F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special RelativityEvidence for Time Dilation: Muons, Atomic Clocks and Accelerators
F3 F2.3 F2.4 A1.6 A1.8 A1.10About 75 minutes
Learning goals- I can explain how cosmic-ray muons reaching Earth's surface provide evidence for time dilation.
- I can use half-life and γ to predict how many muons survive a trip, with and without time dilation.
- I can describe the Hafele–Keating atomic-clock experiment and interpret its results.
- I can analyse decay data from a simulation to find a half-life and test a relativistic prediction.
Words to know
muon
An unstable elementary particle, like a heavy electron (about 207 times the electron's mass), with a mean lifetime of about 2.2 µs in its rest frame.
cosmic rays
High-energy particles from space (mostly protons) that strike the upper atmosphere and create showers of other particles, including muons.
mean lifetime (τ)
The average time an unstable particle exists before decaying. The half-life equals τ × ln 2 ≈ 0.693τ.
half-life
The time for half of a sample of unstable particles to decay.
caesium atomic clock
A clock that counts the very regular microwave frequency absorbed by caesium-133 atoms; the second is defined by it.
experimental uncertainty
The range (±) within which the true value is likely to lie, given the limits of the measurement.
Can we test this?
Time dilation is a bold prediction. If it were wrong, special relativity would collapse. Physics does not accept a theory because it is clever; it accepts it because measurements agree with it. In this lesson you will look at three kinds of evidence: particles from space, atomic clocks on airplanes, and particles in accelerators. You will also analyse decay data yourself, the way an experimenter would.
Muons from the sky
When cosmic-ray protons hit the upper atmosphere (typically 10 to 15 km up), they create showers of particles. Some of these become muons. A muon at rest decays with a mean lifetime of about 2.2 µs, which means a half-life of about 1.52 µs (2.2 µs × 0.693).
Even at nearly the speed of light, a muon living 2.2 µs would travel only about 3.00 × 10⁸ m/s × 2.2 × 10⁻⁶ s ≈ 660 m. Without time dilation, almost none would survive the 10 km or more to the ground. Yet more than a hundred muons pass through every square metre at sea level every second. (Some are passing through you as you read this.)
Remember from Grade 11: in Thermal and Nuclear Energy you used half-life: after n half-lives, the fraction left is (1/2)ⁿ. We use that here.
The mountain experiment
In 1963, David Frisch and James Smith measured the number of fast muons (speeds of about 0.995c) arriving per hour at the top of Mount Washington in New Hampshire, then at a site near sea level about 1907 m lower. Their counts were about 560 per hour at the top and about 410 per hour at the bottom.
- Newtonian prediction: the trip takes 1907 m ÷ (0.995 × 3.00 × 10⁸ m/s) = 6.39 µs. That is 6.39 ÷ 1.52 = 4.20 half-lives. Fraction left: (1/2)^4.20 = 0.0543. Expected: 560 × 0.0543 ≈ 30 per hour.
- Relativistic prediction: γ = 1/√(1 − 0.995²) = 10.0. The muons' own clocks record only 6.39 ÷ 10.0 = 0.639 µs, which is 0.420 half-lives. Fraction left: (1/2)^0.420 = 0.747. Expected: 560 × 0.747 ≈ 420 per hour.
The measured value, about 410, agrees with relativity and is more than ten times the Newtonian prediction.
Simplified data from the 1963 Frisch–Smith experiment. Of about 560 fast muons per hour at the summit, only about 30 per hour should reach sea level if their clocks ran at the normal rate. About 410 per hour arrived, as time dilation predicts.Worked exampleExample 1: What γ do the data imply?
Using the counts 560 per hour (summit) and 410 per hour (sea level), a half-life of 1.52 µs and an Earth-frame travel time of 6.39 µs, find the time that passed in the muons' frame and the value of γ.
- Fraction surviving: 410 ÷ 560 = 0.732.
- Number of half-lives n: (1/2)ⁿ = 0.732, so n = log(0.732)/log(0.5) = 0.450.
