University Physics IV · Special Relativity II · 3.4
Length Contraction & Proper Length
To measure something that is moving you have to catch both ends at once — and "at once" is exactly what relativity refuses to make universal. This lesson turns that one procedural sentence into L = L₀/γ, keeps the transverse directions out of it, and shows that nothing whatever has been done to the rod.
Build the model
Connect the measurement to the mechanism.
A length is not read off a body the way a mass is; it is computed from two events, and the whole of relativity's intervention happens where those two events get chosen. Classically the choice is invisible, because every frame agrees about which pairs of events are simultaneous, so length can be treated as a property the body carries around and nobody is ever forced to notice the convention buried inside it. Once the boost mixes space and time that agreement is gone, and one rod handed to two frames yields two numbers — not because the frames disagree about the rod, but because they are not comparing the same pair of events.
What singles out the rest frame is only this: there, and nowhere else, the answer does not depend on timing at all, which is what makes proper length L₀ the anchor that every other frame's L = L₀/γ is measured against. Three features are worth carrying forward. The shrinkage acts along the relative velocity alone, so a fast sphere is measured as a spheroid rather than as a smaller sphere.
It is exactly symmetric between the two frames, so neither answer is a correction to the other. And it is kinematics rather than mechanics: the atoms of the rod never register that anyone measured it. Send β to zero and γ to one and the classical picture returns unchanged, which is why a quantity resting on a hidden convention went three centuries without anybody having to examine the convention.
- Simple definition
- The length of a moving object is the distance between the positions of its two ends taken at the same instant in the measuring frame, and it comes out shorter than the proper length by the factor 1/γ.
- Example
- A 100 m rocket at β = 0.600 has γ = 1.250, so a lab that marks both ends at one lab instant finds them 80.0 m apart — while the rocket's own crew, at rest with it, keep measuring 100 m.
The only length a frame can claim without a simultaneity convention — every other frame's answer is derived from it.
Metres. The ends' coordinates are constant in time, so no instant has to be chosen.
Without this line 'length' is undefined for a moving body, and any answer at all can be manufactured by delaying one reading.
Δt = 0 holds in the measuring frame only; in S′ those same two readings are not simultaneous.
β = 0.600 → γ = 1.250 and L = 0.800 L₀; β = 0.866 → γ = 2.00 and the rod measures half its proper length.
β = v/c is dimensionless; L and L₀ in metres. Applies along the relative velocity only.
A sphere measures as an oblate spheroid; a gold nucleus at γ ≈ 100 measures a hundredth as thick along the beam as it is across.
Widths in metres, V in m³, ρ in kg m⁻³; mass is invariant, and only the dimension along v is divided by γ.
This is contraction's whole content: 100 m of proper length at β = 0.600 puts the lab's two readings 200 ns apart in the rod's frame.
Seconds, with L₀ in metres. In the object's own frame the leading end is the one read first.
A 70 m airliner at 300 m s⁻¹ loses 3.5 × 10⁻¹¹ m — about a third the width of a hydrogen atom, which is why nobody noticed.
Dimensionless fractional shortening; the next term is ⅛β⁴.
A moving length is undefined until you fix an instant
For an object at rest you can measure at leisure: read one end this morning and the other after lunch, and because neither end has moved the answer is the same. That leisure is what makes proper length L₀ well defined, and it is why L₀ belongs to the object's rest frame alone. Set the object moving and the leisure is gone. Take a 100 m rod passing a lab at β = 0.600, which the lab will find to be 80.0 m long: read its nose at t = 0 and its tail 100 ns later, by which time the tail has advanced 18.0 m, and you report 62.0 m. Read the tail first and the nose 100 ns later and you report 98.0 m. The reading has become a story about your delay rather than about the rod. So the procedure must be pinned down: mark both ends at the same time in the frame doing the measuring, Δt = 0 there. A length is therefore a pair of events, not a property — and the moment simultaneity stops being universal, that pair stops being universal too.
Four lines from the boost — and the slip that reverses it
Put the rod at rest in S′ with its ends at fixed coordinates x′₁ and x′₂, so L₀ = x′₂ − x′₁ at whatever times you like. The lab S reads both ends at one lab time t. Apply x′ = γ(x − vt) to each event and subtract: x′₂ − x′₁ = γ[(x₂ − x₁) − v(t − t)] = γ(x₂ − x₁). The v-term dies because Δt = 0, and what survives is L₀ = γL, so L = L₀/γ = L₀√(1 − β²). The common slip is to start instead from the inverse form x = γ(x′ + vt′) and set Δt′ = 0. That is a legal calculation, but it answers a different question — the lab separation of two events simultaneous in the rod's frame — and it delivers L = γL₀, which is backwards. Choose the form of the transformation whose Δt is the one your apparatus actually zeroed. One check catches the error every time: γ ≥ 1, so the largest number always belongs to the frame in which the object is at rest.
