Chapter 3
Events, Clocks, and the Lattice
When did it happen? Ask the clock that was there.
What this chapter does
Every surprise in relativity is about when and where things happen. Before we can trust any of them, we need an honest way to say when something happened. This chapter builds one, in three steps.
- Synchronize a lattice. Meter rods with a clock at every joint, all set by one flash of light.
- Seeing versus reckoning. Why the moment you see a firework is not the moment it burst, and the one subtraction that fixes it.
- One frame, many people. Why two people standing still in different places see an event at different moments, yet agree on when it happened.
You’ll leave able to say what an event is, how a lattice is synchronized, and what “measure” means for the rest of the course.
Warm-up
Three quick questions from earlier chapters. They wake up exactly the tools this chapter needs.
Light travels 299 792 458 meters in one second. How many meters of time is one nanosecond, \(10^{-9}\) s? From Chapter 1
That’s how many nanoseconds are in one meter of time. Which way does this question go?
Hint
A meter of time is the time light takes to cross one meter. How far does light go in a billionth of a second?
Solution
\(299\,792\,458 \times 10^{-9} \approx 0.2998\). One nanosecond is about 0.3 m of time: a foot of light.
You are sealed in a windowless cabin on a ship sailing smoothly, in a straight line, at a steady speed. Which experiment can tell you the ship is moving? From Chapter 2
Hint
Galileo’s cabin had flies, fish and a dripping bottle; Chapter 2 added light. Did any of them notice the voyage?
Solution
None of them. No experiment inside a cabin moving steadily in a straight line can reveal that motion, not even one with light. This chapter leans on that: a lattice that moves steadily is as good a lab as one that sits still.
A well is 6 m east and 8 m north of a stake. How far apart are they? From Chapter 1 Need a refresher?
That adds the two offsets as if you walked east and then north. How long is the straight line?
Hint
East and north make the two legs of a right triangle.
Solution
\(\sqrt{6^2 + 8^2} = \sqrt{100} = 10\) m. Hold on to this: in a minute we’ll add a third leg, going up.
The puzzle
It’s festival night. Lin stands on the harbor wall with a fast camera. It stamps every flash it catches with the reading on Lin’s watch, to the nanosecond.
A firework bursts over the bakery, 300 m away. Another bursts over the lighthouse, 3 km away by the bay. The camera’s stamps are clear: the bakery flash arrived first, and the lighthouse flash arrived 1200 meters of time later (that is, about 4 microseconds).
“The bakery went first,” says Lin.
Did it?
Predict
Which firework exploded first?
Experiment
To answer Lin, we need something better than eyes. Here is the tool physicists use instead: a lattice of clocks.
Picture a jungle gym of meter rods filling space, with a clock at every joint. The clock in the corner is the reference clock. Every other clock is set to show a reading equal to its distance from the corner, in meters of time, and held there, stopped. A clock 5 m from the corner is set to read 5. At 0 the reference clock starts and sends out a flash. When the flash reaches a waiting clock, that clock starts too. The lab below is a small lattice: five clocks along each edge of the floor, three layers high. Press Play (or Step, with reduced motion) and watch.
A cube of clocks one meter apart, five along each edge of the floor and three layers high. The corner clock flashes at 0; a sphere of light spreads out and starts each preset clock as it arrives. Afterwards every clock reads the same time. The clock 3 m along one edge and 4 m along the other is preset to 5; the far top corner, at (4, 4, 2), is preset to 6.
Describe this lab
An isometric drawing of a lattice of clocks, 5 by 5 on the floor and 3 layers high, one meter apart. x runs to the right, y runs back into the picture and z runs up; the reference clock sits at the front-left corner of the floor. Every clock starts out waiting at its preset, its distance from the corner. Press Play (Step under reduced motion): the reference clock flashes at time 0, a gold sphere of light spreads at one meter per meter of time, and each clock starts when the sphere reaches it. Select a clock by tapping it, with the x, y and z buttons under the drawing, or with the keys (arrow keys move along the floor, Page Up and Page Down change the layer; the selection stops at the edges). The card under the drawing gives the selected clock’s place, its preset, its reading now, and the reading of the reference clock at the instant the flash started it. “Flash again” restarts the synchronization.
Try it · box 1 of 3
- Before you select anything, work out the preset of the floor clock 3 m along x and 4 m along y. Then find it: tap it, or use the x and y buttons under the drawing. Does its card agree with you?
