Chapter 11
Straight Lines Age the Most
The twin paradox, and the principle that runs the universe.
What this chapter does
Twins part on their birthday. One stays home. The other flies to a star and back. When they meet again, who is older? It’s the most famous puzzle in relativity, and this chapter settles it with the interval from Chapter 6 and some careful drawing.
- The puzzle (about 15 minutes). A warm-up, then the trip. You make the best case you can for the traveler, then test it in a lab where you bend her path any way you like.
- Two paths (about 25 minutes). The trip by hand with friendly numbers, a triangle rule that flips in spacetime, a relay of three brothers with no engines, what the traveler calls “now” back home, and birthday cards sent by light.
- Nature, practice and code (about 25 minutes). Atomic clocks flown around the world, muons in a ring, practice, and a function you write that then runs the trip lab.
You’ll leave able to work out how much any traveler ages, leg by leg, and to settle the puzzle for a doubtful friend with one picture.
Warm-up
Three quick ones from earlier chapters. Each is a tool this chapter leans on.
From Chapter 6. A clock coasts at steady speed from the event \((t, x) = (0, 0)\) to the event \((5, 3)\), in years and light-years. How much time does the clock itself record? Need a refresher?
That’s \(\sqrt{5^2 + 3^2}\), the distance on a map. Spacetime’s rule has a different sign in the middle. Which one?
That’s the lab time between the events. The clock moved 3 light-years along the way; doesn’t that cost it something?
Hint
The clock is present at both events, so it reads the interval between them: \(\sqrt{\dt^2 - \dx^2}\).
Solution
\(\tau = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\) years. One clock, present at both events, reads the interval.
From Chapter 8. A clock moving at \(v = 0.8\) records 3 years between two of its own ticks. How much time passes between those two ticks according to Lin’s lattice of clocks?
That divides by \(\g\). Which clock is present at both ticks, the moving one or Lin’s? Which one reads less?
Hint
At \(v = 0.8\), \(\g = 5/3\). The clock present at both events reads the shorter time.
Solution
\(\lab{\dt} = \g\,\inv{\tau} = \tfrac{5}{3} \times 3 = 5\) years. Check with the interval: in 5 years the clock moves 4 light-years, and \(\sqrt{5^2 - 4^2} = 3\).
From Chapter 7. Rae coasts at \(v = 0.6\). On a diagram drawn in Lin’s frame, Rae’s line of “now” (the events her lattice calls simultaneous) is…
Solution
Rocket now-lines drawn in lab coordinates are \(t = v x + t'/\g\), so their slope is \(v = 0.6\). They tilt up toward the direction of motion, mirroring the worldline across the light line.
The puzzle
Lin and Rae are twins. On their twentieth birthday Rae boards a rocket, coasts to a star 3 light-years away at \(v = 3/5\), turns around, and coasts home at the same speed. Lin stays on Earth.
Chapter 8 taught us that moving clocks are measured running slow. So Lin says Rae’s clocks dawdled the whole way. But from Rae’s seat it was Earth that rushed away, then came rushing back. Can’t she say exactly the same about Lin?
Here’s the catch. When Rae steps off the rocket, the twins stand side by side. Two faces, one mirror. Either one of them is older, or neither is. Nature has to pick.
Predict
Play devil’s advocate. Write the strongest argument you can that Rae, the traveler, comes home the older twin. Make it good: in a few minutes you’ll hunt for its flaw.
Experiment
Here is the trip as a spacetime diagram, in years and light-years. Lin’s worldline runs straight up: she stays put. Rae’s worldline runs out to the turnaround and back. Both start at the same event, Depart, and end at the same event, Reunion. The lab’s readouts Home ages (Lin) and Traveler ages (Rae) work like odometers: they add up each twin’s wristwatch time, leg by leg.
One idealization for the whole chapter: we ignore Earth’s gravity. Picture Lin’s home as a space station coasting far from any planet. Gravity waits for the epilogue of the final chapter.
Try it
1. Read the odometers. With the turnaround at \((5, 3)\), how much does each twin age? Then check one leg by hand: the leg table gives \(\dt = 5\) and \(\dx = 3\). Which triangle from Chapter 6 turns those into \(\tau = 4\)?
The trip lab needs JavaScript. In words: the straight worldline from (0, 0) to (10, 0) ages 10 years. Rae’s bent worldline through (5, 3) ages 4 + 4 = 8. Every other bent worldline between the same two events ages less than 10.
Describe this lab
A spacetime diagram with time in years pointing up and distance in light-years pointing right. Home’s worldline is a straight blue line from Depart (0, 0) to Reunion (10, 0), with a tick at every year. The traveler’s dashed vermilion worldline goes from Depart through one or more turning points to Reunion, with a tick at every year of her own wristwatch time. Readouts show how much each twin ages and the difference, plus a table of each leg’s Δt, Δx and wristwatch time τ.
Buttons: “Add a turning point” adds one in the middle of the longest leg; “Remove a turn” removes the turn you last touched. Tapping the diagram adds a turn at that time, kept slower than light. Keyboard: Tab to a turning point, then use the arrow keys to move it by 0.1 (Shift for 1). N adds a turning point; Delete removes the focused one. A leg at or beyond 45° turns red: no clock can ride it, and the totals pause until you fix it.