Along the boost only, and why transverse must be untouched
The boost leaves y and z alone: y′ = y, z′ = z. This is not an aesthetic choice, and a symmetry argument forces it. Let a peg and a hole have equal proper diameters and send the peg through, its motion along the axis. If transverse lengths contracted, the hole's frame would say the peg is narrow and it passes cleanly, while the peg's frame would say the hole is narrow and it jams. Whether the peg scrapes is a single physical fact that no change of frame can rewrite, so the only consistent transverse factor is 1. Consequences follow at once. A sphere of proper radius r measures as an oblate spheroid, squashed to r/γ along v while staying r across; a volume becomes V₀/γ and, since mass is invariant, a density γρ₀. Charge is invariant while the volume holding it is not, so charge density transforms as well — which is exactly why a wire that is neutral in the lab is not neutral in the electrons' frame, and why the force one frame calls magnetic another calls electric.
Contraction is the simultaneity offset, converted
Run the lab's measurement through the boost and watch where the missing metres go. Rod of proper length 100 m, β = 0.600, γ = 1.250, so the lab gets L = 80.0 m by marking the ends at t = 0: events R = (x = 0, t = 0) and F = (x = 80.0 m, t = 0). In the rod's frame those same events sit at x′ = 0 and x′ = 100 m — the rod's full length, as they must, since the marks landed on its ends — but at t′ = 0 and t′ = −200 ns. The offset is exactly Δt′ = vL₀/c² = 0.600 × 100 m/c. So the rod's crew describe the lab's procedure like this: your two markers are 80.0 m apart in your frame, so they span only 64.0 m in mine; you marked my nose 200 ns early, your marker pair then slid 36.0 m before marking my tail, and 64.0 + 36.0 = 100 m is my rod. Nobody made an arithmetic error. The lab's number is short precisely because the readings the lab calls simultaneous are, in the rod's accounting, spread over 200 ns.
Both frames call the other short, and nothing breaks
Contraction is reciprocal: two 100 m rods passing at β = 0.600 each measure the other at 80.0 m. That sounds like a contradiction until you notice that 'shorter than' is not a frame-free relation between rods — it is a claim about a pair of events, and the two crews use different pairs. Follow the frame-free facts instead, which are coincidences. Let the two noses meet at t = 0 as B overtakes A. In A's frame B is 80.0 m long, so B's tail already lies 20.0 m past A's tail, and the tails met 111 ns before the noses did. In B's frame A is the short one, A's tail falls 20.0 m short of B's tail, and the tails meet 111 ns after the noses. The two frames reverse the order of the same two events — which is permitted, because with Δx = 100 m against cΔt = 33.3 m the pair is spacelike separated, and only spacelike pairs can be reordered. No signal outruns light, no causal chain is disturbed, and no contradiction has been created.
Where it shows up, and why it is not a squeeze
The evidence is routine. Muons made at 10 km with β = 0.995 have γ = 10.0: in the ground frame the trip takes 33.5 μs against a dilated lifetime of 22.0 μs, while in the muon's frame the atmosphere is contracted to 1.00 km and crossed in 3.35 μs against the proper lifetime of 2.20 μs. Both give 1.52 lifetimes and a survival fraction of 0.22 — one frame contracts a length, the other dilates a time, and the counter does not care which story you tell. At 7 TeV a circulating proton has γ ≈ 7460, so the LHC's 26.7 km ring measures 3.58 m to it. None of this is a squeeze. Lorentz and FitzGerald first proposed contraction as a real dynamical deformation produced by motion through the aether, and that version returns the same measured lengths, so no single measurement refutes it. What retired it is universality: the same factor applies to steel, to a beam of light and to the ticking of a clock, which makes it a statement about spacetime rather than about the forces holding any one material together. Accelerating a rod up to speed does involve genuine internal forces, but that is a separate, dynamical question.
Change one variable at a time
Make the relationship visible.
Start at β = 0.60 and read 80.0 m against a proper 100 m, then push β to 0.95: the lab bar falls to 31.2 m while the rod's own bar never moves, and the mark offset climbs from 200 ns to 317 ns.
LORENTZ FACTOR γ1.25
LAB LENGTH L80.0 m
SHORTFALL L₀ − L20.0 m
MARK OFFSET Δt′200 ns
Live interpretationLORENTZ FACTOR γ: 1.25. LAB LENGTH L: 80.0 m. SHORTFALL L₀ − L: 20.0 m. MARK OFFSET Δt′: 200 ns
Catch the common trap
Explain before calculating.
