The now a travelling twin can measure

The twin who comes back younger settled the twin paradox by geometry: the stay-at-home twin’s worldline is a straight line, the traveller’s is bent, and in spacetime the straight path between two events is the one along which the most time passes. Everything from an exchange of pulses settled it again by counting flashes, with no coordinates at all. Neither account leaves anything to explain about who ends up older.

What both leave standing is the traveller’s own story. On the outbound leg, by the simultaneity of the traveller’s inertial frame, the home clock runs slow. On the return leg, by the simultaneity of the new frame, it runs slow again. And yet it ends up far ahead. The textbook resolution is that at the turnaround the traveller’s “now on Earth” lurches forward by years. That is correct arithmetic, and it is also a strange thing to say, because nothing lurches: nobody at home notices anything, and the traveller cannot see the lurch either. This essay asks whether a travelling observer has a “now” that is defined by something the traveller can actually do — and finds that there is one, that it never lurches, and that during the turnaround it agrees with the stay-at-home’s.

A clock timed by echo #

The procedure is the one any radar uses. The traveller sends a light pulse towards a distant event at their own time $\tau_1$; it is reflected there and the echo comes back at their time $\tau_2$. Light took as long to go as to return, so the traveller dates the reflection at the middle of the wait,

$$\tau = \frac{\tau_1 + \tau_2}{2},$$

and places it at the distance $c(\tau_2 - \tau_1)/2$. For an inertial observer this is exactly Einstein’s synchronisation: the speed that cannot be measured one way found that the convention of taking the outward and return journeys of light to be equal is what defines simultaneity in an inertial frame, and radar time is that convention, applied by the observer to the echo. Hermann Bondi built the whole of special relativity out of such exchanges, and in 2001 Carl Dolby and Stephen Gull pointed out that the same definition works, unchanged, for an observer who accelerates.

The definition has two virtues and one limitation. It uses only the traveller’s own clock and the round trip of light, so it does not depend on any coordinates or on any frame the traveller happens to be momentarily at rest in. It gives every event the traveller can signal and hear back from exactly one time and one distance. And it says nothing about events the traveller cannot both reach and hear from — events too far away, or too late to be heard before the trip ends — which is honest: the traveller has no way of timing them.

The surfaces of now #

Every event the traveller would time at the same radar time $\tau$ lies on a surface — in a diagram with one space dimension, a line — and the family of those lines is the traveller’s sequence of “nows.” For an inertial observer they are the familiar tilted lines of now is a choice of slicing. For the travelling twin they can be constructed exactly: for each half-delay $\delta$, the pulse that leaves the twin at $\tau - \delta$ and the echo that arrives at $\tau + \delta$ meet at one event on each side of the twin, and the surface is the set of all such meetings.

Early in the trip, the twin’s radar surfaces are the tilted lines of the outbound rest frame — for every event whose echo returns before the turnaround, the twin is inertial for the whole round trip, and radar time is Einstein time in that frame. Late in the trip they are the tilted lines of the return frame. In between, for events whose round trip straddles the turnaround, the surfaces bend: each has a corner where it crosses the light rays going out from the turnaround event, and inside the region those rays bound, on the home side, the surfaces are flat.

Flat in the home frame means they are the home twin’s own lines of simultaneity. That is the surprise in the drawing. For the stretch of the trip during which the traveller’s signals to home straddle the turnaround, the events the traveller times as simultaneous are exactly the events the stay-at-home twin calls simultaneous. The traveller, by echo, agrees with home about what “now” is, for as long as the turnaround sits inside every round trip. It is not a coincidence: for an event on the home side of the turnaround, the pulse that reaches it leaves the outbound leg and the echo arrives on the return leg, and the two legs’ Doppler shifts are reciprocal, so their average is the home frame’s.

