University Physics IV · Special Relativity II · 3.2
Proper time along a worldline
One γ for a whole journey is a special case that almost no real trip satisfies. Chop the worldline into instants instead, read the tick rate at each and add — the same integral then says which of two paths between one pair of events banks more time, and what has to be assumed about the clock before any of it can be trusted.
Build the model
Connect the measurement to the mechanism.
Time dilation as usually met is a statement about one pair of events and one constant velocity: Δt = γΔτ, with a single γ. Almost no real journey is like that. The repair starts from the invariant interval, c²dτ² = c²dt² − dx², which every inertial frame evaluates to the same number. Divide it by dt² and it becomes a rate rather than a ratio: dτ/dt = √(1 − v(t)²/c²), legitimate instant by instant because at each instant there is a momentarily co-moving inertial frame in which the clock is at rest.
Proper time is whatever that rate accumulates to, τ = ∫√(1 − v(t)²/c²) dt, and it belongs to the path through spacetime, not to any observer: different frames assign different coordinates to the same worldline and all return the same τ. Two consequences pay for the machinery. First, τ is the integral of 1/γ, never 1/γ evaluated at an average speed, so the mean-speed shortcut fails the moment the speed changes, and it always fails on the generous side. Second, among all worldlines joining two timelike-separated events, the unaccelerated one banks the most proper time — Minkowski's triangle inequality runs backwards, and every bend subtracts.
The cost is an assumption you cannot derive: that an ideal clock's rate depends on its speed alone and not on its acceleration. That is the clock hypothesis, and it is held up by experiment, not by proof.
- Simple definition
- The proper time along a worldline is the reading of a clock carried along it, accumulated as dt/γ(t) instant by instant, so it depends on the whole speed history rather than on any single γ.
- Example
- A beacon flown at 0.800c for 30.0 s of station time and then held at rest for 20.0 s logs 30.0(0.600) + 20.0 = 38.0 s while the station logs 50.0 s.
Proper time belongs to the worldline, not to a frame, which is why it can be accumulated along a path.
dτ in s; dx, dy, dz in m; c = 2.998 × 10⁸ m s⁻¹. Every inertial frame gets the same dτ from its own dt and dx.
Replaces an accelerating clock by a stack of momentarily co-moving inertial ones, one per instant.
v(t) is the speed in the chosen frame, m s⁻¹; γ(t) ≥ 1, so dτ ≤ dt, with equality only while at rest.
The only version that survives a changing speed: it fixes τ for any timelike path, accelerated or not.
Limits are the coordinate times of the two events in one frame; τ comes out the same in every frame.
Turns most problems into a three-term sum, and shows a leg spent at rest contributing its Δtᵢ in full.
Δtᵢ is leg i's coordinate duration, βᵢ = vᵢ/c its constant speed. Direction never enters; only βᵢ² does.
The triangle inequality reversed: every bend subtracts. It is a maximum, never a minimum.
Δt and Δx are the coordinate separations of the two events, in s and m; equality only for the unbent path.
The one smooth accelerated worldline that integrates in closed form — the benchmark for numerical work.
α is the acceleration felt on board, m s⁻²; t is frame time from rest. c/α = 0.968 yr when α = g.
The interval gives a rate, not a ratio
Start from the one quantity every inertial frame agrees on: c²dτ² = c²dt² − dx² − dy² − dz², where dτ is what a clock present at both ends of that infinitesimal interval reads. Divide through by dt² and take the root: dτ/dt = √(1 − v(t)²/c²) = 1/γ(t). Nothing in that step assumed constant velocity. It is legitimate for an accelerating clock because at each instant there exists a momentarily co-moving inertial frame — one in which the clock is instantaneously at rest — and the interval is invariant, so the frame you compute in cannot change the answer. What comes out is a rate: how fast the carried clock ticks relative to coordinate time at this instant, at this speed. The familiar Δt = γΔτ is only the special case in which v never changes, so the rate can be pulled out of the integral as a constant.