- Only one clock in this lattice has a preset of exactly 6. Which one? Use Pythagoras twice: the top layer is 2 m up. Select it, press Flash again, and watch its card. What does it read while it waits? What does the reference clock read at the instant the flash arrives?
Now spread a lattice like that over Lin’s whole town, with rods every 300 m and every clock synchronized the way you watched above. Lin stands on the harbor wall. Her friend Sam stands near the lighthouse. Neither is moving, and each wears a watch set by the lattice. Three fireworks are about to go off.
A town map, 3600 m by 3000 m, on one lattice of synchronized clocks. Lin stands at (300, 900) m and Sam at (3000, 2700) m. Three fireworks burst:
| Firework | Clock there read | Lin sees it at | Sam sees it at |
|---|---|---|---|
| A Bakery | 2000 | 2300 (300 m away) | 5000 (3000 m away) |
| B Bridge | 1200 | 2700 (1500 m away) | 2949 (1749 m away) |
| C Lighthouse | 500 | 3500 (3000 m away) | 800 (300 m away) |
Describe this lab
A top-down map of a harbor town, 3600 m east to west and 3000 m north to south, covered by a lattice of synchronized clocks every 300 m. Three fireworks: A over the bakery at (600, 900), B over the bridge at (1500, 1800), C over the lighthouse at (2700, 2700). Lin starts at (300, 900) and Sam at (3000, 2700). A time slider, Play, and “Next sighting” run the time from 0 to 5800 m. In “What I see” you learn about a firework only when its light reaches Lin or Sam: a short gold streak marks each arrival, and the table lists when each of them saw each flash. Select a firework (Enter) to draw its distances to Lin and Sam and add them to the table. Give your verdict on the true order with the three menus at the bottom. The verdict also asks when the first firework burst, in meters of time. A right verdict switches to “What the lattice reckons”: the reading of the clock beside each firework, each person’s subtraction, and a replay of the light spreading from each burst. A wrong verdict gets a hint, more detailed each time. Focus Lin or Sam and use the arrow keys to move them (100 m per press, 10 m with Shift). The line under the map gives the gap between the two flashes a person sees closest together; near a tie, they step onto the exact spot. With your code, the light delays come from your clockPreset and the microsecond view from your metersToSeconds.
Try it · box 2 of 3
- Press Play, or Next sighting to go one flash at a time. Write down the order in which Lin sees the three flashes, then Sam’s order. In this view you learn about a firework only when its light reaches someone, as Lin and Sam do.
- Now reckon. Tap a firework: dashed lines show how far it is from Lin and from Sam, and the distances join the table. For each firework, take the time Lin saw it and subtract its distance.
- Put the three in order at the bottom of the lab, and type when the first one burst. A right verdict switches the map to What the lattice reckons, where the clock beside each firework checks your subtractions.
Try it · box 3 of 3
Drag Sam until he sees two flashes at the same moment. (On a keyboard, Tab to Sam and use the arrow keys; hold Shift for 10 m steps.) While you drag, the line under the map shows how far apart his two closest flashes arrive. When it says “Let go”, let go, and Sam steps onto the exact spot. How far apart, in lattice time, did those two fireworks really burst? Stuck? Try Sam at 1000 m east, 600 m north.
Then play freely. Put Lin right under the lighthouse, or stand Sam out by the bay. The seen order changes as you go. Does the lattice order ever change?
What just happened?
Lin’s camera caught the bakery first and the lighthouse last. But the lattice clocks say the lighthouse burst first, at 500, and the bakery last, at 2000. Seeing gave Lin the order exactly backwards.
If you picked the bakery because its flash arrived first, you reasoned exactly like Lin, and that is the trap this chapter is about. If you picked the lighthouse by weighing the distances against the stamps, that is this chapter in one move. If you said nobody could tell, you were half right: the order of arrival can’t tell you, but stamps plus distances can.
It takes one subtraction. Light covers one meter per meter of time, so a flash spends its distance, in meters of time, on the road. The lighthouse is 3000 m from Lin: \(3500 - 3000 = 500\). The bakery is 300 m away: \(2300 - 300 = 2000\). The far firework went first, by 1500 m of time.
Sam stood somewhere else and saw a different order. After his subtractions he gets 500, 1200 and 2000, exactly Lin’s numbers. Two people, two orders of seeing, one order of events.