The arithmetic of the flat stretch #

The flat stretch can be checked with two lines of algebra, using the numbers of the drawings: speed $0.6c$, so $\gamma = 1.25$, turnaround at the twin’s 4 years, which is home’s 5 years at 3 light-years out. A pulse that reaches home when the home clock reads $t$ must have left the outbound leg at home time $t/1.6$, which is the twin’s time $t/2$. Its echo, travelling outward at the speed of light, catches the returning twin at home time $(t + 6)/1.6$, which is the twin’s time $(t + 6)/2$. The radar time is the average of the two:

$$\tau = \frac{1}{2}\left(\frac{t}{2} + \frac{t+6}{2}\right) = \frac{t+3}{2}, \qquad t = 2\tau - 3.$$

So for every home event whose round trip straddles the turnaround — home times from 2 to 8, the twin’s radar times from 2.5 to 5.5 — the home clock advances at exactly twice the twin’s rate, and at the twin’s radar time 4, the moment of turning, it reads 5, which is what the home twin’s own clocks read at the turnaround event. The factor of two is the Bondi factor $k = \sqrt{(1+\beta)/(1-\beta)}$ at $0.6c$, and it appears because both ends of these round trips keep the same pace against the home clock: pulses sent from the receding outbound leg arrive home spaced out by $k$, and echoes from home reach the approaching return leg crowded together by $k$, so the twin’s sending time and receiving time each advance by $1/k$ of a year for every year on the home clock — and so does their average.

The radar distance tells the same story from the other side. On the outbound leg it is $0.6c$ times the twin’s time, the same as the rest frame’s distance. In the flat stretch it is half the difference of the two times above, $\tfrac12\left(\frac{t+6}{2} - \frac{t}{2}\right) = 1.5$ light-years, whatever $t$: by echo, home stays at a fixed distance of a light-year and a half for the three years of the twin’s life around the turnaround, and then closes in. The rest frame would have the distance reach 2.4 light-years at the moment of turning, and in the home frame the twins are 3 light-years apart then. The radar never sees home further than 1.5. Each of these numbers is a different answer to a question that has no observer-independent answer, and only the radar’s is one the traveller can measure.

The rest frames’ nows, which do not fit together #

Set beside the radar surfaces, the textbook recipe looks less like physics and more like a stitching error. At each moment, take the inertial frame in which the traveller is momentarily at rest, and use its simultaneity.

The lines do two things no definition of “now” should. They skip: as the traveller turns, the line pointing at home swings from the home clock’s 3.2 years to its 6.8, and the 3.6 years in between are never “now” for the traveller at all. And they cross: beyond the turnaround, on the far side from home, outbound and return lines intersect, so events there are “now” twice, once on each leg. For an instantaneous turnaround both defects are abrupt; for a smooth turnaround at a finite acceleration they are spread over the time of turning, but they do not go away. The momentary rest frame’s lines cross each other at a distance $c^2/a$ behind an observer with acceleration $a$ — the same distance at which the wall of silence behind a rocket that never stops put the accelerated observer’s horizon — and every event beyond that distance is dated more than once.

Neither defect is a mistake in the arithmetic. The rest-frame recipe is a convention, chosen because it is easy to compute with, and conventions do not have to be well behaved. But it is a convention that an accelerating observer cannot carry out by any measurement: to know where the momentary rest frame’s line meets a distant clock, the traveller must already know the distant clock’s history. Radar time is defined by a measurement the traveller makes, and it has neither defect.

The home clock, by three reckonings #

The clearest comparison is the reading the traveller assigns to the home clock at each moment of their own life.

The rest frame’s reckoning runs at 0.8 of the twin’s rate on both legs — the home clock running slow, as a moving clock should — and jumps between them. The radar reckoning runs at 0.8 until the twin’s 2.5 years, then at twice the twin’s rate until 5.5 years, then at 0.8 again: the home clock’s catching-up is spread over three years of the traveller’s life, centred on the turnaround, and it is continuous throughout. What the traveller actually sees — the reading on the home clock carried by the light arriving at each moment — lags both, running at half the twin’s rate on the way out and twice it on the way back, as the flash-counting argument found, switching at the turnaround itself.

All three reckonings agree at the start and the end, where the twins are together and no convention is needed. They differ only about the distant clock in between, and they differ in instructive ways. The sight reckoning is a fact about light arriving, not a statement about simultaneity. The rest-frame reckoning is a statement about simultaneity that cannot be measured. The radar reckoning is a statement about simultaneity that can be, by the one instrument the traveller has.

Where the catching-up happens #

The rate at which the home clock advances, by radar, is the most direct picture of the asymmetry that makes the twins age differently.

