Muons Don’t Read Textbooks

Warm-up

Three quick ones from earlier chapters. They are the three tools this chapter leans on: the stretch factor, the interval, and the tilted “now”.

The puzzle

Remember the prologue’s muons? They are born high in the air and live about 2.2 microseconds. In 2.2 μs even light covers only 660 m. Yet they rain down through you all day long.

In the early 1960s, David Frisch and James Smith put that to a test. On the summit of Mount Washington, their detector counted fast muons, moving at about \(v = 0.995\): about 563 every hour. Then they counted the same kind of muons with a second detector at sea level, 1907 m lower.

Before you scroll, commit to a number.

Experiment

Here is the mountain, in a box. Each little ring is a muon carrying its own countdown clock. Its lifetime is drawn at random, the way real decays are, and averages 2.2 μs of its own time. Where a muon decays, it leaves a small ×, so the column keeps a record of where they die. The tan triangle behind them is the mountain. The counters show the hourly rates this simple model predicts, scaled so the summit reads 563.

Reading the side panel: the hourly rates are the model’s long-run answer. The “This run” counters start again whenever you change something, and many of the muons they count are still falling, so “reached” stays well below the share the hourly rates predict. Use the hourly rates for the numbers. ≡ marks a number every frame agrees on.

A second puzzle to carry with you

Here is the same idea with friendly numbers, and a twist. Rae’s rocket passes Lin at event O, where both their clocks read 0. Rae moves at \(v = 0.6\). Lin has a lattice: a clock at every meter along \(x\), all synchronized in her frame. Five meters of time later by those clocks, Rae passes the one at \(x = 3\). Call that event A.

Lin can time Rae’s clock with her lattice. But Rae carries a lattice of her own, and she can run the same test on Lin. The lab draws one spacetime twice. In the Lab frame panel, Lin’s clocks stand still. In the Rocket frame panel, Rae’s do. (On a phone, the Lab frame panel is on top.) Press Lin’s check, then Rae’s check.

Now predict before you slide. Lin’s lattice has a clock everywhere, and the lab always follows the one Rae passes at \(t = 5\). At \(v = 0.8\), Rae gets farther in that time, so A is beside the clock at \(x = 4\), and Rae’s own clock reads 3 there. Rae reckons Lin slow by the new stretch factor, \(\g = 5/3\). Where on Lin’s worldline should C land? Make a guess. Then slide to 0.8, press Rae’s check, and read C. Hold on to the puzzle: each frame says the other clock runs slow. We solve it below, in Each frame finds the other slow.

What just happened?

Those were your two lab counters: 421 and 31. And the second Try it? Going from 0.98 to 0.999 barely shortens the fall by Earth’s clocks (6.49 to 6.37 μs), so the Galileo count hardly moves: about 29 to 31 per hour. In the Einstein universe \(\g\) jumps from 5 to 22, the muons’ own time for the fall shrinks more than fourfold, and the count climbs from about 313 to 495.

Back to the Prologue’s puzzle: muons from 10 km up

The Prologue asked you to guess what fraction of a batch of muons, made 10 km up and falling at \(v = 0.995\), reaches the ground. Here is the answer with time stretching counted.

Sagredo is right to worry. In the muon’s frame, Earth’s clocks are the moving ones, so shouldn’t they run slow? That is the same puzzle as the two-frame lab’s. The muon’s own story belongs to Chapter 9. Why the two stories don’t contradict each other is our job, right now.

Which clock was there?

Back to Rae and Lin from the two-frame lab, at \(v = 0.6\): event O, where they pass, and event A = \((t, x) = (5, 3)\), where Rae passes Lin’s lattice clock. How much time passed between O and A? Look at who was there.

• Rae’s wristwatch was at O and at A. It reads 4.
• No clock of Lin’s was at both. The clock at \(x = 0\) was at O; the clock at \(x = 3\) was at A. Lin’s lattice subtracts their readings: \(5 - 0 = 5\).

Check with the interval: \(5^2 - 3^2 = 25 - 9 = 16 = 4^2\). Rae coasts from O to A, and a coasting clock present at both events reads the interval. That reading is the wristwatch time, and physicists call it the proper time. Two clocks, synchronized in a frame where the events happen at different places, always read more.

That is the whole rule. Here it is in shorthand, which is all an equation is:

\[ \lab{\dt} = \g\,\inv{\tau}, \qquad \g = \frac{1}{\sqrt{1 - v^2}}. \]

Here \(\inv{\tau}\) is read by the one clock present at both events, and \(\lab{\dt}\) by two clocks of a frame in which that clock moves at speed \(v\). With Rae’s numbers, \(v = 0.6\) and \(\g = 5/4\), and indeed \(5 = \tfrac54 \times 4\). This is time stretching; textbooks call it time dilation.

How many muons survive

Muons decay at random, but with a steady habit: in each stretch of its own time, every living muon has the same chance of decaying. After one lifetime \(\tau_\mu = 2.2\) μs, a fraction \(e^{-1} \approx 0.37\) is still alive. After two lifetimes, \(0.37 \times 0.37 \approx 0.14.\) After any wristwatch time \(\tau\), the surviving fraction is

\[ \frac{N}{N_0} = e^{-\inv{\tau}/\tau_\mu}. \]

Need a refresher? Powers of \(e\), and why \(e^{-1} \approx 0.37\).

Now the whole muon story, one step at a time. Press “Next step” to reveal each.

