Chapter 9
Shrinking Rulers
Length is a question about “now”, and “now” depends on who’s asking.
What this chapter does
Chapter 8 was about time: a muon’s own clock counts less time for its fall than Earth’s clocks do. This chapter asks the same question about space. How long is something that moves past you? The answer hides in a word from Chapter 7: “now”.
- The puzzle and three labs (about 20 minutes). Measure a moving rod, settle a fight about a pole and a barn, and ride along with a muon.
- What a length is (about 15 minutes). One rule, one formula, and three things the formula doesn’t mean.
- Nature, practice and code (about 25 minutes). Real evidence, problems, and a function that powers the rod lab.
You’ll leave able to say how a 10 m pole can be 6 m in one frame and still 10 m in its own, with nobody wrong.
Warm-up
Three quick ones from earlier chapters. Each is a tool you will use today.
From Chapter 8: Rae’s wristwatch ticks off 3 m of time between two events on her rocket, which moves at \(v = 0.8\) through Lin’s lab. How much time passes between those two events in Lin’s frame?
That divides by \(\g\). Rae’s watch is present at both events, so it reads the shorter time. Which frame’s time is longer?
Hint
Which clock is present at both events? That clock reads wristwatch time, the shortest time between them.
Solution
Rae’s watch is at both events, so \(\tau = 3\). At \(v = 0.8\), \(\g = 5/3\), and Lin’s frame time is \(\g\tau = \tfrac53 \times 3 = 5\) m of time.
From Chapter 7: Rae’s rocket is 10 m long, with a clock at each end, synchronized in her frame. The rocket moves at \(v = 0.6\). In Lin’s lab frame…
Hint
Front clock lags, by \(v\) times the rocket’s own length.
Solution
Clocks synchronized on a rocket of own length \(L\) are out of step by \(vL\) in the lab, with the front one behind: \(0.6 \times 10 = 6\) m of time.
From Chapter 6: two events are \(\dt = 3\) m of time and \(\dx = 5\) m apart in Lin’s frame. Every frame agrees on the number \(\dx^2 - \dt^2\) (Chapter 6’s squared interval with its sign flipped, so it’s positive when the space gap is bigger). What is it?
That’s \(\dt^2 - \dx^2\), the squared interval. This question asks for the same number with the opposite sign. Which is bigger here, Δx or Δt?
That adds the squares. Spacetime subtracts them. What changes?
Hint
\(5^2 - 3^2\).
Solution
\(25 - 9 = 16\). The pair is spacelike, and its proper distance is \(\sqrt{16} = 4\) m. Hold on to that: today it becomes a length.
The puzzle
Lin owns a barn 8 meters long, with a door at each end. Rae owns a pole 10 meters long, and she carries it through the barn at \(v = 4/5\). (She’s fast. Go with it.)
Lin has heard that moving things measure shorter. “At that speed,” she says, “your pole is only 6 meters. I can shut both doors at once with the whole pole inside.”
Rae has heard the same rumor. “Moving things are shorter? You’re the one moving. Your barn is only 4.8 meters. My 10-meter pole will never fit.”
They can’t both be right. Can they?
Predict
Can both doors be shut at once with the whole pole inside?
Experiment
Before we judge the barn, we need to agree on what “length” even means for something that won’t hold still. Here is a rod, Rae’s, 5 meters long in its own frame. It slides along Lin’s ruler at \(v = 0.6\). Lin’s lattice has a clock at every few meters, all synchronized. Your job: measure the rod with it.
Rae’s rod is 5 m long in its own frame. Marked at one instant of Lin’s clocks, at v = 0.6, its ends are 4 m apart.
Describe this lab
A horizontal ruler in meters represents Lin’s lab frame, with synchronized lattice clocks above it that all show the same lab time. Rae’s rod, 5 m long in its own frame, slides to the right at the chosen speed. Buttons (or the keys M and N) leave a chalk mark at the current position of the front or rear end, stamped with the lab time. Play, Pause and a time slider control the lab time. When both marks exist, the lab reports their separation and whether they were made at the same lab time; if not, it warns that the result isn’t a length. An inset spacetime diagram shows the rod’s two ends as dashed worldlines, Lin’s current “now” as a horizontal line, your marks as dots, and optionally Rae’s tilted “now” through the front mark, which meets the rear end at the event Rae calls simultaneous. The Describe button gives a text account and a table of the marks.
Try it
- Freeze the rod anywhere on the ruler. Without moving time, mark the front, then the rear. How long is the rod in Lin’s frame at \(v = 0.6\)? At \(v = 0.8\)?
- Now cheat: mark the front, let the rod run a moment, then mark the rear. What “length” do you get? Can you make it zero?
- Tick “Show Rae’s now-line” after a fair measurement. Which rear event does Rae call simultaneous with your front mark (the readout gives its time and place in Lin’s frame), and how far apart are those two events for her?
Now the barn. This lab starts in the barn frame at \(v = 0.8\). Scrub the time and watch the doors. Then switch to the pole’s frame and watch the same doors again. There, drag the time back to about \(t' = -11\) to see the exit door shut first.
Barn frame, t = 0: both doors shut at once with the 6 m pole inside. Pole frame: the barn is 4.8 m and the exit door shuts 10.67 m of time before the entrance.