A frame is a lattice, not a pair of eyes
Events
An event is something that happens at one place, at one instant. A firecracker going off. A ball touching the floor. Two fingers snapping. It isn’t a thing; it’s a happening, as small and sharp as you can make it. Events are the atoms of relativity, and the rest of this course is about how they’re arranged.
To pin an event down you need four numbers: three to say where (\(x, y, z\)) and one to say when (\(t\)). The question of this chapter is who supplies them, and how.
Quick check. Which of these is an event?
Hint
An event happens at one place, at one instant.
Solution
The burst. The lighthouse is a thing, the light’s trip covers many places and moments, and the festival is a whole evening. The burst, and the light’s arrival at Lin’s camera, are events: each has one \(x, y, z\) and one \(t\).
The lattice
Not a person squinting from far away. A frame of reference is the whole jungle gym: every rod, every clock, all at rest relative to one another. When a firecracker bursts, the clock right beside it notes its reading. The rods say where; the clock that was there says when. That’s the subtitle of this chapter: when did it happen? Ask the clock that was there.
Our drawings show only a few joints. Imagine the rods as fine as you need: a clock every millimeter if the experiment fits on a table, every few hundred meters if it fills a town.
Setting the clocks
The clocks are useless unless they agree. Here is the trick you watched in the first lab, with friendly numbers. Take a clock 3 m from the reference clock and set it to 3, stopped. At 0 the reference clock starts and flashes. Light covers one meter per meter of time, so the flash reaches our clock when the reference reads exactly 3. Our clock starts from 3 at that instant. From then on it keeps step with the reference clock forever.
That works for every clock, as long as its preset is its distance from the reference clock. Distance in a lattice is Pythagoras, used twice. Need a refresher?
- A clock sits 3 m east, 4 m north and 12 m up. Across the floor it is \(\sqrt{3^2 + 4^2} = 5\) m from the corner. East and north are the two legs of a right triangle on the floor.
- Now go up: \(\sqrt{5^2 + 12^2} = 13\) m. The floor distance and the height make a second right triangle, standing up.
- Preset it to 13 m of time. The flash needs 13 m of time to get there, so that is the reading it must start from.
- At reference time 13 the flash arrives; the clock starts at 13. In step. Every clock starts at the moment the reference clock reaches its preset.
Both triangles fold into one line, the shorthand for what we just did:
\[ \text{preset} = \sqrt{x^2 + y^2 + z^2} \]
Why, really?
Why go to all this trouble? Why not set every clock side by side, then carry each one to its joint? Because a carried clock is a moving clock, and nobody has yet promised you that a moving clock keeps the same time as one at rest. (In Chapter 5 we’ll find that it doesn’t.) Light’s speed, though, is the one thing every free-float lab agrees on. That was Chapter 2’s hard-won fact. So we let light do the carrying.
He’s right. One clock can time a round trip, and light’s round-trip speed always comes out 1. But that the trip out and the trip back take equal times is a choice, the one Einstein made in 1905.
Does a choice make the lattice times less real? No. A different choice would set the far clocks a little ahead or a little behind, and so relabel their readings. It could never change what happens: every flash, echo and burst would happen just as before, only with different numbers on the far clocks. Equal times out and back is the simplest choice, and the one everybody uses.
Once it’s made, anyone in the lattice can check any clock with an echo. Send a flash from the reference clock when it reads \(t_1\). A mirror at clock B bounces it back, and it gets home at \(t_2\). Out and back take equal times, so the bounce came exactly halfway: if B is in step, it read \((t_1 + t_2)/2\) at the bounce. If it read anything else, it’s off, and you reset it. Try it:
Quick check. Clock B is 5 m from the reference clock. The reference clock sends a flash when it reads 0. A mirror at B bounces the flash back, and it gets home when the reference clock reads 10. If B is in step, what did B read at the bounce?
That’s when the echo got home. The bounce happened partway through the round trip. Which part?
That’s when the flash left. It still had 5 m to travel.
Hint
With Einstein’s choice, the trip out and the trip back take equal times.
Solution
Halfway: \((0 + 10)/2 = 5\). If B had read 6, it would be one meter of time fast, and you’d reset it. Anyone in the lattice can run this test on any clock, so the readings are facts, not opinions.
