There Is No Universal Now

Warm-up

Three quick questions from earlier chapters. Each is a tool you’ll need in the next hour.

The puzzle

A train glides along a platform at \(v = 0.6\). Lightning strikes both ends of the train, scorching the train and the platform. Lin stands on the platform exactly halfway between the two scorch marks there, and both flashes reach her together. She’s in the middle and light is light, so her lattice of clocks reckons the strikes simultaneous.

Rae sits exactly in the middle of the train, halfway between the scorch marks on the train.

Here’s the question, before we’ve built any tools for it. Expect to wrestle with it. A wrong answer argued carefully is worth more here than a right one guessed.

Experiment

Two labs. The first shrinks the puzzle to one flash inside one railcar. The second is the full thunderstorm.

Now the full storm, which is the puzzle itself: a 20 m train with lightning at both ends. On top is the scene at one moment. Below is a spacetime diagram of the whole story. One time slider drives both.

Free play: tick Show where the strikes slide as you change frame and switch frames again. Each strike slides along a curve on which τ² from the origin never changes (Chapter 6’s hyperbolas). Then try \(v = 0.8\) and \(v = 0\). At 0 the train stands still, and both frames agree on everything.

What just happened?

The most common prediction says Rae “corrects for” rushing toward the front flash, and then agrees with Lin. That correction quietly uses Lin’s frame. In Rae’s frame the train sits still and the platform slides backward. Equal distances, equal speeds, different arrival times: the flashes left at different times.

Galileo’s ship answers Sagredo: nothing Rae can do inside her train tells her it moves (Chapter 2). Being bolted to the Earth makes Lin’s lattice handy, not special.

So “at the same time” was never a fact about two events alone. It’s a fact about two events and a frame, like “to the left of” is a fact about two trees and which way you face. Physicists call this the relativity of simultaneity.

Sort out what the two frames share and what they don’t.

Tilting now

The flash in the middle, with numbers

Back to the railcar. In Rae’s frame it’s 10 m long with the lamp in the middle. The light goes 5 m each way at speed 1, so it reaches both walls at \(\rkt{t'} = 5\).

Now Lin’s frame, where the car moves at 0.6. Here we borrow from Chapter 9: Lin measures the moving car shorter, 8 m instead of 10. That’s \(10/\g = 10/1.25 = 8\). Take it on trust for now; Chapter 9 earns it. And the tilt we are after won’t depend on this length at all: you’ll see it cancel in step 3 below. So in Lin’s frame each wall starts 4 m from the flash.

The rear wall rushes toward the leftward light. Their 4 m gap closes at \(1 + 0.6 = 1.6\) meters per meter of time, so it takes \(4/1.6 = 2.5\). Your turn for the front wall.

Call the two arrivals R and F. In Lin’s coordinates R is at \((t, x) = (2.5, -2.5)\) and F is at \((10, 10)\): 7.5 m of time apart. For Rae they are simultaneous. Draw both frames:

“Now” is a tilted line

Join R and F in Lin’s diagram. Rise: \(10 - 2.5 = 7.5\). Run: \(10 - (-2.5) = 12.5\). Slope: \(7.5/12.5 = 0.6\). That’s the car’s speed. Coincidence? The steps below redo it with letters, to show it isn’t.

Optional If the numbers convinced you, skim steps 1 to 3. Step 4 states the result.