Chop the journey, then add
Split the coordinate time between the two events into slices Δtᵢ short enough that v is effectively constant across each. Each slice advances the carried clock by Δtᵢ√(1 − vᵢ²/c²). Sum and refine, and the sum becomes τ = ∫√(1 − v(t)²/c²) dt, taken between the coordinate times of the two events in whichever single inertial frame you have chosen. Written this way, τ = T⟨1/γ⟩: the time-weighted average of the tick rate, times the coordinate duration. That is a different object from 1/γ evaluated at the average speed, and the gap is not small. Four hours at 0.600c followed by four hours at 0.800c gives 4.00(0.800) + 4.00(0.600) = 5.60 h, while the mean speed 0.700c predicts 8.00(0.7141) = 5.71 h. Because √(1 − β²) is concave in β, the speed-first shortcut always comes out too high.
Piecewise speeds reduce to a table
Most problems hand you a worldline as a few legs of constant speed, and then no calculus is needed: τ = Σ Δtᵢ√(1 − βᵢ²). Build a table with one row per leg — coordinate duration, β, rate factor, product — and add the last column. Three habits keep it honest. Direction never appears, because only β² enters, so an outbound leg and a return leg at the same speed contribute equally. A leg spent at rest in the chosen frame has factor 1 and contributes its full Δtᵢ, which is why parking at a distant station adds real years to both clocks alike. And every Δtᵢ must be a coordinate duration in the same frame; slipping a leg measured in the ship's frame into the sum yields a number that is neither τ nor T.
The straight worldline is the longest
Take two events with a timelike separation and work in the frame where they happen at the same place. The inertial worldline joining them sits still in that frame, so its proper time is the whole coordinate separation there, τ = √(Δt² − Δx²/c²) evaluated in any frame you like. Every other worldline must move in that frame, and each instant of motion multiplies dt by a factor strictly below one. So the unaccelerated path banks the most proper time and every departure from it subtracts. This reverses the Euclidean triangle inequality, and it makes spacetime diagrams read backwards: the bent worldline is plainly the longer line on the page and the shorter time in fact. It also fixes the wording — an inertial worldline extremises proper time as a maximum, which is why nobody writes a principle of least proper time.
One smooth worldline you can integrate
Constant proper acceleration α — what an accelerometer bolted to the floor reads — gives v(t) = αt/√(1 + (αt/c)²) in the launch frame, a speed that climbs towards c without reaching it. Feed it in and, with u = αt/c, the rate factor is 1/√(1 + u²) and the integral closes: τ = (c/α) asinh(αt/c). The scale c/α is worth carrying for α = g: 2.998 × 10⁸/9.81 = 3.056 × 10⁷ s, or 0.9683 yr. Two years of Earth time at 1 g therefore give u = 2.0654, a final speed of 0.900c, and τ = 0.9683 × asinh(2.0654) = 0.9683 × 1.4725 = 1.426 yr. Applying the final γ = 2.295 to the whole span would have returned 2.000/2.295 = 0.872 yr, low by 39%, because the ship spent most of those two years far slower than its final speed.
The clock hypothesis is measured, not proved
All of the above rests on a claim the two postulates do not supply: that an ideal clock's rate depends on its instantaneous speed alone, with no term in acceleration or any higher derivative. That is the clock hypothesis. It is a statement about clocks rather than about spacetime, and it can fail for a real device — a pendulum will not survive 10⁶ g, and any clock with an internal length scale comparable to c²/α is distorted by the acceleration itself. The evidence is good where anyone has looked. Muons circulating in the CERN storage ring at γ ≈ 29.3 carry a transverse proper acceleration of order 10¹⁸ g, and their measured lifetime matched γτ₀ to about two parts in a thousand, leaving no room for an acceleration term of that size. Use it as an experimental input, and name it when you do.
Change one variable at a time
Make the relationship visible.
Push β₁ and β₂ to 0.95 with w = 0: the turning event climbs almost to the crossing of the light lines and the ship's clock reads 3.75 yr against home's 12.00 yr, even though its path is visibly the longer line on the page. Then raise w — the parked leg is vertical and ticks at the full rate.
OUTBOUND γ₁1.67
RETURN γ₂1.67
TRAVELLER CLOCK τ7.20 yr
AGE GAP T − τ4.80 yr
Live interpretationOUTBOUND γ₁: 1.67. RETURN γ₂: 1.67. TRAVELLER CLOCK τ: 7.20 yr. AGE GAP T − τ: 4.80 yr
Catch the common trap
Explain before calculating.
