}\vec\ell,\qquad \nabla\times\vec u = 2\vec\Omega$
Universal phase	$\Delta\phi = \dfrac{4}{\hbar}\left(\dfrac{p}{v}\right)\vec\Omega\cdot\vec A$
Optical fringe shift	$\Delta N = \dfrac{4\,\vec\Omega\cdot\vec A}{\lambda c}$
Ring-laser beat	$\Delta f = \dfrac{4\,\vec\Omega\cdot\vec A}{\lambda L}$
Matter-wave enhancement	$\dfrac{\Delta\phi_{\text{matter}}}{\Delta\phi_{\text{light}}} = \dfrac{mc^2}{\hbar\omega}$
Synchronisation gap	$\delta\tau = \dfrac{2\,\vec\Omega\cdot\vec A}{c^2} = \dfrac{\Delta t}{2}$

:::svg[THE SPEED LEDGER — the owned result, and the base's error, on one plot. Horizontal axis: the speed of the signal sent round the loop, from a 343 m/s sound wave to light. The BLUE curve is the Sagnac TIME delay Δt = 4Ω·A/v², and it is NOT flat — it climbs as 1/v², so sound in the same loop sees a delay 7.6×10¹¹ times larger than light does. The base says this curve is flat. It is not. The RED band is what actually is invariant: the PHASE, Δφ = (4/ħ)(p/v)Ω·A, where p/v is ħω/c² for a photon and m for a massive particle — the momentum-to-speed ratio is the universal Sagnac charge. Divide, and the matter-wave enhancement falls out: mc²/ħω, which for rubidium-87 (80.96 GeV) against a helium-neon photon (1.959 eV) is 4.1×10¹⁰. THE PUNCHLINE: flatten Δt as the base demands and that enhancement becomes mv²/ħω — smaller by c²/v², ten orders of magnitude. The base's intuition card refutes the base's own eq[5] by the entire factor atom interferometry exists to exploit]

Derivation AThe rotating-frame metric in our signature, and the exact delay that follows

The goal. The base's rotating-frame metric has the opposite overall sign, where timelike intervals come out negative. We are locked to $(+,-,-,-)$ , where $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}\mathbf{x}^2$ and $\mathrm{d}\tau = \mathrm{d}s/c$ with no minus sign under the root. So we build it from flat Minkowski space by a coordinate change, in our own signature, and solve the null condition exactly — not to leading order — to ask what the higher-order corrections mean, and whether the base's story is true. It is not, and this derivation convicts it.

Step 1 — flat spacetime, cylindrical, our lock. $\;\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}r^2 - r^2\mathrm{d}\phi_{\text{lab}}^2 - \mathrm{d}z^2$ . A clock at rest has $\mathrm{d}s^2 = c^2\mathrm{d}t^2 > 0$ : timelike is positive.

Step 2 — spin the angular coordinate. Let $\phi$ be the angle read on the platform. A point fixed to it holds $\phi$ constant while its laboratory angle advances at $\Omega$ :

\phi_{\text{lab}} = \phi + \Omega t \qquad\Longrightarrow\qquad \mathrm{d}\phi_{\text{lab}} = \mathrm{d}\phi + \Omega\,\mathrm{d}t

A change of coordinates, not a Lorentz transformation — nothing is boosted; we are relabelling events.

Step 3 — substitute. Only the angular term changes, $r^2\,\mathrm{d}\phi_{\text{lab}}^2 = r^2\mathrm{d}\phi^2 + 2\Omega r^2\,\mathrm{d}\phi\,\mathrm{d}t + \Omega^2r^2\,\mathrm{d}t^2$ , so collecting the $\mathrm{d}t^2$ terms:

\boxed{\;\mathrm{d}s^2 = \left(1 - \frac{\Omega^2r^2}{c^2}\right)c^2\mathrm{d}t^2 \;-\; 2\Omega r^2\,\mathrm{d}\phi\,\mathrm{d}t \;-\; \mathrm{d}r^2 - r^2\mathrm{d}\phi^2 - \mathrm{d}z^2\;}

