University Physics IV · Special Relativity II · 3.1

Proper time and time dilation

Before you write a single γ, name the two events and find the clock that is present at both. That clock reads the proper time; every other inertial frame reads more. This lesson makes the bookkeeping automatic, so dilation stops being a paradox to argue about and becomes a substitution you can check.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Time dilation is not a statement about clocks running slow; it is a statement about which pair of events a clock manages to be present at. Fix two events — a muon is created, that same muon decays. In the muon's own rest frame both happen at one place, Δx′ = 0, so the interval between them is read off a single clock, and that reading is the proper time Δτ.

Any other inertial frame sees the muon move between the two events, so no single clock in that frame is at both; the interval there must be assembled from two clocks standing at different places, synchronised by a convention the moving frame does not share. Feed Δx′ = 0 into the inverse Lorentz transformation t = γ(t′ + vx′/c²) and the whole result drops out in one line: Δt = γΔτ, with γ = 1/√(1 − β²) ≥ 1. Proper time is therefore the shortest reading any inertial frame will report, and it belongs to the event pair rather than to a frame, because it is built from the invariant c²Δτ² = c²Δt² − Δx².

What the model costs is the idea of a universal now. The effect is reciprocal — each frame finds the other's clock slow — and that is consistent only because each frame is comparing one clock against a pair, and the two frames disagree about which distant clocks are synchronised.

Simple definition: Proper time is the interval between two events measured by a single clock that is present at both; time dilation is the statement that every other inertial frame assigns a larger interval, Δt = γΔτ.
Example: A muon travelling at 0.995c lives 2.20 μs on its own clock; γ = 10.0, so laboratory clocks record 22.0 μs between birth and decay and a flight of 6.57 km rather than 657 m.

Proper time: the condition that defines itΔτ = Δt′ with Δx′ = 0

Forces you to name the event pair and its home frame before any γ appears, which is where most solutions go wrong.

Δτ in s. The primed frame is the one where both events happen at the same place — the particle's rest frame.

Lorentz factorγ = 1/√(1 − β²), β = v/c

One number carries the whole size of the effect, and it hugs 1 until β is large: γ = 1.005 at β = 0.10.

β dimensionless, γ ≥ 1 always. β = 0.60 → γ = 1.25; β = 0.866 → γ = 2.00; β = 0.995 → γ = 10.0.

Dilation from the inverse boostΔt = γ(Δt′ + vΔx′/c²) = γΔτ

Shows dilation is one substitution into a transformation you already derived, not a separate postulate to accept.

Δt is the coordinate time in the frame the particle moves through; Δt, Δt′, Δτ in s, v in m s⁻¹.

The invariant that ranks the twoc²Δτ² = c²Δt² − Δx²

Subtracting a positive Δx² forces Δt > Δτ, so proper time is the shortest reading and is frame-independent.

Every term in m². Every inertial frame computes the same value of the right-hand side for the same event pair.

Decay in flightN(t) = N₀ e^{(−t/(γτ₀)}), d = βγcτ₀

Turns dilation into a counting prediction: at β = 0.9994, γ = 28.9 and d = 19.0 km instead of 0.66 km.

τ₀ = 2.20 μs for the muon, t is lab time in s, d the mean lab decay length in m.

Low-speed limitΔt ≈ Δτ (1 + ½β² + ⅜β⁴)

Recovers Newtonian absolute time as β → 0, and says why only atomic clocks and fast particles ever see the effect.

Fractional excess ≈ ½β². At 300 m s⁻¹ that is 5.0 × 10⁻¹³, or 43 ns per day.

Two events first, then a frame

Dilation problems go wrong at the first line, when γ is written before deciding what is being dilated. Fix that by naming the event pair. An event is a point in space and time: the muon is created at (t₁, x₁); the muon decays at (t₂, x₂). Every frame agrees these two things happened, and disagrees only about the coordinates. Now ask a mechanical question — is there an inertial frame in which both events occur at the same place? For a particle that is born, moves, and dies, yes: its own rest frame, where Δx′ = 0 and one clock carried along is present at both. The reading of that clock is the proper time Δτ between those two events, and the phrase 'the proper time' means nothing until the pair is named. It is not a property of a clock in general; the same clock has a different Δτ for a different pair. Two events joined by a light ray have no such frame at all: the invariant gives Δτ = 0 for them, and no clock, however fast, can be present at both.

Impose Δx′ = 0 and the algebra is one line

The boost from the laboratory S to the particle frame S′ reads t′ = γ(t − vx/c²) and x′ = γ(x − vt). Work in that direction and you must supply Δx, the laboratory separation, which is usually one of the unknowns. Use the inverse instead — swap the primes and reverse the sign of v — giving t = γ(t′ + vx′/c²). Take differences between the two events: Δt = γ(Δt′ + vΔx′/c²). Now impose the condition that defines the proper frame, Δx′ = 0. The second term vanishes, Δt′ is Δτ, and Δt = γΔτ. That is the whole derivation. Notice what did the work: not a claim that moving clocks are defective, but the fact that in S the two events are separated in space, so the laboratory interval has to be read off two clocks rather than one. With β = 0.60, γ = 1.25, and a proper interval of 4.00 s appears in the laboratory as 5.00 s.