- Muon-frame time: Δt₀ = 0.450 × 1.52 µs = 0.684 µs.
- γ = Δt/Δt₀ = 6.39 ÷ 0.684 = 9.3.
Δt₀ ≈ 0.68 µs and γ ≈ 9.3, close to the γ = 10 expected for 0.995c. (The real experiment had to correct for muons slowing down in the air; Frisch and Smith found a time-dilation factor of about 9, in agreement with relativity within their uncertainty.)
Simulated decay data for 1000 muons at rest (points) with the curve N = 1000e^(−t/2.2 µs). Half of the muons are left after about 1.5 µs, the half-life.Atomic clocks on airliners
In October 1971, Joseph Hafele and Richard Keating flew four caesium-beam atomic clocks around the world on regular passenger flights, first eastward, then westward, and compared them with clocks at the U.S. Naval Observatory.
Two effects were predicted:
- Special relativity (speed): think from a frame that does not rotate with Earth (centred on Earth's axis). Earth's surface moves eastward. An eastbound plane adds its speed to Earth's, so it moves faster than the ground clocks and should lose time. A westbound plane moves slower than the ground (it partly cancels Earth's rotation) and should gain time relative to the ground clocks.
- General relativity (gravity): clocks higher up, where gravity is slightly weaker, run slightly faster. Both flights gain from this.
Results (ns = nanoseconds):
- Eastbound: predicted −40 ± 23 ns, measured −59 ± 10 ns.
- Westbound: predicted +275 ± 21 ns, measured +273 ± 7 ns.
The measured values agree with the predictions within the experimental uncertainties. Later experiments with better clocks have confirmed the same physics far more precisely. In 2010, physicists at the U.S. National Institute of Standards and Technology even detected time dilation with optical clocks moving at speeds of less than 10 m/s.
Seen from above the North Pole, Earth turns eastward (counterclockwise). An eastbound plane moves faster than the ground in a non-rotating frame and its clock loses time; a westbound plane moves slower and gains time. The measured results agreed with relativity's predictions within the uncertainties.Particle accelerators: daily confirmation
Physicists at accelerators work with fast unstable particles every day, and the lifetimes always follow Δt = γΔt₀. In the 1970s, an experiment at CERN stored muons moving in a ring with γ ≈ 29.3. Their measured lifetime was about 64 µs, about 29 times the at-rest value, in close agreement with time dilation. If time dilation did not happen, beams of pions, kaons and muons would decay long before reaching detectors, and accelerator experiments could not be designed the way they are.
Worked exampleExample 2: Newton vs. Einstein for high-altitude muons
Muons are created 10.0 km above sea level and travel straight down at 0.998c. What fraction reaches sea level (a) according to Newtonian physics and (b) according to special relativity? Use a half-life of 1.52 µs.
- Earth-frame time: Δt = 10 000 m ÷ (0.998 × 3.00 × 10⁸ m/s) = 3.34 × 10⁻⁵ s = 33.4 µs.
- (a) Newton: 33.4 ÷ 1.52 = 22.0 half-lives. Fraction: (1/2)^22.0 = 2.4 × 10⁻⁷ (about 1 in 4 million).
- (b) γ = 1/√(1 − 0.998²) = 15.8. Muon-frame time: 33.4 ÷ 15.8 = 2.11 µs = 1.39 half-lives. Fraction: (1/2)^1.39 = 0.38.
(a) about 2 × 10⁻⁷; (b) about 0.38 (38%). Detectors at sea level count large numbers of muons, as relativity predicts.
Analysing evidence like a physicist (A1.8, A1.10)
- Compare prediction with measurement, including uncertainty. −59 ± 10 ns and −40 ± 23 ns overlap (−49 to −69 vs. −17 to −63), so they agree.
- Look for ratios that rule out alternatives. 410 measured versus 30 predicted by Newton is not a small error; it is a factor of more than 10.
- Identify sources of error. Muons lose energy in the air (so slow down), the detectors are not identical, and counts fluctuate randomly (about ±√N for N counts).
