Physics › Turning points › Time dilation and muon decay
Time dilation and muon decay
Proper time is the interval measured on a clock present at both events, and any observer moving relative to that clock measures a longer one. Time dilation follows t = t₀/√(1 − v²/c²), and cosmic-ray muons are the standard evidence. About a quarter reach the ground, whereas an undilated lifetime would let almost none arrive.
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The consequences of special relativity, part 1 of 2. Part 2 is Length contraction, mass and energy.
Builds on The Michelson-Morley experiment and Radioactive decay and the decay constant.
IN THIS TOPIC
- Use the time dilation equation, and identify which observer measures the proper time.
- Cite muon decay as the evidence, and run the numbers both ways round.
- For Edexcel, say when the relativistic stretching of a particle's lifetime is significant; that board asks for no equation with it.
COMMON MISCONCEPTION
Time dilation is an illusion caused by the travel time of the light reaching the observer.
Time dilation is not an optical effect: it survives after every light travel time has been corrected for. Muons created in the upper atmosphere reach the ground in numbers only a genuinely slower-running moving clock can explain.
Time dilation
Einstein's two postulates cannot both hold while time stays universal. The first thing to go is the idea that all observers share one clock. The proper time t0 between two events is the time measured in the inertial frame where both events happen at the same place, on a clock riding along with them. Any observer moving at speed v relative to that clock measures a longer time:
The root is smaller than one, so t is bigger than t0. A moving clock runs slow, as judged from the frame it moves through. At everyday speeds the correction is invisibly tiny. Near c it is enormous.
Describe the drawing
The computed dilation factor against v over c: the curve hugs one so closely that an airliner's point, marked in amber at the start, is stretched by only three and a half parts in ten to the thirteen, and it is still under five per cent at three tenths of c. Beyond that it climbs steeply towards a dashed vertical wall at v equals c that it can never reach, passing through the marked question speeds: 3.2 at 0.95c, 7.1 at 0.99c and 11.2 at 0.996c, the factor that lets the cosmic-ray muons reach the ground.
WORKED EXAMPLE
A muon's stretched lifetime
A muon at rest lives about t0 = 2.2 μs before decaying. Find its lifetime as measured by a laboratory it passes at 0.99c.
t = t0/√(1 − v2/c2) = 2.2/√(1 − 0.992) = 2.2/0.141 = 16 μs.
The muon's own clock still reads 2.2 μs at decay; the laboratory watches that clock run seven times slow. Proper time belongs to the muon, because both events, birth and decay, happen at the muon.
GUIDED PRACTICE
A dilation read backwards
A beam of pions has a rest-frame lifetime of 26 ns, and a laboratory measures their lifetime in flight as 39 ns. Find the speed of the pions.
Show the working
t/t0 = 39/26 = 1.5, so √(1 − v2/c2) = 1/1.5 = 0.667.
Squaring gives 1 − v2/c2 = 0.44, so v = 0.75c = 2.2 × 108 m s−1. The equation runs in either direction: given the stretch, it hands back the speed.
The muons that should not arrive
Nature runs this experiment constantly. Cosmic rays striking the upper atmosphere create muons about 10 km up, travelling near c. One undilated lifetime at 0.996c covers 650 m, so the 10 km descent lasts fifteen lifetimes, and since decay is exponential, fewer than three muons in every ten million should survive it. Ground-level detectors measure far more than that. Around a quarter of them arrive.
Describe the drawing
A 10 kilometre drop from a dashed creation line to the ground: a short coral bar shows the 650 metres one undilated muon lifetime covers, and a long cyan bar the 7.4 kilometres a lifetime dilated elevenfold covers, which is why a quarter of the muons survive the trip that classical physics says almost none should.
Time dilation closes the gap exactly. At 0.996c the factor is 11.2, the laboratory-frame lifetime stretches to 25 μs, the descent lasts only 1.4 dilated lifetimes, and the predicted survival is 26 per cent, matching what detectors count. Muon showers are therefore a continuous measurement of time dilation.
The stretch is only noticeable when two conditions hold together: the speed is a large fraction of c, and the particle's natural lifetime is short compared with the time it takes to make the crossing. At everyday speeds the dilation exists and is far too small to see. The same factor √(1 − v2/c2) also applies to lengths, and to the relation between mass and energy.
Length contraction has moved to the next part of this lesson.
Mass, energy, and Bertozzi's electrons has moved to the next part of this lesson.
ASSESSMENT FOCUS
- Choose the proper quantity first, and say why. Proper time is measured in the frame where the two events happen at the same place, on the clock riding with them; taking the wrong quantity as t0 inverts the whole calculation.
- Sanity-check the direction of every answer. Moving clocks read less time, so t0 is always the shorter of the two times.
- The muon answer needs its numbers in a chain. Lifetime dilated by the factor, descent time compared with it, survival transformed from negligible to substantial, and the prediction matched against what detectors count.
- This page sits in an AQA option, but one strip of it is core Edexcel. That board asks when the relativistic stretching of a particle's lifetime matters, and says outright that the equations are not required, so the muon section is the part to read. Two situations answer it. Cosmic-ray muons reach the ground when an undilated lifetime would never get them there, and short-lived particles made in accelerators survive long enough to cross a detector and be recorded. Both share the condition worth stating: the speed is a large fraction of c, and the unstable particle's natural lifetime is short compared with the time the trip takes. At everyday speeds the stretch is there and far too small to notice.
CHECK YOURSELF
A spacecraft passes Earth at 0.80c, and its on-board clock records 30 s between two ticks. State which observer measures the proper time and why, calculate the interval Earth measures, and find the speed at which the dilation factor would reach 2.
Show a hint
The root is √(1 − 0.64) = 0.60. For the last part, set the whole factor equal to 2 and solve for v.
Show the answer
Both ticks happen at the ship's clock, at the same place in the ship's frame, so the ship's crew measure the proper time t0 = 30 s.
Earth measures t = 30/0.60 = 50 s. The moving clock runs slow.
A factor of 2 needs √(1 − v2/c2) = 0.5, so v2/c2 = 0.75 and v = 0.87c = 2.6 × 108 m s−1. Nothing slower than about 0.87c dilates by even a factor of two, which is why the effect went unobserved for so long.
A moving clock runs slow by the factor √(1 − v²/c²), and cosmic-ray muons reaching the ground are the standard evidence.
Proper time belongs to the frame where both events happen at the same place.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the time dilation and muon decay questions page.
CHECK YOUR PROGRESS
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- Use the time dilation equation, and identify which observer measures the proper time.
- Cite muon decay as the evidence, and run the numbers both ways round.
- For Edexcel, say when the relativistic stretching of a particle's lifetime is significant; that board asks for no equation with it.
Open the full revision checklist to track your progress across the whole unit.