Why the reciprocity is not a contradiction

Each frame finds the other's clock slow, and that sounds self-defeating until you notice the two measurements are not the same measurement. When S times a clock at rest in S′, it compares one moving clock against a pair of stationary, pre-synchronised clocks standing at different places. When S′ times a clock at rest in S, it does the same thing with its own pair. The asymmetry hides in the synchronisation: two clocks that S has set to agree are, in S′, offset by vΔx/c², the relativity of simultaneity. So the two frames are running different experiments, each internally consistent, and neither result contradicts the other. Every clock face read at an actual meeting is agreed by both. The moment you insist on bringing one clock back to meet the other a second time, something must accelerate, the situation stops being symmetric, and you have the twin problem rather than this one.

Proper time is the shortest, and it belongs to the events

Because γ ≥ 1 for every real v, Δt = γΔτ says the proper time is the smallest interval any inertial frame will assign to the pair. That is not an accident of the algebra; it follows from an invariant. Square the boost and the combination c²Δt² − Δx² comes out the same in every inertial frame. In the proper frame Δx′ = 0, so that invariant is exactly c²Δτ²; in any other frame it is c²Δt² − Δx² with Δx ≠ 0, and subtracting a positive quantity forces Δt > Δτ. Two consequences matter in practice. First, Δτ is a property of the event pair, so there is no 'really right' frame to hunt for. Second, the relation is one-directional numerically: an astronaut's 4.00 s at β = 0.60 is 5.00 s in the laboratory, and a laboratory interval of 5.00 s between those same two events is 4.00 s of proper time. You never multiply by γ twice.

Muons: the measurement that settles it

Cosmic-ray muons are produced high in the atmosphere and decay with a rest-frame mean lifetime τ₀ = 2.20 μs. At 0.995c a muon covers only 0.995 × 3.00 × 10⁸ × 2.20 × 10⁻⁶ = 657 m in one mean lifetime if dilation is ignored, so almost none should survive the roughly 2000 m from a mountain summit to sea level: the surviving fraction would be e^{(−2000/657)} = 0.048. With γ = 10.0 the laboratory mean lifetime is 22.0 μs, the mean decay length becomes 6.57 km, and the surviving fraction is e^{(−2000/6570)} = 0.74 — about fifteen times more muons at the bottom. Rossi and Hall first measured a survival ratio of this kind between two altitudes in Colorado in 1941, and Frisch and Smith repeated it two decades later between the summit of Mount Washington and sea level with muons selected near 0.995c; both results landed on the dilated side. Ride along with the muon and nothing odd happens to its clock: its mean life is still 2.20 μs. What has changed in its frame is the mountain, contracted from 2000 m to 200 m, which at 0.995c takes 0.67 μs to pass — and e^{(−0.67/2.20)} is the same 0.74.

What the formula does and does not cover

Δt = γΔτ is a statement about one pair of events and one constant relative velocity, and three restrictions follow. It does not apply to a clock whose speed changes: for that you integrate, τ = ∫√(1 − v(t)²/c²) dt, which is the next topic. It says nothing about what an observer sees, because light from the two events takes different times to arrive; the visual Doppler factor √((1 + β)/(1 − β)) is a separate calculation from γ, and confusing the two is a standard source of wrong answers about the twin problem. And it is not a claim about the mechanism of any particular clock. A caesium transition, a muon, and a pendulum all stretch by the same γ, because what dilates is the coordinate interval between two events, not the workings of a device. Whether acceleration itself alters a clock's rate — the clock hypothesis — is a separate experimental question, taken up with the worldline integral.

Change one variable at a time

Make the relationship visible.

Interactive model

Speed β = v/c0.80

Proper time Δτ2.2 μs

Push β from 0.20 to 0.95 with Δτ held at 2.2 μs and watch the decay climb the hyperbola while the gap between the two dashed levels opens from 0.05 μs to 4.85 μs — the proper time never moves, and the hyperbola's lowest point, the clock at rest, is why Δτ is the shortest reading.

LORENTZ FACTOR γ1.667

PROPER TIME Δτ2.2 μs

LAB TIME Δt = γΔτ3.67 μs

LAB FLIGHT βγcΔτ0.88 km

Live interpretationLORENTZ FACTOR γ: 1.667. PROPER TIME Δτ: 2.2 μs. LAB TIME Δt = γΔτ: 3.67 μs. LAB FLIGHT βγcΔτ: 0.88 km

Catch the common trap

Explain before calculating.

Misconception check

The common error is to treat γ as a factor you apply to whichever interval happens to be in front of you, so a laboratory time of 22.0 μs gets multiplied again to 220 μs, or a proper time is 'converted' by dividing. The relation runs one way only: Δt = γΔτ, proper time in, coordinate time out. The repair is procedural rather than algebraic — before writing γ, name the two events, ask which frame has both of them at one place, and call that frame's reading Δτ. The second error is believing the moving clock is physically damaged, that acceleration broke it or that the motion disturbed its mechanism. Nothing happened to the clock. What differs is that one frame timed the pair with a single clock while the other had to use two clocks standing at different places — and the two frames do not agree that those two clocks are synchronised.

