University Physics IV · Special Relativity I · 2.8
Time dilation and length contraction
Two of the most quoted results in physics, and each is one line of algebra — provided you say first which frame holds the clock still and which frame marks the rod's two ends at the same instant. Leave that unsaid and the same algebra hands you each observer's clock running slow in the other's frame, with nothing left to tell you why that is not a contradiction.
Build the model
Connect the measurement to the mechanism.
Neither result is a new postulate. The boost relates the coordinates of two events in two frames and, on its own, says nothing about clocks or rods; what turns it into a measurement is the side condition you impose. Ask for the time between two events at the same place in S′ — set Δx′ = 0 — and Δt = γ(Δt′ + vΔx′/c²) collapses to Δt = γΔτ, so the proper time, read by the one clock present at both events, is the shortest interval any frame assigns.
Ask instead for the length of a rod at rest in S′ and you must first decide what length means for a moving object: the separation of its ends marked at one instant of the measuring frame, Δt = 0. Feed that into x′ = γ(x − vt) and the proper length comes out as L₀ = γL, so L = L₀/γ. The two γ's point opposite ways because the two constraints are different, and each result is a claim about a chosen pair of events rather than a property of the clock or the rod.
That is also what dissolves the reciprocity that looks like a contradiction: each frame checks a single clock belonging to the other against two clocks of its own that it synchronised, so the two accounts run over different event pairs and are not mirror images. The cost is the fine print — constant relative velocity, contraction along the boost axis only, transverse lengths untouched, and no claim at all about what a camera would photograph.
- Simple definition
- Proper time is what one clock present at both events reads, and every other inertial frame assigns a longer coordinate time Δt = γΔτ; proper length is a rod's length in its rest frame, and a frame that marks both ends at one instant gets the shorter L = L₀/γ.
- Example
- At β = 0.995 (γ = 10.01) a muon of proper lifetime 2.20 μs survives 22.0 μs of lab time and crosses 6.57 km of atmosphere; in its own frame it lives 2.20 μs and that 6.57 km is contracted to 656 m. One journey, two bookkeepings.
Says when either effect is measurable: γ − 1 rises as β², so 10% of c costs half a percent and 60% of c costs a full 25%.
β and γ dimensionless, γ ≥ 1; β = 0.10 → γ = 1.00504, β = 0.600 → γ = 1.250, β = 0.866 → γ = 2.000
The single clock does the work: Δτ needs one clock and Δt needs two synchronised ones, and γ always sits on the two-clock side.
Δτ in seconds, read by the single clock present at both events; Δt is coordinate time spanning two synchronised clocks
Returns Δt = γΔτ without choosing a frame, and shows why no frame owns the clock that is 'really' right.
Timelike pairs only, cΔt > |Δx|; Δτ is frame-independent, while Δt and Δx separately are not
Divide, do not multiply: the rod's own frame is the only one that can lay a ruler beside it at rest, and no other frame assigns more.
L₀ in metres, the rod's rest-frame length; L is the end-to-end separation marked at one instant of the measuring frame
Names the missing length: in the rod's frame the ends are marked at different times, and the drift in between restores L₀.
Δx = L is the marked separation; the offset is in the rod's own frame; β = 0.600 with L₀ = 12.0 m gives 24.0 ns
A rod lying across the motion is untouched; one at an angle both shortens and swings toward the transverse.
θ₀ is the rod's angle to v in its own frame, θ the lab's; y′ = y and z′ = z, so θ₀ = 90° returns L = L₀
Both results are the boost plus one side condition
The standard-configuration boost, ct′ = γ(ct − βx) and x′ = γ(x − βct), relates the coordinates of one event in two frames. Hand it a pair of events and it returns Δt′ and Δx′ — and that is all. Nothing in it yet mentions a clock running slow or a rod being short, because a measurement is not simply a pair of events: it is a pair of events chosen by a rule. Time dilation chooses the rule Δx′ = 0, meaning the two events happen where one clock is sitting. Length contraction chooses Δt = 0, meaning the two ends are marked at one instant of the measuring frame. Different rule, different answer, out of the same transformation. Write the constraint on the page before the algebra and the γ can never come out on the wrong side.