Try it
- Beat the free particle. Drag the turnaround toward home, or tap Add a turning point (or press N) to add more turns, up to eight. Remove a turn takes one away. Can the traveler ever age more than 10 years? What does she age with a single turn at \((5, 1)\)? (Drop the turn right onto Lin’s line, at \(x = 0\), and you tie at 10. That isn’t a detour, though: it’s Lin’s own path.)
- Race the light. Push the turnaround out toward 45°. What happens to the traveler’s age as her legs tilt toward 45°? What is the interval along a leg at exactly 45°, and why can no wristwatch ride it? (At 45° or past it, the lab marks the leg in red, faster than light, and hides the totals. Watch the trend as you close in.)
Then play. Draw zigzags, lopsided trips, a trip that waits at the star. The So close… board keeps your best tries. Some come close to 10 years. None gets past.
What just happened?
Rae comes home younger: she ages 8 years while Lin ages 10. And no bent path you drew ever beat the straight one.
So the argument has a flaw somewhere, however good it was. Before you read on, hunt for it.
Yours may be worded differently, but most arguments for an older Rae have this shape:
- From Rae’s seat, Earth rushes away and then rushes back.
- On each leg, Rae’s lattice measures Lin’s clocks running slow.
- Rae sits still in one free-float frame for the whole trip, so her view counts exactly as much as Lin’s.
- So Lin should come home the younger one.
Which step breaks?
Hint
Look at Rae’s worldline in the lab. Is it one straight line?
Solution
Step 3. A free-float frame belongs to a straight worldline, and Rae’s is bent at the star. She rides one frame out and another one home, so she has no single frame in which she sat still all along. Lin does.
Here’s the step that breaks. A symmetry argument says “from Rae’s point of view, Earth moves, so Earth’s clocks run slow.” That sentence quietly treats Rae’s point of view as one free-float frame for the whole trip. It isn’t. Rae rides one frame going out and a different frame coming home. Lin rides the same frame from start to finish.
Draw it and the asymmetry jumps off the page. Lin’s worldline is straight. Rae’s is bent. They join the same two events, Depart and Reunion, along different paths. Asking why their ages differ is like asking why two roads between the same two towns have different lengths. They’re different roads.
A fair worry, and a famous one. Hold on to it: in a few minutes we’ll take the push away entirely and see whether anything changes.
Two paths, one pair of events
Let’s do the trip by hand with friendly numbers. Then a picture. Then the shorthand.
Ten years and eight
- Lin’s worldline runs from Depart \((0, 0)\) to Reunion \((10, 0)\). She never moves, so \(\dx = 0\) and her wristwatch time is the whole \(10\) years. For a clock at rest, wristwatch time is plain lab time.
- Rae’s first leg runs from \((0, 0)\) to the turnaround \((5, 3)\): \(\dt = 5\) years and \(\dx = 3\) light-years. At \(v = 3/5\) she covers 3 light-years in 5 years.
- One straight leg, one clock present at both ends, so the wristwatch reads the interval: \(\sqrt{5^2 - 3^2} = \sqrt{16} = 4\) years. That’s Chapter 6: the clock present at both events reads \(\tau\).
- The way home is the mirror image: \(\dt = 5\), \(\dx = -3\), so another \(4\) years. Squaring erases the minus sign. Direction doesn’t matter, only how far.
- Add the legs: Rae ages \(4 + 4 = 8\) years. Lin ages \(10\). Wristwatch time adds up along a worldline, the way an odometer adds up kilometers.
The triangle inequality, reversed
On a map, a detour costs you. Walk from A to C by way of B, and the two sides of the triangle add up to more than the straight side. That’s the triangle inequality: the straight line is the shortest way.
Draw the same triangle in spacetime and the rule flips. The straight side is 10 years. Each bent side is 4 years. In spacetime the detour is shorter, and the straight line is the longest path there is.
Quick check. Same Depart and Reunion, but move the turnaround farther out, to \((5, 4)\). How much wristwatch time does the detour hold?
That’s \(2\sqrt{5^2 + 4^2}\), the map rule. Which sign goes between the squares in spacetime?
That’s the old turn at \((5, 3)\). Each leg is now \(\dt = 5\), \(\dx = 4\). What is \(\sqrt{5^2 - 4^2}\)?
Hint
Each leg has \(\dt = 5\) and \(|\dx| = 4\): a 3-4-5 triangle.
Solution
Each leg: \(\sqrt{25 - 16} = 3\) years, so the detour holds \(3 + 3 = 6\). Farther out on the page, less time on the watch: 10, then 8, then 6.
The reversed rule has a name, the Principle of Maximal Aging: of all the worldlines a clock can follow between two events, the free-float one, the straight one, records the most wristwatch time.