Two features, both load-bearing. The $g_{tt}$ coefficient $\left(1 - \Omega^2r^2/c^2\right)$ is $1/\gamma^2$ for rim speed $\Omega r$ ; it vanishes at the light cylinder $r = c/\Omega$ , beyond which no platform point can be at rest. That is where rim time dilation lives — second order in $\Omega$ . The off-diagonal $g_{t\phi} = -\Omega r^2$ is first order, with no analogue in any static metric — the entire Sagnac effect: a $\mathrm{d}t\,\mathrm{d}\phi$ cross-term means the two senses round the loop are not equivalent.

Step 4 — the null condition, solved exactly. Light on the rim: $r$ , $z$ fixed, $\mathrm{d}s^2 = 0$ . The metric collapses to a quadratic in $\mathrm{d}t$ :

\left(c^2 - \Omega^2 r^2\right)\mathrm{d}t^2 - 2\Omega r^2\,\mathrm{d}\phi\,\mathrm{d}t - r^2\mathrm{d}\phi^2 = 0

Solve it, not expand it:

\mathrm{d}t_\pm = \frac{r\,|\mathrm{d}\phi|}{c \mp \Omega r}

where $\pm$ labels the sense of travel. (As signed roots they read $r\,\mathrm{d}\phi/(c-\Omega r)$ and $-r\,\mathrm{d}\phi/(c+\Omega r)$ ; the second beam has $\mathrm{d}\phi<0$ , so both times are positive.) The chase argument, appearing unbidden out of the geometry: the co-rotating beam advances against a reduced closing speed $c - \Omega r$ ; the counter-rotating one enjoys $c + \Omega r$ . No approximation made.

Step 5 — go round. Integrate each root over a full circuit in its own sense, $t_+ = 2\pi r/(c-\Omega r)$ and $t_- = 2\pi r/(c+\Omega r)$ :

\Delta t = 2\pi r\left[\frac{1}{c-\Omega r} - \frac{1}{c+\Omega r}\right] = \boxed{\;\frac{4\pi\Omega r^2}{c^2 - \Omega^2r^2}\;}

Step 6 — read off the leading order. With $A = \pi r^2$ :

\Delta t = \frac{4\Omega A}{c^2}\cdot\frac{1}{1 - \Omega^2r^2/c^2} = \frac{4\Omega A}{c^2}\,\gamma^2 \;\longrightarrow\; \frac{4\,\vec\Omega\cdot\vec A}{c^2}

Step 7 — be careful about that $\gamma^2$ : the base is not. The base's "exact circular loop" card prints this $\gamma^2$ and explains it as "the $\gamma^2$ factor from rim time dilation." That is false: this derivation used no clock rate. The $\gamma^2$ came from the denominators $c \mp \Omega r$ in step 4 — a closing-speed factor in coordinate time, which a purely Galilean chase calculation would produce too. Real dilation enters only when converting coordinate time to what the detector reads: the beamsplitter is a rim clock, ticking slow by $\gamma$ :

\Delta\tau = \frac{\Delta t}{\gamma} = \frac{4\Omega A}{c^2}\,\gamma

So the observable delay carries $\gamma^{1}$ , not $\gamma^{2}$ — dilation does not supply the second power, it removes one. Verified in verify_aq-sr-sagnac-effect.py: metric rebuilt from Minkowski in $(+,-,-,-)$ , shown to be the exact negative of the base's mostly-plus card; null condition solved in closed form; $\Delta t$ and its $4\Omega A/c^2$ leading term confirmed; $\Delta\tau = \Delta t/\gamma$ confirmed to carry one power of $\gamma$ .

Go deeper · Derivation B · 3 minThe circulation theorem — and why a gyroscope does not need to know where it is pivoting

Derivation BThe circulation theorem — and why a gyroscope does not need to know where it is pivoting

The goal. We want $\Delta t$ for an arbitrary closed loop, on a platform rotating about an arbitrary axis — and to prove, not assert, the two independences the technology rests on: only the enclosed area matters, and the rotation axis position does not matter. The second is the surprising one — it lets a ring-laser gyro be bolted anywhere in an aircraft without knowing the centre of rotation. It falls out of one vector identity and one fact about closed curves.

