Spacetime, by Hand

Chapter 1

The Parable of the Surveyors

Everyone disagrees about coordinates. Everyone agrees about distance.

About 45 minutes · Builds on Chapter 0

What this chapter does

The Prologue promised that Chapters 1 to 7 build the tools for the muon trial in Chapter 8. This chapter builds the first two. There are no rockets in it yet, only two teams of surveyors mapping one small town.

  1. Invariants (about 25 minutes). The surveyors disagree about which way is north, yet they agree exactly on how far apart two places are. A number everyone agrees on is called an invariant. Relativity is built on them.
  2. Meters of time (about 20 minutes). Time measured in the same unit as distance, with the speed of light as the conversion factor. Every chapter after this one uses it.

The puzzle

Once upon a time, the town of Middlemark had two guilds of surveyors. The Daytimers worked by sunlight and found north with a magnetic compass. The Nighttimers worked by lantern light and found north with the North Star. Those two norths differ by about ten degrees, so the two guilds’ maps never quite lined up.

There was one more oddity. By a habit so old that nobody remembered who started it, both guilds measured north–south offsets in miles and east–west offsets in meters.

One morning the mayor sent both guilds the same question: how far is it from the town square to the mill, the one that stands beside the old well?

Two norths: a map of Middlemark with two survey grids A parchment map, drawn to scale. The town square sits at the origin, the church to the north-west, and the well, with the windmill right beside it, to the north-east. A solid blue grid follows magnetic north. A dashed vermilion grid follows the North Star and is turned about 10 degrees counterclockwise. Two compass roses in the corner show the two norths. NORTH EAST NORTH′ EAST′ Town square Church Well Mill N MAGNETIC NORTH STAR 100 m
Plate 1.1 Middlemark, surveyed twice. The Daytimers’ grid (solid blue) follows magnetic north; the Nighttimers’ grid (dashed vermilion) follows the North Star, about 10° away. The mill stands right beside the old well.

Predict

Each guild runs Pythagoras on its own offsets to the mill, north in miles and east in meters. Will the two guilds agree on the distance from the town square to the mill?

  • Yes: a distance is a distance.
  • No: different norths give different answers.
  • Only after they fix something.

Experiment

Here is Middlemark with both grids laid over it. The panel lists each guild’s offsets to the selected landmark (the mill, to start), and the “distance” each guild gets from Pythagoras, miles and all. Turn the Nighttimer grid with the slider. Drag the mill, the well or the church, or tab to them and use the arrow keys. The pale bands are the town’s roads. They stay put, so a landmark you drag leaves its road behind.

With north measured in miles, the guilds disagree: at a 10° tilt the Daytimers compute 300.0 for the mill and the Nighttimers 365.3. With 1609.344 meters per mile, both get 501.9 m, at every angle.

The surveyors’ town map. Two grids, one town. Turn one grid, convert the miles, and watch what both guilds agree on.
Describe this lab

A map of Middlemark with the town square at the origin, a church, a well and a mill. The Daytimer grid is drawn in solid blue and points to magnetic north. The Nighttimer grid is drawn in dashed vermilion and can be turned from −45° to 45° with a slider. A side panel lists each guild’s east and north offsets of the selected landmark, with north in miles, and the distance each guild computes with Pythagoras. A number box sets a conversion factor in meters per mile. At 1609.344 both guilds compute the same distance for every landmark and every angle; until then they disagree. The landmarks can be dragged, or moved with the arrow keys in steps of 10 m.

Try it

Turn the Nighttimer grid until the mill is straight north′ for the Nighttimers, so their east offset reads about 0 m (anything within 2 m counts). From the starting 10°, turn the other way: past 0° and into negative angles. Which angle does it take, and how many miles north′ is the mill then?

Try it

Before you touch the factor, turn the grid back to 0°. Now the guilds agree: both say 300.0. Problem solved? Look at the map. The mill is 300 m east and a quarter mile north of the square. Can it really be only 300 m away? Two guilds can agree by making the same mistake. An invariant has to agree at every angle.

Try it

The guilds disagree because Pythagoras is adding miles to meters. Type a conversion factor, in meters per mile, into the box. Which number makes the two distances agree? (A mile is a bit more than 1600 m.) If they nearly agree but not quite, your factor is close but isn’t a mile yet.

Try it

With the factor fixed, sweep the grid all the way from −45° to 45°, then drag the well somewhere new. Does the distance ever change? Click the church and write down its distance from the square.

When you’re done measuring, play. Put the mill somewhere silly. Turn the grid as far as it goes. See if you can make the guilds disagree again without touching the factor.

What just happened?

With north in miles, the guilds disagreed: at a 10° tilt the Daytimers got 300.0 for the mill and the Nighttimers 365.3. With the factor set to 1609.344 meters per mile, both got 501.9 m, at every angle.

So the honest answer was “only after they fix something”. And the thing to fix was never the norths. Lining up the norths makes the numbers match, but they match on a wrong answer: 300 m for a mill that is 502 m away.

If you answered “yes”, your instinct about the town was sound: the mill really is one fixed distance from the square. What failed was the guilds’ arithmetic, which added miles to meters. Hold on to that instinct. It’s the hero of this chapter.