In Lin’s frame the proof fits in a few lines. Chop any trip into straight legs. Each leg’s wristwatch time is \(\sqrt{\dt^2 - \dx^2}\), and that can never be more than \(\dt\), because something is taken away before the square root. So the legs add up to at most the total \(\dt\), which is Lin’s 10 years. They tie only if every \(\dx\) is zero: only if the traveler never leaves home.
\[ \begin{gathered} \inv{\tau_\text{trip}} \;=\; \sum_i \sqrt{(\lab{\dt_i})^2 - (\lab{\dx_i})^2} \\ \le\; \sum_i \lab{\dt_i} \;=\; \inv{\tau_\text{home}} \end{gathered} \]
Both ends of that line are wristwatch times, so they’re invariant. The conclusion holds in every frame, even frames where Lin is the one moving.
Why, really?
Because moving costs time. Every bit of sideways travel is subtracted inside the square root, so a wristwatch that wanders pays for the trip out of its own reading. The only way to keep the whole budget is to coast straight and spend nothing. Nobody knows why Nature put a minus sign there instead of a plus. But that one sign is the entire twin effect.
Is it the push? Three brothers
Now for Sagredo’s question. Is the turnaround push what makes Rae younger? Let’s remove it.
Meet three brothers. Home stays on Earth. Out is already coasting at \(3/5\) when he flies past Earth, and as he passes he sets his watch to match Home’s: zero. Out never fires an engine. At the star he meets Back, who is already coasting toward Earth at \(3/5\). As they pass, Back sets his watch to Out’s reading: 4 years. Back never fires an engine either.
No clock in the relay ever feels a push. When Back flies past Earth, what does his watch read?
Hint
Out’s leg runs from \((0, 0)\) to \((5, 3)\); Back’s from \((5, 3)\) to \((10, 0)\). What does a coasting clock read along each?
Solution
8. Out’s watch reads \(\sqrt{5^2 - 3^2} = 4\) at the star, Back copies it, and his own leg adds another 4. Home reads 10.
Back passes Earth with \(4 + 4 = 8\) on his watch, while Home’s reads 10: exactly Rae’s result. Nobody accelerated. Every clock coasted the whole time, and copying a number from one watch to another ages nobody. The push was never the cause. The cause is geometry: the relay’s path is bent and Home’s is straight.
Myth
“The twin paradox is solved by the acceleration,” or “it needs general relativity.” The relay has no accelerating clock and gives the same 8 years. In the original story the push has one job: it bends Rae’s path, so it tells you which twin took the detour. The ages come from the lengths of the two paths.
Myth
“Special relativity can’t handle acceleration.” It can. A curved worldline is a chain of very short straight legs, and wristwatch time adds up along it the same way (the Going further section flies a rocket at a steady 1 g). What needs general relativity is gravity, not acceleration.
What Rae calls “now” back home
One loose end. During the outbound leg, Rae’s own lattice measures Lin’s clocks running slow, by the same factor \(5/4\) that Lin measures for Rae’s. That’s the reciprocity of Chapter 8, and it’s true. So how does Lin end up older?
Watch Rae’s line of simultaneity, her line of “now,” as she travels. Where it crosses Lin’s worldline tells you what Rae’s lattice reckons Lin’s clock reads at that moment.
This lab needs JavaScript. In words: on the way out, Rae’s now-line reaches Lin’s clock at 3.2 years when Rae turns. On the way back, her new now-line starts at 6.8 years. The 3.6-year gap is a change of bookkeeping, not a change in Lin.
Describe this lab
A spacetime diagram of the same 10-year trip. A slider sets Rae’s wristwatch time; a marker sits at that point on her worldline, and a dashed vermilion line through it shows the events Rae’s current frame calls “now.” Where the line meets Lin’s worldline, a readout gives Lin’s clock as Rae reckons it. A second switch adds Lin’s own line of now (horizontal). A speed slider changes the trip (always 10 years for Lin). A toggle replaces Rae with the three-brothers relay. Once the marker reaches the turnaround, both turnaround now-lines stay drawn and a bracket on Lin’s worldline marks the gap between them.
Buttons: Play, “Jump to the turn,” “Lin’s now too,” and the Rae / Three brothers switch. Keyboard: the sliders take arrow keys. Space plays or pauses (or steps a year, with reduced motion), T jumps to the turnaround, B toggles the three brothers, L toggles Lin’s line of now.
Try it
1. Scrub Rae’s watch to 2.00 years and tap Lin’s now too. Two numbers appear. Rae’s now-line meets Lin’s clock at 1.60 years: Rae reckons Lin is the slow one. Lin’s own now-line meets Rae at 2.50 years of Lin’s time: Lin reckons Rae is the slow one. Both by the same factor, \(2.50 / 2 = 2 / 1.60 = 5/4\). That is reciprocal slowing, in one picture.
2. Scrub to just before the turn. What does Rae’s now-line say Lin’s clock reads? Tap Jump to the turn (or press T), then nudge one step past. What does it say now?
On the way out, Rae reckons Lin aging slowly: 3.2 years while she ages 4. On the way home, the same again: another 3.2. That’s 6.4 years of slow aging, yet Lin ages 10. The missing 3.6 years sit between Rae’s two now-lines at the turn.