A probe holds 0.400c for 10.0 yr of station time, then 0.900c for a further 10.0 yr of station time, so the station logs 20.0 yr for the journey. What does the probe's own clock read?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA drone's speed in the lab frame is 0.600c for 6.00 s, then zero for 4.00 s while it hovers, then 0.800c for 5.00 s heading back the way it came. Find the reading accumulated by the clock bolted to the drone, and compare it with the lab time.
- Lab time is just the sum of the leg durations: T = 6.00 + 4.00 + 5.00 = 15.00 s.
- Each leg has constant speed, so τ = Σ Δtᵢ√(1 − βᵢ²). Leg 1: √(1 − 0.600²) = √0.640 = 0.800, giving 6.00 × 0.800 = 4.80 s.
- Leg 2 is at rest in the lab frame, factor √(1 − 0) = 1, so it contributes its full 4.00 s. A parked clock keeps lab time exactly.
- Leg 3: √(1 − 0.800²) = √0.360 = 0.600. The reversed direction is invisible because only β² enters, so 5.00 × 0.600 = 3.00 s.
- Add the column: τ = 4.80 + 4.00 + 3.00 = 11.80 s.
Answerτ = 11.80 s against 15.00 s of lab time — the drone's clock is 3.20 s behind, a mean tick rate of 11.80/15.00 = 0.787.
MediumA probe leaves Earth at 0.800c, holds that speed for 5.00 yr of Earth time, parks for 3.00 yr at a station at rest relative to Earth, then flies home at 0.600c. Find the station's distance, the total Earth time, and the probe's clock reading. Then work out what the γ of the round-trip average speed would have predicted, and say why it fails.
- Outbound distance: d = 0.800c × 5.00 yr = 4.00 ly.
- Return leg: Δt₃ = 4.00 ly ÷ 0.600c = 6.667 yr, so the total Earth time is T = 5.00 + 3.00 + 6.667 = 14.667 yr.
- Rate factors: √(1 − 0.800²) = 0.600 outbound, 1 while parked, √(1 − 0.600²) = 0.800 inbound.
- τ = 5.00(0.600) + 3.00(1) + 6.667(0.800) = 3.000 + 3.000 + 5.333 = 11.333 yr.
- Average speed over the round trip is 8.00 ly ÷ 14.667 yr = 0.5455c, whose factor is √(1 − 0.5455²) = 0.8381, predicting 14.667 × 0.8381 = 12.29 yr.
- The shortcut sits 0.96 yr high: one rate factor cannot stand in for three different ones, and the curvature of √(1 − β²) fixes the sign of that error in advance.
Answerd = 4.00 ly, T = 14.667 yr, τ = 11.333 yr — a gap of 3.33 yr. The average-speed shortcut returns 12.29 yr, high by 0.96 yr.
HardA ship starts from rest and holds a constant proper acceleration α = 9.81 m s⁻², felt as 1 g on board, for 1.000 yr of Earth time (3.156 × 10⁷ s). Find its final speed and its clock reading, then test two shortcuts against it: applying the final γ to the whole year, and applying the γ of half the final speed.
- Constant proper acceleration gives v(t) = αt/√(1 + (αt/c)²). Work in the dimensionless u = αt/c = 9.81 × 3.156 × 10⁷ ÷ 2.998 × 10⁸ = 1.0327.
- Final speed: γfinal = √(1 + u²) = √2.0665 = 1.4375, so βfinal = u/γfinal = 1.0327/1.4375 = 0.7184.
- Proper time: τ = (c/α) asinh(u), and c/α = 2.998 × 10⁸ ÷ 9.81 = 3.056 × 10⁷ s = 0.9683 yr.
- asinh(1.0327) = ln(1.0327 + 1.4375) = ln 2.4702 = 0.9043, so τ = 0.9683 × 0.9043 = 0.876 yr. The ship ages 0.876 yr while Earth ages 1.000 yr, a deficit of 45.4 days.
- Shortcut A, final γ throughout: 1.000/1.4375 = 0.696 yr, low by 21%, because the ship was slower than 0.718c for almost the whole year.
- Shortcut B, half the final speed: β = 0.3592 gives √(1 − 0.1290) = 0.933 yr, high by 6.6%. Only the integral tracks a rate that is still changing.
Answerβfinal = 0.718 with γfinal = 1.4375; τ = 0.876 yr against Earth's 1.000 yr, a deficit of 45.4 days. The shortcuts give 0.696 yr and 0.933 yr.