Step 1 — the local light cone on moving space. Let $\vec u(\vec x)$ be the platform material's velocity at $\vec x$ . Over a segment $\mathrm{d}\vec\ell$ , to first order in $u/c$ , the coordinate time for light to cross it in the $\pm$ sense is

\mathrm{d}t_\pm = \frac{|\mathrm{d}\vec\ell|}{c} \pm \frac{\vec u\cdot\mathrm{d}\vec\ell}{c^2}

This is the $g_{0i}$ cross-term of the previous derivation, written locally. (Check: on the rim $\vec u\cdot\mathrm{d}\vec\ell = \Omega r^2\,\mathrm{d}\phi$ , and expanding $r\,\mathrm{d}\phi/(c \mp \Omega r)$ to first order gives exactly $r\,\mathrm{d}\phi/c \pm \Omega r^2\mathrm{d}\phi/c^2$ . It agrees.)

Step 2 — go round, and subtract. The path-length pieces $|\mathrm{d}\vec\ell|/c$ are identical for the two senses and cancel identically. What survives is pure circulation:

\boxed{\;\Delta t = \frac{2}{c^2}\oint \vec u\cdot\mathrm{d}\vec\ell\;}

Already a theorem: the Sagnac delay is a line integral of the velocity field of space around the loop — nothing said about circles.

Step 3 — Stokes. $\;\oint\vec u\cdot\mathrm{d}\vec\ell = \int_S (\nabla\times\vec u)\cdot\mathrm{d}\vec A$ .

Step 4 — rigid rotation has constant vorticity. With $\vec u = \vec\Omega\times\vec r$ ,

\nabla\times(\vec\Omega\times\vec r) = 2\vec\Omega \qquad\Longrightarrow\qquad \Delta t = \frac{2}{c^2}\int_S 2\vec\Omega\cdot\mathrm{d}\vec A = \frac{4\,\vec\Omega\cdot\vec A}{c^2}

Shape-independence, for free. The integrand is constant, so the integral is the area. Bend the loop into a square, an ellipse, a fibre coil: all that survives is the enclosed vector area $\vec A$ , and only its component along $\vec\Omega$ . A $4\times4$ m square ring laser and a circle of equal area read the same rotation — and winding a fibre $N$ times multiplies the answer by $N$ .

Step 5 — axis independence, the good one. Move the axis through some point $\vec d$ . Every point of the platform now has velocity $\vec u = \vec\Omega\times(\vec r - \vec d)$ , and the circulation splits:

\oint\vec u\cdot\mathrm{d}\vec\ell = \oint\left(\vec\Omega\times\vec r\right)\cdot\mathrm{d}\vec\ell \;-\; \oint\left(\vec\Omega\times\vec d\right)\cdot\mathrm{d}\vec\ell

The second term has a constant vector dotted into $\mathrm{d}\vec\ell$ , so it comes straight out of the integral:

\oint\left(\vec\Omega\times\vec d\right)\cdot\mathrm{d}\vec\ell = \left(\vec\Omega\times\vec d\right)\cdot\oint\mathrm{d}\vec\ell = 0 \qquad\text{because}\qquad \oint\mathrm{d}\vec\ell = \vec 0

The sum of the displacement vectors around any closed curve is zero — that is the entire proof. The offset $\vec d$ vanishes, not approximately, not to leading order, but identically, for any loop and any offset.

In an aeroplane. A ring-laser gyro sits in an airframe simultaneously yawing near its centre of mass, rolling about its long axis, and carried round the Earth's axis six thousand kilometres away. It is at the centre of none of those rotations and need not be: it responds to $\vec\Omega$ alone. It is a pure vorticity meter — stronger than "it measures rotation", trusted without knowing the pivot.