Describe this lab
A barn 8 m long in its own frame, with an entrance door on the left and an exit door on the right, and a pole 10 m long in its own frame, moving right through it. A frame switch toggles between the barn frame (Lin’s lattice) and the pole frame (Rae’s lattice). A time slider, a Play button and a speed slider from 0.05 to 0.95 control the scene. In the barn frame both doors shut at t = 0, when the pole is centered, and reopen a moment later. A mini spacetime diagram shows the doors’ worldlines as solid lines, the pole’s ends as dashed lines, and the door events as dots, with the current frame’s “now” line. A list orders the events by time in the chosen frame; readouts give the pole’s length in the barn frame, the barn’s length in the pole frame, and how much earlier the exit door shuts in the pole frame. An option “Exit door stays shut” lets the pole hit the exit door and shows the stop signal running back along the pole no faster than light.
Try it
In the pole frame, how long before the entrance door shuts does the exit door shut? Read it from the event list. Then find the slowest speed at which the pole fits in the barn frame.
One more. In Chapter 8, muons born 10 km up reached the ground because, in Earth’s frame, their clocks run slow. Ride along with a muon instead. Its own clock is perfectly normal. So what changed?
Before you flip the lab: in the muon’s frame its clock ticks at the normal rate. Compared with Earth’s frame, what fraction of the muons reach the ground?
Hint
Is “this muon reached the detector” a fact about one event?
Solution
The same fraction. A muon reaching the detector is an event, and all frames agree on events. The muon’s frame explains it differently, and the lab shows how.
In Earth’s frame, muons at v = 0.995 cross 10 km of air with slow clocks, and 21.8% survive. In the muon’s frame the air is 998.7 m thick, the muon’s clock is normal, and the same 21.8% survive.
Describe this lab
A column of atmosphere 10 km high with a ground detector. Cohorts of 36 muons, drawn as small rings that empty like countdown clocks for each muon’s own lifetime, are made at the top and fall at the chosen speed. The fall is replayed in slow motion, and a legend under the column gives the rate. A frame switch shows either Earth’s frame, where the muons’ clocks run slow and a new cohort sets off every 1.2 seconds, so several are in the air at once, or the muon’s frame, where the muon clocks tick normally and the atmosphere, contracted along the motion, rushes up past them with the detector, one cohort at a time. A muon whose own clock passes its random lifetime leaves a faint × where it decayed; in the muon’s frame the marks ride up with the air. Survivors make the detector glow and collect on it as gold dots, and a running tally counts arrivals and decays. A chart beside the column shows the fraction still alive, and a teal readout shows the fraction that reaches the ground, which is the same in both frames. Keys while the lab has focus: F switches frame, Space plays or pauses (when motion is reduced, it drops one cohort and shows where each of its muons ended up).
Try it
At \(v = 0.995\), switch to the muon’s frame. How thick is the atmosphere now, to the nearest meter? What fraction of muons reach the ground in each frame?
The big percentage is what the rules predict. The run’s own counts catch up slowly: the first muons land after a few seconds, and muons still in the air aren’t counted yet.
What just happened?
Both frames were right, about different questions. In the barn frame, both doors are shut at the same instant with the 6 m pole inside. In the pole frame, the barn is 4.8 m and the pole never fits, but the doors are never shut at the same time: the exit door shuts 10.67 m of time before the entrance door, and swings open again before the pole reaches it.
Look at what the two frames disagree about. Not about any single event. Every door slam, every end passing every door, happens in both stories at a definite place on the diagram. They disagree about which events happen at the same time.
“Fine,” says Sagredo, “but which is it really? Is the pole 10 meters, or 6?”
Here is the answer, and the rest of the chapter unpacks it. “The pole is inside the barn” is a claim about two events at once: the rear end is inside at the same moment the front end is. So it has a hidden “now” in it. Change your “now”, and you change the answer. Ten meters is the pole’s own length: its length in the frame where it sits still. Every frame agrees on that number. Six meters is its length in the barn’s frame. Asking which is “really” its length is asking “in which frame?”, and Nature has no favorite.
The rod lab told the same story in miniature. Marks made at one lab time give 4 m for a 5 m rod. Marks made at different times give nonsense, and Rae’s own “now” gives 5.
What a length is
Both ends, same time
Measuring a parked car is easy: lay a tape along it and read both ends whenever you like. A moving car is harder. Mark where the front bumper is, and by the time you get to the back bumper, the car has moved on.
So here is the rule. It is the whole chapter in one sentence: the length of a moving thing, in your frame, is the distance between where its two ends are at the same time in your frame. “The same time” means the same reading on your lattice clocks. That’s exactly what the rod lab made you do.
Quick check. A train rolls past Lin’s platform. Her lattice marks where the front of the train is at \(t = 0\), and where the rear is at \(t = 2\) m of time. Is the distance between the two marks the train’s length in Lin’s frame?
Hint
Did the train move between the two marks?
Solution
No. Between \(t = 0\) and \(t = 2\) the train moved, so the gap between the marks mixes its length with its motion. It is the rod lab’s “cheat” again.
And Chapter 7 already told us the catch. Lin’s “same time” and Rae’s “same time” are different sets of events. So they will measure different lengths. The only question is how different.
A length from a clock
We can get the number with no new physics at all. Rae’s rod is 5 m long and moves at \(v = 3/5\). Lin stands beside one of her lattice clocks; call it C.
