Experimental Evidence: Muon Decay Revision Notes | A Level AQA Physics

1. The Puzzle: Particles That Should Not Exist at Ground Level

Every second, cosmic rays — high-energy protons from deep space — slam into the upper atmosphere roughly 15 km above your head. These collisions produce showers of particles, including muons. Here is the puzzle: muons are unstable. They decay with a mean lifetime of about $2.2 \times 10^{-6}$ s. Even travelling at close to the speed of light, classical physics predicts they should disintegrate long before reaching the ground. Yet detectors at sea level record them in their thousands. Something is wrong with the classical picture.

This is not a minor discrepancy — it is a factor of roughly ten times more muons arriving than Newtonian physics allows. The resolution is one of the most elegant confirmations of Einstein's special relativity: time dilation. From the Earth's frame of reference, the muon's internal clock runs slow, extending its apparent lifetime enough to survive the journey.

2. The Classical Prediction — and Why It Fails

Start with what Newtonian physics would predict. A muon is created at an altitude of approximately 15 km and travels downward at roughly $v = 0.98c$, where $c = 3.0 \times 10^8$ m/s. The mean lifetime in the muon's own rest frame is $t_0 = 2.2 \times 10^{-6}$ s. Classically, the distance the muon could travel before decaying is speed multiplied by lifetime.

Worked example: Classical Distance Prediction for Muon Travel

A muon travels at $v = 0.98c$. Its rest-frame mean lifetime is $t_0 = 2.2 \times 10^{-6}$ s. Calculate the maximum distance it could travel classically before decaying.

1. $d = v \times t_0$ — Classically, distance equals speed multiplied by the time available — the rest-frame lifetime.
2. $v = 0.98 \times 3.0 \times 10^8 = 2.94 \times 10^8 \text{ m/s}$ — Convert the fractional speed to SI units before substituting.
3. $d = 2.94 \times 10^8 \times 2.2 \times 10^{-6}$ — Both values are now in SI units (m/s and s), so the result will be in metres.
4. $d = 2.94 \times 2.2 \times 10^{8-6} = 6.468 \times 10^2 \approx 647 \text{ m}$ — The muon can travel only about 647 m classically — far short of the 15 000 m to ground level.
5. $647 \text{ m} \ll 15\,000 \text{ m}$ — The classical prediction says muons should decay about 23 times before reaching the ground. Yet they arrive. The classical model is clearly inadequate.

Answer: Classically, a muon could travel only ~647 m, far less than the 15 km altitude of production. Classical physics predicts virtually no muons should reach sea level.

The numbers make the problem stark. Classically, a muon would need to survive roughly 23 half-lives to cover 15 km — the probability of that is vanishingly small. The fact that muons do arrive in large numbers demands a different framework.

3. Time Dilation — The Relativistic Resolution

Special relativity introduces a crucial correction: time is not absolute. A clock moving relative to an observer ticks more slowly than a clock at rest relative to that observer. The muon carries its own internal 'clock' — its decay process. From the Earth's frame, that clock runs slow, so the muon appears to live much longer than $2.2 \times 10^{-6}$ s.

The proper time $t_0$ is the time measured in the frame where the muon is at rest — the muon's own frame. The dilated time $t$ is what an Earth-based observer measures. Because the muon moves at $v = 0.98c$ relative to Earth, $\gamma$ is significantly greater than 1, stretching the observed lifetime considerably.

Worked example: Relativistic Lifetime and Distance for a Muon at 0.98c

A muon travels at $v = 0.98c$ relative to Earth. Its proper mean lifetime is $t_0 = 2.2 \times 10^{-6}$ s. Calculate (a) the Lorentz factor $\gamma$, (b) the dilated mean lifetime observed from Earth, and (c) the distance the muon can travel in that dilated lifetime.

