Muon Decay (VCE SSCE Physics): Revision Notes

Muon Decay

Introduction: muons and the particle zoo

Muons are small fundamental particles that were first discovered in 1936 by American physicist Carl Anderson. He detected them when cosmic rays collided with atoms in Earth's atmosphere. Anderson had previously won a Nobel Prize for discovering the positron, another elementary particle.

By the 1950s, physicists had observed so many new particles from cosmic ray collisions and particle accelerator experiments that they called them the "particle zoo." Over time, scientists worked out which particles were truly fundamental (unable to be broken down further) and which were made up of smaller particles. They organised these fundamental particles into a classification system based on their properties, which became known as the Standard Model.

The Standard Model groups particles into three main categories:

• Quarks (purple boxes) - make up protons and neutrons
• Leptons (green boxes) - includes electrons, muons, and neutrinos
• Bosons (orange/red/yellow boxes) - force carrier particles

Muons belong to the lepton family, appearing in the second generation alongside the muon neutrino. Organising knowledge into patterns and classifications like this is a key scientific technique, and many discoveries have come from attempting to fill gaps in such systems.

Evidence of special relativity: muon decay

What are muons?

Muons are small fundamental particles classified as a type of lepton. They have several important properties:

• Mass: 207 times the mass of an electron (but still much smaller than atoms)
• Mean lifetime when stationary: $2.20 \times 10^{-6}$ s (proper time)
• Formation: Created when cosmic rays (mainly high-energy protons) collide with atoms in the upper atmosphere
• Height of formation: Approximately 5000 m above Earth's surface
• Velocity: Travel at around $0.995c$ (where $c$ is the speed of light)

Despite their extremely short lifetimes, muons are frequently detected hitting Earth's surface. This observation provides compelling evidence for Einstein's special theory of relativity.

The classical physics problem

According to classical (Newtonian) physics, we can calculate how far a muon should travel during its lifetime using the basic distance formula:

$s = tv$

Substituting the known values:

$s = 2.20 \times 10^{-6} \times 0.995 \times 3.00 \times 10^{8}$

$s = 656.7 \text{ m}$

Solution from Earth's reference frame: time dilation

The answer lies in the effects of special relativity, specifically time dilation. Understanding this requires thinking carefully about different reference frames.

When muons are measured in a laboratory, they are stationary relative to the scientist observing them. In this case, both the muon and scientist agree on the decay time ($2.20 \times 10^{-6}$ s) because there is no relative motion. This measurement is called proper time ($t_0$) - the time interval measured in the reference frame where the events occur at the same location.

However, when muons are created in the upper atmosphere, they travel at $0.995c$ relative to a scientist on Earth. Because the muon is moving in the Earth observer's reference frame, time dilation occurs. The muon still experiences its own proper lifetime of $2.20 \times 10^{-6}$ s, but the Earth-based scientist measures a longer, dilated lifetime.

We can calculate the dilated lifetime using the time dilation formula:

$t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Substituting values:

$t = \frac{2.20 \times 10^{-6}}{\sqrt{1-0.995^2}}$

$t = 2.20 \times 10^{-5} \text{ s}$

The scientist on Earth observes the moving muon's lifetime to be $2.20 \times 10^{-5}$ s - ten times longer than the laboratory measurement. This extended lifetime explains why muons reach Earth's surface.

Recalculating the distance with the dilated time:

$s = tv$

$s = 2.20 \times 10^{-5} \times 0.995 \times 3.00 \times 10^{8}$

$s = 6575 \text{ m}$

From Earth's reference frame, muons can travel 6575 m, which is greater than the 5000 m needed to reach the surface.

Solution from the muon's reference frame: length contraction

A fundamental principle of special relativity is that all inertial reference frames are equally valid. This means the phenomenon must also make sense from the muon's perspective.

From the muon's reference frame, its own lifetime is still $2.20 \times 10^{-6}$ s (proper time). The muon doesn't experience time dilation in its own frame. So how does it travel the required distance?

The answer is length contraction. In the muon's reference frame, Earth is travelling towards it at $0.995c$. Due to this relative motion, the distance between the muon's creation point and Earth's surface becomes contracted.

To calculate the contracted length, we first need the Lorentz factor ($\gamma$):

$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$

$\gamma = \frac{1}{\sqrt{1-0.995^2}} = 10.01$

The contracted length is then:

$L = \frac{L_0}{\gamma}$

Where $L_0 = 5000$ m is the proper length (measured in Earth's frame):

$L = \frac{5000}{10.01} = 499 \text{ m}$

From the muon's frame of reference, the distance to Earth's surface is only 499 m. The muon can easily cover this distance in its proper lifetime of $2.20 \times 10^{-6}$ s when travelling at $0.995c$.

Being clear on your reference frame

Remember!