On both legs, far from the turnaround, the rate is $1/\gamma$: the traveller sees the home clock running slow, exactly as relativity’s symmetry between two inertial observers requires. Near the turnaround the rate is well above one, and for long enough to make up the difference. The stretch over which it is high is set by the light-travel time to home and back: at $0.6c$ the traveller is three light-years from home at the turnaround, so a pulse sent home 1.5 years before turning round, by the traveller’s clock, returns 1.5 years after, and the catching-up occupies that whole interval. A faster trip goes further and catches up longer and harder. The area under each curve is the home clock’s total elapsed time, and since the slow stretches are the same fraction of the trip as the twins’ symmetry says they must be, it is the catching-up stretch that holds the whole of the difference.

There is nothing here the traveller would describe as a lurch. A traveller who timed the home clock by radar throughout would report that it ran slow, then fast for a while around the turnaround, then slow again — and that the fast stretch began before the traveller turned round. That last point sounds paradoxical and is not: the radar time of a distant event depends on when its echo returns, and an echo that returns after the turnaround carries the effect of the turnaround in it, whenever the pulse was sent. Radar time is not a real-time display of distant events. It is a dating of them, assigned after the fact, from the round trip.

None of this involves the acceleration directly. The clock that does not feel the turn found that a clock’s rate depends only on its speed, not on its acceleration, however violent; here the acceleration enters only through the shape of the worldline, which decides which round trips straddle which part of the journey. The radar description of the trip would be the same if the turnaround took an hour or a microsecond, except within a light-hour of the turnaround itself. What decides the home clock’s catching-up is not how the traveller turns but how far away home is when the traveller does it.

What the definition does and does not settle #

Radar time is not the only possible definition of an accelerated observer’s simultaneity, and it is not privileged by any law of physics — the speed that cannot be measured one way found that even inertial simultaneity is a convention, chosen for convenience. What singles radar time out is that it is operational, that it reduces to Einstein’s convention for inertial observers, and that it is well behaved for any observer whose worldline is timelike: its surfaces never cross, and every event the observer can exchange signals with is dated once. Those properties have made it the preferred definition in discussions of what an accelerated observer “sees” — including whether such an observer detects particles that an inertial observer does not, the effect the heat hidden in a Doppler shift followed, where the choice of simultaneity decides which modes count as positive frequency. For a rotating observer the radar definition has the problem how big now is found for every definition: round trips around a rotating ring do not close, and the radar now of an observer on a turntable is consistent only over a patch of limited size.

It is also, unremarkably, how distant clocks are actually read. Laser ranging to the reflectors left on the Moon times the round trip of a pulse and places the reflection at the midpoint, and the tracking of spacecraft across the solar system — whose velocities change continually as they fall round the Sun — is done by two-way radio ranging, with the spacecraft’s events dated by the midpoint of the exchange and the general-relativistic delays of the path added on. Nobody tracking a probe uses the probe’s momentary rest frame to decide what time it is at home. The twin who times home by echo is doing what navigators do.

The figures assume one space dimension, an instantaneous turnaround and flat spacetime. In three dimensions the radar surfaces are three-dimensional and the construction is the same for every direction. For a turnaround with finite acceleration the corners of the surfaces are rounded but the qualitative picture does not change. In curved spacetime, radar time still makes sense wherever light can make round trips, and it is one of the few ways of defining “now” for an observer near a massive body that does not depend on coordinates. The domain of the drawings is a twin moving at constant speed on two legs in flat spacetime, with the events dated those the twin can exchange signals with during the trip.

Still open: what “now” should mean far from any observer #

Radar time gives each observer a now that extends only as far as the observer’s signals can reach and return during the observer’s lifetime. Beyond that, for events the observer can see but never echo, or echo but never see, it says nothing. Whether there is any natural way to extend a local observer’s now further — over a galaxy, or over a cosmological horizon, where the universe’s expansion means signals sent now will never return — and whether such an extension has any physical content beyond convenience, is still argued. In cosmology the practical answer is to use the frame of the cosmic background radiation, which is a choice about the universe rather than about any observer, and the radar definition shows how much that choice depends on there being a privileged set of observers to make it.

The traveller’s story, at least, is settled. Dating each event by the midpoint of an echo’s round trip gives an accelerating observer a now that never jumps and never crosses itself: for a twin at 0.6c turning after 4 years, the home clock reads 0.8 of the twin’s time until 2.5 years, runs at twice the twin’s rate until 5.5, then at 0.8 again, reaching 10 when the twin returns at 8 — and while the turnaround sits inside the round trip, the twin’s radar now is home’s own. The years the textbook traveller skips are not skipped by the one measurement the traveller can make; they are spent, on the home clock, during the long echo of the turnaround.