1. $\frac{v^2}{c^2} = (0.98)^2 = 0.98 \times 0.98 = 0.9604$ — Square the fractional speed. Always expand the multiplication explicitly: $0.98 \times 0.98 = 0.9604$.
2. $1 - 0.9604 = 0.0396$ — This small value under the square root is what makes $\gamma$ large at high speeds.
3. $\sqrt{0.0396} \approx 0.199$ — Check: $0.199 \times 0.199 = 0.0396$ (to 3 s.f.). This is the denominator of $\gamma$.
4. $\gamma = \frac{1}{0.199} \approx 5.0$ — At $0.98c$, the Lorentz factor is approximately 5. Time in the Earth frame runs about 5 times faster than in the muon's frame — equivalently, the muon's clock runs 5 times slower as seen from Earth.
5. $t = \gamma t_0 = 5.0 \times 2.2 \times 10^{-6} = 1.1 \times 10^{-5} \text{ s}$ — From Earth's frame, the muon appears to live for $1.1 \times 10^{-5}$ s — five times longer than its rest-frame lifetime.
6. $d = v \times t = 2.94 \times 10^8 \times 1.1 \times 10^{-5} = 3.234 \times 10^3 \approx 3230 \text{ m}$ — Even with time dilation, this is a mean distance. Many muons in the high-energy tail of the distribution travel significantly further. At higher energies, $\gamma$ is larger and survival to sea level is well within reach.
7. $3230 \text{ m (relativistic)} \gg 647 \text{ m (classical)}$ — The relativistic prediction is about 5 times larger — consistent with $\gamma \approx 5$. High-energy muons with even larger $\gamma$ values readily survive the full 15 km.

Answer: $\gamma \approx 5.0$; dilated lifetime $\approx 1.1 \times 10^{-5}$ s; relativistic travel distance $\approx 3230$ m for mean-energy muons. Higher-energy muons with larger $\gamma$ survive the full 15 km to sea level.

4. The Experimental Evidence — What Measurements Actually Show

The muon decay experiment is not a thought experiment — it has been performed repeatedly with real detectors. The key measurement involves placing particle detectors at two different altitudes, for example at the top of a mountain and at sea level, and counting the muon flux (number of muons per unit area per unit time) at each location.

If muons obeyed classical physics, the count at sea level would be negligible — the vast majority would have decayed during the journey. What experimenters actually find is that the sea-level count is far higher than the classical prediction, and it matches the relativistic prediction using the time-dilation formula with the measured muon speed. This agreement between theory and experiment is the direct evidence for time dilation.

The agreement between the relativistic prediction and the experimental count is not approximate — it is quantitatively precise. Physicists can measure the muon speed using time-of-flight techniques, calculate $\gamma$, predict the dilated lifetime, and then verify that the observed survival fraction matches the exponential decay law using that dilated lifetime. The match is excellent.

5. Two Frames, One Event — Consistency of the Relativistic Picture

One of the most instructive aspects of the muon problem is that it can be analysed from two different reference frames, and both give the same physical result — the muon reaches the ground. This consistency is a hallmark of a correct relativistic treatment.

In the Earth's frame: the muon travels 15 km, but its lifetime is dilated by $\gamma$, so it survives long enough to arrive. In the muon's rest frame: the muon's lifetime is the proper value $t_0 = 2.2 \times 10^{-6}$ s, but the distance to the ground is length-contracted by the same factor $\gamma$, reducing the apparent distance to a manageable value. Both frames agree on the outcome — the muon reaches the detector — but they attribute it to different relativistic effects.

6. Why This Evidence Is Compelling

The muon decay experiment is particularly powerful as evidence for special relativity for several reasons. The muon speed can be measured independently, so $\gamma$ is not a free parameter — it is calculated from a measured quantity. The rest-frame lifetime of the muon is also independently known from laboratory measurements of muons at rest. The prediction is therefore fully determined before the altitude-comparison experiment is run.

Furthermore, the effect is large. At $v = 0.98c$, $\gamma \approx 5$, so the predicted survival rate differs from the classical prediction by a factor of roughly five in the lifetime — not a subtle correction but a dramatic, unambiguous discrepancy that no experimental uncertainty can explain away. The data consistently and quantitatively support the relativistic prediction.

7. Exam Technique and Common Mistakes

AQA questions on muon decay typically ask you to either calculate a dilated lifetime and compare it with the classical prediction, or to explain why muons are detected at sea level using the language of time dilation. Both types reward precise use of terminology and a clear logical chain.