Did Lin suddenly age 3.6 years while Rae turned around? No. Nothing happened to Lin at all. Rae switched frames, so she switched which far-away events she calls “now.” It’s a change of bookkeeping, like switching from one map grid to another: the town doesn’t move. Switch on Three brothers in the lab. The last now-line the relay borrows from Out meets Lin at 3.2; Back’s first meets her at 6.8. Out keeps coasting, Back was coasting all along, and nobody jumps anywhere. The two brothers disagree about distant simultaneity, exactly as Chapter 7 said two frames would. And the reunion ages, 10 and 8, don’t depend on anyone’s now-line at all.
Pitfall
“Earth’s twin suddenly ages at the turnaround.” Lin’s clocks tick steadily: ten ticks in ten years, one per year, no lurch. The “jump” lives in Rae’s bookkeeping, in the tilt of her now-line, not in Lin’s life. And the reunion ages, 10 and 8, don’t depend on anyone’s now-line at all.
Birthday cards
Now-lines are bookkeeping. Here is something that isn’t. Each twin sends a flash of light on every birthday, by her own wristwatch, and each counts the flashes she receives. Counting needs no lattice and no convention: a flash either reaches you or it doesn’t. So every observer must agree on the counts.
Count before you read on. Using Plate 11.4 (or the live lab just below, whose tally stays covered until you answer), how many birthday cards does Rae receive during the whole trip, and how many does Lin receive?
Two counting rules. A card sent at the reunion counts: it is caught the instant it is sent. A card caught exactly at the turnaround counts as an outbound card.
Close. Did you count the card Lin sends at the reunion itself? By the rule above, it counts: it arrives the instant it leaves.
Eight is how many cards Rae sends. Count the rings on Rae’s path in the left panel. How many are there?
Ten is how many cards Lin sends. Count the rings on Lin’s worldline in the right panel. How many are there?
Hint
Count rings, not lines. A ring marks where a card is caught. Left panel: rings on Rae’s path. Right panel: rings on Lin’s.
Hint
Lin’s tenth birthday happens at the reunion, with Rae standing right there. That card is caught the moment it is sent.
Solution
Rae catches 2 of Lin’s cards on the way out (at Rae’s years 2 and 4) and 8 on the way home: \(2 + 8 = 10\). Lin catches Rae’s 4 outbound cards at her years 2, 4, 6 and 8, then the 4 homebound cards at 8.5, 9, 9.5 and 10: \(4 + 4 = 8\). Each twin receives exactly one card per birthday of the other.
The birthday-card lab needs JavaScript. In words: at v = 0.6, Rae receives 2 + 8 = 10 of Lin’s cards and Lin receives 4 + 4 = 8 of Rae’s.
Describe this lab
The same trip diagram with gold 45° light lines: each one is a birthday flash from one twin, drawn until it reaches the other. A panel tallies the cards each twin receives. It is covered until you answer the counting exercise above or press “Show the tally.” A switch shows both twins’ cards, only Lin’s, or only Rae’s. Keyboard: Tab to the turnaround and use the arrow keys; F toggles the flashes.
Now look at the rates. On the way out, Rae ages 4 years and catches only 2 of Lin’s cards, one every 2 years (the second arrives exactly at the turnaround, so it counts as outbound). On the way home she ages 4 more and catches 8: two a year. Lin catches Rae’s outbound cards one every 2 years for 8 years, then the homebound ones two a year, crammed into her last 2 years.
Everything runs on one ratio. Moving apart, cards arrive at half the rate they’re sent; moving together, at twice the rate. Call the ratio you just counted \(k\). At \(v = 3/5\), \(k = 2\). It’s the Doppler factor, and Chapter 12 builds it properly.
And see where the asymmetry hides. Rae switches from slow cards to fast cards at her turnaround, halfway through her trip. Lin can’t see the turnaround until its light reaches her, at year 8, so she spends 8 of her 10 years in slow cards. The twins’ experiences are not mirror images, and the light tells on them.
Real numbers
Rae’s \(v = 3/5\) is \(179\,875\,475\) m/s. The star is \(3\) light-years \(= 2.838 \times 10^{16}\) m away. The age gap of 2 years is \(63\,115\,200\) s, counting Julian years of \(365.25\) days.
Deep dive: how big is the bookkeeping jump?
Keep Lin’s trip at 10 years and let Rae fly at any speed \(v\). She turns at \(t = 5\), \(x = 5v\). Her outbound now-line has slope \(v\), so it meets Lin’s worldline \(x = 0\) at \(t = 5 - v \cdot 5v = 5(1 - v^2)\). Her homebound now-line has slope \(-v\) and meets it at \(5(1 + v^2)\). The jump is the difference:
\[ 5(1 + v^2) - 5(1 - v^2) = 10\,v^2 . \]
At \(v = 0.6\) that’s \(3.6\) years. Add Rae’s two slow stretches, \(2 \times 5(1 - v^2) = 6.4\), and the jump: \(6.4 + 3.6 = 10\). The books balance. The bookkeeping jump is exactly the part of Lin’s life that neither of Rae’s frames ever called “now” during the coasting.
That’s the whole twin paradox: two paths, one minus sign, and a flashlight to keep everyone honest. Has anybody done it for real?
Ask Nature
Twins are hard to come by. Clocks aren’t. You met both of these experiments in Chapter 8 as tests of time stretching. Here they are again, read as twin experiments: each one is a round trip that ends in a reunion.
