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Physics IV course sequence · Interactive lesson

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.

UPIV-02.03 · Lesson 3 of 9 in this unit. Position is not a completion score.
Starting model: The postulates and clock synchronisationSpacetime diagram of the synchronising signal: distance along the baseline runs right, clock A's reading runs up, and events off A's worldline are placed with the ε = ½ chart. Light leaves A, reflects at clock B 240 m away, and is back 1.60 μs later on A's clock. The open circle marks the reading 0.80 μs the rule hands B at the reflection, and the dashed line through it and the reflection event is what the chosen ε calls simultaneous. One-way speeds implied: 1.00 c out, 1.00 c back.ε = 0.50: light out at 1.00 c, back at 1.00 cdashed line: the events this ε calls simultaneoust₁ = 0set B to 0.80 μst₂ = 1.60 μsclock Aclock Bround trip on A: 1.60 μs for every ε
Start with a picture. Change the model in the lab below.
1

Build the model

One speed is measured. Half of it is stipulated.

The key idea

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.

The complete physical model and assumptions

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.

1The 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.

2Why 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.

3Einstein'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.

4Three 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.

5What 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.

6Where 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.

2

Key equations

Read the units, assumptions and limits before substituting.

Einstein synchronisation

tB = t₁ + ε(t₂ − t₁), with ε = ½

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

Two-way speed — the measurable one

c̄ = 2L/(t₂ − t₁)

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

One-way speeds under a chosen ε

c₊ = c/(2ε)c₋ = c/(2(1 − ε))1/c₊ + 1/c₋ = 2/c

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

Michelson–Morley fringe shift

Δn = 2Lv²/(λc²)

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

Ives–Stilwell Doppler test

λ± = λ₀√((1 ∓ β)/(1 ± β))√(λ₊λ₋) = λ₀(λ₊ + λ₋)/2 = γλ₀

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

What the three experiments each fix

MM: β − δ = ½KT: β − α = 1IS: α = −½

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
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.50
0.2 to 0.8; steps of 0.05.
240 m
150 to 300 m; steps of 30.

Try this. 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.

Live results

OUTWARD c₊
1.00 c
RETURN c₋
1.00 c
ROUND TRIP ON A
1.60 μs
B IS SET TO
0.80 μs

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics IV modelSpacetime diagram of the synchronising signal: distance along the baseline runs right, clock A's reading runs up, and events off A's worldline are placed with the ε = ½ chart. Light leaves A, reflects at clock B 240 m away, and is back 1.60 μs later on A's clock. The open circle marks the reading 0.80 μs the rule hands B at the reflection, and the dashed line through it and the reflection event is what the chosen ε calls simultaneous. One-way speeds implied: 1.00 c out, 1.00 c back.ε = 0.50: light out at 1.00 c, back at 1.00 cdashed line: the events this ε calls simultaneoust₁ = 0set B to 0.80 μst₂ = 1.60 μsclock Aclock Bround trip on A: 1.60 μs for every ε
Spacetime diagram of the synchronising signal: distance along the baseline runs right, clock A's reading runs up, and events off A's worldline are placed with the ε = ½ chart. Light leaves A, reflects at clock B 240 m away, and is back 1.60 μs later on A's clock. The open circle marks the reading 0.80 μs the rule hands B at the reflection, and the dashed line through it and the reflection event is what the chosen ε calls simultaneous. One-way speeds implied: 1.00 c out, 1.00 c back.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Synchronisation choice ε: 0.50; Baseline A to B: 240 m

Baseline 0.50; choose 0.2 to 0.8, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The commonest error is to say that Michelson–Morley measured the speed of light in two directions and found it the same. It measured no one-way speed at all. Light went out and came back, and the interferometer compared the total times of two closed paths, so an anisotropy that slows light one way and speeds it up by the compensating amount on the return leaves the fringes exactly where they were. That is not a loophole a better interferometer could close; it is the reason ε = ½ has to be a definition. The second, related error is to treat the null result as proof of relativity: Lorentz's aether theory, with a real contraction by 1/γ along the motion, predicts precisely the same null result, which is why Kennedy–Thorndike and Ives–Stilwell had to be done at all.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Clock 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

A 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?

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

In 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

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?
7

Review & continue

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.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.