Seeing is late news
You already reckon like a lattice. In a storm you see lightning, count three seconds, and hear thunder. Sound covers about 343 m per second, so you say the strike was about a kilometer away, and that it happened three seconds before you heard it. You subtracted the travel time without thinking.
Light is almost a million times faster than sound, so its delay hides from us. But it is never zero. Everything you see is a little old: the bakery by 300 m of time, the Sun by eight minutes, a distant star by centuries. Physicists keep two words apart, and so will we. To see an event is to receive its light. To reckon it, or measure it, is to read the lattice clock that was there. (To reckon is an old word for to work out.) They are linked by one line:
\[ \begin{gathered} \lit{t_\text{seen}} = \lab{t_\text{event}} + \text{distance} \\ \text{so} \quad \lab{t_\text{event}} = \lit{t_\text{seen}} - \text{distance} \end{gathered} \]
Think of the lattice clock’s reading as a postmark, the date stamped on a letter when it is mailed. Seeing is opening the letter. The postmark doesn’t change however long the letter spends in the mail, and it doesn’t change with who opens it.
Quick check. Someone 900 m from the bridge sees its firework when their watch reads 2100. When did it burst, by the lattice?
That’s when the news arrived. How long was the light on the road?
That adds the travel time. Did the burst come before or after the sighting?
Hint
Seen time minus distance.
Solution
\(2100 - 900 = 1200\): the reading of the clock beside the bridge in the town lab.
Myth
“An event happens when I see it.” It happens when the clock beside it says so. When you see it depends on how far away you stand.
The same mix-up hides inside a famous phrase: “a moving clock looks slow.” In this course, measure means the lattice record, after every light delay has been subtracted. What your eyes see of a moving clock includes those delays, and an approaching clock actually appears to run fast, because each later tick has less distance to travel to reach you. We’ll measure moving clocks in Chapter 5 and see them in Chapter 12.
Lin and Sam share one frame
Lin saw A, B, C. Sam saw C, B, A. Do they live in different frames, with different times? No. They are at rest relative to each other, so their watches are clocks of the same lattice. Each sees a different order, each subtracts, and each gets the same burst times to the last meter of time. Where you stand changes what you see. Only how you move could change what you reckon, and Lin and Sam move the same way: not at all.
A frame is a lattice, not a pair of eyes.
Quick check. Sam walks 500 m farther from the lighthouse and stands still there. Then the lighthouse firework bursts. Compared with watching from his old spot, what happens to the burst time he reckons?
Hint
Burst time = seen time − distance. Which of those two change, and by how much?
Solution
It doesn’t change. He sees the flash 500 later, but he also subtracts a distance 500 bigger. Standing still anywhere in the lattice, he is still one of its clocks.
Pitfall
“Every observer has their own frame.” Not if they share a lattice. Two people standing still a kilometer apart, a pigeon on the roof, the clock on the far wall: one frame. A second frame needs a second lattice moving through the first, and that is where the real surprises live (Chapter 7).
Real numbers
One meter of time is 3.3356 ns, so 300 m of time is 1.0007 μs and 3000 m of time is 10.007 μs. So the lighthouse light spent about 10 μs on the road to Lin. (The puzzle’s 4 μs is something else: the gap between the two arrivals.) Thunder from the lighthouse would reach her about 8.7 s late: almost a million times worse.
Deep dive: why not just time light one way?
To time a flash from clock A to clock B, you read A when it leaves and B when it arrives. That only works if A and B already agree, and making them agree is the very thing we are trying to do. Every way of timing light one way runs into the same circle. So in 1905 Einstein took the echo test as the definition of “in step”: B is in step if it reads \((t_1 + t_2)/2\) at the bounce. In Chapter 4 the echo becomes a way to locate events by radar.
That’s the whole toolkit: events, a lattice, presets, and one subtraction. Next, a lattice that is quietly working in your pocket.
Ask Nature
Is a lattice of clocks a fantasy? You carry the receiving end of one.
Ask Nature
The GPS satellites circle Earth on orbits of radius about 26 562 km. Each carries atomic clocks, deliberately adjusted to keep the time of one imaginary lattice. That lattice is centred on Earth but does not turn with it, so its clocks can be set by our flash rule. (On a turning lattice they can’t all be; the deep dive at the end of the chapter says why.)
The satellites are not clocks of that lattice: they move, and they move relative to one another. They are its messengers. Each one broadcasts, over and over, “I am here, and the lattice time was this when I sent this message.”