1. Say the walls start a distance \(a\) from the flash in Lin’s frame and move at \(v\). The rear wall closes on the leftward light at \(1 + v\), so R happens at \(t_R = \dfrac{a}{1+v}\), \(x_R = -\dfrac{a}{1+v}\). Same arithmetic as \(4/1.6\), with letters. Leftward light has \(x = -t\).
2. The front wall flees the rightward light at \(1 - v\), so F happens at \(t_F = \dfrac{a}{1-v}\), \(x_F = \dfrac{a}{1-v}\). Rightward light has \(x = t\): F is on the 45° line.
3. The slope from R to F is \[ \frac{t_F - t_R}{x_F - x_R} = \frac{\frac{a}{1-v} - \frac{a}{1+v}}{\frac{a}{1-v} + \frac{a}{1+v}}. \] Multiply top and bottom by \((1-v)(1+v)/a\): \[ \frac{(1+v) - (1-v)}{(1+v) + (1-v)} \] \[ = \frac{2v}{2} = v. \] The \(a\) cancels. The length of the car never mattered.
4. R and F are simultaneous for Rae. Put more lamps anywhere in her car, each midway between its own pair of walls, at any spacing: each pair of arrivals she reckons simultaneous is joined by a line of slope \(v\). Join them all up and her “now” is a straight line of slope \(v\). Rae’s lines of “now” have slope \(v\) in Lin’s diagram. Step 3 never used \(a\) or where the flash was, so every such pair gives the same slope.

Now tilt them yourself. Each event gets a dashed line through it: the rocket’s “now” at that event.

Which line? The intercept

A slope doesn’t pick out one line. Which line of slope \(v\) is Rae’s “now” when her own clock reads, say, 5? The one through the event on her worldline where her clock reads 5.

Numbers first, at \(v = 0.6\). From Chapter 5, Lin reckons Rae’s clock reads 5 at lab time \(1.25 \times 5 = 6.25\). By then Rae is at \(x = 0.6 \times 6.25 = 3.75\). A line of slope 0.6 through that event drops 0.6 for every meter you go left. Back at \(x = 0\) it has dropped \(0.6 \times 3.75 = 2.25\), so it crosses the \(t\) axis at \(6.25 - 2.25 = 4\). That crossing is the intercept, and it is \(5/1.25\): her clock reading divided by \(\g\).

Optional The same with letters (the result is the formula just below). Write the line as \(t = vx + b\), where \(b\) is the intercept. Rae’s worldline is \(x = vt\). To find where they meet, put Rae’s worldline \(x = vt\) into the line: \(t = v^2 t + b\). Solve for \(t\): \(t = b/(1 - v^2) = \g^2 b\). Her clock there reads \(\tau = t/\g = \g b\). So \(b = \tau/\g\), and her “now” at her time \(\rkt{t'}\) is

\[ \lab{t} = v\,\lab{x} + \frac{\rkt{t'}}{\g}. \]

Check it on R and F. Their line has slope 0.6 and passes through R at \((t, x) = (2.5, -2.5)\), so its intercept is \(b = 2.5 - 0.6 \times (-2.5) = 2.5 + 1.5 = 4\). That makes Rae’s clock read \(1.25 \times 4 = 5\) there: the same 5 the wall clocks read.

Turn the formula around and it hands you any event’s rocket time. An event at \((t, x)\) sits on the line with intercept \(t - vx\), so Rae’s lattice gives it

\[ \rkt{t'} = \g\,(\lab{t} - v\,\lab{x}). \]

Two strikes, one formula

Take two events simultaneous for Lin: A at \(x = 0\) and B at \(x = \dx\), both at \(t = 0\). Rae passes A, and her clock reads 0 there, so A has \(t' = 0\). For B, \(t' = \g(0 - v\dx)\). So

\[ \rkt{\dtp} = -\g\, v\, \lab{\dx} \]

Here \(\dx\) is the distance measured on the platform. This shortcut is only for a pair that is simultaneous in the lab.

The minus sign is the punchline. The event farther along Rae’s direction of motion happens earlier for her.

1. Strikes at \(x = \pm 150\) m at \(t = 0\); train at \(v = 0.6\). Front minus rear: \(\dx = 300\) m. Measure Δx from the rear event to the front event, so a negative Δt′ means “front first”.
2. \(\g v = 1.25 \times 0.6 = 0.75\). The friendly-number row for 0.6: γ = 5/4, γv = 3/4.
3. \(\dtp = -0.75 \times 300 = -225\) m of time. Negative: the front strike comes first on the train.
4. In seconds: \(225 \times 3.336\un{ns} \approx 750\un{ns}\). One meter of time is about 3.336 ns.

