University Physics IV · Special Relativity II · 3.5

Contraction Paradoxes & Rigidity

Length contraction only looks paradoxical while you are still asking which frame is right. The cure is to stop scoring lengths against each other and start tabulating events. Do that and the barn shuts, the pole survives, rigid bodies leave physics, and a camera stops agreeing with a ruler.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Length contraction is not a claim about a rod; it is a claim about a procedure. To assign a length you must locate both ends at one time, and “at one time” is exactly what two frames disagree about, so a length is a joint statement about an object and a convention. Contraction paradoxes are built by hiding that second half. The pole-and-barn story asks whether a 5.0 m pole really fits a 3.0 m barn, as though fit were a property of the pole and not a relation between two door-closing events.

Replace the question with an event table — where and when each door shuts, in each frame — and the contradiction evaporates: the frames disagree about which door shut first, and agree that no door and no part of the pole ever share an event. What the resolution costs is the rigid body. A rod that kept its shape while you pushed it would move its far end at the instant you struck the near one, and that pair — simultaneous, separated in space — is reordered by a boost, so some frame watches the far end leap before the hammer lands. Rigidity is not an idealisation that improves with stiffer steel; it is forbidden.

What survives is Born rigidity, a class of motions so tightly constrained that the driver keeps almost no freedom. And one cost more: the contracted length is what a ruler returns, never what a lens returns, because the light reaching a camera at one instant left the near and far parts of the object at different times.

Simple definition: A contraction paradox is an apparent contradiction manufactured by comparing lengths across frames instead of comparing events, and it dissolves as soon as each frame writes out its own list of what happened, where and when.
Example: At β = 0.866 a 5.0 m pole measures 2.50 m in the barn frame and clears a 3.0 m barn by 0.50 m; in the pole's frame that barn is 1.50 m long and its far door shuts 17.3 ns before the near one.

Measured length against proper lengthL = L₀/γ (both ends located at one t)

Names the hidden premise: change whose “one time” you used and the number changes with it.

L₀ (m) is the rest-frame length, L (m) the measured one; only the dimension along the boost changes.

Offset between the two door-closingsΔt′ = −γβD/c for events with Δt = 0, Δx = D

Turns the paradox into a number: 17.3 ns across a 3.0 m barn at β = 0.866.

D (m) is the barn's proper length, β = v/c, and Δt′ (s) the gap the pole's frame reports; the far door leads.

Clearance at each doorgap = γs/2 = (γD − L₀)/2, with s = D − L₀/γ

Both frames report clearance, so neither frame ever closes a door on wood.

s (m) is the barn-frame spare room; the identical gap appears at the near door and at the far door.

Barn displacement between the closingsΔx′ = γβ²D

4.5 m for a 3.0 m barn at β = 0.866, which is three times that barn's own contracted length.

Δx′ (m) is the ground the barn covers in the pole's frame while the two doors shut in turn.

The stiffness ceiling on any materialv_sound = √(B/ρ) ≤ c ⇒ B ≤ ρc²

Steel's 1.6 × 10¹¹ Pa sits nine orders below it, and even the ceiling leaves a finite delay.

B (Pa) is the bulk modulus and ρ (kg m⁻³) the density; for steel the ceiling is 7.0 × 10²⁰ Pa.

Terrell–Penrose apparent rotationsin α = β, apparent width = L(cos α + sin α)

At β = 0.60 the turn is 36.9° and the two projections add to 1.40 L, wider than the cube at rest.

α is the apparent turn of a small distant object photographed broadside; L (m) is its proper side.

A fit is a relation between events, not a length

Run a pole of proper length L₀ = 5.0 m through a barn of proper length D = 3.0 m with a door at each end, at β = 0.866 so that γ = 2. The barn crew measures the pole at L₀/γ = 2.50 m and sees 0.50 m of room to spare. The pole crew measures the barn at D/γ = 1.50 m and sees 3.5 m of pole hanging out of it. Both statements are correct measurements, and asking which describes the real fit is the wrong question. The claim that the pole fits is shorthand for a claim about two events: the near door shuts, the far door shuts, and no part of the pole is at either door when it does. Events are frame-independent — either a door met wood or it met air, and no boost can change that. Lengths are not. So put the two lengths aside and write the event table.

The barn frame's ledger

Put the near door at x = 0 and the far door at x = D = 3.0 m, and slam both at t = 0 with the pole centred, so its tail is at 0.25 m and its nose at 2.75 m. With v = βc = 2.596 × 10⁸ m s⁻¹, the tail crossed the near door at t = −0.25/v = −0.96 ns and the nose reaches the far door at t = +0.96 ns, so the pole is wholly enclosed for a 1.93 ns window. Both doors are shut, both are shut at the same instant, and neither touches the pole. Nothing here is strange, which is why this is the frame everyone trusts — and exactly why it settles nothing on its own. A paradox is only resolved once the other frame's ledger is written too.

