University Physics IV · Special Relativity I · 2.3
The postulates and clock synchronisation
Two postulates look like two experimental facts. One is a principle you extend on faith, and half of the other is a convention you choose. This is where you learn which parts of “light goes at c” were measured, which were stipulated, and which experiment pins down which.
Build the model
Connect the measurement to the mechanism.
The second postulate is usually read as a bare experimental fact, and it is not one. To measure how fast light goes from A to B you need clocks at A and B that already agree on what “the same moment” means, and to make them agree you must send a signal between them whose one-way speed is precisely what you were trying to find. The circle cannot be broken, so Einstein cut it: he stipulated that the outward and return legs of a round trip take equal times — the ε = ½ choice — and everything experiment can actually reach turns out to be a round trip, or a ratio read at one place.
What survives as physics is the two-way speed 2L/Δt read on a single clock: isotropic (Michelson–Morley), independent of the laboratory's own velocity (Kennedy–Thorndike), and accompanied by a time-dilation factor γ (Ives–Stilwell). What is convention is the split of that round trip into two legs, since any ε between 0 and 1 reproduces every experiment ever performed, at the price of anisotropic coordinates in which Maxwell's equations lose their symmetry. The cost of the postulates is therefore not that they are unproven; it is that one of them is half a definition, and every conclusion you draw about events happening “at the same time” somewhere else inherits that definition rather than measuring it.
- Simple definition
- Einstein synchronisation defines two separated clocks to agree when the outward and return legs of a light signal's round trip between them are assigned equal durations — the ε = ½ choice — which makes the one-way speed of light c by stipulation rather than by measurement.
- Example
- A pulse leaves clock A, reflects off clock B and returns 2.00 μs later, so B stands 300 m away and is told to read 1.00 μs at the reflection. Choosing ε = 0.25 instead sets it to 0.50 μs, so light runs out at 2c and back at 2c/3 — and nothing timed on one clock can tell the two accounts apart.
Fixes what “at the same time” means at a distance. t₁ and t₂ are read off A's clock; the ½ is chosen, not found.
t₁, t₂ emission and echo read on A's clock in s; tB the reading handed to B; ε dimensionless, 0 < ε < 1
The second postulate in testable form: c̄ = 2.998 × 10⁸ m s⁻¹ whatever the direction, the source, or the frame.
L the proper distance in m; both times read on the one clock at A, so no synchronisation enters
Experiment fixes the harmonic mean and the split is the convention: ε → 0 sends the outward speed to infinity and the return speed to c/2, and no single clock notices.
c₊ outward and c₋ return, both in m s⁻¹; ε = ½ is the only value that makes them equal
The shift expected on turning the apparatus through 90°: about 0.4 fringe for an 11 m path at Earth's orbital speed, and none was seen.
L arm length in m, λ vacuum wavelength in m, v the frame's speed through a putative preferred frame in m s⁻¹
A classical moving-source law puts the geometric mean at λ₀/γ and the midpoint on λ₀ itself, so the residual measures γ and nothing else.
λ± the head-on and tail-on lines in m; β = v/c of the beam; λ₀ the rest wavelength in m
Solving all three returns α = −½, β = ½, δ = 0 — the Lorentz transformation and nothing else — and ε never enters.
α, β, δ are the v²/c² coefficients of the Mansouri–Sexl time, longitudinal and transverse factors; β here is not v/c
The first postulate is a promise about every law
Galilean relativity already said that mechanics cannot tell a uniformly moving cabin from a stationary one. Einstein's first postulate widens that to every law of physics, and the widening is the whole content: Maxwell's equations must hold unchanged in the moving cabin, with the same μ₀ and ε₀, so c = 1/√(μ₀ε₀) = 2.998 × 10⁸ m s⁻¹ must come out as the same number in every inertial frame. That is a promise, not a measurement — it asserts that no experiment whatever, optical or mechanical, will single out a rest frame, which is why a null aether-drift result supports it without proving it. The promise costs something at once: if c is frame-independent while lengths and times behave classically, velocity addition breaks, so something in the kinematics has to give. What gives is the assumption that “now” is the same everywhere.
Why a one-way speed cannot simply be measured
Speed is distance over elapsed time, and an elapsed time between two places needs two clocks that already agree. Try to make them agree. Set them side by side and carry one to B: transport changes its reading by an amount you can only compute from a theory of moving clocks, which is what you were trying to test. Send a light signal instead: the correction depends on the one-way flight time, the very unknown. Every route closes the same loop. A round trip escapes it, because emission and reception happen at the same place on the same clock, so 2L/(t₂ − t₁) needs no second clock and no convention at all. That asymmetry is why every genuine test in this topic — interferometers, optical resonators, Doppler shifts — is built from closed light paths or from frequency ratios read at one point.