Ask Nature: clocks around the world
In 1971 Joseph Hafele and Richard Keating flew cesium atomic clocks around the world on ordinary airliners, once eastward and once westward, then compared them with clocks that had stayed home at the U.S. Naval Observatory. Clocks for twins.
Picture a lattice of clocks centered on Earth’s center that doesn’t spin with Earth. (Imagine floating far above the North Pole, not turning. Below you, Earth spins eastward and carries the ground clock with it.) In that frame the ground clock is already moving east. An eastbound plane adds its speed, so it moves faster through that frame than the ground clock. Every bit of its trip subtracts more inside \(\sqrt{\dt^2 - \dx^2}\), so it should age less. A westbound plane subtracts its speed, moves slower than the ground clock, and should age more.
Here are the totals, in nanoseconds gained by the flying clocks over the ground clocks (a minus sign means the flying clock lost time). The predicted total adds a second effect to the speed part: clocks higher above Earth tick faster. That’s gravity, which belongs to general relativity, and it is why the westbound total is so large.
| Eastward | Westward | |
|---|---|---|
| Speed part | −184 ± 18 | +96 ± 10 |
| + Gravity part | +144 ± 14 | +179 ± 18 |
| = Predicted | −40 ± 23 | +275 ± 21 |
| Measured | −59 ± 10 | +273 ± 7 |
Predicted and measured agree within their uncertainties. We discuss only the speed part here, and it is the same leg-by-leg sum you just did: the clock that moved faster through that frame aged less.
J. C. Hafele and R. E. Keating, “Around-the-World Atomic Clocks: Predicted Relativistic Time Gains” and “…: Observed Relativistic Time Gains,” Science 177, 166–168 and 168–170 (1972).
Ask Nature: muons on a merry-go-round
At CERN in the 1970s, muons were sent round and round a storage ring by magnets, at \(\g = 29.33\). As Chapter 8 showed, they lived 64.4 μs by the lab’s clocks, \(\g\) times a resting muon’s 2.2 μs, within about 0.1%.
The lab is the stay-at-home twin. Each muon travels a looping path through spacetime and comes back to the same place younger. And the magnets pushed the muons sideways, hard, all the way round, yet their clocks showed only what their speed predicts. The ages depend on the path, not on the push.
J. Bailey et al., “Measurements of relativistic time dilatation for positive and negative muons in a circular orbit,” Nature 268, 301–305 (1977).
Practice
Three tiers. Friendly numbers first, so the arithmetic never hides the idea. Short on time? Do the ◇ tier now; the ◆ and ★ tiers keep for a second visit.
◇ Back of the envelope
A star is 3 light-years away. Rae flies there at \(v = 0.6\) and straight back. How many years does Rae age?
That’s Lin’s age: the lab time. Whose wristwatch rides the bent path?
That multiplies Lin’s 10 years by \(\g\). Does the traveling clock read more than the one at home, or less?
Hint
Each leg takes \(\dt = 3/0.6 = 5\) years of lab time and covers \(\dx = 3\).
Solution
Each leg: \(\sqrt{5^2 - 3^2} = 4\) years. Two legs: \(8\) years. (Or: Lin’s 10 years divided by \(\g = 5/4\).)
The variant: a star 4 light-years away, at \(v = 0.8\), there and back. How many years does the traveler age?
That’s \(10 \times \g\). The traveler’s path is the bent one. Should her total be bigger or smaller than Lin’s 10?
That’s how long the trip takes for Lin: \(2 \times 4/0.8\). What does the traveler’s own watch read?
Hint
Each leg: \(\dt = 4/0.8 = 5\), \(\dx = 4\). Which Pythagorean triple has a long side of 5 and a side of 4?
Solution
Each leg: \(\sqrt{5^2 - 4^2} = 3\) years. Total: \(6\) years, against Lin’s \(10\). At \(v = 0.8\), \(\g = 5/3\), and \(10 / \tfrac{5}{3} = 6\).
On her way out at \(v = 0.6\), Rae catches one of Lin’s yearly cards every how many of her own years? That ratio is the Doppler factor \(k\).
Hint
In Plate 11.4 (left panel), Rae ages 4 years on the way out. How many rings sit on her outbound leg?
Solution
Rae ages 4 outbound years and catches 2 cards: one every 2 years, so \(k = 2\). On the way home it flips: 8 cards in 4 years, two a year, a rate of \(k = 2\) times the sending rate.
What makes the twins’ ages differ at the reunion?
Solution
Different paths. Lin’s worldline is straight; Rae’s is bent at the star. Wristwatch time is the length of a path in spacetime, and the straight one is longest. The relay of three brothers, with no acceleration anywhere, gives the same 8 years.
Tick every statement that is true. All the worldlines run between the same two events.
Hint
Any worldline that isn’t straight ages less than the straight one. Is a trip that pauses at the star straight?
Solution
True: the first, second and fourth. Any bend, jog or pause bends the worldline and costs wristwatch time; legs near 45° contribute almost nothing. The push damages nothing: the relay shows the same result with no push at all.