If you blamed the different norths, you saw something real: turning the grid did change the answers. But it changed them only because the units were mixed. Once north was in meters, you could turn the grid as far as you liked and the distance never moved.

Each guild is right about its own offsets. When the mill sits straight north′ for the Nighttimers, its east′ offset really is zero; for the Daytimers it really is 300 m east. “East” means something different on each map.

The real trouble was the units. Pythagoras adds the squares of two legs, and that only works when both legs are in the same unit: 0.25² square miles plus 300² square meters is nonsense. Fix the units, and something remarkable shows up. The guilds still disagree about east and about north. About distance they agree, to the last millimeter, however far you turn the grid.

Hold on to Sagredo’s question. By the end of the chapter you’ll be able to answer it yourself.

What a turn can’t change

Friendly numbers first

Let’s leave the miles behind for a moment and measure everything in meters. Take the well. The Daytimers put it 300 m east and 400 m north of the square. That’s a 3-4-5 triangle scaled up by 100, so the well is 500 m away. Need a refresher?

Now turn the Nighttimer grid by an angle that keeps the arithmetic friendly. Lay a 5 m stick along east, then swing it counterclockwise until its tip sits 4 m east and 3 m north of its base: 3-4-5 again. It has turned about 36.87°. The share of the stick still along east, 4 out of 5 or 0.8, is called the cosine of that angle. The share along north, 3 out of 5 or 0.6, is its sine. Sine and cosine refresher Here’s how a Nighttimer finds the well.

  1. East′ offset: \(\rkt{x'} = 300 \times 0.8 + 400 \times 0.6 = 240 + 240 = 480\un{m}\). Both Daytimer legs lean partly along the new east′ direction. Each casts a “shadow” on the east′ axis: the east leg its cosine share, the north leg its sine share. The lab below draws these shadows.
  2. North′ offset: \(\rkt{y'} = -300 \times 0.6 + 400 \times 0.8 = -180 + 320 = 140\un{m}\). Turning the grid counterclockwise swings north′ toward the west, so walking east now counts slightly against north′.
  3. Squares: \(480^2 + 140^2 = 230\,400 + 19\,600 = 250\,000\). Completely different legs, but we add their squares anyway, just as Pythagoras says.
  4. Distance: \(\sqrt{250\,000} = 500\un{m}\). The Daytimers got the same from \(300^2 + 400^2 = 90\,000 + 160\,000\). The legs changed beyond recognition. The sum of their squares didn’t move. Look at the Nighttimers’ triangle: 480, 140, 500. Divide by 20 and it’s 24-7-25, a completely different right triangle with the same long side.
One hypotenuse, two pairs of legs The well, 500 m from the square. The Daytimer legs are 300 m east and 400 m north, drawn in solid blue. The Nighttimer grid is turned 36.87 degrees; its legs are 480 m and 140 m, drawn in dashed vermilion. Both pairs of legs end at the same teal hypotenuse of 500 m. 300 400 480 140 500 ≡ square well

At a 36.87° turn the well’s legs are 300 and 400 for the Daytimers, 480 and 140 for the Nighttimers, and the distance is 500 m for both.

One hypotenuse, two pairs of legs. Turn the Nighttimer grid and watch both guilds’ triangles share one long side.
Describe this lab

A map with the square at the origin and a chosen stake: the well at (300 m, 400 m) or the church at (−250 m, 350 m). The Daytimer right triangle has solid blue legs along east and north. The Nighttimer grid can be turned from −45° to 45°; its right triangle has dashed vermilion legs along east′ and north′. Both triangles share the teal hypotenuse from the square to the stake. Below the map, a ledger shows each guild’s two squared legs as a stacked bar: the split changes as the grid turns, while the total, the squared distance, stays the same. Readouts list both guilds’ offsets, the distance, and the sum x + y, which is not invariant. Buttons jump to 0°, 10° and the 3-4-5 turn of 36.87°. A Show shadows switch drops dotted lines from the Daytimer corner onto the east′ axis, splitting the Nighttimer east′ leg into the east leg’s share, x cos φ, and the north leg’s share, y sin φ. Keys: [ and ] turn the grid by 1°, with Shift by 10°; T jumps to the 3-4-5 turn.

Try it

Press 3-4-5 turn and check the worked example: the Nighttimer legs should read 480 and 140. Press Show shadows to see the 480 split into its two shares, 240 from each Daytimer leg. Then turn the grid slowly and watch the ledger. Which bars grow and shrink? Which total never moves? Keep an eye on \(x + y\), too.

The shorthand

Now the symbols, as shorthand for what we just did. Call the turn angle \(\varphi\) (the Greek letter phi). A stake at Daytimer offsets \((\lab{x}, \lab{y})\) has Nighttimer offsets

\[ \begin{gathered} \rkt{x'} = \lab{x}\cos\varphi + \lab{y}\sin\varphi \\ \rkt{y'} = -\lab{x}\sin\varphi + \lab{y}\cos\varphi \end{gathered} \]

Every coordinate changes. One combination doesn’t:

\[ \inv{d^2} = \lab{x}^2 + \lab{y}^2 = \rkt{x'}^2 + \rkt{y'}^2 \]

For two stakes anywhere, use the differences of their offsets: \(d^2 = \Delta x^2 + \Delta y^2\). The proof that this works for every angle takes three lines; it’s in the Deep dive at the end of this section.