The pole frame's ledger — same events, different order

Boost the two closings with t′ = γ(t − vx/c²). The near door, at (t = 0, x = 0), gives t′ = 0. The far door, at (t = 0, x = 3.0 m), gives t′ = −γβD/c = −17.3 ns. In the pole's frame the far door shuts first, by 17.3 ns, and reopens long before it reaches the nose. Positions complete the picture. At t′ = −17.3 ns the far door sits at x′ = γD = 6.0 m and the pole's nose at 5.5 m, so the door shuts on 0.50 m of air. Nothing about the pole moves in this frame, so it is the door that eats that gap, arriving at the nose 1.93 ns later. Over the full 17.3 ns the barn sweeps γβ²D = 4.5 m back along the pole — further than its own proper length — and at t′ = 0 the near door shuts at x′ = 0 with the pole's tail at 0.50 m. The clearance γs/2 = 0.50 m is identical at both doors. Two orderings, one event list, no contact anywhere.

Rigidity dies with simultaneity

A rigid body keeps fixed separations no matter what you do to it. Strike one end of a rod of proper length L and its far end must start moving at the same instant: two events with Δt = 0 and Δx = L. Boost that pair and Δt′ = −γβL/c, so some frame sees the far end move before the hammer landed. Rigidity is therefore not an idealisation that gets better with stiffer steel; it is causally forbidden, and the bound shows up in material constants. The longitudinal signal speed √(B/ρ) can never exceed c, capping B at ρc² = 7.0 × 10²⁰ Pa for steel at ρ = 7.8 × 10³ kg m⁻³. Real steel manages 1.6 × 10¹¹ Pa — more than nine orders short — giving a sound speed of 4.5 km s⁻¹, or 1.5 × 10⁻⁵ c. Strike one end of a 1.00 m bar and the other end learns of it 0.22 ms later, some 66 000 light-crossing times.

Born rigidity, and Bell's thread

What survives is Born rigidity: motion in which every element holds a constant proper distance from its neighbours. It exists, but the constraint bites hard — fix one element's worldline and every other element's is already settled. A rod boosting along its own length must be driven harder at the tail than at the nose, with proper acceleration a(X) = c²/X measured from the Rindler horizon, so a rod of proper length L₀ cannot be pushed Born-rigidly with a nose acceleration above c²/L₀. Bell's spaceships show the cost of ignoring this. Two rockets separated by L in the launch frame run identical acceleration programmes, so their launch-frame separation stays exactly L — but each rocket's own frame measures the gap as γL, which grows without bound, and a taut thread between them snaps. Identical acceleration is not rigid motion; it is a programme that stretches the object.

What a camera sees is not what a ruler reports

Contraction is what a ruler reports when both ends are located at one time. A camera does something else: it collects light that arrives together, and light from an object's far parts left earlier. Photograph a cube of side L passing broadside at β. Light from the receding back corner travels an extra L, taking L/c, during which the cube advances βL — so that side face appears with width βL, beside a front face foreshortened to L√(1 − β²). Those two widths are precisely what an unsquashed cube turned through α, with sin α = β, would project. That is the Terrell–Penrose result: a small distant object photographs as rotated rather than flattened, and Penrose showed a sphere's outline stays a perfect circle at any speed. None of it softens contraction: the apparent turn is what finite light-travel time adds on top of a side that really does measure 0.800 L at β = 0.60.

Change one variable at a time

Make the relationship visible.

Interactive model

Speed β = v/c0.866

Pole proper length L₀5.0 m

Raise β and watch the lower picture stretch: the two closings pull further apart in time and the barn sweeps further between them. Then raise the pole's proper length until the spare room turns negative, and the doors start shutting on wood in both frames.

LORENTZ FACTOR γ2.00

POLE IN BARN FRAME2.50 m

SPARE ROOM IN BARN0.50 m

FAR DOOR LEADS BY17.3 ns

Live interpretationLORENTZ FACTOR γ: 2.00. POLE IN BARN FRAME: 2.50 m. SPARE ROOM IN BARN: 0.50 m. FAR DOOR LEADS BY: 17.3 ns

Catch the common trap

Explain before calculating.

Misconception check

The usual escape is to declare one frame honest — the pole really is 2.50 m long, because the pole is the thing that was accelerated — and to treat the pole frame's account as bookkeeping. That fails twice. It fails because both frames are inertial throughout the episode: the transformation knows nothing about which object was once pushed, and it is symmetric. And it fails on its own terms, because both frames agree the pole survives; if one were simply mistaken, the two would have to disagree about whether wood was struck, and they never do. The mirror-image error runs the other way. Told that a photograph shows a rotation rather than a squash, students conclude contraction is merely an appearance. It is not. Locate both ends at one barn-frame time and you get 2.50 m, and the doors shut on 0.50 m of air. Camera and ruler answer different questions, and both answer honestly.