Same for every observer, or does it depend on the frame?
- Rae’s age at the reunion
- The number of birthday cards Lin receives
- What Lin’s clock reads “at the moment” Rae turns around
- The distance from Earth to the star
- The wristwatch time along Rae’s homebound leg
- How fast Lin’s clocks tick, as Rae’s lattice measures them on the way out
Hint
Things you can read off one clock, or count at one place, are invariant. Things that need a “now” far away, or a lattice’s rulers, depend on the frame.
Solution
Invariant: Rae’s age at the reunion, Lin’s card count, and the wristwatch time of a leg. Frame-dependent: what “now” at a distance means, the distance to the star (it’s contracted in Rae’s frame) and the measured tick rate of a moving clock.
◆ Pencil and paper
Find the wristwatch time from \((0, 0)\) to \((10, 0)\) through the turnaround \((5, 1)\). Need a refresher?
That’s \(2\sqrt{5^2 + 1^2}\): the map length. Which sign goes between the squares in spacetime?
Hint
Each leg has \(\dt = 5\) and \(|\dx| = 1\).
Hint
\(\sqrt{25 - 1} = \sqrt{24} \approx 4.899\). There are two such legs.
Solution
\(2\sqrt{5^2 - 1^2} = 2\sqrt{24} \approx 9.80\) years. A small detour costs only a fifth of a year: near the straight line, the penalty is tiny.
Finish the worked example. This time Rae flies to the star at \(3/5\), waits there 3 years, then flies home at \(3/5\). Her worldline is \((0, 0) \to (5, 3) \to (8, 3) \to (13, 0)\).
- Leg 1: \(\dt = 5\), \(\dx = 3\), so \(\tau = 4\).
- Leg 2 (waiting): \(\dt = 3\), \(\dx = 0\), so \(\tau = 3\).
- Leg 3: \(\dt = \,?\), \(\dx = \,?\), so \(\tau = \,?\)
- Total for Rae: \(?\) Lin, who stays home, ages 13.
That’s Lin’s total. Rae’s legs add to less. What are they?
Hint
Leg 3 runs from \((8, 3)\) to \((13, 0)\): \(\dt = 5\), \(\dx = -3\).
Solution
Leg 3: \(\sqrt{5^2 - 3^2} = 4\). Total: \(4 + 3 + 4 = 11\) years, against Lin’s 13. The gap is still 2 years: waiting at the star costs both twins the same 3 years, so only the moving legs make a difference.
A 10-year round trip that turns around at the halfway time, \(t = 5\). Drag the turnaround (keep it at \(t = 5\)) so the traveler ages exactly 6 years.
No pointer? Tab to the turnaround and use the arrow keys (Shift moves by 1). The odometer shows the traveler’s age as you go.
Hint
Each leg must age 3 years, and each leg lasts 5 years of lab time. Which 3-4-5 triangle fits?
Solution
Put the turn at \(x = 4\) (or \(-4\)): each leg is \(\sqrt{5^2 - 4^2} = 3\), so \(3 + 3 = 6\). That’s \(v = 0.8\): the variant trip to a star 4 light-years away, from the second ◇ exercise.
On her 4th birthday, at the turnaround \((5, 3)\), Rae flashes a card toward Earth. At what time on Lin’s clock does it arrive?
That’s when the card is sent. It still has 3 light-years to cover. How long does light take?
Hint
Light covers 1 light-year per year, so 3 light-years take 3 years.
Solution
\(t = 5 + 3 = 8\). So Lin spends 8 of her 10 years receiving Rae’s slow outbound cards, and only the last 2 receiving fast ones. That lopsided split is the asymmetry, written in light.
The same 10-year trip, but at \(v = 0.8\): the turnaround is at \((5, 4)\). Just before the turn, Rae’s now-line has slope \(0.8\). Just after, it has slope \(-0.8\). Where does each line meet Lin’s worldline \(x = 0\)?
Hint
A line through \((5, 4)\) with slope \(s\) is \(t - 5 = s\,(x - 4)\). Set \(x = 0\).
Solution
Before: \(t = 5 - 0.8 \times 4 = 1.8\). After: \(t = 5 + 0.8 \times 4 = 8.2\). The bookkeeping jump is \(6.4 = 10 \times 0.8^2\) years. Check the books: Rae reckons Lin aging \(1.8\) on each leg, and \(1.8 + 6.4 + 1.8 = 10\).
Put the three-brothers argument in order.
- Out, already coasting, passes Earth and sets his watch to Home’s reading: 0.
- Out coasts to the star; his watch reads 4 when he meets Back.
- Back, coasting homeward, copies Out’s reading of 4 as they pass.
- Back passes Earth reading 8, while Home reads 10.
- No clock accelerated, so the 2-year gap comes from the bent path, not from a push.
Hint
Follow the watch reading as it travels: set, carried, copied, carried home, compared.
Solution
Set to 0 at Earth, carried to 4 by Out, copied by Back, carried home to 8, compared with Home’s 10, and only then the conclusion.
★ Head-scratchers
Birthday cards for the variant trip from the second ◇ exercise: a star 4 light-years away at \(v = 0.8\), where \(k = 3\). How many cards does the traveler receive, and how many does home receive?