In Rae’s frame, clock C slides along her 5 m rod at 0.6. It takes \(5 \div 0.6 = 8.33\) m of time to get from her front end to her rear end, as measured by Rae’s two end clocks (two clocks, so this is lattice time).
But clock C is present at both events: passing the front, then passing the rear. So C reads the wristwatch time between them, which is shorter by \(\g = 5/4\): \(8.33 \div 1.25 = 6.67\) m of time.
In Lin’s frame, then, the rod slides past C at 0.6 for 6.67 m of time. So its length in her frame is \(0.6 \times 6.67 = 4\) meters. Not 5.
Time stretching forced length shrinking. They aren’t two separate effects. They are one fact about spacetime, seen from two sides.
The picture
Here is the same rod as a spacetime diagram in Lin’s frame. The rod’s two ends trace two parallel dashed worldlines, and the strip between them is the rod’s whole history. Lin’s “now” at \(t = 0\) is the horizontal axis. It cuts the strip at R and F, which are 4 apart. Rae’s “now” through F is tilted, as every rocket now-line is. It meets the rear end at R′, and R′ and F are the two ends at the same time for Rae.
Now check the 4 a second way, with the one thing every frame agrees on. We’ll follow one pair of events, R and F in Plate 9.1, through both frames. Click through one step at a time.
- Lin’s marks. Write events as (t, x). Lin’s two marks are R = (0, 0) and F = (0, 4): both ends at \(t = 0\). So \(\dt = 0\) and \(\dx = 4\). Like your fair marks in the rod lab: one lab time, 4 apart. Here we make them at \(t = 0\) (scrub the lab to \(t = 0\) to get exactly these).
- Lin’s number, the one every frame agrees on. R and F have more space than time between them: they are a spacelike pair. For a pair like that, Chapter 6’s invariant is \(\dx^2 - \dt^2\), the squared proper distance. For Lin it is \(4^2 - 0^2 = 16\). Every frame must get 16 for this pair. It’s the squared interval with its sign flipped, so it comes out positive. Its square root, 4, is the proper distance.
- Rae’s clocks. For Rae, R and F don’t happen at the same time. Chapter 7’s rule for a pair that is simultaneous in the lab gives \(\dtp = -\g v \dx = -1.25 \times 0.6 \times 4 = -3\). So F happens 3 m of time before R, by Rae’s clocks. This is Warm-up 2’s front-clock lag: \(1.25 \times 0.6 \times 4 = 0.6 \times 5\), \(v\) times the rod’s own length. At one lab time, Rae’s front clock reads 3 behind her rear clock.
- Rae’s rulers. In Rae’s frame the rod sits still. Its rear end stays at \(x' = 0\) and its front end stays at \(x' = 5\), the rod’s own length, at every moment. So any event on the rear end and any event on the front end are 5 apart for Rae, whenever each one happens. For R and F, \(\dxp = 5\). It’s the parked car from the start of this section: when a thing sits still, you may read its two ends at different times.
- Rae’s number. For Rae, \(\dxp^2 - \dtp^2 = 5^2 - 3^2 = 25 - 9 = 16\). That is Lin’s 16, as it must be, so Lin’s 4 passes the check. Rae’s 5 is \(5 \div 4 = 1.25\) times Lin’s 4, and 1.25 is \(\g\).
Rae’s own length over Lin’s length is the stretch factor again.
One pair of events, two frames. Lin writes (t, x); Rae writes (t′, x′). Event Lin Rae R (0, 0) (0, 0) F (0, 4) (−3, 5) Squared proper distance 16 − 0 = 16 25 − 9 = 16
The shorthand
The check came down to one line of arithmetic: \(25 - 9 = 16\). Now say it with letters, so it works for any rod at any speed. Call the rod’s own length \(\inv{L_0}\) (Rae’s 5) and its length in Lin’s frame \(\lab{L}\) (Lin’s 4). Each number in the check becomes letters:
| Number | Where it came from | Letters |
|---|---|---|
| 16 | Lin’s length, squared: \(4^2\) | \(\lab{L}^2\) |
| 25 | Rae’s length, squared: \(5^2\) | \(\inv{L_0}^2\) |
| 9 | Rae’s time gap, squared: \(3^2\), with \(3 = 1.25 \times 0.6 \times 4\) | \((\g v \lab{L})^2\) |
So \(25 - 9 = 16\) becomes:
\[ \inv{L_0}^2 - (\g v \lab{L})^2 = \lab{L}^2 \]
Move the 9 across, \(25 = 16 + 9\), and pull out \(\lab{L}^2\). In numbers that’s \(25 = 16 \times 1.5625\), and 1.5625 is \(1.25^2\), which is \(\g^2\):
\[ \inv{L_0}^2 = \lab{L}^2\,(1 + \g^2 v^2) = \g^2\,\lab{L}^2 \]
Deep dive: why 1 + γ²v² = γ² at every speed
Check it at \(v = 0.6\): \(1 + 1.5625 \times 0.36 = 1 + 0.5625 = 1.5625\), which is \(1.25^2\). For any speed, square \(\g = 1/\sqrt{1 - v^2}\) to get \(\g^2 (1 - v^2) = 1\), which is \(\g^2 - \g^2 v^2 = 1\). Add \(\g^2 v^2\) to both sides.