The front clock lags

Now turn it around and stand on the rocket. Rae synchronizes clocks along her rocket of proper length \(L\): the rear at \(x' = 0\), the front at \(x' = L\). In Rae’s frame it’s Lin who moves, at \(-v\). The same construction says Lin’s lines of “now” have slope \(-v\) in Rae’s diagram.

If the next two paragraphs go by too fast, try the Two lattices lab just below first: freeze it, read the moving row, then come back for the why.

Here is the key step. One of Lin’s now-lines, drawn in Rae’s diagram, is one single instant of Lin’s. So where it crosses Rae’s rear clock and Rae’s front clock tells you what those two clocks read at that one Lin-instant.

Take the Lin-instant whose now-line passes the rear clock as it reads 0. That line is \(t' = -v x'\). At the front, \(x' = L\), it gives \(t' = -vL\). So at one instant of Lin’s, the rear clock reads 0 and the front reads \(-vL\). The front clock lags by \(vL\). A 10 m rocket at 0.6: 6 m of time, the gap you read off the car clocks in the first lab.

\[ \text{lag} = v\,L \]

Lin gets the same number from her side. She measures the rocket shorter, \(L/\g\), and the strike formula gives \(\g v \times L/\g = vL\). The \(\g\) factors cancel, which is why the rule has no \(\g\).

Which orders can flip?

When does a rocket find B happening before A? When its now-line through A passes above B. Its now-lines have slope \(v\); the line from A to B has slope \(\dt/\dx\). So the order flips once \(v > \dt/\dx\). For the pair from the warm-up, \(\dt = 3\) and \(\dx = 5\): simultaneous at \(v = 0.6\), reversed above it.

But \(v\) is always less than 1. A flip needs \(\dt/\dx < 1\), so \(|\dx| > |\dt|\): a spacelike pair. For a timelike or lightlike pair no rocket is fast enough.

Sidebar: the tachyon telephone

Unless someone builds a faster-than-light signal. Call it a tachyon telephone. At \(t = 0\), Lin sends a question that is instantaneous in her frame to a station 10 light-years away. A rocket passing the station at 0.8 relays the reply, instantaneous in its frame: along its now-line, slope 0.8, back toward Lin. At \(x = 0\) that line is at \(t = 0.8 \times (0 - 10) = -8\). The reply arrives eight years before Lin asked.

Relativity doesn’t ban tachyons by decree. It says you can’t have both a working tachyon telephone and cause before effect. Pick one. No experiment has ever caught a faster-than-light signal. The closest call was Chapter 6’s OPERA neutrinos, which seemed to arrive about 60 ns early. The main culprit turned out to be a loose fiber-optic connector in the timing system.

That’s the whole machine: a tilted line, its intercept, and a lag.

Ask Nature

Relative simultaneity sounds like a curiosity for trains near light speed. It’s in your pocket.

Practice

Three tiers: ◇ back of the envelope, ◆ pencil and paper, ★ head-scratcher. Use the friendly numbers: at 0.6, \(\g = 5/4\) and \(\g v = 3/4\); at 0.8, \(\g = 5/3\) and \(\g v = 4/3\).

Want every equation from this chapter on one page? It’s on the formula card.

Build it

Two small functions. When they pass, your code takes over in the labs above: your simultaneityLine draws Rae’s “now” in the lightning lab, and your frontClockLag sets the car clocks in the flash lab.

function rectangle(w, h) {
  if (w < 0 || h < 0) throw new RangeError('no negative sides');
  return { area: w * h, perimeter: 2 * (w + h) };
}

Checkpoint

Five scored items, then one written answer. Feedback waits until you submit.

Explain it to Sagredo

Going further

Two head-scratchers (★) for the road, then some reading.

Reading: Rachel Scherr, Peter Shaffer and Stamatis Vokos, “The challenge of changing deeply held student beliefs about the relativity of simultaneity”, American Journal of Physics 70, 1238 (2002), a study of exactly the struggle most people have at the start of this chapter. Einstein’s own popular book, Relativity: The Special and the General Theory (German 1916, English 1920), tells the train-and-embankment story in his words.