A number that comes out the same in every grid is called an invariant. A number that changes when you turn the grid, like the east offset, is grid-dependent. Later in the course the observers will be moving instead of turning, and the word becomes frame-dependent. Chapter 6 puts turning and moving side by side.

Quick check, no calculator needed. Imagine turning the Nighttimer grid a full 90° counterclockwise, further than the labs go, so that \(\cos\varphi = 0\) and \(\sin\varphi = 1\). Where is the well, \((300\un{m}, 400\un{m})\), for the Nighttimers now?

Hint

Put \(\cos\varphi = 0\) and \(\sin\varphi = 1\) into the shorthand. Half the terms vanish.

Solution

\(\rkt{x'} = 300 \times 0 + 400 \times 1 = 400\un{m}\) and \(\rkt{y'} = -300 \times 1 + 400 \times 0 = -300\un{m}\). After a quarter turn, east′ points north and north′ points west, so the well’s 400 m north becomes 400 m east′, and its 300 m east becomes 300 m against north′. The distance is still \(\sqrt{400^2 + 300^2} = 500\un{m}\).

Why, really?

Why should distance survive the turn? Because the distance belongs to the stakes, and the grid belongs to the surveyors. Hammer two stakes into a field and stretch a rope between them. The rope has no idea which way anybody’s compass points. East and north are questions you ask the map. Distance is a question you ask the field.

Myth

“Relativity says everything is relative.” It says nearly the opposite. The guilds disagree about east and north, and both are right. What makes their maps trustworthy is the one thing they agree on. Relativity is the same kind of theory: the time and the distance between two events change from frame to frame, and the physics lives in the quantities that don’t. This whole course is a hunt for them.

The guild mistake

The guild mistake A right triangle with an east leg of 300 m and a north leg of 0.25 mile. Squaring and adding the two numbers gives about 300.0001, marked with a red cross: it adds square meters to square miles. Converting first, 0.25 mile is 402.3 m, and the distance is 501.9 m, marked with a check. 300 m 0.25 mi ? square mill Guild arithmetic √(300² + 0.25²) ≈ 300.0001 square meters plus square miles Convert first 0.25 × 1609.344 = 402.3 m √(300² + 402.3²) = 501.9 m
Plate 1.2 The guild mistake: Pythagoras can’t add miles to meters. Convert one leg first, and the distance comes out the same for every guild.

Plate 1.2 puts the whole fix on one page. The number that does the work, 1609.344 meters per mile, is a conversion factor. It’s exact: a mile is defined as 1609.344 meters.

Pitfall

Units are not physics. Miles and meters measure the same kind of thing, and the number 1609.344 says nothing about Middlemark. It records a habit of whoever invented the mile. Here’s a test worth remembering: if two quantities can always be converted into each other with one fixed factor, they are probably one quantity wearing two costumes.

A Daytimer apprentice reports: “The bridge is 0.1 mile north and 120 m east of the well, so it’s \(\sqrt{0.1^2 + 120^2} \approx 120\) m away.” How far is the bridge from the well, really?

Hint

Convert the north leg to meters with 1609.344 meters per mile, then use Pythagoras.

Solution

\(0.1 \times 1609.344 = 160.9\un{m}\). Then \(\sqrt{160.9^2 + 120^2} \approx 200.7\un{m}\). The apprentice was off by 80 m, all from adding square miles to square meters.

Time is the other guild

Now the big idea, and the reason a parable about maps opens a book about relativity.

We measure space in meters and time in seconds. That feels like the order of nature, but it’s a guild habit too. Space was surveyed by people with rulers; time was surveyed by people with clocks. Each guild picked a unit that suited its tools, and for centuries nobody suspected the two were measuring parts of one thing.

There is a conversion factor between them, and Nature hands it to us: the speed of light. Light crosses 299 792 458 meters every second. A conversion factor that changed from one observer to the next would be useless. This one doesn’t change. Every free-floating observer measures the same speed for light, however fast they move. That is a fact about Nature, not a habit, and Chapter 2 shows how we know it (and what free-floating means). So we can measure time in meters. One meter of time is the time light takes to cross one meter. One second is 299 792 458 meters of time.

\[ \begin{gathered} t\,[\text{m}] = c \cdot t\,[\text{s}] \\ c = 299\,792\,458\ \text{m/s} \end{gathered} \]

Call it the sacred factor of the space-and-time surveyors. Use it, and time and space share one unit, so they can finally sit in the same formula. Forget it, and you make the guild mistake: adding seconds to meters.

And here’s the deeper half of the analogy. The guilds found 1609.344 by demanding one thing: that distance come out the same in every grid. In Chapter 6 you’ll meet \(c\) the same way. It is the one factor that makes a certain combination of time and space come out the same for every moving observer. That combination looks like Pythagoras with a minus sign: time squared minus distance squared.