Take the square root: \(\inv{L_0} = \g \lab{L}\), just as \(5 = 1.25 \times 4\). Divide both sides by \(\g\), and you have the shorthand for “mark both ends at the same time”: Need a refresher?
\[ \lab{L} = \frac{\inv{L_0}}{\g} \qquad \text{(along the motion only)} \]
Physicists call \(\inv{L_0}\) the proper length: the length in the frame where the thing is at rest, its own length. In Plate 9.1 it is the gap from R′ to F, the two ends at one moment of Rae’s clocks. So a proper length is a proper distance too: for two events on the rod’s ends that are simultaneous in the rod’s own frame, the proper distance is the proper length. (R and F have a proper distance too, 4, but they aren’t simultaneous for Rae, so their 4 is Lin’s length, not the rod’s own.) It’s teal because every frame agrees on it. The shrinking is called Lorentz contraction. Don’t let “proper” fool you: the other lengths aren’t improper. They’re answers to a different question. From here on I’ll say proper length.
Pitfall
Multiply or divide? The proper length is always the longest one. If your moving rod comes out longer than its proper length, you multiplied by \(\g\) where you should have divided.
Quick check. A rocket is 12 m long in its own frame and passes Earth at \(v = 0.8\). How long is it in Earth’s frame?
That multiplies by \(\g\). The moving rocket should come out shorter than its own 12 m.
Hint
At \(v = 0.8\), \(\g = 5/3\). Divide the proper length by \(\g\).
Solution
\(L = 12 \div \tfrac53 = 12 \times 0.6 = 7.2\) m.
They can, for the same reason each frame measured the other’s clocks slow in Chapter 8. Each measures the other’s ruler with her own “now”. Lin’s two marks and Rae’s two marks are different pairs of events, so there’s no contradiction. A contradiction needs one universal now, and there isn’t one.
Pole and barn, settled
Back to the barn, with numbers. At \(v = 4/5\), \(\g = 5/3\), so in the barn frame the pole is \(10 \div \tfrac53 = 6\) m. Both doors shut at \(t = 0\): two events 8 m apart at the same barn time.
In the pole frame those same two events are not simultaneous. Chapter 7’s rule gives \(\dtp = -\g v \dx = -\tfrac53 \cdot \tfrac45 \cdot 8 = -10.67\) m of time. The exit door shuts 10.67 m of time before the entrance door. In the pole frame the pole sits still and the short barn slides onto it. The exit door shuts before it has reached the front end, then swings open again. Then the exit slides past the front end. Only later does the entrance slide past the rear end and shut behind it.
Every door event happens in both stories. Only the order of some pairs differs, and only for pairs too far apart for light to connect, the spacelike pairs of Chapter 6. So in no frame does a door ever hit the pole: a collision would be an event, and every frame would have to agree on it.
Why, really?
Because “the pole is inside the barn” is a sentence with a hidden “now” in it. A sentence about one event, like “the exit door shut” or “the front end passed the exit”, is true in every frame. A sentence about two events at once inherits its frame’s choice of “at once”. Length is that second kind of sentence. So is “fits”.
Quick check. In the pole frame the barn is only 4.8 m long, and its exit door shuts while the barn is sliding onto the 10 m pole. Does the exit door ever hit the pole?
Hint
Answer it in the barn frame first: are the doors ever shut while part of the pole is in a doorway? Then remember that a collision is an event.
Solution
No. In the barn frame the doors are shut only around \(t = 0\), while the 6 m pole sits inside with room to spare, so nothing touches. A hit would be an event, and events are the same in every frame. In the pole frame the same fact shows up as timing: the exit door shuts and opens again before it reaches the front end.
Only along the motion
Does a fast rocket also get thinner? No, and here is a proof with no algebra. Two rings, each 1 m across in its own frame, fly toward each other along a shared axis. Suppose moving things shrank sideways too. Then Lin would say Rae’s ring is the smaller one and passes inside hers. Rae, with exactly the same right, would say Lin’s ring passes inside hers.
Give each ring a fringe of wet paintbrushes around its rim, bristles pointing inward. If one ring slips inside the other, the outer ring’s brushes paint a stripe on the inner ring’s rim, and the inner ring’s brushes touch nothing. Afterward anyone can check which ring is striped, so that’s a record every frame must agree on. Lin’s story stripes Rae’s ring; Rae’s story stripes Lin’s. Neither frame is special, so neither story can be the true one. The rings stay the same size and meet rim to rim: each ring’s brushes touch the other’s rim equally, so neither ring ends up inside.
Myth
“Moving things shrink in every direction.” Only the length along the motion changes. Heights and widths across the motion are the same in every frame.
Quick check. A hoop flies at high speed along its own axis, like a ring tossed over a peg. In the lab frame, does its diameter contract?
Hint
Which way does the hoop move, and which way does its diameter point?
Solution
No. Contraction happens only along the motion. The two-rings argument shows that a sideways shrink would make two frames disagree about which ring passes inside the other, a disagreement about events, which is impossible.
Nothing gets squeezed
Is Rae’s rod under strain? Ask Rae. In her frame it sits still at its proper length, as relaxed as it was on the workbench. A strain gauge glued to it reads zero, and every frame must agree on what a gauge reads, because a reading is an event. Contraction isn’t something done to the rod. It’s a fact about which events Lin’s lattice calls “now”.
