Drawing Spacetime

Warm-up

Three quick questions from earlier chapters. Each one wakes up an idea this chapter is about to draw.

The puzzle

Three short stories, all starting at the same moment in the same spot. Lin stays put. Rae leaves Lin’s side at 3/5 of light speed. And at the moment Rae leaves, Lin flashes a lamp the way Rae is going.

Below are three lines, drawn the way this whole course draws things: time runs up the page, position runs to the right, and both are measured in meters. The whole picture covers 8 meters of time, about 27 nanoseconds, so these are quick stories. Line B runs straight up the time axis. Nobody told the lines which story they belong to.

Experiment

This is a spacetime diagram you can draw on. Each dot is an event: one place, one instant, like a firecracker going off. Click twice and the two dots join into a worldline, the history of one thing. Keep clicking to add more legs. Switch Click adds to Event to drop a single happening, like a firecracker, that belongs to no worldline. The board follows one rule of the universe: no clock’s worldline may lean flatter than light’s 45°.

Drawing is easy when you already know where everything is. But how does Lin find out where and when something happened far away? She can’t see it until light brings the news. The oldest trick is radar: send a pulse, wait for the echo, and think.

Then find the last two targets, the Asteroid and the Probe.

What just happened?

A dot is one instant. Lin lives through every instant of the story, and at each one she is somewhere: at \(x = 0\). Stack those events up the page and you get a vertical line. A worldline is not the road a thing travels. It is the thing’s whole history, one event per moment. A map of where Lin spent the story would be a dot. A spacetime diagram of it is a line, because time is on the page.

Here is the same thing as a table. Every story has a row, and every row has a slope:

Notice: the steepest line is the slowest thing. Vertical is standing still.

And the radar room? The bounce never happens when the echo arrives. It happens exactly halfway between sending and hearing. Rae’s echoes came home at 4, 8 and 12: each one at 4 times its send time. (Why 4? The optional Deep dive at the end of the next section works it out.) With a send at 3, the Comet was 9 m out at \(t = 0\), met your pulse 5 m out at \(t = 8\), and was only 2.5 m away when the echo got home at 13. An echo tells Lin where a target was at the bounce, not where her lattice puts it when the echo lands. Next we’ll see why the bounce sits exactly halfway.

Reading the picture

Light draws the 45° lines

Start with the flash. In 1 meter of time, light goes 1 meter. In 2, it goes 2. In 7, it goes 7. On a page where a meter of time and a meter of space are the same length, those points line up at exactly 45°.

Every clock goes less far than light in the same time, so a clock’s worldline climbs more steeply than 45°. The speed limit has become an angle you can see. A flash spreads both ways, so it draws a V: one 45° line each side.

Radar: when and where, from two clock readings

Lin sends a pulse at 1 on her clock. The echo comes home at 7. She never sees the target; she only has two readings of her own clock. That is enough.

The round trip took \(7 - 1 = 6\). By the rule Lin’s lattice clocks were set with in Chapter 3, light takes equal times out and back, so it spent half of that, 3, on each half of the trip. It bounced at \(1 + 3 = 4\) on Lin’s clocks, and in 3 m of time it covered 3 m. The bounce was at \(t = 4,\ x = 3\).

In symbols, with \(t_1\) for the send and \(t_2\) for the echo, here is the shorthand for what we did:

\[ \lab{t} = \frac{t_1 + t_2}{2}, \qquad \lab{x} = \frac{t_2 - t_1}{2}. \]

Here the pulse went toward \(+x\). An echo tells Lin how far away the bounce was, not which way; she knows that because she knows where she aimed. A pulse sent toward \(-x\) bounces at \(\lab{x} = -(t_2 - t_1)/2\).

The picture says the same thing as the algebra. Draw a 45° line up from the send and a 45° line down from the echo. Where they cross is the event. That works for any target, moving or not.

1. A pulse leaves Lin at \(t_1 = 4\) and its echo comes home at \(t_2 = 10\). Find the round trip: \(10 - 4 = 6\). The round trip is all Lin can measure directly.
2. Split it: 3 going out, 3 coming back. Lin’s lattice is synchronized on the rule that light takes equal times out and back (Chapter 3), so the two halves count as equal.
3. When: \(t = 4 + 3 = 7\), which is \((4 + 10)/2\). The bounce happened 3 after the send, which is also 3 before the echo.
4. Where: in 3 m of time, light covers 3 m, so \(x = 3\), which is \((10 - 4)/2\). One meter of distance per meter of time: that is what 45° means.