Time dilation: hold one clock still, then read Δt = γΔτ
Let a clock ride in S′. Two of its ticks are two events at the same place in S′, so Δx′ = 0. The inverse boost, Δt = γ(Δt′ + vΔx′/c²), then collapses to Δt = γΔt′ = γΔτ, where Δτ is the proper time — the interval read by that one clock. Since γ ≥ 1, every other inertial frame assigns more. At β = 0.866, γ = 2.000, so a 1.00 s tick aboard occupies 2.00 s of lab time. Notice the asymmetry hiding inside the procedure: the clock's own frame needs one clock, while the lab needs two clocks at different places, synchronised beforehand. The dilation is not a fault in the moving clock. It is a comparison between one clock and a pair, and the pair's synchronisation is exactly what depends on the frame.
Length contraction: the measurement is a simultaneity rule
A moving rod cannot be measured by laying a ruler beside it at leisure; you have to fix where both ends are, and 'both ends at once' is where the frame dependence enters. Put the rod at rest in S′ with its ends at x′ = 0 and x′ = L₀. In S, mark the ends at one lab time t. Apply x′ = γ(x − vt) to each mark: 0 = γ(xA − vt) and L₀ = γ(xB − vt). Subtract, and the vt cancels: L₀ = γ(xB − xA) = γL, so L = L₀/γ. Take L₀ = 15.0 m at β = 0.800, where γ = 1.667: the lab marks 9.00 m. Had you instead marked the ends simultaneously in S′, you would have recovered 15.0 m, because that is the rod's own rule. The rod did not change. The rule did.
Reciprocity is not a contradiction
Each frame says the other's clocks run slow and the other's rods are short, and both are right. The apparent paradox comes from imagining a single symmetric comparison when there are two different comparisons in play. When the lab says the ship's clock is slow, it is checking one ship clock against a succession of lab clocks it has synchronised. When the ship says the lab's clocks are slow, it is checking one lab clock against a succession of ship clocks it has synchronised. The two procedures use different event pairs, and the word doing the work — synchronised — is precisely what the two frames disagree about. Nothing here is a race with a winner. A real asymmetry needs something more, such as one twin changing frames, which is Unit 3's business.
The missing length is the simultaneity offset in metres
The deficit L₀ − L is not lost; it is the offset spent as distance. Work in the rod's frame and count what the lab's two markers do there. They are at rest in the lab a distance L apart, so the rod's frame sees them separated by L/γ = L₀/γ², and it sees them fire βL₀/c apart in time, the leading one first. In the interval between those firings the pair slides along the rod by βc · βL₀/c = β²L₀. Add the pieces: L₀/γ² + β²L₀ = L₀(1 − β²) + β²L₀ = L₀, exactly, at every speed. Put numbers on it with L₀ = 20.0 m at β = 0.600: the lab marks 16.0 m, the firings are 40.0 ns apart in the rod's frame, the markers stand 12.8 m apart there, and they drift 7.20 m in between, restoring 20.0 m. Neither frame has lost anything. One of them credits a drift the other's rule concealed.
Where the two corollaries stop
Four limits. First, both were derived between inertial frames in constant relative motion; an accelerating clock needs the proper time integrated along its worldline, τ = ∫√(1 − β²) dt, which is Unit 3's tool. Second, contraction acts along the boost axis only — y and z are untouched, so a rod at angle θ₀ in its rest frame measures L₀√(cos²θ₀/γ² + sin²θ₀) and swings toward the transverse by tan θ = γ tan θ₀. Third, contraction is not a squeeze: no force acts, no stress appears, and the rod's proper length is exactly what it was. Fourth, it is not what a camera records. A photograph gathers light that left different parts of an object at different times, so a small fast object images as rotated rather than flattened, and a fast sphere keeps a circular outline — the Terrell–Penrose effect. Contraction is a statement about coordinates assigned by a synchronised frame, not about an image.
Change one variable at a time
Make the relationship visible.
Push β from 0.20 to 0.80 and watch the filled dots close up while the open dot swings far right and high: the rod's frame never disputes the marks, only which pair counts as one measurement. At L₀ = 30 m and β = 0.80 the lab reads 18.0 m.
LORENTZ FACTOR γ1.250
LAB LENGTH L = L₀/γ19.20 m
LENGTH LOST L₀ − L4.80 m
MARK OFFSET IN ROD FRAME48.0 ns
Live interpretationLORENTZ FACTOR γ: 1.250. LAB LENGTH L = L₀/γ: 19.20 m. LENGTH LOST L₀ − L: 4.80 m. MARK OFFSET IN ROD FRAME: 48.0 ns
Catch the common trap
Explain before calculating.
