The Interval: Hero of Spacetime

Warm-up

Three quick questions from earlier chapters. Each one is an ingredient of today’s hero.

The puzzle

Remember the surveyors? Two guilds mapped one town with two different norths. They argued about every east and every north, and agreed perfectly on every distance.

Now the surveyors bring clocks. Two firecrackers go off, at events O and B. Lin, in the lab, writes B down as \((t, x) = (5, 3)\): 5 m of time after O and 3 m to the right. Rae’s rocket moves at 3/5 of light speed, and it passes right through both firecrackers. So for Rae, both bangs happen at her own place, \(x' = 0\). She writes the same pair as \((4, 0)\).

Experiment

Two labs, one gesture. The first is the surveyors’ town: turning the Nighttimers’ grid changes every east and every north. The second is a spacetime diagram with O and B. Its speed slider redraws everything as a rocket moving at that speed would measure it. Physicists call that a boost.

The two sliders are tied: move one and the other follows, so a turn and a boost happen side by side. We linked them by hand, so that the special turn (about 53°) and the special boost (\(v = 0.6\)) arrive together. No law of nature connects an angle to a speed. The bar above the labs keeps score. Watch what changes, and what doesn’t.

Play a little more. Drag B onto a gold 45° line, say \((4, 4)\), and boost. It slides, but it never leaves the line. What does the panel say about \(\tau^2\) there?

Sorting spacetime

Every event around O belongs to one of three families. One number decides which: τ² = \(t^2 - x^2\), worked out from O, the same τ² as in the boost machine’s O → B panel.

• Positive: timelike. Time wins, and the event is inside O’s light cone.
• Zero: lightlike. A tie, and the event is on the cone.
• Negative: spacelike. Space wins, and the event is outside the cone.

Sort eight events into the bins. When all eight are in, a speed slider unlocks.

Then play. Hide the light cone and sort by the numbers alone, or press “Shuffle new events” for a fresh set.

What just happened?

We call \(t^2 - x^2\) the squared interval and write it \(\inv{\tau^2}\), said “tau squared”. When \(\tau^2\) is positive, its square root \(\tau\) is the interval itself: for O and B, \(\tau^2 = 16\) and \(\tau = 4\). \(\tau^2\) is invariant: every free-float frame gets the same value, and so the same \(\tau\). It is the hero of this course. Time stretching, shrinking rulers, the twins: each is this one number, seen from another side. Even \(E = mc^2\) comes from the same minus sign (Chapter 14 shows how).

The sorter told the same story in another way. If \(\tau^2\) is the same for everyone, then so is its sign. So no boost can ever move a card to another bin. Sagredo’s question is a good one, and the “Why, really?” box below answers it.

One number everybody agrees on

Numbers first

Line up the two stories. The Daytimers put the well at (300, 400). Turn the Nighttimers’ east axis until it points straight at the well, and they put it at (500, 0). Different pairs, same sum of squares: \(300^2 + 400^2 = 500^2 + 0^2\).

Lin puts B at (5, 3). Rae, whose rocket passes through both firecrackers, puts it at (4, 0). Different pairs, same difference of squares: \(5^2 - 3^2 = 4^2 - 0^2\).

Rae’s pair is special, just like the Nighttimers’ was. She is present at both events, so for her the 4 is a plain clock reading: her wristwatch ticks off 4 m of time between O and B. That is the wristwatch time of Chapter 5. Physicists call it proper time, \(\tau\). So the interval \(\tau\) has a meaning you can hold in your hand: it is the time on the watch of anyone who rides straight from one event to the other.

The interval can do real work on its own, with no light clock in sight. Watch it rebuild the stretch factor.

1. Lin’s clock ticks twice at her own place, 10 m of time apart. In her frame: \(\dt = 10\), \(\dx = 0\), so \(\tau^2 = 100\). A clock at rest in a frame is the simplest pair of events there is.
2. Rae’s rocket moves at 0.6, so in Rae’s frame Lin moves at 0.6 the other way. Between the ticks Lin travels \(0.6\,\dtp\). Distance is speed times time, in any frame.
3. The interval doesn’t care whose notebook it is: \(\dtp^2 - (0.6\,\dtp)^2 = 100\). Since \(0.6^2 = 0.36\), that is \((1 - 0.36)\,\dtp^2 = 100\), so \(0.64\,\dtp^2 = 100\). This is the whole trick. One equation, one unknown.
4. So \(\dtp^2 = 156.25\) and \(\dtp = 12.5\). Lin moved \(0.6 \times 12.5 = 7.5\). Check: \(12.5^2 - 7.5^2 = 156.25 - 56.25 = 100\). Always check by putting the numbers back in.
5. Rae measures 12.5 between ticks that Lin’s own clock calls 10. The ratio is \(1.25\): the \(\g\) of Chapter 5, rebuilt from the interval alone. The light clock wasn’t a trick of light. It was the interval all along.

The picture: a circle with a minus sign

Look at the left half first. Every point 500 m from the square sits on one circle, and its rule is east² + north² = 500². Turning the grid changes a landmark’s two numbers, but never which circle it sits on.

Now the right half. Every event with \(t^2 - x^2 = 16\) sits on one curve too. It’s the same recipe: square one number, square the other, combine them. The only change is the sign in the middle, and that one sign changes the shape. Flip it yourself.