Your phone receives the message a little later. From a satellite straight overhead, about 20 200 km up, the signal is about 67 ms on the road. The phone multiplies each delay by light’s speed to get a distance, and with four satellites it solves for four unknowns: \(x\), \(y\), \(z\) and \(t\). The fourth unknown is there because the phone doesn’t trust its own cheap clock. It doesn’t see where it is. It reckons, from postmarks.
A timing error of 10 ns is 3 m of position, which is why the satellite clocks must be so good. Why adjust them at all? Because of the rest of this course. Left alone in orbit they would drift ahead of clocks on the ground by about 38 μs a day. Chapter 8 explains why orbiting clocks drift. For now, the point is that they must be corrected.
N. Ashby, “Relativity in the Global Positioning System”, Living Reviews in Relativity 6 (2003), 1. Read it free
Practice
Three tiers, easiest first: ◇ back of the envelope, ◆ pencil and paper, ★ head-scratchers. Do the first tier, then pick what interests you. All times are in meters of time unless a question says otherwise.
◇ Back of the envelope
A firecracker explodes at \(t = 0\), \(x = 0\). When does its flash reach the lattice clock at \(x = 6\) m?
Hint
Light covers one meter in one meter of time.
Solution
Six meters at one meter per meter of time: it arrives at \(t = 6\) m of time.
The same flash: how many nanoseconds after the explosion does it reach \(x = 6\) m?
That multiplies by 0.3, which converts nanoseconds into meters of time. Which way are we going here?
Close. You used 3 ns for each meter of time, but one meter of time is 3.3356 ns. Use the more exact number.
Hint
One meter of time is 3.3356 ns.
Solution
\(6 \times 3.3356\un{ns} \approx 20.0\un{ns}\).
A rocket moves at \(v = 0.6\). What is that in meters per second? Need a refresher?
That is the speed as a fraction of light’s. How many meters per second is light itself?
Close: that is the speed in kilometers per second. The question asks for meters per second.
Hint
\(v\) is a fraction of light speed, and light does 299 792 458 m/s.
Solution
\(0.6 \times 299\,792\,458 \approx 1.80 \times 10^8\) m/s. You can type it as 1.8e8.
Lin and Sam stand 1 km apart on the same station platform. Neither is moving. Are they in the same frame?
Hint
Their watches can both be clocks of one lattice. Is either of them moving relative to it?
Solution
Yes. A frame is a lattice of clocks at rest relative to one another; Lin and Sam are two of its clocks. They see events at different moments, and they reckon them at identical ones. Rae, passing on a rocket, or a passenger on an express train gliding through, would not share it: each moves relative to the platform, so each carries a different lattice (Chapter 7).
Seeing or reckoning? Sort each one.
- The moment the flash reaches Sam’s camera
- The reading of the clock beside the firework when it burst
- Thunder arriving three seconds after the lightning
- The time stamp a GPS satellite puts on its message
- Starlight that left its star long ago, arriving tonight
- The seen time minus the distance
Hint
Ask: is this the moment something arrives somewhere, or the reading of the clock where it happened?
Solution
Seeing (receiving the signal, whether light or sound): the flash at Sam’s camera, the thunder, the starlight. Reckoning: the clock beside the firework, the satellite’s time stamp, and seen time minus distance (that subtraction turns a sighting into a reckoning).
◆ Pencil and paper
What preset does the lattice clock at \((x, y, z) = (3, 4, 12)\) m need?
That adds the three offsets, as if the flash had to walk along the rods. Does light follow the rods?
Hint
Do Pythagoras on the floor first, then once more going up.
Solution
Floor: \(\sqrt{9 + 16} = 5\). Up: \(\sqrt{25 + 144} = 13\). Preset 13 m of time.
A proton from a particle accelerator flies from one lattice clock to the next, 3 m away. It leaves when the first clock reads 0 and arrives when the second clock reads 15 m. What is its speed, as a fraction of light speed?
That divides time by distance. Speed is distance divided by time.
Hint
Speed is distance divided by time, and here both are in meters.
Hint
Notice you used two different clocks, one at each end. That only works because they are synchronized.
Solution
\(v = 3 / 15 = 0.2\): a fifth of light speed, an ordinary speed for a proton from an accelerator. The two readings come from two different clocks, which is fine because the lattice keeps them in step.