Hint
Outbound, the traveler ages 3 years and receives home’s cards at a third of the rate; homebound, 3 years at three times the rate.
Hint
Shortcut: each twin receives one card per birthday of the other. How old is each twin at the reunion?
Solution
Traveler: \(3 \times \tfrac13 + 3 \times 3 = 1 + 9 = 10\), one per home birthday. Home: the traveler’s turnaround light reaches Earth at \(t = 5 + 4 = 9\), so home gets \(9 \times \tfrac13 = 3\) slow cards, then \(1 \times 3 = 3\) fast ones in the last year: \(6\), one per traveler birthday.
A friend says: “The traveler felt acceleration, so special relativity can’t handle this. You need general relativity.” Write a rebuttal.
Model answer
Special relativity handles any worldline, bent or curved: chop it into short straight legs and add up \(\sqrt{\dt^2 - \dx^2}\) along it. The three-brothers relay gets the same 8 years with no accelerating clock at all, so acceleration can’t be the cause. In the twin story, the push only tells us which path is the bent one. General relativity is needed for gravity, not for acceleration.
- The interval sums along any path, bent or curved
- The three-brothers relay has no accelerating clock and the same result
- Acceleration only tells you which path is bent
Stuck on a square root? The toolkit has a refresher, and the formula card has every equation from this chapter.
Build it
Time to teach the computer the odometer. Your function receives a whole trip, a list of events, and adds up the wristwatch time leg by leg. Once it passes, the trip lab at the top of this chapter runs on your code.
Code you need
An array is a list in square brackets. Count from zero: stops[0] is the first item, and stops.length is how many there are. A for loop repeats a block while a counter i climbs; i++ adds 1 each time round.
const stops = [2, 5, 9];
for (let i = 1; i < stops.length; i++) {
console.log(stops[i] - stops[i - 1]); // 3, then 4
}Starting at i = 1 and looking at stops[i - 1] and stops[i] visits every neighboring pair. For a trip, each pair is one leg.
Build it: wristwatchTime(path)
Add up the wristwatch time of every leg. Your properTime and intervalSquared from Chapter 6 are already loaded (or the reference ones, if you borrowed them). The tests check Rae’s 8-year trip, a stay-at-home 10, a small detour, two light-speed legs that add to zero, refusals for legs that run backward in time or faster than light, and one property: 100 random trips, none of which may beat the straight line.
Check each leg before you add it. With dt = b.t - a.t and dx = b.x - a.x, the leg runs backward in time if dt < 0, and faster than light if Math.abs(dx) > dt. Refuse those the way you did in Chapter 2: if (dt < 0) throw new RangeError('backward in time');. Do the checks first: properTime returns NaN for a faster-than-light leg instead of refusing it.
When your function passes, scroll up to the first lab: its chip says it’s powered by your wristwatchTime(). Curious? Break your function on purpose, choose “Run the lab with my code anyway,” and see whether a bent trip can beat the free particle in that universe.
Checkpoint
Six items, five of them scored. Feedback waits until you submit.
A star is 12 light-years away. A traveler flies there at \(v = 12/13\) and straight back, while Earth ages 26 years. How many years does the traveler age?
That multiplies Earth’s 26 years by \(\g\). Is the bent path longer or shorter?
Solution
Each leg: \(\dt = 13\), \(\dx = 12\), so \(\tau = \sqrt{169 - 144} = 5\). Two legs: \(10\) years. (A 5-12-13 triangle, with \(\g = 13/5\).)
Three brothers: Out and Back each coast at constant speed the whole time, and Back copies Out’s watch reading as they pass at the star. Back passes Earth showing 8 years while Home shows 10. What does this show?
Solution
The bent path. The relay has no acceleration anywhere and still gives 8 against 10, so acceleration is not the cause, and special relativity handles it with a sum of intervals.
At the turnaround, Rae’s now-line swings, and her reckoning of Lin’s clock jumps from 3.2 to 6.8 years. What happens to Lin during the turnaround?
Solution
Nothing happens to Lin. The jump is in Rae’s choice of simultaneity: her outbound frame and her homebound frame slice spacetime differently. With the three brothers it is two different people’s now-lines, and nobody jumps.
Every path below runs from \((0, 0)\) to \((10, 0)\). Order them from the largest total interval to the smallest. (Where a clock can ride the path, that total is its wristwatch time.)
- Stay home: no turnaround at all
- One turn at \((5, 1)\)
- One turn at \((5, 3)\)
- One turn at \((5, 4)\)
- Two 45° legs through \((5, 5)\): no clock can ride them
Solution
10, then \(2\sqrt{24} \approx 9.80\), then 8, then 6, then 0. The farther the path bends from straight, the less time it holds. The 45° path has interval exactly 0: only light travels it, and light carries no wristwatch.
Lin flashes a card on her 1st birthday, at \((1, 0)\). Place the event C where that flash reaches Rae.
No pointer? Type t and x into the boxes under the diagram.