A ladder of light A ladder with six rungs, not to scale. Each rung pairs a time with the distance light covers in that time. 1 nanosecond is 0.30 m, a foot of light. 3.34 nanoseconds is 1 meter of time. 1 second is 299 792 458 m. 1.28 seconds takes light from Earth to the Moon. 499 seconds takes it from the Sun to Earth. 1 year is 1 light-year. TIME METERS OF TIME 1 ns 3.34 ns 1 s 1.28 s 499 s ≈ 8.3 min 1 year 0.30 m a foot of light 1 m one meter of time 299 792 458 m the sacred factor 3.84 × 10⁸ m Earth to Moon 1.50 × 10¹¹ m Sun to Earth 9.46 × 10¹⁵ m one light-year × 3.34 × 300 million × 1.28 × 390 × 63 000 not to scale: each rung is a bigger jump than it looks

Seconds times 299 792 458 gives meters of time. Between the rungs: 1 μs is 299.8 m, about three soccer fields, and 1 ms is 299.8 km, a three-hour highway drive.

Plate 1.3 A ladder of light. Each rung pairs a time with the distance light covers in it: the same quantity in two guilds’ units. Slide or type any time, and read it in meters of time. The top rung is one light-year.
Describe this lab

A vertical ladder on a logarithmic scale, from 1 nanosecond at the bottom to 1 year at the top, with rungs at 1 ns, 1 meter of time (3.34 ns), 1 microsecond, 1 millisecond, 1 s (the sacred factor), 1.28 s (Earth to Moon), 499 s (Sun to Earth) and 1 year (one light-year). A gold marker shows the chosen time. A slider moves the marker; a text box accepts any time in seconds, such as 0.8 or 1e-9; buttons jump to each rung. Readouts give the time, the same time in meters of time, and the nearest rung. Once your secondsToMeters passes its tests in the Build it section, the conversion is computed by your code.

Try it

Jump to the nanosecond rung: how long a ruler is one nanosecond? Then type the length of one heartbeat, about 0.8 seconds, into the box. How many meters of time is one beat of your heart? The line under the big number tells you how much of the way to the Moon that is.

Why, really?

Why is the sacred factor such a monster? Because the second and the meter were picked to fit human bodies. A second is about a heartbeat; a meter is about a stride. Light was never built to our scale. In one heartbeat it covers about 300 million strides. The big number isn’t telling you that light is strange. It’s telling you that we are slow.

Deep dive: why no turn can change \(x^2 + y^2\)

Square the two Nighttimer offsets:

\[ \begin{aligned} \rkt{x'}^2 &= x^2\cos^2\varphi + y^2\sin^2\varphi \\ &\quad + 2xy\cos\varphi\sin\varphi \\[4pt] \rkt{y'}^2 &= x^2\sin^2\varphi + y^2\cos^2\varphi \\ &\quad - 2xy\sin\varphi\cos\varphi \end{aligned} \]

Add them. The two \(2xy\) terms on the second lines cancel, and what’s left groups into \(x^2(\cos^2\varphi + \sin^2\varphi)\) \(+\; y^2(\sin^2\varphi + \cos^2\varphi)\). On a circle of radius 1, Pythagoras says \(\cos^2\varphi + \sin^2\varphi = 1\). So \(\rkt{x'}^2 + \rkt{y'}^2 = x^2 + y^2\), for every angle.

Notice what did the work: the cross terms \(\pm 2xy\cos\varphi\sin\varphi\) canceled because the two axes are perpendicular. Keep that in mind when you meet the rebel surveyor in Going further, at the end of the chapter.

So far it’s all been arithmetic and argument. Has anybody actually bet on the sacred factor?

Ask Nature

Measuring time and distance in one unit isn’t a classroom trick. Since 1983 it has been, quietly, the law of the land.

Ask Nature

In 1983 the world’s measurement authority redefined the meter as the distance light travels in vacuum in 1/299 792 458 of a second. Since then the speed of light is exactly 299 792 458 m/s, by definition, with no error bar.

Before 1983, labs measured \(c\) in meters per second. By the 1970s their laser measurements had become so precise that the biggest uncertainty was in the definition of the meter itself. So the rules were flipped. The speed of light became a fixed conversion factor, and “measuring the speed of light” turned into “measuring the meter”. In effect, the two guilds had merged.

A committee can fix a number, but it can’t make Nature keep it. The bet was safe because of what experiments kept finding: every careful measurement gives light the same speed, however the lab is moving. Chapter 2 is about that measured fact.

17th General Conference on Weights and Measures (CGPM), Resolution 1, 1983.

Engineers felt the sacred factor long before 1983. The computer pioneer Grace Hopper used to hand out pieces of wire about 30 cm (11.8 inches) long. Each was a “nanosecond”: the farthest light, and so any signal, can travel in a billionth of a second. Her point was practical. A computer can’t work faster than signals can cross it, so its wires had better be short.

A nanosecond of light against a ruler A gold wire lies along a 30 centimeter ruler. It is 29.98 cm long: the distance light travels in one nanosecond. 0 10 20 30 cm light in 1 ns: 29.98 cm
Plate 1.4 A nanosecond, measured with a ruler: the wire Grace Hopper handed out.

Real numbers

1 meter of time is 3.3356 ns, and 1 ns is 0.2998 m: about a foot of light.