Same recipe, one sign flipped: east² + north² = 500² draws a circle, and \(t^2 - x^2 = 16\) draws a hyperbola. The minus sign is the whole difference. It also explains the shape. On a light line, \(t\) and \(x\) are the same size, so \(t^2 - x^2\) is 0 there. A curve where \(t^2 - x^2\) is 16 can never touch a line where it is 0. So it bends away from O and runs alongside the light lines without ever reaching them.

A hyperbola has two halves, or branches: one in O’s future, one in its past. Boosting slides B along its own branch, the bold curve in the boost machine. So the hyperbola is the spacetime circle, drawn with a minus sign. The circle collects every point that is the same distance from the center. The hyperbola collects every event that is the same wristwatch time from O.

The shorthand

Here is everything above in one line. The colors are the code of the course: blue for Lin’s lab, vermilion for Rae’s rocket, teal for what everyone agrees on.

\[ \inv{\tau^2} = \lab{\dt^2} - \lab{\dx^2} = \rkt{\dtp^2} - \rkt{\dxp^2} \]

Space has three directions, and each one gets a minus sign:

\[ \inv{\tau^2} = \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 \]

For example, a firecracker at \((t, x, y) = (13, 5, 12)\) from O gives \(169 - 25 - 144 = 0\). That pair sits exactly on the light cone, though no single coordinate shows it. Need a refresher?

Don’t take one example’s word for it. Here are four notebooks for the same pair O and B. Check the last column yourself.

Future, past, elsewhere

A squared number can’t be negative, but \(\tau^2\) can, because it is a difference. Its sign sorts every pair of events into one of three families, and every observer agrees on the family.

• Timelike, \(\tau^2 > 0\). Time wins. A clock can ride from one event to the other, and \(\tau\) is what it reads. For A in the plate, \(\tau^2 = 8\): a clock that rides there from O arrives with \(\sqrt 8 \approx 2.8\) on its face.
• Lightlike, \(\tau^2 = 0\). A tie. Only a flash of light connects the two events. These events make up O’s light cone.
• Spacelike, \(\tau^2 < 0\). Space wins. Getting from O to E means covering 4 m of space in 1 m of time. Nothing can, not even light. E is elsewhere.

This is causality in one picture. O can affect only events in its future cone (inside or on it). Only events in its past cone can affect O. Elsewhere is cut off in both directions.

Now the order of events. Take a timelike pair: B at (5, 3). As you boost, B slides along the upper branch of its hyperbola. That half never comes down to \(t' = 0\), so every rocket says B came after O. Now a spacelike pair: B at (3, 5). Its hyperbola lies on its side, and it crosses the line \(t' = 0\). At \(v = 0.6\) the two events are simultaneous, and faster rockets put B first. You saw both in the lab.

Proper distance

For a spacelike pair, \(\tau^2\) is negative and no wristwatch can measure it. Flip the sign and you get a positive number: \(\sigma^2 = \dx^2 - \dt^2\). Its square root \(\sigma\) is the proper distance. It is the distance between the events in the one frame where they happen at the same time.

For B at (3, 5): \(\sigma^2 = 25 - 9 = 16\), so \(\sigma = 4\). Your boost machine agreed: at \(v = 0.6\) it put B at \(t' = 0\), \(x' = 4\). Wristwatch time and proper distance are the same invariant, read in the two families where each one makes sense.

Light has no rest frame

One family is odd. Every lightlike pair has \(\tau = 0\). Does that mean light’s own clock stands still?

That’s the physics of the chapter: one number, three families, and a guarantee about cause and effect. Next question: has Nature ever broken the guarantee?

Ask Nature

If anything ever crossed from O into its elsewhere, the guarantee would fail. A message could reach a rocket that measures it arriving before it was sent. So physicists keep looking for anything faster than light. In more than a century, nothing has ever been caught outside the light cone. Once, it looked as if something had.

Light’s side of the bargain has been tested as well, by the CERN experiment you met in Chapter 2. In 1964, photons from particles that were themselves moving at about 0.99975 of light speed were timed in flight. They still moved at light speed, to within about one part in 10 000. A source that is moving doesn’t tilt its own light cone. That’s why everyone can draw the same cone.

Practice

Three tiers: ◇ back of the envelope, ◆ pencil and paper, ★ head-scratcher. Units are meters of time and meters unless the problem says years and light-years.

Stuck on a square root? The math toolkit has a refresher, and the formula card has this chapter’s equations on one page.

Build it

Three small functions, and the labs above start running on your code. Once they pass, the boost machine’s τ² column and its timelike, lightlike and spacelike tags, and the sorter’s verdicts, all come from you.

function weather(celsius) {
  if (celsius > 25) return 'hot';
  else if (celsius < 5) return 'cold';
  else return 'mild';
}

const pool = { width: 10 };
// pool has no depth: gives 2
const depth = pool.depth ?? 2;

Checkpoint

Five graded items, plus one written answer at the end. Feedback waits until you submit.

Explain it to Sagredo

Going further

Reading. Taylor and Wheeler’s Spacetime Physics builds the whole subject on the interval, and the surveyors’ parable comes from it; see eftaylor.com. Feynman’s chapter on space-time in volume I of The Feynman Lectures on Physics treats the same idea with rotations, free to read at feynmanlectures.caltech.edu.