Solution
The flash runs along \(x = t - 1\). Rae’s outbound path is \(x = 0.6\,t\). They meet where \(t - 1 = 0.6\,t\), so \(t = 2.5\), \(x = 1.5\). Rae’s watch there reads \(\sqrt{2.5^2 - 1.5^2} = 2\): Lin’s first card reaches Rae on Rae’s 2nd birthday, which is \(k = 2\) at work.
In two or three sentences: why does the twin who stays home age more?
Model answer
Both twins travel between the same two events, but along different worldlines: Lin’s is straight and Rae’s is bent. Wristwatch time is the spacetime length of a worldline, and because of the minus sign the straight one is the longest. So Lin ages 10 years to Rae’s 8.
- Same two events, different worldlines
- Wristwatch time is measured along the path
- The straight path has the most
Explain it to Sagredo
Sagredo isn’t done. “Fine, Rae comes home younger. But from her rocket, Earth was the thing moving the whole time. Why doesn’t Lin come out younger?” Answer him.
Model answer
Because the two trips aren’t mirror images. Lin’s worldline is straight from Depart to Reunion; Rae’s bends at the star. There is no single free-float frame in which Rae sits still for the whole trip: she rides one frame out and another one back, and switching frames swings her line of “now.” Wristwatch time is the length of a path through spacetime, and with spacetime’s minus sign the straight path is the longest. So Lin ages 10 years and Rae 8. The push at the turnaround only marks which path is bent: three coasting brothers get the same answer with no push at all.
- Names the two different paths between the same two events
- Says the straight (free-float) path has the most wristwatch time
- Explains why Rae can’t claim one frame for the whole trip, or uses the brothers to rule out acceleration
What’s invariant here?
- The wristwatch time along any worldline between two events, \(\inv{\tau} = \sum_i \sqrt{(\dt_i)^2 - (\dx_i)^2}\). Every frame computes the same total.
- Maximal aging: the straight, free-float worldline has the most wristwatch time of all. Detours in spacetime are shortcuts.
- The birthday-card counts: Rae receives 10, Lin receives 8. Counting flashes needs no convention.
- Not invariant: Rae’s “now” at home. It swings at the turnaround, and that swing is bookkeeping, not aging.
Still strange?
It should be: the twin who works hard to travel ends up with less time, and the straight line is the long way round. Next chapter we stop measuring and start looking, and the \(k\) you counted turns out to set the color shift of starlight.
Going further
Optional extras for when the main idea feels solid.
Deep dive: a rocket at a steady 1 g
Special relativity happily flies a rocket that never stops accelerating. Suppose the crew feels a steady push \(a\), the same as standing on Earth: \(1\,g\), which in years and light-years is \(a = 1.0323\) ly/y². Starting from rest, after the crew’s own watch reads \(\tau\), the lab finds them at
\[ \lab{t} = \frac{\sinh(a\tau)}{a}, \qquad \lab{x} = \frac{\cosh(a\tau) - 1}{a}, \qquad v = \tanh(a\tau). \]
Chapter 13 builds \(\sinh\), \(\cosh\) and \(\tanh\) from \(e^x\) with plain arithmetic Need a refresher?. For now, the numbers: when the crew’s watch reads 1 year, the rocket moves at \(v = 0.7748\), it is 0.564 light-years out, and the Earth-frame lattice clock it is passing reads 1.187 years. The crew’s worldline is a curve, a hyperbola, and its wristwatch time is still the sum of tiny straight legs. Nothing here needed gravity.
Careful with that 1.187. It compares the crew’s watch with a lattice clock far from home, so it depends on the frame, exactly like Chapter 8’s slow clocks. The invariant question needs a reunion: fly out under 1 g, turn, fly back, and the crew comes home younger, because theirs is the bent worldline.
★★ At what steady speed must Rae fly out and back so that she ages exactly half as much as Lin?
Hint
Half as much means \(\g = 2\). Which friendly triangle gives \(\g = 2\)?
Solution
\(\g = 2\) means \(\sqrt{1 - v^2} = 1/2\), so \(v^2 = 3/4\) and \(v = \sqrt{3}/2 \approx 0.866\). That’s the 30-60-90 triangle: a leg with \(\dt = 2\) and \(\dx = \sqrt{3}\) has wristwatch time \(\sqrt{4 - 3} = 1\).
★★ Use the deep dive. A crew accelerates at 1 g for one year of its own time, starting from rest at Earth. At that moment, what does the Earth-frame lattice clock beside the rocket read? Use \(a = 1.0323\) ly/y² and a calculator with sinh.
Hint
\(\lab{t} = \sinh(a\tau)/a\) with \(\tau = 1\).
Solution
\(\sinh(1.0323)/1.0323 \approx 1.2257/1.0323 \approx 1.187\) years. The crew’s watch reads 1 year; the Earth-frame lattice clock beside them reads about 1.19. That comparison depends on the frame: on the crew’s own momentary now-line, the clock back on Earth reads only \(1.187 - 0.7748 \times 0.5636 \approx 0.75\) years. Only a reunion settles who is younger.
For more, Edwin Taylor and John Wheeler’s Spacetime Physics (2nd edition, 1992) treats the twins in depth. Their Exploring Black Holes (2000) adds gravity and uses the principle of maximal aging to tell a thrown stone where to fall.