- Use more than one independent test. Particles, clocks and accelerators are very different experiments that all agree.
Just for funAbout one cosmic-ray muon passes through an area the size of your palm roughly every second. Thanks to time dilation, you are a muon detector that never needs charging; you just cannot read the display.
Hands-on activitySimulate muon decay with coins
This simulation (F2.4) shows why time dilation makes such a dramatic difference. You need: 50 coins (any kind, or 50 paper squares marked X on one side), a cup or shoebox, a pencil and paper.
Each coin is a muon. One shake-and-toss of the box is one half-life: coins that land heads up have decayed; remove them.
- Newtonian muons: the trip lasts 10 half-lives in the muons' own frame. Start with 50 coins. Toss, remove heads, record how many remain. Repeat for 10 tosses (or until none remain). Record the number left after each toss.
- Relativistic muons (γ = 10): the same trip lasts only 1 half-life in the muons' frame. Start again with 50 coins and toss once. Record how many remain.
- Repeat each run three times and average.
- Plot 'muons remaining' against 'half-lives elapsed' for the Newtonian run. On the same graph, mark the relativistic result.
What to look for: Newtonian runs usually reach zero after 5 to 7 tosses (expected after 10: 50 ÷ 1024 ≈ 0.05 coins). The relativistic run leaves about 25. The graph halves at each step, just like the decay curve above.
Analyse: (a) Why do your runs not give exactly 25, 12.5, 6.25...? (Random variation: with small numbers the fluctuations are large, about ±√N.) (b) How would using 500 coins change your results? (c) Explain in two sentences how this models the Frisch–Smith experiment.
Your turn
1Why is it surprising that muons created 10 km up are detected at sea level?
Their mean lifetime is only 2.2 µs. Even at nearly c they would travel only about 660 m in that time without time dilation, so almost none should reach the ground.
2A muon has a mean lifetime of 2.2 µs. Find its half-life.
t½ = 0.693 × 2.2 µs = 1.52 µs.
3Muons travel 3.0 km at 0.99c (Earth frame). How long does the trip take in Earth's frame, and how many half-lives is that?
Δt = 3000 ÷ (0.99 × 3.00 × 10⁸) = 1.01 × 10⁻⁵ s = 10.1 µs, which is 10.1 ÷ 1.52 = 6.6 half-lives.
4For the muons in the previous question, how many half-lives pass in the muons' frame? (γ = 7.09)
10.1 ÷ 7.09 = 1.42 µs, which is 1.42 ÷ 1.52 = 0.94 half-lives.
5What fraction survives in each case from the last two questions?
Newton: (1/2)^6.6 = 0.010 (1%). Relativity: (1/2)^0.94 = 0.52 (52%).
6In the Hafele–Keating experiment, why did the eastbound clocks lose time from the speed effect?
In a non-rotating frame, the eastbound plane's speed adds to Earth's rotation, so it moved faster than the ground clocks. Faster clocks run slower, so the flying clock lost time.
7Do the eastbound results (predicted −40 ± 23 ns, measured −59 ± 10 ns) agree? Explain using the uncertainties.
Yes. The predicted range is −63 to −17 ns and the measured range is −69 to −49 ns. The ranges overlap (−63 to −49 ns), so they agree within uncertainty.
More practice: warm-up
1Where are cosmic-ray muons created?
High in the atmosphere (typically 10 to 15 km up) when cosmic-ray protons collide with air molecules.
2What did Frisch and Smith compare?
The number of fast muons per hour at the top of Mount Washington and near sea level.
3What kind of clocks did Hafele and Keating use?
Caesium-beam atomic clocks (four of them).
4Besides speed, what other effect changed the flying clocks' rates?
Gravity (general relativity): clocks at higher altitude run slightly faster.
5After 3 half-lives, what fraction of muons remains?
More practice: on your own
1Using the decay graph, estimate the number of muons left after 3 µs and check with N = 1000e^(−t/2.2).
From the graph: about 250. Formula: 1000e^(−3/2.2) = 1000 × 0.256 = 256.