Quick check. Rae glues a strain gauge to her rod, which coasts past Lin’s lab at \(v = 0.9\). The gauge would read any squeeze in the rod. What does it read?
Hint
Answer it in Rae’s frame, where the rod is at rest. Then remember that a gauge reading is the same for everyone.
Solution
Zero. A coasting rod feels no stress. Its contraction in Lin’s frame comes from Lin’s choice of simultaneous events, not from any force.
Myth
“Length contraction squeezes the atoms.” A rod coasting at steady speed feels no stress at all. Stress appears only when something forces a proper length to change: try “Exit door stays shut” in the barn lab, or read about Bell’s spaceships below.
Myth
“A perfectly rigid pole would stop all at once.” If the exit door stays shut, then in the barn frame the front end stops when it hits. For the rear to stop at that same instant, the news “stop!” would have to cross the pole instantly, faster than light. It can’t. The rear keeps coming until the news arrives, and the pole crumples. In relativity there are no perfectly rigid bodies: no material can pass a push along faster than light. Real materials are far slower still: the news travels at the speed of sound, about 5 km/s in steel. So every real pole gives a little when it’s stopped.
Measured is not seen
One more warning about the word “shorter”. Everything in this chapter is measured with a lattice: each clock notes what is beside it at its own reading. A camera does something else. It collects the light that arrives at its lens together, and that light left different parts of the object at different times. Work it through, and a fast sphere photographs as a circle, not a squashed ellipse, and a fast cube looks rotated. That surprise is Chapter 12’s.
Myth
“Length contraction is a visible squashing.” What you see includes light delay. James Terrell and Roger Penrose showed, independently, that a fast sphere still photographs with a circular outline.
Deep dive: Bell’s spaceships
Two identical spaceships sit 100 m apart in a launch frame, with a thread tied between them, exactly long enough. At a signal that is simultaneous in the launch frame, both fire identical engines with identical programs. Does the thread break?
In the launch frame, the two ships have the same speed at every launch-frame instant, so the gap between them stays 100 m. But the thread is now moving at \(v\), so its length in the launch frame should be \(100/\g\), shorter than the gap. A thread can’t be both. It stretches, and then it snaps.
In the ships’ frame, after the engines stop, the gap is \(\g \times 100\) m: about 167 m at \(v = 0.8\). How did it grow? The launch frame’s simultaneous ignitions are not simultaneous for the ships. It’s the same rule as the rod: for the ships, the front one fired first, because it is ahead along the motion (Chapter 7). So it pulled ahead. Two frames, one verdict: the thread breaks. Dewan and Beran posed the puzzle first; John Bell made it famous.
That’s the theory: one rule about simultaneous marks, one formula, and three things it doesn’t mean. Next: does Nature agree?
Ask Nature
The muons of Chapter 8 already ran this experiment for us. Earth’s frame explained their survival with slow clocks. The muon’s frame has to explain the same count without them, and it does.
Ask Nature
A muon is made 10 km up and falls at \(v = 0.995\). In Earth’s frame its clock runs slow by \(\g = 10.0125\), and 21.8% of such muons reach the ground.
In the muon’s frame its clock is normal: it lives about 2.2 μs, like any muon at rest. But the atmosphere rushes up at 0.995, and its thickness along the motion is \(10\,000 \div 10.0125 =\) 998.7 m. Crossing that takes 1004 m of time, which is 3.35 μs, about 1.5 lifetimes, and \(e^{-3.35/2.2}\) is again 21.8%. Need a refresher?
Same muons, same count, two explanations. Muon lifetime 2.1969811 μs: Particle Data Group. Survival numbers computed with this course’s reference library.
Myth
“The muon lives longer in its own frame.” No. In its own frame it lives about 2.2 μs, like every muon at rest. What changes, for the muon, is the thickness of the air.
Ask Nature
At the Relativistic Heavy Ion Collider on Long Island, which ran from 2000 to 2026, gold nuclei collided at about \(\g \approx 100\) (more precisely, about 107). In the collider’s frame each nucleus was flattened along the beam to about a hundredth of its thickness at rest, a pancake, while its width across the beam was unchanged. Models of these collisions start from the pancakes. The RHIC experiments’ summary papers, Nuclear Physics A 757 (2005).
Real numbers
The pole’s speed, \(v = 0.8\), is \(0.8 \times 299\,792\,458 \approx 2.40 \times 10^8\) m/s. In the barn frame the 6 m pole takes \(6 \div 0.8 = 7.5\) m of time, about 25.0 ns, to slide past one point. The pole-frame door gap of 10.67 m of time is \(10.67 \times 3.3356\un{ns} \approx 35.6\un{ns}\).
Practice
Three tiers. The first few take a line each; the last one deserves a cup of tea. Fractions such as 32/3 are fine as answers. Short on time? Do the ◇ tier and any two ◆ items; the rest are extra practice.
◇ Back of the envelope
A rod is 5 m long in its own frame and moves at \(v = 0.6\) along its length. How long is it in the lab frame?
That multiplies by \(\g\). Is a moving rod longer or shorter than its proper length?
Hint
At \(v = 0.6\), \(\g = 5/4\). Divide the proper length by \(\g\).
Solution
\(L = L_0/\g = 5 \div 1.25 = 4\) m.