Sagredo has found a real soft spot. Every experiment in Chapter 2 timed light on a round trip. Nobody can time a one-way trip without a clock at each end, and setting those two clocks already needs a rule. In Chapter 3, Lin set her lattice with exactly this one: light takes equal time each way. So the halfway point is not something she measured; it is part of what “synchronized” means for her lattice. Lin could have set her clocks by another rule, say that light takes less time going out than coming back. Her clocks would then reckon light faster one way than the other. But every experiment we can do times only a round trip, and a round trip comes out the same under either rule. So no experiment could tell the two rules apart, and we pick the rule that treats both directions the same. With that rule, radar and the lattice clocks always agree.

Slope is speed

Rae’s worldline passes through \((t = 10, x = 6)\). In 10 m of time she covered 6 m: her speed is \(6/10 = 0.6\). Any two events on a straight worldline give the same answer. From now on we write an event as \((t, x)\), time first, because time is the axis that goes up. From \((5, 3)\) to \((10, 6)\): 3 across in 5 up, still 0.6.

The shorthand: \[ v = \frac{\dx}{\dt}. \] A vertical line has \(\dx = 0\), so \(v = 0\). Light has \(\dx = \dt\), so \(v = 1\). A line that leans the other way has negative \(\dx\), so \(v\) is negative: the traveler is heading toward \(-x\). The sign shows the direction; the size shows how fast. Need a refresher?

If a line leans flatter than 45°, \(|\dx|\) is bigger than \(\dt\) and the speed comes out above 1. That is the leg the sketchpad drew dotted and flagged “faster than light”. Nothing that carries a clock has ever been caught doing it.

One story, four ways

Here is a longer story. Rae leaves Lin at \(t = 0\) and flies at 0.6 toward a beacon 3 m away. She waits there for 2 m of time, then flies home at 0.6. Here it is three more ways.

As equations: \(x = 0.6\,t\) going out, \(x = 3\) at the beacon, and \(x = 3 - 0.6\,(t - 7)\) coming home. In the last one, \((t - 7)\) is how long she has been flying home since she left the beacon at \(t = 7\), and \(0.6\,(t - 7)\) is how far she has come back. At \(t = 12\) it gives \(x = 3 - 0.6 \times 5 = 0\): home. The plate marks the home leg −0.6 because \(x\) is shrinking. The sign gives the direction; the 0.6 gives how fast.

Story, table, diagram, equations: four views of one history, each with its own job.

• Story: what happens, in plain words.
• Table: the exact time and place of each event.
• Diagram: the whole trip at a glance. Every speed is a lean, and you can check the light limit by eye.
• Equations: where Rae is at any moment, even one no row lists. At \(t = 11\), the last one gives \(x = 0.6\).

Use whichever view makes the problem easy, and check it against another: the table puts B at (5, 3), and \(0.6 \times 5 = 3\).

That’s the whole toolkit: events as dots, histories as lines, light at 45°, speed as slope, and radar to find the dots. Next, real radar aimed at the Moon.

Ask Nature

Lin’s radar room is not a cartoon. Observatories still play it, with the Moon as the target.

Practice

Three tiers: ◇ back of the envelope, ◆ pencil and paper, ★ head-scratcher. Sketch a little diagram for every one. It is faster than it sounds.

◇ Back of the envelope

◆ Pencil and paper

★ Head-scratcher

Every answer here can be checked on the formula card: one line for speed, two lines for radar.

Build it

Now teach the computer both tricks. Each function is a few lines, and each task below says in full what the starter code’s comments say, so on a phone you needn’t scroll the editor sideways to read them. Once a function passes its tests, a lab above starts running on it: the radar room gets a button that locates bounces with your radarEvent, and the sketchpad measures its legs with your worldlineSpeed.

function middle(a, b) {
  if (b < a) {
    throw new RangeError('b < a');
  }
  return {
    mid: (a + b) / 2,
    half: (b - a) / 2,
  };
}

Checkpoint

Five scored items plus one written explanation. Feedback waits until you submit, so take your time and sketch. Two items are marked “Key idea”, and both need to be right to pass.

Explain it to Sagredo

Going further

Reading. Hermann Bondi’s Relativity and Common Sense (1964) builds a large part of relativity from radar echoes and nothing else; the hidden factor \(k = 2\) from the Deep dive is where his method starts. Taylor and Wheeler’s Spacetime Physics (the authors’ page) is where much of this course comes from, and its early chapters are full of diagrams like these.