Special relativity · Accelerated observers

Twin Clocks and Radar Time

Which moment at home does a travelling twin call now, and why does the turnaround matter? The travelling twin comes home younger because the bent path through spacetime accumulates less proper time; neither twin sees the other's clock jump. Which home moment counts as "now" is a convention. Radar time, the average of an echo's send and return readings, gives one smooth answer through the turn, while hopping between momentary rest frames makes home time leap.

Send a twin out and back, then compare how each twin dates events at the other end: radar echoes give a smooth answer, momentary rest frames jump at the turn.

A's clock when the trip ends $10\,\mathrm{yr}$

B's clock when the trip ends $6\,\mathrm{yr}$

B's radar time for the home event $3\,\mathrm{yr}$

B receives its light at $4.33\,\mathrm{yr}$

Model assumptions

One space dimension, with light speed set to one light-year per year. Observer A rests at the origin. B follows piecewise-constant proper acceleration, integrated in B's own time with fourth-order Runge-Kutta steps of at most 0.02 years; a sudden turn is a jump in rapidity. Radar time follows Dolby and Gull (2001): the average of B's clock readings when an echo leaves and when it returns. Momentary rest-frame simultaneity is drawn only for comparison. The model has no gravity, no rocket fuel budget and no optics of real images.

Green is radar time and purple the momentary rest frame; the right panel counts the flashes each twin receives.

More numbers and the saved record

Proper acceleration while not coasting

\infty

A's yearly flashes that reach B

10

B's yearly flashes that reach A

6

Last home time whose light reaches B

\infty

The JSON record holds B's sampled worldline, the radar pair for the home event and every flash with its reception time.

Start with the essentials

Focus question: Which moment at home does a travelling twin call now, and why does the turnaround matter?
One-sentence intuition: Proper time along each path decides who is younger. Radar time is a smooth, operational way for B to call a home event simultaneous; momentary rest frames are not.

Core mathematical model

Radar time and distance

\tau=\tfrac12\,(\tau^{+}+\tau^{-}),\qquad \rho=\tfrac12\,c\,(\tau^{+}-\tau^{-})

B sends an echo at its own reading $\tau^{-}$ and receives it back at $\tau^{+}$ . Dolby and Gull use this average for any observer, accelerated or not.

Proper time along a path

\tau=\int \sqrt{1-v(t)^2/c^2}\;\mathrm{d}t

With a sudden turn at eight tenths of light speed and ten years at home, B's clock gains six years.

Momentary rest-frame "now" at home

t_{\mathrm{home}}=t_B-\frac{v\,x_B}{c^2}

Changing v at the turn changes this home time abruptly. In the sudden-turn preset it leaps from 1.8 to 8.2 years.

Constant proper acceleration from rest

x(t)=\frac{c^2}{a}\left(\sqrt{1+\frac{a^2t^2}{c^2}}-1\right),\qquad t_{\mathrm{horizon}}=\frac{c}{a}

Light leaving home after the horizon time never catches B. With a equal to half a light-year per year squared, that time is two years.

Common difficulties

The turn is not where the years go

Typical misconceptionB's missing years are lost during the acceleration.

Better mental modelShrink the turnaround toward zero: the age gap does not vanish but approaches the value set by the two coasting legs. The difference is the length of each path in proper time.

Simultaneity is a convention

Typical misconceptionThere is one true home event happening "now" for B.

Better mental modelDistant simultaneity needs a rule. Radar time is one operational rule that works for accelerated observers without jumps; it agrees with Einstein synchronization for events whose echo leaves and returns during the same coasting leg.

What the model leaves out

Typical misconceptionThe diagram includes gravity or the rocket's engine.

Better mental modelOne space dimension, with light speed set to one light-year per year. Observer A rests at the origin. B follows piecewise-constant proper acceleration, integrated in B's own time with fourth-order Runge-Kutta steps of at most 0.02 years; a sudden turn is a jump in rapidity. Radar time follows Dolby and Gull (2001): the average of B's clock readings when an echo leaves and when it returns. Momentary rest-frame simultaneity is drawn only for comparison. The model has no gravity, no rocket fuel budget and no optics of real images.

Run the experiment

01
Count the years at reunion
Start with the sudden-turn preset and let the trip play.
What to observe: A's clock shows ten years and B's shows six. The flash staircases end at the same pair of numbers.
02
Date a home event by radar
Drag the round home event up and down A's line.
What to observe: B's radar time changes smoothly; the home event at five years gets radar time three, the middle of B's trip.
03
Compare the two rules for "now"
Watch the purple and green markers on A's line during the turn, then try the gradual-turn preset.
What to observe: The purple rest-frame marker leaps across years of home time at a sudden turn, while the green radar marker moves on smoothly.
04
Find the horizon
Choose Endless acceleration hides home and drag the home event past two years.
What to observe: Its light never reaches B, so the radar time and reception time become infinite.

Twin Clocks and Radar Time

Two clocks, one reunion

Core mathematical model

Radar time and distance

Proper time along a path

Momentary rest-frame "now" at home

Constant proper acceleration from rest

Common difficulties

The turn is not where the years go

Simultaneity is a convention

What the model leaves out

Run the experiment

Count the years at reunion

Date a home event by radar

Compare the two rules for "now"

Find the horizon

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Relativity as measurement: clocks, light signals and four-momentum