Muon Lifetime Experiment (AQA A Level Physics)

Revision Note

Dan Mitchell-Garnett

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Muon Lifetime Experiment

  • Muon decay experiments provide experimental evidence for time dilation and length contraction

  • Muons are unstable subatomic particles that are around 200 times heavier than an electron and are produced in the upper atmosphere (as a result of pion decays produced by cosmic rays)

  • Muons travel at 0.98c  and have a half-life of 1.6 µs (or mean lifetime of 2.2 µs)

    • The distance they travel in one half-life is therefore around 470 m

  • Muons are created around 10 km above the surface

    • From the value above calculated with Newtonian physics, very few muons are expected to reach the surface

    • Roughly 21 half-lives pass over the course of 10 km

    • However, in practice we find that a considerably larger number of muons can be detected on the Earth's surface

  • The detection of the muons is a result of time dilation (or length contraction, depending on the viewpoint of the observer)

Muon Decay From Time Dilation

  • From the muon's own reference frame, its half-life is 1.6 μs

  • For an observer on Earth, however, the muon is travelling close to the speed of light so its half life is dilated

  • We can see this from the time dilation equation

increment t space equals space gamma increment t subscript 0

  • Where:

    • The gamma factor, gamma space equals space fraction numerator 1 over denominator square root of 1 space minus space open parentheses 0.98 close parentheses squared end root end fraction space equals space 5

    • increment t = the half-life measured by an observer on Earth

    • increment t subscript 0 = the proper time for the half-life measured in the muon's inertial frame

  • Therefore, in the reference frame of an observer on Earth, the muons have a lifetime of

increment t space equals space 5 space cross times space 1.6 space equals space 8 space µ straight s

  • According to the observer on Earth, the muons travel 10 km at a speed of 0.98c, taking 34 μs

    • This is 4.3 half-lives as seen by the observer on Earth

    • This is far fewer half-lives than the Newtonian prediction of 21 half-lives, so this correctly explains why a larger number of muons reach the Earth's surface before decaying

Muon Decay From Length Contraction

  • From the Earth's reference frame, the muons are seen to cover a distance of 10 km

  • According to the muon's reference frame, however, length is contracted so the distance they travel is shorter

  • We can see this from the length contraction equation

L space equals space fraction numerator space L subscript 0 over denominator gamma end fraction

  • Where:

    • L subscript 0 = the proper length for the distance measured by an observer on Earth (stationary relative to that 10 km path)

    • L space=  the distance measured from the muon's reference frame (moving relative to the 10 km path)

  • Therefore, in the reference frame of the muons, they only have to travel a distance:

L space equals space fraction numerator 10 space 000 over denominator 5 end fraction space space equals space 2000 space straight m

  • To travel this distance takes a time of fraction numerator 2000 over denominator 0.98 c end fraction = 6.8 µs, according to the muon's reference frame

  • This is 4.3 half-lives again, so a significant number of muons remain undecayed at the surface

Proper time and length

  • Proper time is the time measured in the reference frame which is stationary relative to the event being timed

    • Here, that event is the muon decaying

    • The muon is stationary relative to itself, so proper time is measured in the muon's reference frame

  • Proper length is the length measured in the reference frame stationary relative to the distance being measured

    • Here, that distance is the path from where the muon is created to the Earth's surface

    • The observer on Earth is stationary relative to that distance, so proper length is measured in the Earth's reference frame

Worked Example

Muons are created at a height of 4250 m above the Earth’s surface. The muons move vertically downward at a speed of 0.980c relative to the Earth’s surface. The gamma factor for this speed is 5.00. The half-life of a muon in its rest frame is 1.6 µs.

(a) Estimate the fraction of the original number of muons that will reach the Earth’s surface before decaying, from the Earth's frame of reference, according to:

 (i) Newtonian mechanics

 (ii) Special relativity

(b) Demonstrate how an observer moving with the same velocity as the muons, accounts for the answer to (a)(ii).