2A pion beam has particles with γ = 5.0 and a rest lifetime of 26 ns. What is their lifetime in the lab, and roughly how far do they go (v ≈ 0.98c)?
Lab lifetime: 5.0 × 26 = 130 ns. Distance ≈ 0.98 × 3.00 × 10⁸ × 1.30 × 10⁻⁷ ≈ 38 m.
3A student says the muon result could be explained if muons simply live longer than we think. How do lab measurements of muons at rest answer this?
Muons stopped in a detector (at rest in the lab) are measured to have a mean lifetime of 2.2 µs. Only moving muons last longer, and by exactly the factor γ, so it is their motion, not a wrong lifetime, that explains the result.
4Why did the westbound clocks show the larger gain?
Both the speed effect and the gravity effect made the westbound clocks gain time (they moved slower than the ground clocks in the non-rotating frame and were higher up). For the eastbound clocks the two effects worked in opposite directions.
5In a 1-hour count, 410 muons are detected. Estimate the random uncertainty in this count.
About ±√410 ≈ ±20 counts per hour.
Challenge
1Muons are produced at 12 km altitude and travel straight down at 0.9990c. Using a half-life of 1.52 µs, find the surviving fraction (a) with no time dilation and (b) with time dilation.
Δt = 12 000 ÷ (0.9990 × 3.00 × 10⁸) = 4.004 × 10⁻⁵ s = 40.0 µs. (a) 40.0/1.52 = 26.3 half-lives; (1/2)^26.3 ≈ 1.2 × 10⁻⁸. (b) γ = 1/√(1 − 0.998001) = 1/√0.001999 = 22.37. Δt₀ = 40.0/22.37 = 1.79 µs = 1.18 half-lives; (1/2)^1.18 ≈ 0.44.
2The CERN storage-ring muons had γ ≈ 29.3. Predict their lab lifetime using a rest lifetime of 2.197 µs and compare with the measured value of about 64 µs.
Δt = 29.3 × 2.197 = 64.4 µs, in close agreement with the measured value of about 64 µs.
3Design improvement: suggest two ways to make the coin simulation a better model of the Frisch–Smith experiment, and one limitation that would remain.
Improvements: use many more 'muons' (e.g., 500 coins or a computer simulation) to reduce random fluctuations; use a die or spinner to model a fraction of a half-life (the real muons experienced 0.42 half-lives, not 1); repeat many runs and average. Limitation: the simulation assumes time dilation instead of testing it; it can show the size of the difference between predictions but cannot by itself prove which is correct. Only real measurements can do that.
Check your understandingQuiz
1. Why do so many cosmic-ray muons reach Earth's surface?
2. A muon's trip takes 6.0 µs in Earth's frame and γ = 10. How much time passes for the muon?
3. In the Hafele–Keating experiment, which flight's clocks gained the most time?
4. A measured value is 410 ± 20 and the prediction is 420. What can you conclude?
5. Muons stored in a ring at CERN had γ ≈ 29.3. About how long did they live in the lab?
6. In the coin simulation, what does one toss of the box represent?
For parents and teachersBig idea (F3, F2.4): time dilation is not just theory. Muons from cosmic rays, atomic clocks flown around the world and particles in accelerators all behave exactly as Δt = γΔt₀ predicts.
How to help: Do the coin simulation together: one of you tosses, the other records. Ask your teen to explain why the 'relativistic' run leaves so many more coins. Then ask them to explain the mountain experiment in their own words using the bar graph.
Common difficulties:
- Using the mean lifetime (2.2 µs) where the half-life (1.52 µs) is needed, or the reverse. With (1/2)ⁿ, use the half-life.
- Forgetting to divide by γ to get the time in the muon's frame.
- Solving (1/2)ⁿ = fraction: n = log(fraction) ÷ log(0.5).
- Deciding whether results 'agree' without looking at the uncertainty ranges.
- Thinking the eastbound plane 'flies against Earth's rotation'. It flies with it, so it moves faster in a non-rotating frame.