A ship passes Earth at β = 0.80 and, moving at constant velocity, later passes Mars, 2.4 × 10¹¹ m away in the Earth–Mars rest frame. Take the two events to be the two passings. Which clock reads the proper time between them, and what does it read?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA pion is created in a laboratory and decays 3.5 × 10⁻⁸ s later, having travelled in a straight line at a constant 0.90c. Find γ and the proper lifetime this particular pion experienced.

Name the pair: creation and decay. The pion is at one place in its own rest frame for both, so that frame owns the proper time, and the laboratory's 3.5 × 10⁻⁸ s is the coordinate time Δt.
β = 0.90, so γ = 1/√(1 − 0.90²) = 1/√0.19 = 1/0.4359 = 2.294.
Δt = γΔτ runs proper-to-coordinate, so invert it: Δτ = Δt/γ = 3.5 × 10⁻⁸ / 2.294 = 1.53 × 10⁻⁸ s.
Check the direction before stopping: Δτ must be the smaller number, and 15.3 ns < 35 ns. The laboratory also records a flight of 0.90 × 3.00 × 10⁸ × 3.5 × 10⁻⁸ = 9.45 m, while that same laboratory distance is 9.45/2.294 = 4.12 m in the pion's frame.

Answerγ = 2.29 and Δτ = 1.53 × 10⁻⁸ s (15.3 ns): the pion's own clock reads less than half the laboratory's 35 ns, and the 9.45 m laboratory flight is only 4.12 m in the pion's frame.

MediumA beam of charged pions with rest-frame mean lifetime τ₀ = 26.0 ns is produced at β = 0.980 and sent down a 40.0 m flight tube. Find γ, the mean decay length in the laboratory, and the fraction of the beam reaching the far end. Compare that fraction with what a calculation ignoring dilation would predict.

γ = 1/√(1 − 0.980²) = 1/√0.0396 = 1/0.1990 = 5.03.
The proper lifetime belongs to the pion: τ₀ = 26.0 ns is read by a clock riding with it. The laboratory mean lifetime is the dilated one, γτ₀ = 5.03 × 26.0 = 131 ns.
Mean decay length in the laboratory: d = βc(γτ₀) = 0.980 × 3.00 × 10⁸ × 1.307 × 10⁻⁷ = 38.4 m.
Survival over 40.0 m: N/N₀ = e^{(−40.0/38.4)} = e^(−1.041) = 0.353, so about 35% of the beam gets through.
Ignore dilation and the decay length would be βcτ₀ = 0.980 × 3.00 × 10⁸ × 26.0 × 10⁻⁹ = 7.64 m, giving e^{(−40.0/7.64)} = e^(−5.23) = 0.0053. Beamlines are built on that factor of 66.

Answerγ = 5.03, mean laboratory decay length 38.4 m, and about 35% survival to 40.0 m. Ignoring dilation gives a decay length of 7.64 m and a survival of 0.53% — 66 times fewer pions at the far end.

HardFrame S′ moves at β = 0.60 along the x-axis of S. Clocks A at x = 0 and B at x = 3.00 × 10⁸ m are synchronised in S. A single clock C, at rest in S′, passes A when both read zero, and later passes B. Find what C and B read at the second event, then redo the bookkeeping in S′ — where C is at rest and B is the moving clock — and identify the term that stops the two accounts contradicting each other.

In S the two events are 3.00 × 10⁸ m apart, so Δt = 3.00 × 10⁸/(0.60 × 3.00 × 10⁸) = 1.667 s. B is synchronised with A in S, so B reads 1.667 s at the meeting.
C is present at both events, so C owns the proper time: γ = 1/√(1 − 0.36) = 1/0.80 = 1.25, and Δτ = Δt/γ = 1.667/1.25 = 1.333 s.
Cross-check in S′, where C is at rest and the A–B ruler sweeps past, contracted to 3.00 × 10⁸/1.25 = 2.40 × 10⁸ m. At 0.60c that takes 2.40 × 10⁸/(1.80 × 10⁸) = 1.333 s, matching C's own reading exactly.
During those 1.333 s, S′ says the moving clock B advances only 1.333/1.25 = 1.067 s. But B displays 1.667 s at the meeting, so 0.600 s is unaccounted for.
The missing term is simultaneity. In S′, at the instant C passes A, B is not reading zero: A and B sweep past in the −x′ direction, so the trailing clock B leads by vΔx/c² = (1.80 × 10⁸ × 3.00 × 10⁸)/(9.00 × 10¹⁶) = 0.600 s. Then 0.600 + 1.067 = 1.667 s, exactly what B shows.
So both frames agree on every clock face read at an actual meeting while disagreeing about rates. S timed one clock (C) against two (A and B); S′ timed one clock (B) against C plus a synchronisation offset it does not share.

AnswerC reads 1.333 s and B reads 1.667 s. In S′ that same 1.667 s is 0.600 s of simultaneity offset plus 1.067 s of B's dilated ticking, so the two frames predict identical clock faces at the meeting.