Lin sees the far firework, 3000 m away, when her watch reads 3500 m. When did it explode, by the lattice?
That’s when the news arrived. How long was the light on the road before it got to Lin?
That adds the travel time. Did the burst happen before or after Lin saw it?
Hint
The light spent 3000 m of time traveling.
Solution
\(\lab{t_\text{event}} = \lit{t_\text{seen}} - \text{distance} = 3500 - 3000 = 500\).
Finish this worked example. Sam stands at the lattice origin. A firework bursts 900 m east and 1200 m north of him, and Sam sees it when his watch reads 2000 m.
- Distance from Sam to the firework: \(\sqrt{900^2 + 1200^2} = \;?\)
- The light spent that many meters of time on the road.
- Burst time by the lattice: \(2000 - \text{distance} = \;?\)
That adds the travel time. Did the burst happen before or after Sam saw it?
Hint
900 and 1200 are 3 and 4, times 300.
Solution
\(\sqrt{810\,000 + 1\,440\,000} = \sqrt{2\,250\,000} = 1500\) m: a 3-4-5 triangle scaled by 300. Burst time \(2000 - 1500 = 500\).
A GPS signal is 0.0700 s on the road from satellite to phone. How far away is the satellite, in kilometers?
That’s in meters. The question asks for kilometers.
Hint
Turn seconds into meters of time first: multiply by 299 792 458. That is the distance in meters.
Solution
\(0.0700 \times 299\,792\,458 \approx 2.0985 \times 10^7\) m, about 20 985 km.
★ Head-scratchers
Lin stands at \(x = 0\) and Sam at \(x = 1000\) m, on a straight road. A firework bursts somewhere on the road between them. Lin sees it at 700 and Sam sees it at 1300. Where was it, and when did it burst?
Hint
Call the place \(x\) and the time \(t\). Lin’s sighting says \(t + x = 700\). What does Sam’s say?
Hint
Sam’s distance to the firework is \(1000 - x\), so \(t + 1000 - x = 1300\). Subtract one equation from the other.
Solution
\(t + x = 700\) and \(t - x = 300\). Adding: \(2t = 1000\), so \(t = 500\). Then \(x = 200\). Two sightings from known places pin the event down completely. That is GPS in one dimension.
Sam sees two fireworks burst at exactly the same moment. Did they burst at the same moment, by the lattice?
Hint
Burst time = seen time − distance. The seen times are equal. What about the distances?
Solution
Only if they’re equally far from Sam. Burst time is seen time minus distance; with the same seen time, the farther firework burst earlier.
Keep the formula card nearby: the seen-time line and the preset rule are both on it.
Build it
Two small functions this time. The first sets the clocks of any lattice; the second turns meters of time back into seconds. Once they pass, the town lab above runs on them: your clockPreset computes every light delay, and your metersToSeconds drives its microsecond view.
Code you need
A parameter can have a default value, used whenever the caller leaves it out:
function greet(name = 'friend') {
return 'Hello, ' + name;
}
greet('Lin'); // 'Hello, Lin'
greet(); // 'Hello, friend'Big numbers are written without commas or spaces: 299792458. Tiny ones can use e: 1e-9 means \(1 \times 10^{-9}\).
Build it: clockPreset(x, y, z)
Return the preset for the lattice clock at \((x, y, z)\) meters: its distance from the reference clock. A clock on the floor is often given as clockPreset(6, 8), so y and z default to 0. The tests try clocks behind the origin too: a preset is a distance, never negative. If you forget the = 0 defaults, clockPreset(30) gets y = undefined, and arithmetic with undefined gives NaN.
Build it: metersToSeconds(m)
In Chapter 1 you turned seconds into meters of time. Now undo it: one second is 299 792 458 m of time, so how many seconds is m meters of time? The tests check tiny answers, like one meter of time being 3.3356 ns, to nine significant digits.
Checkpoint
Six items. Feedback waits until you submit.
A firework bursts 1200 m from Sam. He sees it when his watch reads 2000 m. When did it burst, by the lattice?
That’s when the light reached Sam. How long had it been traveling?
That adds the travel time. The burst came before the sighting, not after.
Solution
\(2000 - 1200 = 800\) m of time.
Lin’s lattice says a firework burst at 500. Sam sits on a bench 2 km away from Lin, at rest relative to her. What burst time does Sam reckon?