Einstein's rule, and Reichenbach's free parameter
Clock A emits at its own reading t₁ and receives the echo at t₂, so B stands L = c(t₂ − t₁)/2 away, a distance that needs only A's clock. Einstein's instruction is to set B to read tB = t₁ + ½(t₂ − t₁) at the moment of reflection. Reichenbach's observation is that the ½ is doing work no observation can justify, so write tB = t₁ + ε(t₂ − t₁) with 0 < ε < 1. The outward speed is then c₊ = L/[ε(t₂ − t₁)] = c/(2ε) and the return is c₋ = c/(2(1 − ε)): ε = 0.9 has light crawling out at 5c/9 and racing back at 5c. Notice what refuses to move: 1/c₊ + 1/c₋ = 2/c for every ε, so the round trip, the only interval one clock can time, is untouched. Einstein's value is chosen because it alone treats every direction alike — B setting A by the same rule agrees with A setting B — and because in the coordinates it defines Maxwell's equations keep their isotropic form.
Three experiments, three different questions
Michelson–Morley compares two round trips at right angles in one laboratory. Rotating the apparatus should shift the fringes by Δn = 2Lv²/(λc²), which for L = 11.0 m, λ = 589 nm and v = 30.0 km s⁻¹ is 0.374 fringe; nothing moved beyond 0.010. That tests isotropy of the two-way speed and nothing else. Kennedy–Thorndike deliberately makes the arms unequal and watches for months: with ΔL = 16 cm and green light, a change of order (30 km s⁻¹)² in v² as the seasons turn would move the pattern by about 3 × 10⁻³ fringe, and none appeared. It tests whether the two-way speed depends on the frame's velocity. Ives–Stilwell views the light from a fast hydrogen beam head-on and tail-on at once: relativity requires the geometric mean of the two shifted wavelengths to be exactly λ₀, whereas a classical moving-source law returns λ₀/γ. It tests the time-dilation factor on its own.
What the three fix together, and what they cannot touch
Put a general test theory between the experiments and the conclusion. Mansouri and Sexl allow a preferred frame and write the transformation with three unknown functions: a time factor a = 1 + αv²/c², a longitudinal length factor b = 1 + βv²/c², and a transverse factor d = 1 + δv²/c², where α, β and δ are coefficients, not velocities. Special relativity is the single point α = −½, β = +½, δ = 0. Michelson–Morley constrains β − δ = ½, Kennedy–Thorndike constrains β − α = 1, and Ives–Stilwell constrains α = −½: three equations, three unknowns, one solution. That is why no single null result establishes relativity — Lorentz's contracted aether satisfies the first and fails the second. And the synchronisation parameter ε appears in none of the three, because every observable in them is a closed path or a frequency ratio read at one place.
Where the convention bites, and where the rule breaks
Sort your results before quoting them. Anything built from an invariant — proper time along a worldline, the interval, a two-way speed, a Doppler ratio, a measured particle lifetime — carries no trace of ε and is safe to report. Anything that names a time at a distant place — “both ends of the rod at once”, a coordinate one-way speed, which twin is older “right now” — is partly the convention you adopted. Einstein synchrony also has a hard boundary: it is defined for inertial frames only. Carried around a rotating frame it fails to close, and synchronising station to station around the equator accumulates a Sagnac gap 2ωA/c², which for A = πR² = 1.28 × 10¹⁴ m² and ω = 7.29 × 10⁻⁵ rad s⁻¹ is 207 ns, or 62 m of position error. The repair is not a better convention but a single declared coordinate time for the whole rotating system.
Change one variable at a time
Make the relationship visible.
Slide ε off ½ and watch the dashed line tilt: at ε = 0.20 the assigned speeds split to 2.50 c out and 0.63 c back, while the round-trip reading on A's clock never moves. Widening the baseline scales the whole diagram and changes nothing about that.
OUTWARD c₊1.00 c
RETURN c₋1.00 c
ROUND TRIP ON A1.60 μs
B IS SET TO0.80 μs
Live interpretationOUTWARD c₊: 1.00 c. RETURN c₋: 1.00 c. ROUND TRIP ON A: 1.60 μs. B IS SET TO: 0.80 μs
Catch the common trap
Explain before calculating.
Two clocks 450 m apart have been set by Einstein's round-trip rule with ε = ½. A pulse is then sent one way across the baseline, and its recorded flight time gives a speed of 2.998 × 10⁸ m s⁻¹. What has the experiment established about the one-way speed of light?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyClock A emits a light pulse at its own reading t₁ = 0 and receives the reflection from clock B at t₂ = 3.00 μs. Find the distance to B, and the reading Einstein's rule tells B to display at the reflection. Then repeat with ε = 0.75, and give the two one-way speeds that choice assigns.
- The round trip is timed on one clock, so it needs no synchronisation: L = c(t₂ − t₁)/2 = 2.998 × 10⁸ × 1.50 × 10⁻⁶ = 450 m.
- Einstein's rule with ε = ½: tB = t₁ + ½(t₂ − t₁) = 1.50 μs. Both legs are then assigned 450 m ÷ 1.50 μs = 3.00 × 10⁸ m s⁻¹, which is c — by construction, not by measurement.