How long is the 10 m pole in the barn frame at \(v = 0.8\)?
That’s \(10 \times \g\). The pole is moving in the barn frame. Should it come out longer or shorter?
Hint
\(\g = 5/3\) at \(v = 0.8\).
Solution
\(10 \div \tfrac53 = 6\) m.
How long is the 8 m barn in the pole frame?
That multiplies by \(\g\). In the pole frame, which object is moving?
That’s the pole in the barn frame. Now turn the question around: the barn moves past the pole.
Hint
In the pole frame the barn moves at 0.8. Its proper length is 8 m.
Solution
\(8 \div \tfrac53 = 4.8\) m. The contraction is mutual: each frame measures the other’s object short.
A spaceship moving at \(v = 0.8\) measures 3 m long in the lab frame. How long is it in its own frame?
That divides by \(\g\) a second time. The 3 m is already the contracted length. Which length is the longest?
Hint
The proper length is the longest. Undo the division: \(L_0 = \g L\).
Solution
\(L_0 = \g L = \tfrac53 \times 3 = 5\) m.
A train moves along the track at \(v = 0.6\). Sort each quantity: as measured in the platform frame, does it shrink compared with the train’s own frame, or stay the same?
- The length of a carriage
- The height of a carriage
- The gap between two windows along the side
- The width of the train, side to side
- The height of a passenger standing up
Hint
Ask of each one: does it point along the track?
Solution
Along the motion: the carriage length and the window gap shrink by \(\g = 1.25\). Across the motion: the height, the width and a standing passenger’s height are unchanged.
◆ Pencil and paper
Finish this worked example. A shuttle is 13 m long in its own frame and passes a space station at \(v = 5/13\).
- First, \(1 - v^2 = 1 - \tfrac{25}{169} = \tfrac{144}{169}\), so \(\sqrt{1 - v^2} = \tfrac{12}{13}\).
- So \(\g = \) (fill in).
- The station measures the shuttle as \(L = L_0/\g = \) (fill in) m.
That’s \(\sqrt{1 - v^2}\), which is \(1/\g\). The stretch factor is never below 1.
That multiplies by \(\g\). The moving shuttle should come out shorter than 13 m.
Hint
\(\g\) is one over \(\sqrt{1 - v^2}\).
Solution
\(\g = 13/12 \approx 1.083\), and \(L = 13 \div \tfrac{13}{12} = 12\) m. It’s the 5-12-13 triangle again.
In the barn frame both doors slam shut at the same instant, at the moment the 10 m pole (moving at \(v = 0.8\)) is wholly inside the 8 m barn. In the pole frame, when does the exit door close relative to the entrance door? Give \(t'_\text{exit} - t'_\text{entrance}\) in meters of time. A negative answer means the exit door closes first.
Right size, wrong sign. In the pole frame the barn moves backward. For a lab-simultaneous pair, is the event farther along +x earlier or later in the rocket frame?
That’s \(-v\,\dx\), without the \(\g\). The rule from Chapter 7 is \(\dtp = -\g v \dx\).
That’s one door’s time, measured from the barn’s middle. The question asks for the gap between the two doors, which are 8 m apart.
That uses the pole’s 6 m. The two door events are 8 m apart in the barn frame. Which distance separates the doors?
Hint
The two door events are simultaneous in the barn frame and 8 m apart.
Hint
Chapter 7: a lab-simultaneous pair separated by \(\dx\) has \(\dtp = -\g v \dx\).
Solution
\(\dtp = -\g v \dx = -\tfrac53 \times \tfrac45 \times 8 = -\tfrac{32}{3} \approx -10.67\) m of time. The exit door closes first, by about 35.6 ns.
A new pole is 20 m long in its own frame, the barn 16 m, and \(v = 0.6\). The doors again shut simultaneously in the barn frame. Find the pole’s length in the barn frame and \(t'_\text{exit} - t'_\text{entrance}\).
That multiplies by \(\g\). The pole is the moving thing in the barn frame.
That’s \(-v\,\dx\) without \(\g = 5/4\).
That uses the pole’s 20 m. The door events are as far apart as the doors.
Hint
At \(v = 0.6\), \(\g = 5/4\). The pole just fits.
Solution
Pole: \(20 \div 1.25 = 16\) m, a perfect fit. Doors: \(\dtp = -\tfrac54 \times 0.6 \times 16 = -12\) m of time.
What is the slowest speed at which the 10 m pole fits, just, inside the 8 m barn in the barn frame? (Check your answer with the barn lab.)
At 0.8 the pole is 6 m: it fits with room to spare. Can it go slower and still fit?
Hint
You need \(10/\g = 8\), so \(\g = 5/4\).
Hint
Which friendly speed has \(\g = 5/4\)?
Solution
\(\g = 10/8 = 5/4\), which is \(v = 3/5 = 0.6\). Check: \(1/\sqrt{1 - 0.36} = 1/0.8 = 1.25\).
Pole and barn at \(v = 0.8\). In the barn frame both doors shut, for an instant, at \(t = 0\), when the 6 m pole sits in the middle of the 8 m barn with 1 m to spare at each end. Put these events in order of time in the pole frame, earliest first.
- The exit door shuts (and springs open again).
- The pole’s front end passes out through the exit.
- The pole’s rear end passes in through the entrance.