Step 6 — the ledger the base gets wrong. Redo step 1 for a signal of speed $v$ rather than $c$ : the flight time over a segment is $|\mathrm{d}\vec\ell|/v$ , and the drag term picks up $v$ where the light calculation picked up $c$ :

\Delta t = \frac{2}{v^2}\oint\vec u\cdot\mathrm{d}\vec\ell = \frac{4\,\vec\Omega\cdot\vec A}{v^{2}}

Or by the chase argument: solve $t_\pm = 2\pi r/(v \mp \Omega r)$ and subtract, giving exactly $4\pi\Omega r^2/(v^2 - \Omega^2r^2)$ . Either way, $\Delta t \propto 1/v^2$ — never signal-speed independent. For sound at $343$ m/s the delay in a given loop is $(c/343)^2 = 7.6\times10^{11}$ times the optical one. Verified in verify_aq-sr-sagnac-effect.py: $\oint(\vec r\times\mathrm{d}\vec r) = 2\vec A$ , $\oint\mathrm{d}\vec\ell = \vec0$ , $\nabla\times(\vec\Omega\times\vec r) = 2\vec\Omega$ integrated symbolically; displaced-axis circulation shown in closed form equal to $2\vec\Omega\cdot\vec A$ with every $\vec d$ component dropping out, numerically frozen against a 50% axis shift; general- $v$ delay solved with its $1/v^2$ leading term confirmed.

Limiting cases

The signal-speed limit — what is actually invariant. The time is not universal. What is? A wave of wavenumber $k = p/\hbar$ and group speed $v$ accumulates, over an extra path length $\Delta s = v\,\Delta t$ , a phase $\Delta\phi = k\,\Delta s$ :

\Delta\phi = \frac{p}{\hbar}\cdot v\cdot\frac{4\,\vec\Omega\cdot\vec A}{v^2} = \frac{4}{\hbar}\left(\frac{p}{v}\right)\vec\Omega\cdot\vec A

The speed has not cancelled; it has combined with the momentum into the single ratio $p/v$ — that is the universal Sagnac charge.

Photon: $p/v = \hbar\omega/c^2 = E/c^2$ , giving $\Delta\phi = 4\omega\,\vec\Omega\cdot\vec A/c^2 = \omega\,\Delta t$ . The optical formula, recovered.
Massive particle: $p = mv$ , so $p/v = m$ , giving $\Delta\phi = 4m\,\vec\Omega\cdot\vec A/\hbar = 8\pi m\,\vec\Omega\cdot\vec A/h$ . The atoms' speed has vanished from the answer.

The bridge is $p/v = E/c^2$ — the relation making $E = mc^2$ readable as "mass is the momentum-to-speed ratio". Divide, and the geometry, $\hbar$ , and the factor-of-two loop convention cancel, since each is the same for both waves:

\frac{\Delta\phi_{\text{matter}}}{\Delta\phi_{\text{light}}} = \frac{m}{\hbar\omega/c^2} = \frac{mc^2}{\hbar\omega}

Convention-free, and enormous: a $^{87}$ Rb atom has $mc^2 = 80.96$ GeV, a helium–neon photon has $\hbar\omega = 1.959$ eV, and the ratio $\mathbf{4.1\times10^{10}}$ is why cold-atom gyroscopes exist. Flatten $\Delta t$ and the ratio becomes $mv^2/\hbar\omega$ — off by $c^2/v^2$ , about $10^{10}$ : the base's intuition card destroys its own eq[5] by exactly that margin.