Solution
500. Sam is at rest relative to Lin, so they share one lattice and every event gets the same time from both. Where Sam sits changes only when he sees it.
What preset does the lattice clock at \((6, 6, 7)\) m need?
Solution
\(\sqrt{36 + 36 + 49} = \sqrt{121} = 11\).
Lin and Sam are at rest in one lattice, far apart. Sort each quantity: does it depend on where you stand, or is it the same for all of them?
- The time a flash is seen
- The time a firework burst
- The order in which flashes reach your eyes
- The order in which the fireworks burst
- Your distance to a firework
Solution
Seen times, seen order and distances depend on where you stand. Burst times and burst order are the lattice’s, the same for everyone at rest in it.
How many microseconds is 3000 m of time?
Solution
\(3000 / 299\,792\,458\) s \(\approx 1.0007 \times 10^{-5}\) s \(\approx 10.0\) μs.
In one or two sentences: why are “when did I see it?” and “when did it happen?” different questions?
Model answer
Light takes time to travel, one meter per meter of time, so I see an event later than it happened, by its distance from me. When it happened is the reading of the lattice clock that was there, the same for everyone in my lattice.
- Mentions the light’s travel time
- Says the delay grows with distance
- Names the clock at the event as the one that answers “when”
Explain it to Sagredo
Sagredo has a theory. “Lin saw the bakery first. Sam saw the lighthouse first. So each of them lives in a different frame, with a different time. Everyone has their own frame.” Answer him.
Model answer
They saw different orders only because their distances to the fireworks differ, and light takes time to travel. Subtract each distance from each seen time and Lin and Sam get the same burst times, because they are at rest relative to each other and share one lattice of synchronized clocks. A frame is a lattice, not a pair of eyes; only moving relative to the lattice would put someone in a different frame.
- Blames the different orders on light’s travel time
- Shows the subtraction gives the same burst times
- Says a frame is the lattice, and only relative motion changes it
What’s invariant here?
- The event itself. A firework bursts once, at one place, at one instant, whoever sees it and whenever.
- Light’s speed, 1 meter per meter of time, which is what lets one flash set every clock.
- Shared, but not invariant: its lattice time. Shared means the same for everyone at rest in this lattice. Invariant means the same in every lattice, moving or not. Seen time minus distance gives everyone at rest here the same burst time, wherever they stand. A lattice moving through this one will not agree (Chapter 7).
- Not even shared: when you see it. That depends on where you stand.
Still strange?
So far, everyone who agrees about motion agrees about time; that is the comfortable part. In Chapter 7 a second lattice glides through this one, its clocks synchronized by exactly the same honest flash, and the two lattices will disagree about which events happen at the same time.
Going further
Deep dive: synchronizing clocks on a spinning planet
The lattices in this course are free-float: they don’t rotate. The ground under your feet does. Suppose you sync a ring of clocks around the equator by passing light signals eastward from clock to clock, all the way around. When the signal comes home, the last clock and the first disagree, by about 207 ns. Earth turned a little while the light was on its way, so the eastward trip was a little longer than the westward one would have been. It is called the Sagnac effect, and the GPS system corrects for it every day. The mismatch grows with the area the loop encloses. Around a loop the size of Lin’s town it is at most about \(2 \times 10^{-14}\) s, a few millionths of a meter of time, so her town lattice is free-float far beyond anything our lab could notice. Strictly, a rotating lab isn’t free-float, and its clocks can’t all be synchronized by our rule: over a town that doesn’t matter, over a planet it does.
A firework is fired at exactly lattice time 0, but nobody knows where. Lin, at \((0, 0)\), sees it at 500. Sam, at \((800, 0)\) m, also sees it at 500. It burst north of the line between them. Where?
Hint
Both are 500 m from the firework. Where are all the points equally far from Lin and from Sam?
Hint
Halfway along: \(x = 400\). Now a right triangle with hypotenuse 500 and one leg 400.
Solution
Equal distances put it on the line \(x = 400\). Then \(\sqrt{400^2 + y^2} = 500\) gives \(y = 300\). Known burst time plus two sightings fixes the place: that is how GPS uses its satellites, turned inside out.
Reading: chapter 2 of Edwin Taylor and John Wheeler’s Spacetime Physics (2nd edition, 1992) builds the lattice of clocks in loving detail. The authors share it free at eftaylor.com.