Proper time is what a moving clock records

A clock does not read coordinate time. It counts the time along its own path through spacetime, called proper time. Between the same two events, a clock that travels records less than one that stays put. Every observer agrees on that reading, and muons circling a storage ring confirm it.

Sandbox Physics · 2026-10-29

A clock reads the interval along its path

Follow a clock moving at constant velocity from one event to another. In its own rest frame it does not move, so $\Delta x'=0$ and the time it shows is $\Delta\tau=\Delta t'$ . Chapter 3 showed that the squared interval is the same in every frame. In A’s frame, therefore, the clock’s reading is the square root of the interval between the two events.

With $\Delta x=v\,\Delta t$ this becomes $\Delta\tau=\Delta t\sqrt{1-v^2}=\Delta t/\gamma$ . A clock moving at 0.6 of light speed records 0.8 years for every year of A’s coordinate time. For a path made of several straight pieces, add the pieces.

\Delta\tau=\sqrt{(\Delta t)^2-(\Delta x)^2}=\frac{\Delta t}{\gamma},\qquad \gamma=\frac{1}{\sqrt{1-v^2}}

Worked example

A round trip at 0.6 of light speed

Clock B leaves A at $t=0$ , travels out at $v=0.6$ for 5 years of A’s time, turns, and comes back at the same speed for another 5 years. On each leg $\gamma=1.25$ , so B records 4 years per leg.

At the reunion A’s clock reads 10 years and B’s reads 8. B’s ticks are spaced 1.25 years apart in A’s coordinates, so the tick from Chapters 1 and 2, the one B reads as 4 years, happens at $t=5$ , just as B turns. Both clocks are at the same place at the end, so the 2-year difference involves no light delay and no choice of simultaneity.

\tau_B=2\times5\,\mathrm{yr}\times\sqrt{1-0.6^2}=8\,\mathrm{yr},\qquad t_A=10\,\mathrm{yr}

The straight path records the most

Fix two events at the same place, 10 years apart. A clock that stays put records 10 years. Any clock that leaves and returns spends part of the time with $\Delta x\neq0$ , and each such piece subtracts from the square root. In spacetime, unlike in a flat plane, the straight line between two timelike-separated events is the longest path, measured by the time a clock records.

This reading is a count of ticks at given events, so every observer agrees on it. In the Lab, moving the drawing frame changes the coordinate time of a tick, while the number B shows there stays the same.

Muons on a circle keep the predicted time

A muon is an unstable particle. At rest its mean life is $\tau_0=2.1969811\,\mu\mathrm{s}$ (PDG 2024). In the 1977 CERN storage-ring measurement by Bailey and colleagues, muons circulated with $\gamma=29.33$ . The abstract, which is all we have read, reports a positive-muon lifetime of $64.419(58)\,\mu\mathrm{s}$ in the laboratory.

The proper-time prediction is $\gamma\tau_0=64.437\,\mu\mathrm{s}$ . It differs from the measurement by 0.018 microseconds, about a third of the quoted uncertainty. The muons move on a circle, so they are accelerating the whole time, yet their lifetime follows the speed alone through the factor $\gamma$ . The abstract states agreement with the expected dilation to about 2 parts in a thousand at 95 per cent confidence.

\gamma\tau_0=29.33\times2.1969811\,\mu\mathrm{s}=64.437\,\mu\mathrm{s},\qquad\tau_{\mathrm{lab}}=64.419\pm0.058\,\mu\mathrm{s}

Predict, change, explain

Predict B’s total reading if both legs run at 0.8 of light speed for 5 years of A’s time each.

In the light-signal Lab, set “First-leg velocity” to

0.8

and “Second-leg velocity” to

-0.8

, then open “More numbers and the saved record”.

Explain why the orange ticks on B’s worldline spread further apart as the speed grows, while A’s ticks stay one year apart.

The clock and twin Labs set light speed to one light-year per year, as these chapters do. The collision Lab uses PDG 2024 masses; the particle collider Lab draws its events from a teaching generator.

Try explaining it yourself

1. B travels at $v=0.8$ and records 3 years on each leg. How long is A’s wait?

Read an explained answer

Each leg lasts $3\times\tfrac53=5$ years of A’s time, so A waits 10 years while B records 6.

2. A muon beam has $\gamma=10$ . What mean life do you expect in the laboratory?

Read an explained answer

About $22.0\,\mu\mathrm{s}$ : ten times the rest mean life of 2.197 microseconds.

3. Which worldline from $(t,x)=(0,0)$ to $(10,0)$ records the most proper time?

Read an explained answer

Staying at rest, which records 10 years. Every detour adds spatial separation to some pieces and reduces the square root.

Sources to read alongside

Particle Data Group, 2024 particle listings: proton, neutral pion, charged kaon and muon

Masses used in the course: proton 938.27208816 MeV, neutral pion 134.9768 MeV, charged kaon 493.677 MeV; muon mean life 2.1969811 microseconds. The pion, kaon and muon values come from the matching 2024 listing files.

Bailey et al., Measurements of relativistic time dilatation for positive and negative muons in a circular orbit, Nature 268, 301 (1977)

Abstract only: the storage-ring Lorentz factor 29.33 and the positive-muon lifetime 64.419(58) microseconds. The full paper was not read, so the course compares only these two quoted numbers.

A clock reads the interval along its path

A round trip at 0.6 of light speed

The straight path records the most

Muons on a circle keep the predicted time

Predict, change, explain

Try explaining it yourself

1. B travels at v=0.8v=0.8v=0.8 and records 3 years on each leg. How long is A’s wait?

2. A muon beam has γ=10\gamma=10γ=10. What mean life do you expect in the laboratory?

3. Which worldline from (t,x)=(0,0)(t,x)=(0,0)(t,x)=(0,0) to (10,0)(10,0)(10,0) records the most proper time?

Sources to read alongside

1. B travels at $v=0.8$ and records 3 years on each leg. How long is A’s wait?

2. A muon beam has $\gamma=10$ . What mean life do you expect in the laboratory?

3. Which worldline from $(t,x)=(0,0)$ to $(10,0)$ records the most proper time?