The medium limit — Harress's paradox. There is a genuine speed-invariance here, likely what the base was reaching for: fill the loop with a co-rotating medium and the delay does not change. Light in glass of index $n$ travels at $c/n$ ; if $\Delta t \propto 1/v^2$ , filling the ring with $n=1.5$ glass should multiply the delay by $n^2 = 2.25$ . It does not, because a co-rotating medium drags the light. Fresnel's drag coefficient (Fizeau, 1851; one line from relativistic velocity addition) gives light in a medium moving at $u$ a lab speed $c/n \pm u\left(1 - 1/n^2\right)$ . The co-rotating beam chases a splitter receding at $u$ , the counter-rotating one meets one approaching at $u$ , so the closing speeds are $c_\pm^{\text{eff}} = c/n \mp u/n^2$ and

\Delta t = 2\pi r\left[\frac{1}{c/n - u/n^2} - \frac{1}{c/n + u/n^2}\right] = \frac{4\pi r\,u}{c^2 - u^2/n^2} \;\longrightarrow\; \frac{4\,\vec\Omega\cdot\vec A}{c^2}

Every $n$ has cancelled. The $n^2$ the slowing puts into the delay is taken back out, exactly, by the $1/n^2$ the drag puts into the closing speed; the residue sits at $O(u^2/n^2c^2)$ . This is why a fibre-optic gyroscope, its light spending its whole life inside glass at $c/1.47$ , obeys the vacuum formula. But the cancellation requires a medium that co-rotates with the loop. A beam of free atoms is dragged by nothing; no Fresnel coefficient repays the $1/v^2$ , so it stands at full strength. Medium-independence is real; signal-speed independence is not. Collapsing the first into the second is the base's error.

The non-rotating limit. $\vec\Omega = 0 \Rightarrow \Delta t = 0$ ; only the component of $\vec\Omega$ along the loop's area vector contributes, so a loop in a plane containing the axis reads zero.

Common trapThe base's metric is mostly-plus — and its "exact"

\gamma^2

is not what it says it is

The base's eq[4] prints $\mathrm{d}s^2 = -\left(1 - \Omega^2r^2/c^2\right)c^2\mathrm{d}t^2 + 2\Omega r^2\mathrm{d}\phi\,\mathrm{d}t + \mathrm{d}r^2 + r^2\mathrm{d}\phi^2 + \mathrm{d}z^2$ — mostly-plus, $(-,+,+,+)$ , self-consistent in its own convention (machine-checked: the exact negative of ours, cross-term included) but not ours — copy it into our lock and every timelike interval comes out imaginary. Separately, its "circular loop, exact" card writes $\Delta t = (4\pi R^2\Omega/c^2)\gamma^2$ — algebraically right — and calls the factor "the $\gamma^2$ from rim time dilation." It is not: step 7 produces that $\gamma^2$ from the closing-speed denominators without invoking a clock rate. It is a coordinate-time factor; real dilation removes a power — the rim detector reads $\Delta\tau = \Delta t/\gamma$ , going as $\gamma^1$ .

:::warning[The base's intuition card refutes the base's own eq[5] — by $c^2/v^2$ ] The base states, in its tldr and intuition card, that the Sagnac delay is "independent of the propagation speed of the signal" — that "any signal with a well-defined speed shows the same $4\vec\Omega\cdot\vec A/c^2$ timing asymmetry."

The delay goes as $1/v^2$ . Solve $t_\pm = 2\pi r/(v\mp\Omega r)$ ; $v$ never leaves. The base cannot afford the error: its own eq[5] asserts a matter-wave enhancement of $mc^2/\hbar\omega$ , a number that only exists because the atom's $\Delta t$ exceeds light's by $c^2/v^2$ . Impose the intuition card and it collapses to $mv^2/\hbar\omega$ : ten orders smaller, and cold-atom gyroscopy has no reason to exist. The base has generalised a true statement — $\Delta t$ is independent of the refractive index of a co-rotating medium, since Fresnel drag cancels $n$ — into a false one about signal speed generally. A dragged medium repays the debt; a free particle beam does not.

