12.3.6 Muon Lifetime Experiment

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 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 is a product of time dilation (or length contraction, depending on the viewpoint of the observer)

Muon Decay From Time Dilation

According to the muon's reference frame, time is dilated so its half-life is longer
We can see this from the time dilation equation

$∆ t = γ ∆ t_{0}$

Where:
- The gamma factor, $γ = \frac{1}{\sqrt{1 - {(0.98)}^{2}}} = 5$
- $∆ t$ = the half-life measured by an observer on Earth
- $∆ t_{0}$ = 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

$∆ t = 5 \times 1.6 = 8 µ s$

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 surface

Muon Decay From Length Contraction

According to the muon's reference frame, length is contracted so the distance they need to travel is shorter
We can see this from the length contraction equation

$L = \frac{L_{0}}{γ}$

Where:
- $L_{0}$ = the proper length for the distance measured in the muon's inertial frame
- $L$ = the distance measured by an observer on Earth
Therefore, in the reference frame of the muons, they only have to travel a distance:

$L = \frac{10 000}{5} = 2000 m$

To travel this distance takes a time of $\frac{2000}{0.98 c}$ = 6.8 µs which is about 4.3 half-lives again, so a significant number of muons remain undecayed at the surface

1-5-12-muon-decay-1

Muon decay from the Earth's and a muon'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:

(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

$t i m e = \frac{d i s t a n c e}{s p e e d} = \frac{h}{v}$

$t = \frac{4250}{0.98 \times (3.0 \times 10^{8})} = 1.45 \times 10^{- 5} s$

Step 3: Calculate the number of half-lives

$\frac{1.45 \times 10^{- 5}}{1.6 \times 10^{- 6}} = 9$ half-lives

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

$\frac{1}{2^{9}} \times 100 % = 0.2 %$

(a) (ii) Special relativity

Step 1: List the known quantities

Time for the muon to travel, $∆ t_{0}$ = 1.45 × 10^–5 s

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

$∆ t = γ ∆ t_{0}$

$∆ t = 5 \times (1.6 \times 10^{- 6}) = 8 \times 10^{- 6}$

Step 3: Calculate the number of half-lives

$\frac{1.45 \times 10^{- 5}}{8 \times 10^{- 6}} = 1.8$ half-lives

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

$\frac{1}{2^{1.8}} \times 100 % = 29 %$

(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 $(\frac{9}{1.8})$ = 5 OR length contraction occurs

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

$L = \frac{L_{0}}{γ}$

$L = \frac{4250}{5} = 850 m$

Step 3: Calculate the time to travel

time taken = $\frac{d i s t a n c e}{s p e e d} = \frac{850}{0.98 \times (3 \times 10^{8})} = 2.9 \times 10^{- 6}$

Step 4: Calculate the number of half-lives

$\frac{2.9 \times 10^{- 6}}{1.6 \times 10^{- 6}} = 1.8$ half-lives (same as (a)(ii))

Exam Tip

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
$s p e e d = \frac{d i s t a n c e}{t i m e}$

Calculating half-lives through $\frac{1}{2^{n u m b e r o f h a l f - l i v e s}}$ is a common way to calculate the number of muons remaining:

After 1 half-life, $\frac{1}{2}$ the original muons remain
After 2 half-lives, $\frac{1}{4}$ or $\frac{1}{2^{2}}$ of the original muons remain
After 3 half-lives, $\frac{1}{8}$ or $\frac{1}{2^{3}}$ of the original muons remain, and so on

AQA A Level Physics

Revision Notes

Muon Lifetime Experiment

Muon Decay From Time Dilation

Muon Decay From Length Contraction

Worked example

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Author: Dan MG