- With ε = 0.75 the instruction becomes tB = 0.75 × 3.00 μs = 2.25 μs, so the outward leg is assigned 2.25 μs and the return leg the remaining 0.75 μs.
- Outward: c₊ = 450 ÷ (2.25 × 10⁻⁶) = 2.00 × 10⁸ m s⁻¹ = 2c/3. Return: c₋ = 450 ÷ (0.75 × 10⁻⁶) = 6.00 × 10⁸ m s⁻¹ = 2c.
- Check the invariant: 1/c₊ + 1/c₋ = 3/(2c) + 1/(2c) = 2/c, exactly as for ε = ½. The round-trip reading on A is the same 3.00 μs either way, so no measurement separates the two accounts.
AnswerL = 450 m, and B is set to 1.50 μs with ε = ½. With ε = 0.75 it is set to 2.25 μs, light runs out at 2.00 × 10⁸ m s⁻¹ and back at 6.00 × 10⁸ m s⁻¹, and the round trip is unchanged.
MediumA Michelson interferometer has effective arm length L = 11.0 m and uses sodium light, λ = 589 nm. Predict the fringe shift on rotating it through 90° if the laboratory moves at v = 30.0 km s⁻¹ through a preferred frame. The observed shift stayed below 0.010 fringe: what bound on v does that set, and what does the null result leave open?
- The classical prediction for a 90° rotation is Δn = 2Lv²/(λc²) — second order in v/c, because the first-order terms cancel between the outward and return legs of each arm.
- Numerator: 2Lv² = 2 × 11.0 × (3.00 × 10⁴)² = 1.98 × 10¹⁰ m³ s⁻². Denominator: λc² = 589 × 10⁻⁹ × 8.988 × 10¹⁶ = 5.294 × 10¹⁰ m³ s⁻².
- So Δn = 1.98 × 10¹⁰ ÷ 5.294 × 10¹⁰ = 0.374 fringe, close to the 0.4 fringe Michelson expected and forty times the 0.010 fringe the 1887 apparatus could resolve.
- Invert for the bound: v = √(Δn λ c² / 2L) = √(0.010 × 589 × 10⁻⁹ × 8.988 × 10¹⁶ ÷ 22.0) = √(2.41 × 10⁷) = 4.9 × 10³ m s⁻¹.
- What stays open: both arms are closed paths, so a one-way anisotropy compensated on the return leg is invisible. A real contraction of the parallel arm by 1/γ also nulls the shift exactly, which is how Lorentz's aether survived this experiment.
AnswerΔn = 0.374 fringe predicted; a limit of 0.010 fringe bounds the laboratory's speed through any preferred frame at about 4.9 km s⁻¹, six times below Earth's orbital speed — but it constrains only the isotropy of the two-way speed.
HardIn an Ives–Stilwell experiment, hydrogen atoms in a canal-ray beam each carry 20.0 keV of kinetic energy and radiate the Hβ line, rest wavelength λ₀ = 486.13 nm; the beam is viewed head-on and tail-on at once. Take the emitter's rest energy as mₚ c² = 938.272 MeV, neglecting the electron's share. Find β, the two Doppler-shifted wavelengths, and the displacement of their midpoint from λ₀. Compare with the classical moving-source prediction, and state the precision the measurement demands.
- Energy first: γ − 1 = K/(mₚ c²) = 2.00 × 10⁴ ÷ 9.38272 × 10⁸ = 2.1316 × 10⁻⁵, so γ = 1.000021316 — a 21 ppm effect.
- β = √(1 − γ⁻²) = 6.5292 × 10⁻³, that is v = 1.957 × 10⁶ m s⁻¹.
- Relativistic longitudinal Doppler: λ± = λ₀√((1 ∓ β)/(1 ± β)). Here √((1 − β)/(1 + β)) = 0.993492, giving λ₊ = 482.966 nm approaching and λ₋ = 489.314 nm receding — a first-order splitting of 2βλ₀ = 6.348 nm.
- Midpoint: (λ₊ + λ₋)/2 = 486.1404 nm = γλ₀, displaced above λ₀ by λ₀(γ − 1) = 0.0104 nm. Equivalently √(λ₊λ₋) = 486.130 nm = λ₀ exactly.
- Classically λ± = λ₀(1 ∓ β) = 482.956 and 489.304 nm: the midpoint sits exactly on λ₀ and the geometric mean is λ₀/γ = 486.1196 nm. The two predictions differ by 0.0104 nm, one part in 4.7 × 10⁴.
- That displacement is 1.6 × 10⁻³ of the splitting, so each line centroid must be located to better than a thousandth of the gap. Hence the test is run as a ratio of the two shifted lines, and it probes γ alone — no synchronisation choice enters it.
Answerβ = 6.529 × 10⁻³; λ₊ = 482.966 nm and λ₋ = 489.314 nm; the midpoint lies at 486.1404 nm, 0.0104 nm above λ₀, where the classical law puts it exactly on λ₀. That needs the centroids to 1.6 × 10⁻³ of the 6.35 nm splitting.