:::warning[The base's matter-wave card is half of what its own eq[5] requires] The base's limits card gives $\Delta\phi = (4\pi m/h)\,\vec\Omega\cdot\vec A$ ; its eq[1] gives the optical phase as $(4\omega/c^2)\,\vec\Omega\cdot\vec A$ . Dividing gives $mc^2/2\hbar\omega$ — half the $mc^2/\hbar\omega$ that eq[5] and the card's own note both assert. Two independent derivations — the lab-frame path-length argument above, and the rotating-frame Lagrangian term $m\,\vec u\cdot(\vec\Omega\times\vec r)$ integrated round the loop — both give $\Delta\phi_{\text{matter}} = 4m\,\vec\Omega\cdot\vec A/\hbar = 8\pi m\,\vec\Omega\cdot\vec A/h$ , reproducing eq[5]'s ratio exactly. Eq[5] is right; the limits card is wrong by two. The likely origin is a geometry swap: $4\pi m/h$ is the two-arm coefficient, each path traversing the loop once. In a Sagnac ring both beams go all the way round in opposite senses, doubling the factor. Trust the ratio, not the coefficient: $mc^2/\hbar\omega$ is convention-free since the geometric factor cancels in the division.

Worked examples

Figure

THE THREE MEASUREMENTS, RECOMPUTED FROM SCRATCH — the HISTORY=PHYSICS check. Every published number here was re-derived from apparatus geometry alone, using ΔN = 4Ω·A/(λc) and nothing else. SAGNAC 1913: an 860 cm² loop at 2 rev/s in indigo light. At λ = 436 nm the formula gives a one-sense fringe shift of 0.0331 — but Sagnac reported 0.07, twice that. His figure is the REVERSAL shift, clockwise against counter-clockwise: 2 × 0.0331 = 0.066. What settles it — read 0.07 as one-sense and you must invert to λ = 206 nm, deep ultraviolet, impossible on a 1913 bench; read it as the reversal and you get 412 nm, an ordinary indigo. MICHELSON AND GALE 1925: a rectangle of evacuated pipe, 2010 by 1113 feet, at Clearing, Illinois, 41.77° north, enclosing 207 836 m². At λ = 570 nm the formula predicts 0.2363 fringes; their paper computed 0.236 ± 0.002 and observed 0.230 ± 0.005. WETTZELL G: a 4 × 4 m square ring laser in Bavarian bedrock at 49.15° north, 632.8 nm — not fringes but a beat note between the counter-propagating cavity modes, Δf = 4Ω·A/(λL) = 348.6 Hz, an audible tone whose pitch IS the spin of the planet

Sagnac's own turntable, and why he could see it. His 1913 loop enclosed $860$ cm² (equivalent radius $16.5$ cm, perimeter $L = 1.04$ m). At two revolutions per second the rim moves at $2.08$ m/s, giving $\beta = \Omega r/c = 6.9\times10^{-9}$ . Yet the fringe shift is

\Delta N = \frac{4\Omega A}{\lambda c} = \frac{2L\beta}{\lambda} = 2\times\left(2.38\times10^{6}\right)\times\left(6.9\times10^{-9}\right) = 0.033

because the lever is $L/\lambda$ : the loop is two and a half million wavelengths long, and that multiplies the tiny $\beta$ back up into something visible through an eyepiece. A second-order effect in the same instrument would give $\Delta N \sim L\beta^2/\lambda \approx 1.1\times10^{-10}$ fringes — a ratio of $2/\beta = 2.9\times10^{8}$ . The whole difference between 1887 and 1913: Michelson and Morley hunted a $\beta^2$ effect with $\beta \approx 10^{-4}$ , fighting for hundredths of a fringe; Sagnac, with $\beta$ ten thousand times smaller, got a bigger signal, because his was linear. The deep consequence: a first-order effect cannot test the isotropy of light. At first order the delay is a geometric consequence of the loop failing to close, and Newton, every ether theory and relativity predict $4\Omega A/c^2$ alike. So the Sagnac effect is not evidence for an ether, nor a violation of relativity. What Sagnac established is that rotation is absolute, detectable locally from inside.

The Earth as the turntable. Send light round the equator. The enclosed area is $\pi a^2 = 1.278\times10^{14}$ m², and with $\Omega = 7.292\times10^{-5}$ rad/s the eastbound-versus-westbound difference is

\Delta t = \frac{4\Omega A}{c^2} = \mathbf{414.8}\ \text{ns}

with the one-way correction half that, $207.4$ ns — enormous for a system resolving nanoseconds.