On average, sunlight takes about 499 s to reach us; across exactly one astronomical unit it takes 499.005 s. That’s \(1.496 \times 10^{11}\) meters of time, and the Sun is on average \(1.496 \times 10^{11}\) m away: one number, in either guild’s unit.

Laser pulses bounced off reflectors left on the Moon come back after about 2.5 s. Corrected to the centres of Earth and Moon, the round trip averages 2.56 s, from about 384 000 km away. So the Moon sits about 1.28 light-seconds from us.

Practice

Three tiers: ◇ for a quick estimate, ◆ for pencil and paper, ★ for a head-scratcher. Every item has a hint and a full solution, and fractions or square roots like sqrt(2) are fine as answers.

How long is 1 second, in meters of time? A round figure like 3e8 is fine.

Hint

One meter of time is the time light takes to cross one meter. How many meters does light cross in a whole second?

Hint

Light’s speed is 299 792 458 meters per second, about \(3 \times 10^8\). Anything within 1% of that counts here.

Solution

Light crosses 299 792 458 m in one second, so \(1\un{s} = 299\,792\,458\) meters of time, about \(3 \times 10^8\). Since 1983 this is exact, by definition.

What is 1 meter of time, in nanoseconds?

Hint

One meter of time is \(1/299\,792\,458\) of a second.

Hint

A nanosecond is \(10^{-9}\) s. How many of those fit in \(1/299\,792\,458\) s?

Solution

\(1/299\,792\,458\un{s} = 3.336 \times 10^{-9}\un{s} = 3.336\un{ns}\). A meter of time is about three and a third nanoseconds. To three figures that’s 3.34 ns, the number the ladder shows.

Sort each quantity. Is it the same for every guild (invariant), or does it depend on the grid?

  • The distance between two stakes
  • The north offset of the well
  • The east offset of the well
  • The length of the bridge
  • The angle between two roads at a crossing
  • The direction “straight north”
Hint

For each card, ask: would a surveyor with a turned grid get a different number?

Solution

Invariant: the distance between stakes, the bridge’s length (a distance between its two ends) and the angle between two roads (turning the grid turns both roads’ directions by the same amount, so their difference stays). Depends on the grid: the north offset, the east offset and the direction “straight north”, which is exactly what the two guilds disagree about.

Which of these numbers are pure conversion factors, guild habits rather than facts about the world? Tick all that apply.

  • 1609.344 meters per mile
  • 12 inches per foot
  • 299 792 458 meters of time per second
  • 501.9 m from the square to the mill
  • The 10° between magnetic north and the North Star
Hint

Ask of each number: could you change it just by agreeing to use a different unit?

Solution

The first three. Redefine the mile, the inch or the second and those numbers change, while nothing in the world does. The distance to the mill and the angle between the two norths are facts about Middlemark: the guilds can only describe them in different units.

One caution. The number 299 792 458 is a habit of whoever defined the second and the meter. That there is a speed every observer agrees on is not a habit. It’s a fact about Nature, and Chapter 2 shows how we know.

The mill is 0.25 mile north and 300 m east of the square. How far is it from the square, in meters? Need a refresher?

Hint

You cannot add miles² to meters².

Hint

One mile is exactly 1609.344 m. Convert the north offset to meters first.

Solution

North: \(0.25 \times 1609.344 = 402.336\un{m}\). Then \(\sqrt{300^2 + 402.336^2} = \sqrt{90\,000 + 161\,874} = \sqrt{251\,874} \approx 501.9\un{m}\).

The well is at Daytimer offsets \((\lab{x}, \lab{y}) = (300\un{m}, 400\un{m})\). The Nighttimer axes are turned 10° counterclockwise. Find the well’s Nighttimer offsets \(\rkt{x'}\) and \(\rkt{y'}\).

Hint

Use the shorthand: \(x' = x\cos\varphi + y\sin\varphi\) and \(y' = -x\sin\varphi + y\cos\varphi\).

Hint

\(\cos 10° \approx 0.9848\) and \(\sin 10° \approx 0.1736\). (Set your calculator to degrees.)

Solution

\(\rkt{x'} = 300(0.9848) + 400(0.1736) = 295.4 + 69.5 = 364.9\un{m}\).
\(\rkt{y'} = -300(0.1736) + 400(0.9848) = -52.1 + 393.9 = 341.8\un{m}\).

Using the Nighttimer offsets you just found for the well, how far is the well from the square in the Nighttimer grid?

Hint

Square each Nighttimer offset, add, then take the square root.

Solution

\(\sqrt{364.9^2 + 341.8^2} = \sqrt{133\,152 + 116\,827} = \sqrt{249\,979} \approx 500\un{m}\). The tiny shortfall is rounding: the exact answer is 500 m, the same as the Daytimers’ \(\sqrt{300^2 + 400^2}\).

Back to the well at \((300\un{m}, 400\un{m})\), with the Nighttimer grid turned 10° counterclockwise. Is the plain sum of the two offsets, \(x + y\), an invariant?

  • Yes: both guilds get the same sum.
  • No: the sum changes when the grid turns.
  • Only if both guilds use meters.
Hint

Use the Nighttimer offsets of the well at 10° from the turned-well exercise above: \(\rkt{x'} = 364.9\), \(\rkt{y'} = 341.8\). Add them, and compare with \(300 + 400\).