A 5.0 m pole passes through a 3.0 m barn at β = 0.866, so γ = 2. In the barn frame both doors slam shut simultaneously at t = 0 and reopen at once, and the pole is untouched. What does the pole's own frame report?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA pole of proper length 12.0 m is to be enclosed, however briefly, inside a barn of proper length 7.2 m with a door at each end. Find the minimum speed β that makes this possible, then find the length the barn crew measures for the pole at β = 0.90 and the room it leaves at each end.

Enclosure in the barn frame needs the measured pole no longer than the barn: L₀/γ ≤ D, so γ ≥ 12.0/7.2 = 1.667.
Invert γ = 1/√(1 − β²): β = √(1 − 1/γ²) = √(1 − 1/1.667²) = √(1 − 0.360) = √0.640 = 0.800.
At β = 0.90, γ = 1/√(1 − 0.81) = 1/√0.19 = 2.294.
L = 12.0/2.294 = 5.23 m, so the spare room is s = 7.2 − 5.23 = 1.97 m, or 0.98 m at each end with the pole centred.

Answerβ ≥ 0.800 for any enclosure at all; at β = 0.90 the barn crew measures 5.23 m of pole and 1.97 m of spare room, 0.98 m at each end.

MediumFor that same 12.0 m pole and 7.2 m barn at β = 0.90 (γ = 2.294), the barn crew slams both doors simultaneously at t = 0. Working in the pole's frame, find the time between the two closings, how far the barn travels between them, and the clearance each door has.

Both closings have Δt = 0 and Δx = D = 7.2 m in the barn frame, so Δt′ = γ(Δt − vΔx/c²) = −γβD/c = −(2.294)(0.90)(7.2)/(2.998 × 10⁸) = −4.96 × 10⁻⁸ s. The far door shuts 49.6 ns before the near door.
In that interval the barn moves Δx′ = v|Δt′| = γβ²D = (2.294)(0.81)(7.2) = 13.4 m — more than four times its own contracted length.
Spare room in the barn frame is s = 7.2 − 12.0/2.294 = 1.969 m, so each door clears the pole by γs/2 = (2.294)(1.969)/2 = 2.26 m. The same number follows from (γD − L₀)/2 = (16.52 − 12.0)/2 = 2.26 m.
Sanity check in the pole's frame: the barn is only D/γ = 3.14 m long against a 12.0 m pole, so the pole is never enclosed there — yet each door still shuts on 2.26 m of clear air, which is the only thing both frames had to agree about.

AnswerThe closings are 49.6 ns apart, the barn covering 13.4 m between them, and each door shuts with 2.26 m of clearance.

HardA steel bar of proper length 1.00 m has ρ = 7.8 × 10³ kg m⁻³ and bulk modulus B = 1.6 × 10¹¹ Pa. (a) Struck on one end, how long before the far end moves, and how many light-crossing times is that? (b) Show that no bulk modulus makes the bar rigid. (c) The same steel is cast as a 1.00 m cube and flies past a distant camera at β = 0.60, broadside. Compare what a ruler measures with what the photograph shows.

Longitudinal signal speed vₛ = √(B/ρ) = √(1.6 × 10¹¹ / 7.8 × 10³) = √(2.05 × 10⁷) = 4.53 × 10³ m s⁻¹, so the far end waits 1.00/(4.53 × 10³) = 2.21 × 10⁻⁴ s = 0.221 ms.
Light crosses 1.00 m in 3.34 ns, so that delay is 6.6 × 10⁴ light-crossing times. Stiffening the steel cannot close the gap: vₛ ≤ c caps B at ρc² = 7.0 × 10²⁰ Pa, more than nine orders above steel's value, and even that ceiling still leaves 3.34 ns.
Rigidity demands zero delay, and zero is what no modulus buys. Far end and struck end moving together is a pair with Δt = 0, Δx = 1.00 m, and any boost gives Δt′ = −γβ(1.00 m)/c ≠ 0 — a frame in which the far end moved first. Rigidity is forbidden, not merely unavailable.
Ruler, at β = 0.60: γ = 1/√(1 − 0.36) = 1.25, so the side along the motion measures 1.00/1.25 = 0.800 m.
Camera: light from the receding back corner left L/c = 3.34 ns earlier, in which time the cube advanced βL = 0.600 m, so that face appears 0.600 m wide beside a front face foreshortened to L√(1 − β²) = 0.800 m.
Those are the projections of an unsquashed cube turned through α = arcsin 0.60 = 36.9°. The photograph spans 0.600 + 0.800 = 1.40 m — wider than the cube at rest — while the measurement is 0.800 m.

Answer0.221 ms, about 6.6 × 10⁴ light-crossing times; no modulus works, since B ≤ ρc² = 7.0 × 10²⁰ Pa still leaves 3.34 ns while rigidity needs zero; the ruler reads 0.800 m while the photograph spans 1.40 m as a 36.9° turn.