Question

The muons created by cosmic rays in the upper atmosphere rain down more-or- less uniformly on the earth's surface, although some of them decay on the way down, with a half-life of about \(1.5 \mu\) s (measured in their rest frame). A muon detector is carried in a balloon to an altitude of \(2000 \mathrm{m}\), and in the course of an hour detects 650 muons traveling at \(0.99 c\) toward the earth. If an identical detector remains at sea level, how many muons should it register in one hour? Calculate the answer taking account of the relativistic time dilation and also classically. (Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive.) Needless to say, the relativistic answer agrees with experiment.

Short answer

Relativistically, 420 muons are detected; classically, 31 are detected.

Step-by-step solution

Calculate the Time Dilation Factor

Use the time dilation formula in special relativity to find out how time dilates for the muons.The Lorentz factor \( \gamma \) is given by:\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \( v = 0.99c \).Substitute and calculate:\[ \gamma = \frac{1}{\sqrt{1 - (0.99)^2}} \approx 7.09 \]The half-life of the muons in the earth's frame is:\[ t_{1/2} = \gamma \cdot t'_{1/2} = 7.09 \times 1.5 \mu s = 10.635 \mu s \]

Determine the Time Taken to Reach Sea Level

Compute the time it takes for the muons to travel 2000 m at speed \( 0.99c \).The velocity \( v \) of muons is \( 0.99c \), so:\[ t_{travel} = \frac{d}{v} = \frac{2000 \, m}{0.99 \times 3 \times 10^8 \, m/s} \approx 6.73 \mu s \]

Calculate the Number of Half-Lives and Survivors Relativistically

Determine the number of half-lives that passed during travel in the observer's frame, and calculate the number of muons reaching sea level.The number of half-lives \( n \) is:\[ n = \frac{t_{travel}}{t_{1/2}} = \frac{6.73 \mu s}{10.635 \mu s} \approx 0.633 \]The fraction of muons that survive is \(2^{-n}\):\[ 2^{-0.633} \approx 0.646 \]Thus, out of 650 muons, the survivors are:\[ 650 \times 0.646 \approx 420 \text{ muons} \]

Calculate the Classical Prediction

Classically calculate the number of muons that would reach sea level, ignoring relativistic effects.Classically, the half-life remains \(1.5 \mu s\).The number of half-lives in classical time is:\[ n = \frac{t_{travel}}{t_{1/2}} = \frac{6.73 \mu s}{1.5 \mu s} \approx 4.49 \]The fraction of muons that survive classically is:\[ 2^{-4.49} \approx 0.048 \]Thus, classically, the survivors are:\[ 650 \times 0.048 \approx 31 \text{ muons} \]

Key concepts

Muon Decay

Muons are fascinating subatomic particles that are similar to electrons but much heavier. They are produced when cosmic rays collide with particles in Earth's atmosphere. These muons rain down on the Earth's surface. However, muons have a property called decay, meaning they eventually change into other particles, which is part of their natural lifecycle.
Muons have a specific half-life, which is the time it takes for half of any given number of them to decay. In their rest frame, this half-life is about 1.5 microseconds. But, when muons move at high speeds, as they do when they travel towards Earth, we must consider relativistic effects to understand how long they "live" from an observer's perspective on Earth.
This is where time dilation becomes crucial. Relativistic effects such as time dilation extend the muon's half-life from an Earth observer's view, allowing more muons to reach the Earth's surface before decaying.

Lorentz Factor

The Lorentz factor, represented by the Greek letter \(\gamma\), is a key component in relativistic physics, especially in explaining time dilation. Time dilation refers to the phenomenon where time appears to move slower for an object traveling close to the speed of light compared to a stationary observer.
The formula for the Lorentz factor \( \gamma \) is:

• \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)

where \(v\) is the velocity of the object, and \(c\) is the speed of light. This factor becomes significant when the object's speed is a large fraction of the speed of light.
In the case of muons traveling at \(0.99c\), the Lorentz factor \(\gamma\) is calculated to be approximately 7.09. This implies that the muons' internal clocks (half-life) run about 7 times slower from our perspective on Earth. Therefore, their half-life in our frame of reference is much longer than in their own rest frame.

Half-Life Calculations

Understanding and calculating the half-life of particles like muons can be essential when studying their behavior. In the context of both classical and relativistic perspectives, it's crucial to calculate how many muons survive after traveling a certain distance.
Classically, the half-life calculation is straightforward. If the time taken by the muons to reach sea level is divided by their half-life, it gives the number of half-lives that have elapsed. For muons traveling at relativistic speeds, however, this half-life is significantly extended due to time dilation.

• Classically, with a half-life of \(1.5 \mu s\), more half-lives will pass compared to the relativistic scenario.
• Relativistically, the observed half-life becomes \( \gamma \cdot t'_{1/2} \), which is about \(10.635 \mu s\) for a velocity of \(0.99c\).

Relativistically, fewer half-lives mean more muons survive the trip to sea level compared to classical predictions, aligning better with experimental observations.