Solution

No. Daytimers: \(300 + 400 = 700\un{m}\). Nighttimers: \(364.9 + 341.8 = 706.7\un{m}\). The sum depends on the grid. Only \(x^2 + y^2\) is the same for both: \(250\,000\un{m^2}\).

Finish the worked example. The church is at Daytimer offsets \((-250\un{m}, 350\un{m})\). The Nighttimer grid has the 3-4-5 turn: \(\cos\varphi = 0.8\), \(\sin\varphi = 0.6\). The first step is done for you.

  1. \(\rkt{x'} = (-250)(0.8) + (350)(0.6) = -200 + 210 = 10\un{m}\).
  2. \(\rkt{y'} = -(-250)(0.6) + (350)(0.8) = \;?\)
  3. Distance from the Nighttimer offsets: \(\sqrt{10^2 + \rkt{y'}^2} = \;?\)

Hint

Two minus signs multiply to a plus: \(-(-250) = +250\).

Hint

To check your distance, compute the Daytimers’ \(\sqrt{250^2 + 350^2}\). It must match.

Solution

\(\rkt{y'} = 150 + 280 = 430\un{m}\). Distance: \(\sqrt{10^2 + 430^2} = \sqrt{100 + 184\,900} = \sqrt{185\,000} \approx 430.1\un{m}\). Check: \(\sqrt{250^2 + 350^2} = \sqrt{62\,500 + 122\,500} = \sqrt{185\,000}\). Same.

The Daytimers say the mill is 300 m east of the square. With their grid turned −36.7°, the Nighttimers say it is 0 m east. Who is right?

  • The Daytimers: magnetic north is the standard.
  • The Nighttimers: the North Star doesn’t wander.
  • Both: “east” means something different on each grid.
  • Neither, until they settle on one north.
Hint

Is “300 m east” a fact about the mill alone, or about the mill and a grid?

Solution

Both. An east offset is a fact about the mill and a grid. Change the grid and the offset changes, while the mill stays put. The distance, 501.9 m, is a fact about the mill and the square alone, so both guilds agree on it.

More practice: four more items (church to well, the Sun, the Moon, the guild fix in order)

So far every distance was measured from the square. How far is the church, at Daytimer offsets \((-250\un{m}, 350\un{m})\), from the well at \((300\un{m}, 400\un{m})\)? Then check it the Nighttimer way: at the 3-4-5 turn the church is at \((10, 430)\) and the well at \((480, 140)\).

Hint

Use the differences of the offsets: \(d^2 = \Delta x^2 + \Delta y^2\), with \(\Delta x = 300 - (-250)\) and \(\Delta y = 400 - 350\).

Solution

Daytimers: \(\Delta x = 550\), \(\Delta y = 50\), so \(d = \sqrt{302\,500 + 2500} = \sqrt{305\,000} \approx 552.3\un{m}\). Nighttimers: \(\Delta x' = 480 - 10 = 470\), \(\Delta y' = 140 - 430 = -290\), so \(d = \sqrt{220\,900 + 84\,100} = \sqrt{305\,000}\). Same, as it must be.

Sunlight takes 499.0 s to reach Earth. What is that in meters of time?

Hint

Multiply by the sacred factor, 299 792 458 meters of time per second. Scientific notation like 1.5e11 is fine.

Solution

\(499.0 \times 299\,792\,458 \approx 1.496 \times 10^{11}\) meters of time. Light covers one meter per meter of time, so that is also the Sun’s distance in meters.

Light takes about 2.56 s to go from Earth’s centre to the Moon’s centre and back. How far away is the Moon, in meters?

Hint

The 2.56 s covers the trip there and back.

Solution

One way takes 1.28 s, which is \(1.28 \times 299\,792\,458 \approx 3.84 \times 10^{8}\) meters of time. So the Moon is about \(3.84 \times 10^8\un{m}\), or 384 000 km, away.

A new apprentice must compute the distance to the mill from “0.25 mile north, 300 m east”. Put the steps of the guild fix in order.

  1. Notice that the two legs are in different units.
  2. Multiply the north offset by 1609.344 meters per mile.
  3. Square each leg, now both in meters, and add.
  4. Take the square root: the distance, the same for every guild.
Hint

Pythagoras is the last thing you do, and it needs matching units.

Solution

Spot the mismatch, convert, square and add, take the root: \(0.25 \to 402.336\un{m}\), then \(\sqrt{300^2 + 402.336^2} \approx 501.9\un{m}\).

Want more? The formula card has every equation from this chapter on one page, and the Math Toolkit has more practice with conversion factors.

Build it

Time to teach the computer what the guilds learned. Two small functions, a few lines each. Once a function passes its tests, it moves into your Toolbox, and the labs above start running on your code: look for the purple chip in their toolbars.

Code you need

An object bundles several named values inside curly braces. You read one of them with a dot: cat.legs means “the legs of cat”. A const line gives a value a name you can use on the lines below it.

const cat = { name: 'Tom', legs: 4 };
const pawsInARoom = cat.legs * 3;   // 12

In this course a map point is an object like {x: 300, y: 400}. So if p is the well, p.x is 300 and p.y is 400. Later, an event will be an object like {t: 5, x: 3}.