Answer:

Part (a) (i) Newtonian mechanics

Step 1: List the known quantities

  • Height of muon creation above Earth's surface, h = 4250 m 

  • Speed of muons, v = 0.980c

  • Lifetime of muon, t = 1.6 µs = 1.6 × 10–6 s

Step 2: Calculate the time to travel for the muon

t i m e space equals fraction numerator space d i s t a n c e over denominator s p e e d end fraction space equals fraction numerator space h over denominator v end fraction

t space equals fraction numerator space 4250 over denominator 0.98 space cross times space open parentheses 3.0 space cross times space 10 to the power of 8 close parentheses end fraction space equals space 1.45 space cross times space 10 to the power of negative 5 end exponent space straight s

Step 3: Calculate the number of half-lives

fraction numerator 1.45 space cross times space 10 to the power of negative 5 end exponent over denominator 1.6 space cross times space 10 to the power of negative 6 end exponent end fraction space equals space 9 half-lives

Step 4: Calculate the fraction of the original muons that arrive

1 over 2 to the power of 9 space cross times space 100 space percent sign space equals space 0.2 space percent sign

Part (a) (ii) Special relativity

Step 1: List the known quantities

  • Time for the muon to travel, increment t subscript 01.45 × 10–5 s

Step 2: Calculate the time travelled in the muons rest frame

increment t space equals space gamma increment t subscript 0

increment t space equals space 5 space cross times space open parentheses 1.6 space cross times space 10 to the power of negative 6 end exponent close parentheses space equals space 8 space cross times space 10 to the power of negative 6 end exponent

Step 3: Calculate the number of half-lives

fraction numerator 1.45 space cross times space 10 to the power of negative 5 end exponent over denominator 8 space cross times space 10 to the power of negative 6 end exponent end fraction space equals space 1.8 half-lives

Step 4: Calculate the fraction of the original muons that arrive

1 over 2 to the power of 1.8 end exponent space cross times space 100 space percent sign space equals space 29 space percent sign

Part (b) 

Step 1: Analyse the situation

  • An observer moving with the same velocity as the muons will measure the distance to the surface to be shorter by a factor of open parentheses fraction numerator 9 over denominator 1.8 end fraction close parentheses = 5 OR length contraction occurs

Step 2: Calculate the distance travelled in the muon's rest frame

L space equals space fraction numerator space L subscript 0 over denominator gamma end fraction

L space equals fraction numerator space 4250 over denominator 5 end fraction space equals space 850 space straight m

Step 3: Calculate the time to travel

time taken = fraction numerator d i s t a n c e over denominator s p e e d end fraction space equals space fraction numerator 850 space over denominator 0.98 space cross times space open parentheses 3 space cross times space 10 to the power of 8 close parentheses end fraction space equals space 2.9 space cross times 10 to the power of negative 6 end exponent

Step 4: Calculate the number of half-lives

fraction numerator 2.9 space cross times space 10 to the power of negative 6 end exponent over denominator 1.6 space cross times space 10 to the power of negative 6 end exponent end fraction space space equals space 1.8 spacehalf-lives (same as (a)(ii))

Examiner Tips and Tricks

Remember that it is the observer on Earth that viewed the muons' lifetime or half-life as longer (time dilation), whilst it is the muons' reference frame that views the distance needed to travel as shorter (length contraction).

Always do a sense check with your answer, you must always end up with a longer time or shorter distance for the muons to be observed on the Earth's surface.

Any exam questions on this topic will only use the following equations:

  • Time dilation

  • Length contraction

  • s p e e d space equals fraction numerator space d i s t a n c e over denominator t i m e end fraction

Calculating half-lives through 1 over 2 to the power of n u m b e r space o f space h a l f minus l i v e s end exponent is a common way to calculate the number of muons remaining:

  • After 1 half-life, 1 half the original muons remain

  • After 2 half-lives, 1 fourth or 1 over 2 squared of the original muons remain

  • After 3 half-lives, 1 over 8 or 1 over 2 cubedof the original muons remain, and so on

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Dan Mitchell-Garnett

Author: Dan Mitchell-Garnett

Expertise: Physics Content Creator

Dan graduated with a First-class Masters degree in Physics at Durham University, specialising in cell membrane biophysics. After being awarded an Institute of Physics Teacher Training Scholarship, Dan taught physics in secondary schools in the North of England before moving to Save My Exams. Here, he carries on his passion for writing challenging physics questions and helping young people learn to love physics.