IBPhysicsDPHLRevision NotesSpace, Time & MotionGalilean & Special RelativityMuon Lifetime Experiment

Muon Lifetime Experiment (DP IB Physics): Revision Note

Ashika

Written by: Ashika

Reviewed by: Leander Oates

Updated on

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

    • 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 a mean lifetime of 2.2 µs)

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

  • A considerable number of muons can be detected on the Earth's surface, which is about 10 km from the distance they are created 

    • Therefore, according to Newtonian Physics, very few muons are expected to reach the surface, as this is about 21 half-lives!

  • The detection of the muons on the Earth's surface is a product of time dilation (or length contraction, depending on the viewpoint of the observer)

Muon decay from time dilation

  • According to the reference frame of an observer on Earth, time is dilated, so the muon's half-life is longer

  • 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

  • The time to travel 10 km at 0.98c is 33 µs, or 4.1 half-lives, so a significant number of muons remain undecayed at the Earth's surface

Muon decay from length contraction

  • According to an observer in the muon's reference frame, 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 in the muon's inertial frame

    • L space=  the distance measured by an observer on Earth

  • 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 an observer in the muon's reference frame

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

1-5-12-muon-decay

Muon decay from the Earth's reference frame and a muon's reference frame

Worked Example

Muons are created at a height of 4250 m above the Earth’s surface. They move vertically downward at a speed of 0.980c relative to the Earth.

The muon has a half-life of 1.6 μs in its rest frame.
For this speed, the Lorentz factor is γ = 5.00.

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 percentage of the original number of muons that reach the Earth’s surface before decaying, as measured in the Earth’s frame of reference, assuming:

(i) Newtonian mechanics

(ii) Special relativity

(b) Show that the result in (a)(ii) can also be obtained by considering measurements made in the muon’s rest frame.

Answer:

(a)(i) Newtonian mechanics

Step 1: List the known quantities:

  • Height above Earth’s surface, h = 4250\ \text{m}

  • Speed of muons, v = 0.980c = 2.94\times10^8\ \text{m s}^{-1}

  • Muon half-life (no relativistic effects), t_{1/2} = 1.6\times10^{-6}\ \text{s}

Step 2: Calculate the travel time of the muons:

t = \frac{h}{v} = \frac{4250}{2.94\times10^8} = 1.45\times10^{-5}\ \text{s}

Step 3: Calculate the number of half-lives elapsed:

n = \frac{t}{t_{1/2}} = \frac{1.45\times10^{-5}}{1.6\times10^{-6}} \approx 9

Step 4: Calculate the fraction of muons that survive:

\text{Fraction} = \left(\frac12\right)^9 = 1.95\times10^{-3}

(a)(ii) Special relativity

Step 1: List the known quantities:

  • Earth-frame travel time, t = 1.45\times10^{-5}\ \text{s}

  • Proper half-life of muon, increment t subscript 0 equals 1.6 cross times 10 to the power of negative 6 end exponent space text s end text

  • Lorentz factor, \gamma = 5.00

Step 2: Calculate the dilated half-life in the Earth frame:

t equals gamma increment t subscript 0 equals 5.00 left parenthesis 1.6 cross times 10 to the power of negative 6 end exponent right parenthesis equals 8.0 cross times 10 to the power of negative 6 end exponent space text s end text

Step 3: Calculate the number of half-lives elapsed:

n equals fraction numerator increment t subscript 0 over denominator t end fraction equals fraction numerator 1.45 cross times 10 to the power of negative 5 end exponent over denominator 8.0 cross times 10 to the power of negative 6 end exponent end fraction almost equal to 1.8

Step 4: Calculate the fraction of muons that survive:

percent sign space equals space open parentheses 1 half close parentheses to the power of 1.8 end exponent equals space 0.29 space cross times space 100 space equals space 29 space percent sign

(b) Show that the result in (a)(ii) can also be obtained by considering measurements made in the muon’s rest frame:

Step 1: List the known quantities:

  • Proper half-life of muon, increment t subscript 0 equals 1.6 cross times 10 to the power of negative 6 end exponent space text s end text

  • Proper distance to Earth’s surface, L_0 = 4250\ \text{m}

  • Lorentz factor, \gamma = 5.00

  • Relative speed of Earth toward muon, v = 0.980c = 2.94\times10^8\ \text{m s}^{-1}

Step 2: Calculate the length-contracted distance:

L = \frac{L_0}{\gamma} = \frac{4250}{5.00} = 850\ \text{m}

Step 3: Calculate the travel time required in the muon frame:

t' = \frac{L}{v} = \frac{850}{2.94\times10^8} = 2.9\times10^{-6}\ \text{s}

Step 3: Calculate the number of half-lives elapsed:

n equals fraction numerator t apostrophe over denominator increment t subscript 0 end fraction equals fraction numerator 2.9 cross times 10 to the power of negative 6 end exponent over denominator 1.6 cross times 10 to the power of negative 6 end exponent end fraction almost equal to 1.8

Step 4: Calculate the fraction of muons that survive:

percent sign space equals open parentheses 1 half close parentheses to the power of 1.8 end exponent equals space 0.29 space cross times 100 space equals space 29 space percent sign

Examiner Tips and Tricks

Remember that it is the observer on Earth who views 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

  • speed space equals space distance over time

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Ashika

Author: Ashika

Expertise: Physics Content Creator

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

Leander Oates

Reviewer: Leander Oates

Expertise: Physics Content Creator

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.