Build it: distance(p, q)

Return the straight-line distance between two map points p and q. Besides a few fixed cases, the tests run the surveyors’ own check: they turn both points by the same random angle, a hundred times over, and your distance must not change. A dare: first define dx = q.x - p.x and dy = q.y - p.y, then return Math.abs(dx) + Math.abs(dy), the distance a taxi drives on a grid of streets. It fails the fixed 3-4-5 test, which is no surprise: the taxi drives 7 blocks, not 5. Now read the turning test’s message. Why does turning the two points change a taxi’s answer, and what does that tell you about whether a taxi’s distance can be an invariant? Once your real distance passes, the One hypotenuse lab measures with your function.

Build it: secondsToMeters(s)

Convert a time in seconds into meters of time, using the sacred factor. One test is the Sun: 499.005 seconds of light travel should come out as \(1.496 \times 10^{11}\) meters of time. Another asks about a time two seconds in the past: negative times convert too. Once it passes, the light ladder runs on your function.

Checkpoint

Six items: five are scored, and the written one you check yourself against the model answer. Feedback waits until you submit, so answer them all first.

A stake sits at Daytimer offsets \((600\un{m}, 800\un{m})\) from the square. A guild whose grid is turned 25° measures its own offsets and uses Pythagoras. What distance do they get?

Solution

Distance is invariant, so there’s no need to turn anything: \(\sqrt{600^2 + 800^2} = 1000\un{m}\), in every grid.

A friend says: “Relativity means everything depends on your point of view.” Based on the surveyors, what’s the best reply?

  • “Right: the guilds couldn’t agree on anything.”
  • “Not quite: coordinates depend on your point of view, but some quantities are the same for everyone (for the surveyors, distance), and those are what the theory is about.”
  • “Right, but only for things moving near light speed.”
  • “Wrong: there’s one true frame, and everything else is a mistake.”
Solution

Coordinates depend on your grid (later, on your frame), and that’s fine. The heart of the theory is the invariants: distance for the surveyors, and in Chapter 6 the spacetime interval.

Invariant, or does it depend on the grid?

  • The distance from the church to the well
  • The church’s north offset
  • The angle at which two roads cross
  • The area of the town square
  • Which of two landmarks lies farther east
Solution

Invariant: the distance, the crossing angle and the area (it’s built from distances). Depends on the grid: the north offset, and even which landmark is farther east, since turning the grid can swap them.

A camera flash lasts 1 microsecond (\(10^{-6}\) s). How long is that in meters of time?

Solution

\(10^{-6} \times 299\,792\,458 \approx 299.8\) meters of time: about three soccer fields of light.

A stake is 0.5 mile north and 400 m east of the square. How far away is it, in meters?

Solution

\(0.5 \times 1609.344 = 804.672\un{m}\). Then \(\sqrt{804.672^2 + 400^2} \approx 898.6\un{m}\).

In one or two sentences: why does this course measure time in meters?

Model answer

Because seconds and meters are two guilds’ units for parts of one thing, and the speed of light is a fixed conversion factor between them, like meters per mile. With one unit for both, time and distance can go into the same formula without the guild mistake.

  • Calls the speed of light a conversion factor
  • Compares it to miles and meters
  • Says it lets time and space share formulas

Explain it to Sagredo

Sagredo hasn’t forgotten his question from earlier, and now it’s your turn to answer it.

Sagredo’s question from earlier: “One of those two norths must be the real north. Why not pick that one and use it?” Answer him in a few sentences.

Model answer

There is no real north to pick. “North” is a choice of grid, like which way you hold a map, and nothing in Middlemark marks one grid as the true one. So both guilds are right about their own offsets. Picking one north wouldn’t have fixed anything either: at 0° the guilds agreed on 300 m, a wrong answer, because they were still adding miles to meters. What’s real, in the sense that every guild must agree on it, is the distance between two stakes, whichever north you use. Relativity says the same about time and space: the time and the distance between two events depend on your frame, but one combination of them, the interval, doesn’t.

  • Says there is no real north: each guild is right in its own grid
  • Names the invariant: distance
  • Points ahead: the same holds for time and space

What’s invariant here?

  • Distance. Turn the grid any way you like: \(\inv{d^2} = \lab{\Delta x^2} + \lab{\Delta y^2} = \rkt{\Delta x'^2} + \rkt{\Delta y'^2}\).
  • Depends on the grid: the east and north offsets, and the direction “straight north”.
  • Pythagoras needs matching units. Miles become meters with 1609.344; seconds become meters of time with \(c = 299\,792\,458\).
  • 1 meter of time ≈ 3.34 ns. 1 ns ≈ 30 cm of light.
  • Once they pass, distance(p, q) and secondsToMeters(s) live in your Toolbox.

Still strange?

Here’s the part we haven’t earned yet. Distance survives a turn thanks to a plus sign in Pythagoras; in spacetime the invariant will have a minus sign, and that one minus sign is where all of relativity’s strangeness lives.