1. Which clock was there? The muon’s: it is present at its birth and at its arrival. Earth’s frame needs two clocks, 10 km apart. This is the habit, and it comes first. It tells us the muon’s reading is the short one.
2. Lab time for the fall: 10 000 m at \(v = 0.995\) takes \(\lab{\dt} = 10\,000 / 0.995 \approx 10\,050.25\) m of time. In meters of time, time = distance ÷ speed, with no \(c\) to carry around.
3. Wristwatch time: the muon’s clock reads the short time, so divide by \(\g = 10.0125\): \(\inv{\tau} = 10\,050.25 / 10.0125 \approx 1003.8\) m of time. The interval agrees: \(\sqrt{10\,050.25^2 - 10\,000^2} \approx 1003.8\). A coasting clock present at both events reads the interval, and the interval is the same in every frame.
4. Convert to microseconds. One meter of time is 3.3356 ns, so 1003.8 m of time is about 3.348 μs. That is \(3.348 / 2.197 \approx 1.52\) lifetimes. 3.3356 ns is the time light takes to cross one meter.
5. Survival: \(e^{-1.524} \approx 0.218\). Without stretching, the muon would have to live through the lab’s 33.52 μs, which is 15.26 lifetimes, and \(e^{-15.26} \approx 2.4 \times 10^{-7}\). The decay clock is the muon’s own, so we feed the muon’s time into the decay law.

Each frame finds the other slow

Now Sagredo’s worry, and the puzzle from the two-frame lab. Lin reckons Rae’s clock slow: Rae’s 4 against Lin’s 5, between O and A. Rae, using her lattice, reckons Lin’s clock slow by the same factor. How can both be true?

Look at which events each of them compares. Lin compares Rae’s reading at A with the lattice clock right there at A. Rae needs to know what Lin’s clock reads “now”, at the moment her own clock reads 4. Her line of now through A is tilted, with slope \(v = 0.6\), as in Chapter 7. Before you read on, find where it lands.

So Rae’s now through A meets Lin not at B, where Lin reads 5, but at C, where Lin reads 3.2. Rae’s lattice has a clock of its own right beside Lin at C, synchronized with Rae’s, and it reads 4 there: that is what “now” means for Rae. So Rae’s two clocks (hers at O, her lattice clock at C) say 4 passed, while Lin’s one clock, present at both O and C, says 3.2. Rae: “Lin is slow, by \(4/3.2 = 1.25\).”

They are comparing different pairs of events. Lin’s pair is (O, A), and the one clock present at both is Rae’s. Rae’s pair is (O, C), and the one clock present at both is Lin’s. Each time, the single clock reads less than the other frame’s two. No contradiction: two honest answers to two different questions. The whole disagreement lives in “now”, and “now” belongs to a frame.

The two-frame lab draws both pairs. Lin’s check lights up O and A; Rae’s check lights up O and C.

Small speeds: a shortcut

Airliners and satellites are slow, as light goes, and for them \(\g\) is 1 plus a sliver your calculator may round away. Compare \(\g\) with \(1 + v^2/2\):

Each time \(v\) shrinks by 10, the sliver shrinks by 100, and the shortcut gets better. So for small speeds, \(\g \approx 1 + v^2/2\). The fraction of time a moving clock loses is \(1 - 1/\g\), and that is about \(v^2/2\) too: dividing by \(1 + v^2/2\) takes away about \(v^2/2\). So a moving clock loses a fraction of about \(v^2/2\) of the time, as reckoned by the frame it moves through. No calculus: the table is the argument. Need a refresher?

That’s the idea, from walking pace to muons. Next: the experiments that pinned it down.

Ask Nature

Time stretching has been measured from almost light speed down to a bicycle’s pace. Here is the staircase, fastest first.

In 1971, Joseph Hafele and Richard Keating flew atomic clocks around the world on airliners, once eastward and once westward, and compared them with clocks that stayed on the ground. Predict first.

The bottom step: in 2010, optical clocks in a lab resolved time stretching at speeds below 10 m/s, about the speed of a fast cyclist. C. W. Chou et al., Science 329, 1630 (2010)

Practice

Every exercise starts with the habit: which clock is present at both events? Answers are in meters of time unless the question says otherwise. Use \(\g = 10.0125\) for \(v = 0.995\) and \(\tau_\mu = 2.1969811\) μs.

◇ Back of the envelope

◆ Pencil and paper

★ Head-scratcher

Want every formula of this chapter on one page? The formula card has them, ready to print.

Build it

Two functions this time. The first is the habit, written as code. The second is the whole muon story in four lines. Both may call gamma(v): yours from Chapter 5 if it passed, or the reference one.

function price(n, each = 3) {       // each is 3 unless you pass another
  const total = n * each;            // a named value
  return total > 20 ? total * 0.9 : total;   // big orders get 10% off
}
Math.exp(1)   // 2.718…, which is e

Checkpoint

Six items: five are graded, and in the last one you explain in your own words. Feedback waits until you submit. To pass, get at least four of the five graded items right, including the one marked as a key idea.

Explain it to Sagredo

Going further

Two harder problems and some reading. All optional.

Reading

• D. H. Frisch and J. H. Smith, “Measurement of the relativistic time dilation using μ-mesons”, American Journal of Physics 31, 342 (1963). Short, clear, and full of honest corrections.
• N. Ashby, “Relativity in the Global Positioning System”, Living Reviews in Relativity 6, 1 (2003). Every term in the GPS budget, worked out.
• E. F. Taylor and J. A. Wheeler, Spacetime Physics, 2nd ed. (1992), the chapters on the muon and the twin paradox.