- The entrance door shuts.
Hint
Put \(x = 0\) at the entrance door, so the exit is at \(x = 8\). Then use \(t' = \g(t - v x)\) with the barn-frame times: the rear end enters at \(t = -1.25\), the front end leaves at \(t = 1.25\).
Solution
With \(t' = \tfrac53(t - 0.8x)\): exit shuts at \(-10.67\); front out at \(\tfrac53(1.25 - 6.4) = -8.58\); rear in at \(\tfrac53(-1.25) = -2.08\); entrance shuts at \(0\). The pole is never wholly inside in this frame, and no door ever touches it. (With \(x = 0\) at the barn’s middle instead, every \(t'\) shifts by 5.33; the order is the same.)
A muon falls at \(v = 0.995\) through 10 km of atmosphere (Earth’s frame). How thick is the atmosphere in the muon’s frame, in meters?
That multiplies by \(\g\). In the muon’s frame, the atmosphere is the moving thing.
Hint
\(\g = 10.0125\) at \(v = 0.995\).
Solution
\(10\,000 \div 10.0125 = 998.7\) m.
Rae’s rod is 5 m long in its own frame and moves at \(v = 0.6\). The dashed lines are its rear and front ends in Lin’s frame, and F = (0, 4) is the front end at \(t = 0\). Place R′: the event on the rear end that Rae’s lattice calls simultaneous with F. The dotted line is Rae’s now-line through F. Give R′ in Lin’s (t, x): Rae’s −3 from the step-through is in her own coordinates.
No pointer? Type t and x into the boxes under the diagram.
Hint
Rae’s now-lines have slope \(v\): through F, they satisfy \(t - 0 = 0.6\,(x - 4)\).
Hint
The rear end is the line \(x = 0.6\,t\). Substitute it into the now-line and solve for \(t\).
Solution
\(t = 0.6(0.6t - 4) = 0.36t - 2.4\), so \(0.64t = -2.4\) and \(t = -3.75\), \(x = -2.25\). Check: from R′ to F, \(\dx = 6.25\) and \(\dt = 3.75\), and \(6.25^2 - 3.75^2 = 25\). Rae’s length is 5, as it should be.
A ball 1 m across flies past a camera at \(v = 0.8\). The camera takes a very short exposure. The ball’s outline in the photo is…
Hint
Does a camera record where every part of the ball is at one instant of its frame?
Solution
A circle. The photo is made of light that arrives at the lens together, and that light left the near and far parts of the ball at different times. Light delay and contraction combine so that a sphere still shows a circular outline. Chapter 12 builds the camera.
A steel pole 1 light-second long floats in space. You shove one end. When does the far end start to move?
Hint
A push is a compression wave, which is sound. How fast is sound in steel, compared with light?
Solution
Much later. The shove travels along the steel as a sound wave at roughly 5 km/s, so it needs \(3 \times 10^8 \div 5000 \approx 60\,000\) s, about 17 hours. Even an ideal material couldn’t beat light: a perfectly rigid rod would be a faster-than-light telegraph.
The two ends of the 10 m pole carry clocks synchronized in the pole frame. In the barn frame at \(v = 0.8\), the front clock reads behind the rear one. By how much?
6 is Warm-up 2’s value, for a different rocket at a different speed. Redo it with this item’s numbers: the pole is 10 m and \(v = 0.8\).
That’s \(\g v L\) with the proper length: one factor of \(\g\) too many. Use \(vL\) with the proper length, 10 m, or \(\g v\,\dx\) with the ends’ barn-frame separation, 6 m. Both give the same lag.
Hint
Front clock lags by \(vL\), where \(L\) is the proper length.
Solution
\(vL = 0.8 \times 10 = 8\) m of time. This lag is the clock-side of the same story: the pole’s clocks disagree with the barn’s about “now”.
★ Head-scratcher
Bell’s spaceships. Two identical ships sit 100 m apart, joined by a thread. At a signal simultaneous in the launch frame, both fire identical engines and reach the same speed together, in that frame. Does the thread break? Explain in both frames.
Model answer
Yes. In the launch frame the two ships move identically, so their gap stays 100 m. But a relaxed thread moving at \(v\) would measure only \(100/\g\) in that frame, so the thread must be stretched to span the gap, and it breaks. In the ships’ frame, once they coast, the gap is \(\g \times 100\) m, because the front ship fired first in their frame and pulled away. Both frames agree on the event that matters: the snap.
- The launch-frame gap stays fixed at 100 m.
- The thread “wants” to contract to \(100/\g\), so it is stretched.
- In the ships’ frame the gap grows to \(\g \times 100\) m.
Stuck on a step? The formula card has every equation from this chapter on one page.
Build it
Now teach your toolbox to shrink rulers. This one is short, because you already built the hard part: gamma, in Chapter 5.
Code you need
A function can call another function. Inside contract, you may call gamma(v): the sandbox loads your passing gamma, or the reference one if you borrowed it.
function double(x) { return 2 * x; }
function quadruple(x) { return double(double(x)); }Build it: contract(L0, v)
Return the length, along the motion, of a rod whose proper length is L0 when it moves at speed v. The tests check the pole (6 m), the barn (4.8 m), the 5 m rod (4 m), the muon’s atmosphere (998.7 m), and that contract(1, 1) throws an error (calling gamma(1) already does this): nothing with a length can ride a light beam.