A spacecraft of proper length 90.0 m passes a station at β = 0.800, so γ = 1.667. One station clock, sitting at a single marker, records 0.225 μs between the nose reaching it and the tail reaching it. What interval do the crew's own clocks assign to those same two events?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA spacecraft has proper length 25.0 m and proper width 8.00 m, and a clock aboard ticks once every 1.00 s of ship time. It passes a station at β = 0.600. Find the length and the width the station measures, and the interval station clocks record between two ticks.
- γ = 1/√(1 − 0.600²) = 1/√0.640 = 1/0.800 = 1.250.
- The 25.0 m is a proper length — measured where the ship is at rest — and it lies along the motion, so the station divides by γ: L = 25.0/1.250 = 20.0 m.
- The width is transverse to the boost. The transformation leaves y and z alone, so the station measures 8.00 m, unchanged.
- The two ticks happen at the same place aboard, so 1.00 s is a proper time and the station multiplies: Δt = γΔτ = 1.250 × 1.00 = 1.25 s. The two γ's point opposite ways because the constraints differ — Δt = 0 for the length, Δx′ = 0 for the clock.
AnswerLength 20.0 m; width unchanged at 8.00 m; 1.25 s between ticks on station clocks. γ divides the length and multiplies the time.
MediumA rod of proper length 12.0 m flies past a laboratory at β = 0.600. The lab marks the positions of both ends at one instant of lab time and gets 9.60 m. In the rod's own frame those two marking events are not simultaneous. Find their separation in time, and show that the rod's frame still measures 12.0 m.
- γ = 1/√(1 − 0.360) = 1.250, so the lab's marked length is L = 12.0/1.250 = 9.60 m — which is what it got.
- The marks are simultaneous in the lab: Δt = 0 and Δx = 9.60 m. Transform: Δt′ = γ(Δt − vΔx/c²) = −γβΔx/c = −(1.250)(0.600)(9.60)/(2.998 × 10⁸) = −2.40 × 10⁻⁸ s. In the rod's frame the front end is marked 24.0 ns before the rear.
- In the rod's frame it is the lab's marking apparatus that is contracted: its two markers are 9.60 m apart in the lab, so here they are 9.60/1.250 = 7.68 m apart.
- Between the two firings the marker pair drifts along the rod by vΔt′ = 0.600 × 2.998 × 10⁸ × 2.40 × 10⁻⁸ = 4.32 m.
- The rod's frame adds them: 7.68 m of marker separation plus 4.32 m of drift = 12.0 m = L₀. Both frames agree on where the marks landed; they disagree only about 'at the same time'.
AnswerThe marks are 24.0 ns apart in the rod's frame, front end first; 7.68 m of marker separation plus 4.32 m of drift restores the full 12.0 m.
HardA beam of charged pions, proper mean life 26.0 ns, is produced at one end of a 42.0 m flight tube and travels its length at β = 0.950. What fraction reaches the far end? Do the calculation in the laboratory frame and again in the pions' frame, and say what the naive answer would have been.
- γ = 1/√(1 − 0.950²) = 1/√0.0975 = 1/0.31225 = 3.203, and v = 0.950 × 2.998 × 10⁸ = 2.848 × 10⁸ m s⁻¹.
- Lab flight time: Δt = 42.0/(2.848 × 10⁸) = 1.475 × 10⁻⁷ s = 147.5 ns. But the decay law runs on the pion's own clock, so convert with the Δx′ = 0 constraint: Δτ = Δt/γ = 147.5/3.203 = 46.0 ns.
- Survival fraction: N/N₀ = exp(−Δτ/τ) = exp(−46.0/26.0) = exp(−1.769) = 0.170, so 17.0% arrive.
- The same number from the beam's frame: the tube is contracted to L = 42.0/3.203 = 13.1 m and sweeps past at 0.950c, taking 13.1/(2.848 × 10⁸) = 4.60 × 10⁻⁸ s = 46.0 ns — the same proper time, as it must be.
- The naive calculation feeds the lab's 147.5 ns straight into the decay law: exp(−147.5/26.0) = exp(−5.673) = 0.0034, or 0.34%. That is 49 times too few, and it is the error of pairing a lifetime measured in one frame with a time measured in another.
Answer17.0% survive. The proper time along the flight is 46.0 ns, not the lab's 147.5 ns; feeding the lab time into the decay law gives 0.34%, low by a factor of 49.