Going further

The synchronisation reading generalises: $\delta\tau = 2\vec\Omega\cdot\vec A/c^2$ , half the Sagnac delay, is a holonomy — a mismatch accumulated by transporting a convention round a circuit and back to the start. Seen that way the family resemblance is clear: the rotation from composing Lorentz boosts around a closed circuit in velocity space (`aq-sr-wigner-rotation`, `aq-sr-thomas-precession`) is the same species of animal. And the circulation theorem outruns the gyroscope it was built for — $\Delta t = (2/c^2)\oint\vec u\cdot\mathrm{d}\vec\ell$ holds for any velocity field.

Is the Sagnac effect a special-relativistic effect, if Newtonian physics predicts the same answer?

It is kinematic, and at leading order theory-blind: the chase argument uses no relativity, and Newton, Lorentz's ether and Einstein all deliver $\Delta t = 4\vec\Omega\cdot\vec A/c^2$ — not a test of SR. What is relativistic is the interpretation: the delay is not an ether wind but the failure of Einstein synchronisation to close on a non-inertial platform, and the platform's clocks are dilated by $\gamma$ , so the proper-time delay carries $\gamma^1$ . Genuine relativity enters at second order.

The formula has

c^2

in it. How can the delay not depend on the speed of the signal?

It does depend on it: $\Delta t = 4\vec\Omega\cdot\vec A/v^2$ , and $c$ appears only because the signal is usually light. The genuine invariance — over-generalised into the false one — is that the refractive index of a co-rotating medium drops out, since Fresnel drag repays what the slowing costs. Light in glass at $c/1.47$ sees the same $\Delta t$ as light in vacuum; a beam of free atoms at $300$ m/s does not.

If matter waves are ten billion times more sensitive, why is every gyroscope in every aircraft still optical?

Sensitivity per unit area is not the whole cost function. The advantage is real — cold-atom gyroscopes achieve remarkable stability with tiny enclosed areas — but an atom interferometer needs a vacuum chamber, laser cooling and a duty cycle, while a fibre coil needs a spool of glass and a diode. An optical gyro recovers by brute force what it lacks in per-particle leverage: wind the fibre $N$ times for effective area $NA$ , with $N$ in the tens of thousands. Where the atom wins decisively is drift: a mass is a constant of nature, a laser's wavelength is not.

Why does the enclosed area matter and not the length of the path?

The delay is a circulation, and Stokes turns circulations into fluxes: $\Delta t = (2/c^2)\oint\vec u\cdot\mathrm{d}\vec\ell = (2/c^2)\int(\nabla\times\vec u)\cdot\mathrm{d}\vec A$ , and rigid rotation has constant vorticity $2\vec\Omega$ everywhere — a constant integrand integrates to the area. See it by pinching the loop into a figure-of-eight: the two lobes enclose opposite vector areas and the signal cancels, though the path length is unchanged. Length buys nothing; enclosed area is everything.

How does the Sagnac effect relate to the relativistic clock corrections in GPS?

Separate corrections doing separate jobs; conflating them is a common error. The clock budget — the satellite's velocity dilation, its gravitational blueshift, the net factory detune — is a property of each clock, applied once, belonging to `aq-sr-gps-relativistic-corrections`, the canonical owner. The Sagnac correction is a property of each signal path, applied per transmission, and forces the system to do its arithmetic in a non-rotating frame. Neither substitutes for the other.

Sagnac thought he had proved the ether exists. What did he actually prove?

That rotation is absolute — a rotating frame is detectable from inside, with a local experiment. Relativity has always agreed; SR is a theory about inertial frames and never claimed a spinning platform equivalent to a stationary one. Sagnac's error was treating "the two beams take different times" as though they needed a medium to take different times through. They need nothing of the sort: in the inertial lab frame both travel at $c$ , one simply having further to go. He measured a real, frame-independent fact about his turntable, and drew from it a conclusion his apparatus could not support.