If your answers come out longer than L0, look back at the Pitfall above. Once it passes, the rod lab in the Experiment section draws the rod at the length your contract computes, and says so on a chip.
Checkpoint
Six items: five are checked when you submit, and the sixth you check yourself against a model answer. Answer what you can, then press Submit; blank items count as “Not yet”.
A rocket is 13 m long in its own frame and passes Earth at \(v = 12/13\). How long is it in Earth’s frame?
That multiplies by \(\g\). Is the moving rocket longer or shorter than its proper length?
Solution
At \(v = 12/13\), \(\g = 13/5\). \(L = 13 \div \tfrac{13}{5} = 5\) m.
A cube, 2 m on each edge in its own frame, moves at \(v = 0.6\) parallel to one edge. Its dimensions in the lab frame are…
Solution
Along the motion: \(2 \div 1.25 = 1.6\) m. The two edges across the motion stay 2 m.
Measured fact, or myth?
- A moving rod’s length in a frame comes from marking both ends at the same time in that frame.
- A rod coasting at high speed is under compressive stress.
- A fast sphere photographs as a squashed ellipse.
- A perfectly rigid rod could pass a push from end to end instantly.
- Each frame measures the other frame’s rulers as shorter.
Solution
Facts: the simultaneous-marks definition, and the mutual contraction. Myths: stress in a coasting rod (there is none), a squashed photo (a sphere photographs round), and perfect rigidity (no signal outruns light).
Rae’s rod is 5 m long in its own frame and moves at \(v = 0.6\). Its rear end passes \(x = 0\) at \(t = 0\) (the dashed line). Place the event where Lin’s lattice finds the front end at \(t = 0\).
No pointer? Type t and x into the boxes under the diagram.
Solution
At \(t = 0\) the rod measures \(5 \div 1.25 = 4\) m in Lin’s frame, so the front end is at \((t, x) = (0, 4)\).
An 8 m barn, a 10 m pole at \(v = 0.6\), and doors that shut at the same time in the barn frame. Find \(t'_\text{exit} - t'_\text{entrance}\) in the pole frame.
Right size, wrong sign: which door shuts first for the pole?
That’s \(-v\,\dx\) without \(\g\).
That’s one door’s time, measured from the barn’s middle. The question asks for the gap between the two doors, 8 m apart.
Solution
\(\dtp = -\g v \dx = -1.25 \times 0.6 \times 8 = -6\) m of time: the exit door shuts first.
In two sentences: why do Earth and the muon agree on how many muons reach the ground, even though they explain it differently?
Model answer
Whether a given muon reaches the ground is a single event (or its absence), and every frame agrees on events. Earth explains it with the muon’s slow clock over 10 km; the muon explains it with its normal clock over 998.7 m of contracted air, and both give the same wristwatch time, 3.35 μs.
- Says the count is about events, which all frames share
- Names slow clocks for Earth and short air for the muon
- Mentions the shared wristwatch time
Explain it to Sagredo
Sagredo isn’t satisfied. “In one frame the pole fits. In the other it doesn’t. So relativity says nothing is real.” Answer him.
Model answer
Every event is real, and every event happens in both frames: each door slam, each end passing each door. What the frames disagree about is which events count as “at the same time”. “The pole is inside the barn” is a claim about two ends at the same time, so it depends on the frame the way “now” does. The frame-free facts are the events and the intervals between them, including the pole’s proper length, 10 m, which every frame agrees on.
- Says the door events happen in every frame
- Names simultaneity as the thing that differs
- Names an invariant: the proper length, or the interval
What’s invariant here?
- The proper length \(\inv{L_0}\): the distance between the ends at the same time in the rod’s own frame. Every frame agrees on it, because it is a proper distance (Chapter 6): any frame that works out \(\dx^2 - \dt^2\) for that pair of events gets \(\inv{L_0^2}\).
- Every single event. A door slam is a door slam in every frame.
- Lengths across the motion.
- Frame-dependent: the length along the motion, \(\lab{L} = \inv{L_0}/\g\), because it depends on which events count as “now”.
Still strange?
A pole that fits and doesn’t fit sounds like a riddle with a trick. The trick is that “at the same time” was never a fact about the world, only about a lattice of clocks, and Nature has never agreed to pick one lattice for us.
Going further
★★ Bell’s spaceships, in numbers. The two ships start 100 m apart and finish coasting at \(v = 0.8\), with the launch-frame gap still 100 m. How far apart are they in their own frame?
That divides by \(\g\). The 100 m is the contracted gap, measured in the launch frame, where the ships are moving.
Hint
In the launch frame the ships are moving and the gap is 100 m. So 100 m is the contracted length of a gap whose proper length you want.
Solution
Gap in the ships’ frame \(= \g \times 100 = \tfrac53 \times 100 \approx 166.7\) m. A thread with a proper length of 100 m can’t span it.
Reading: Taylor and Wheeler’s Spacetime Physics sets the pole-and-barn and many cousins of it as problems, with the diagrams to settle them. John Bell tells the spaceship story in his essay How to teach special relativity, reprinted in Speakable and Unspeakable in Quantum Mechanics.
Next chapter, one machine, the Lorentz transformation, will do in one step everything this chapter did with clocks, now-lines and the interval.