A rocket of proper length 20.0 m passes a station at β = 0.800, so γ = 1.667. The station crew mark the positions of its nose and its tail at the same station instant. Which account of that measurement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA metal bar of proper length 2.40 m and proper width 0.10 m is flown lengthways past a laboratory at v = 0.600c. Find γ, the length the laboratory measures, and the width the laboratory measures.
- β = v/c = 0.600, so 1 − β² = 1 − 0.360 = 0.640 and γ = 1/√0.640 = 1/0.800 = 1.250.
- The proper length belongs to the bar's rest frame: L₀ = 2.40 m. The laboratory is not that frame, so it gets L = L₀/γ.
- L = 2.40/1.250 = 1.92 m — a shortfall of 0.48 m, or 20.0% of the proper length.
- The width is perpendicular to v, and y′ = y under the boost, so the laboratory still measures 0.10 m. The bar measures as a different shape, not as a smaller copy of itself.
Answerγ = 1.250, so L = 1.92 m along the motion while the width stays 0.10 m; the bar measures 20.0% shorter than at rest.
MediumA spacecraft of proper length 120 m passes a station marker at v = 0.800c. (a) How long does the station measure the craft to be? (b) How long, by station clocks, does the craft take to pass the marker? (c) How long does that same passage take by the craft's clocks, and do the two answers agree?
- γ = 1/√(1 − 0.800²) = 1/√0.360 = 1/0.600 = 1.667, so the station measures L = 120/1.667 = 72.0 m.
- (b) In the station frame the craft's 72.0 m sweeps past a fixed marker at v = 0.800c = 2.398 × 10⁸ m s⁻¹, so t = 72.0 ÷ (2.398 × 10⁸) = 3.00 × 10⁻⁷ s = 300 ns.
- (c) The craft's crew watch the marker run the whole proper length 120 m at 0.800c: t′ = 120 ÷ (2.398 × 10⁸) = 5.00 × 10⁻⁷ s = 500 ns. Nothing is contracted here — the moving object is the marker, and a point has no length.
- Cross-check: both events happen at the marker, one place in the station frame, so 300 ns is the proper time between them and the craft's frame must read γ × 300 ns = 1.667 × 300 = 500 ns. The two accounts are one fact told twice, and they agree.
AnswerL = 72.0 m; the passage takes 300 ns by station clocks and 500 ns by ship clocks, the two tied together by γ = 1.667.
HardA thin rod of proper length 2.00 m lies at 60.0° to the x-axis in its own rest frame. It flies along the x-axis at v = 0.800c past a laboratory, which marks both ends at one laboratory instant. Find the length and the orientation the laboratory measures, and state how far apart in time the two marking events are in the rod's frame.
- γ = 1/√(1 − 0.800²) = 1/√0.360 = 1/0.600 = 1.667. Only the component along v is affected, so resolve the rod before applying anything.
- Rest-frame components, along v (the x-direction) and across it (the y-direction): L₀∥ = 2.00 cos 60.0° = 1.00 m and L₀⊥ = 2.00 sin 60.0° = 1.732 m.
- The laboratory marks both ends at Δt = 0, so its along-v separation is L∥ = L₀∥/γ = 1.00 × 0.600 = 0.600 m, while L⊥ = 1.732 m is untouched because y = y′.
- L = √(0.600² + 1.732²) = √(0.360 + 3.000) = √3.360 = 1.83 m. That is shorter than 2.00 m but not by 1/γ, because only part of the rod lay along v.
- The rod also tilts: tan θ = L⊥/L∥ = 1.732/0.600 = 2.887, so θ = 70.9° against a proper 60.0°. Equivalently tan θ = γ tan θ₀, and the rod swings toward the transverse direction because only the along-v leg shrank.
- The two marking events are simultaneous in the laboratory and Δx = 0.600 m apart there, so in the rod's frame they are Δt′ = γvΔx/c² = vL₀∥/c² = 0.800 × 1.00 m/c = 2.67 ns apart, the leading end marked first.
- Limits: θ₀ = 0 returns L = 2.00/1.667 = 1.20 m with no tilt, and θ₀ = 90.0° returns the full 2.00 m with no change at all.
AnswerL = 1.83 m at 70.9° to the x-axis, against a proper 2.00 m at 60.0°; the along-v leg falls from 1.00 m to 0.600 m while the transverse 1.732 m is unchanged, and the two laboratory marks are 2.67 ns apart in the rod's frame.