Going further

This section is optional, and it’s a preview. It shows you a trap you’ll meet in Chapter 7, and Chapter 6 shows the way around it. Middlemark has a third guild, the rebels. Their first axis points east, like the Daytimers’. But their second axis leans 30° away from north, toward east, so their two axes meet at 60° instead of at a square corner. To reach a stake, a rebel walks some distance along the first axis, then turns and walks some distance parallel to the second. Then they square the two numbers and add them, like everyone else. Try their problem before you read on.

★★ A stake is 3 units along the rebels’ first axis, then 5 units parallel to their second axis. The two axes meet at 60°. How far is the stake from their origin, in a straight line?

Hint

Draw it. Walk 3 units east. Then walk 5 units along a line tilted 60° up from east. The two legs and the straight line home make a triangle, but not a right triangle.

Hint

Turn the slanted 5 into ordinary east and north pieces, then use Pythagoras. A line tilted 60° from east sends exactly half its length east (a 60° corner is a corner of an equilateral triangle). So the 5 gives 2.5 east. Its north piece follows from Pythagoras: \(\sqrt{5^2 - 2.5^2} = \sqrt{18.75} \approx 4.33\).

Solution

East: \(3 + 2.5 = 5.5\). North: \(\sqrt{18.75} \approx 4.33\). Then \(d^2 = 5.5^2 + 18.75 = 30.25 + 18.75 = 49\), so \(d = 7\). The rebels’ rule gives \(\sqrt{34} \approx 5.83\), which is wrong on a skewed grid. The shortcut for any two axes that meet at 60°: \(d^2 = a^2 + b^2 + 2ab\cos 60° = 9 + 25 + 15 = 49\).

Tried it? Show where the 7 comes from, and the extra 15

Start by splitting the slanted 5 into ordinary east and north pieces, with the cosine and sine shares from earlier: \(5\cos 60° = 2.5\) east and \(5\sin 60° \approx 4.33\) north. (Exactly, the north piece squared is \(5^2 - 2.5^2 = 18.75\).) So the stake is \(3 + 2.5 = 5.5\) east and 4.33 north, and \(5.5^2 + 18.75 = 49\): the distance is 7.

Where does the extra 15 come from? Look at the east total, \(3 + 2.5\). Squaring it gives \(3^2\), plus \(2.5^2\), plus one more piece: \(2 \times 3 \times 2.5 = 15\). The \(2.5^2\) and the north piece squared, 18.75, add up to 25, which is the \(5^2\) the rebels expected. That’s Pythagoras on the slanted leg alone. The 15 is left over. It’s built from both offsets at once, the 3 and the 5, so it’s called the cross term. On a grid with square corners the 5 would have no east share, the cross term would be zero, and plain Pythagoras would be back. (You met a cross term in the Deep dive, too. There it canceled, because the Nighttimers’ axes are perpendicular.)

The rebel surveyor’s skewed grid A grid whose two axes meet at 60 degrees. A stake is reached by walking 3 units along the first axis and then 5 units parallel to the second. The straight-line distance from the origin is 7, drawn in teal. The sum of the squares of the two offsets, 34, falls short of 7 squared, 49. The missing 15 is the cross term. 3 5 7 ≡ 60° first axis second axis 3² + 5² = 34 7² = 49 = 34 + 15
Plate 1.5 The rebels’ grid. Walk 3 along one axis and 5 along the other, and you end up 7 away: \(7^2 = 49 = 3^2 + 5^2 + 15\). The 15 is the cross term, \(2 \cdot 3 \cdot 5 \cos 60°\).

The rebel surveyor. Back to the rebels, whose two axes meet at 60°. They compute distances as \(x^2 + y^2\) from their own offsets. Is \(x^2 + y^2\) still invariant for them? Will they agree with the other two guilds?

Model answer

No. On skewed axes the two offsets aren’t the legs of a right triangle, so \(x^2 + y^2\) isn’t the squared distance, and the rebels won’t agree with the other two guilds. The correct rule has one more piece, made from both offsets multiplied together, called the cross term: for axes that meet at 60°, \(d^2 = x^2 + y^2 + 2xy\cos 60°\). A plain sum of squares only works when the axes meet at a right angle. And since the distance stays put while the cross term changes as their grid turns, their \(x^2 + y^2\) changes too: it isn’t an invariant.

  • Says no: \(x^2 + y^2\) is not the distance for skewed axes
  • Mentions the missing cross term, or that the offsets no longer form a right triangle
  • Connects the rule to how the grid is drawn (perpendicular axes)

Why look at a crooked grid at all? Because you’ll meet one again. In Chapters 7 and 10 you’ll draw a rocket’s time axis and space axis on the same page as Lin’s, and they lean toward each other, a lot like the rebels’ axes. When you see that lean, don’t reach for a cross term. The lean comes from drawing spacetime on flat paper, which can’t show it without distorting it. Flat world maps have the same trouble: on many of them Greenland looks about as big as Africa, which is really 14 times bigger. Nobody thinks Greenland grew. You can’t check this next part yet, so take it as a preview: the number every observer agrees on in spacetime has no cross term. It has a minus sign where Pythagoras has a plus, and Chapter 6 shows where that minus sign comes from.

Read further: the surveyors first appeared in chapter 1 of Edwin F. Taylor and John Archibald Wheeler, Spacetime Physics (2nd edition, W. H. Freeman, 1992), still the clearest book on special relativity built around invariants.