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Relativity as measurement · A course with experiments

L3 · Chapter 6

Radar time for an observer who accelerates

The light-echo rule of Chapter 2 never asked whether the observer was moving steadily. Any observer can send a pulse, catch the echo and average the two readings of their own clock. Dolby and Gull applied this to the travelling twin in 2001. The result is a home-time assignment with no jump, and for a rocket that never stops accelerating, a region of home events it can never date.

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Radar time needs only one clock

Let an observer carry a clock that reads τ\tau along any worldline. For an event, take τ−\tau^-, the latest reading at which light can leave the observer and still reach the event, and τ+\tau^+, the earliest reading at which light from the event can arrive. Dolby and Gull define radar time and radar distance from these two numbers.

Nothing in the definition refers to coordinates or to a rest frame, so the answer does not depend on how the diagram is drawn. Each event has one earliest arrival and one latest departure, so radar time is single-valued. For an inertial observer it is exactly Einstein’s synchronization from Chapter 2.

τ=12(τ++τ−),ρ=12(τ+−τ−)\tau=\tfrac12\left(\tau^++\tau^-\right),\qquad \rho=\tfrac12\left(\tau^+-\tau^-\right)

Worked example

The home event at year 5, seen from the turning twin

Take the trip of Chapter 5: speed 0.8, turnaround at t=5, x=4t=5,\ x=4 when B’s clock reads 3. Light that reaches home at t0=5t_0=5 left B on the way out at t=25/9t=25/9, when B read τ−=5/3\tau^-=5/3. Light leaving home at t0=5t_0=5 catches B on the way back at t=65/9t=65/9, when B reads τ+=13/3\tau^+=13/3.

B therefore dates A’s fifth birthday at 3 years, the moment of B’s own turnaround, at a radar distance of 4/3 light-years. The comoving shortcut instead jumped from 1.8 to 8.2 years at that moment. With radar time, home events between t0=1t_0=1 and t0=9t_0=9 are spread evenly over B’s clock, one third of a year of B’s time per home year, across the turnaround.

τ=12 ⁣(53+133)=3 yr,ρ=12 ⁣(133−53)=43 ly,τ(t0)=3+t0−53  (1≤t0≤9)\tau=\tfrac12\!\left(\tfrac53+\tfrac{13}3\right)=3\,\mathrm{yr},\qquad \rho=\tfrac12\!\left(\tfrac{13}3-\tfrac53\right)=\tfrac43\,\mathrm{ly},\qquad \tau(t_0)=3+\tfrac{t_0-5}{3}\ \ (1\le t_0\le 9)

A gradual turn changes almost nothing far away

A real rocket turns with finite acceleration. Dolby and Gull let the twin accelerate uniformly for a short stretch of proper time on each side of the turnaround and recompute the radar surfaces. Far to either side of the turn, in the regions where the outgoing pulse leaves on one leg and the echo returns on the other, the surfaces are unchanged. In the future and past regions they are only relabelled slightly, because the turn takes a little longer.

This is the opposite of the comoving picture, in which a brief acceleration swings the simultaneity lines around and sends distant events backwards and forwards. With radar time, a brief turn has a brief effect.

A rocket that never stops leaves home behind a horizon

Now let B start at rest at home and keep a constant proper acceleration aa for ever. Its path bends towards the light line x=t−1/ax=t-1/a without reaching it. Light leaving home at tet_e travels along x=t−tex=t-t_e and catches B only if te<1/at_e<1/a.

Home events at or after t=1/at=1/a never send B a signal, so they have no τ+\tau^+ and no radar time: they lie beyond B’s horizon. With a=0.5 ly/yr2a=0.5\,\mathrm{ly/yr^2} the horizon cuts home off after 2 years. One standard gravity is about 1.03 ly/yr21.03\,\mathrm{ly/yr^2}, so a rocket at 1 g stops receiving news from home about 0.97 years after launch, if it never stops accelerating.

Dolby and Gull put the accelerating observer’s turning point at x=1/ax=1/a at time zero. In that placement the radar time takes the form below, and the radar coordinates cover only the wedge x>∣t∣x>|t|. The formula in our placement follows by shifting xx by 1/a1/a.

xB(t)=1+a2t2−1a,te<1aτ=12aln⁡x+tx−t\begin{gathered}x_B(t)=\frac{\sqrt{1+a^2t^2}-1}{a},\qquad t_e<\frac1a\\ \tau=\frac1{2a}\ln\frac{x+t}{x-t}\end{gathered}

Where classical radar stops

The horizon above is pure geometry of light lines. It says nothing about particles. Whether a detector that accelerates through empty space registers anything is a question for quantum field theory, known as the Unruh effect. The review by Crispino, Higuchi and Matsas, of which we have read only the abstract, describes it as showing that the particle content of a field theory depends on the observer. Reaching that conclusion takes the field theory that comes next on this path.

An experiment to think through

Predict, change, explain

  1. Predict whether the radar assignment of home time keeps its jump when the turnaround becomes gradual.
  2. In the twin Lab, compare “Sudden turnaround: 10 years against 6” with “Gradual turnaround”. Then choose “Endless acceleration hides home” and drag the round home event up A’s line past the time shown as “Last home time whose light reaches B”.
  3. Explain why “B receives its light at” turns into infinity once the home event passes that time, using the home light lines that lie on or behind B’s asymptote.
Count the signals each twin receives

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. On the sudden-turn trip, what radar time does B give A’s eighth birthday, t0=8t_0=8?

Read an explained answer

It lies in the range 1≤t0≤91\le t_0\le 9, so τ=3+(8−5)/3=4\tau=3+(8-5)/3=4 years. Directly: light from B leaves at τ−=8/3\tau^-=8/3 and the echo returns at τ+=16/3\tau^+=16/3.

2. A rocket accelerates from home at a=1 ly/yr2a=1\,\mathrm{ly/yr^2} for ever. After which home time can no signal from home reach it?

Read an explained answer

After t=1/a=1 yrt=1/a=1\,\mathrm{yr}. Light leaving later runs parallel to the rocket’s asymptote and never catches up.

3. Why can radar time never give one event two different times?

Read an explained answer

An event has exactly one earliest reception on the observer’s clock and one latest emission, so their average is a single number. Overlapping simultaneity lines come from patching together different rest frames; the radar definition uses one clock on one worldline throughout.

Sources to read alongside

  1. Dolby & Gull, On Radar Time and the Twin “Paradox”, arXiv:gr-qc/0104077v2 (2002); American Journal of Physics 69, 1257 (2001)

    Read in full (nine pages, Figures 1–10). The course reproduces the radar-time formulas of the immediate turn-around, the region formulas of the gradual turn-around and the uniformly accelerating case; the implicit formulas for the four corner regions and the coordinate maps of Figures 8–9 are not reproduced.

  2. David Tong, Quantum Field Theory, University of Cambridge Part III lecture notes

    Course page and chapter list checked: Lorentz invariance in Chapter 1, causality and relativistic normalisation in Chapter 2. A next step after this course, not a source for its calculations.

  3. Crispino, Higuchi & Matsas, The Unruh effect and its applications, arXiv:0710.5373; Reviews of Modern Physics 80, 787 (2008)

    Abstract only. The review treats the quantum response of accelerated observers; the classical horizon in this course does not derive that effect.

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