Question

In the frame in which they are at rest, the number of muons at tiroe \(r\) is given by $$ N=N_{0} e^{-\nu / \tau} $$ where \(N_{0}\) is the number at \(r=0\) and \(\tau\) is the mean lifetime 2.2 \mus. (a) If muons are produced at a height of \(4.0 \mathrm{~km}\), beading toward the ground at \(0.93 \mathrm{c}\). what fraction will survive to reach the ground? (b) What fraction would reach the ground if classical mechanics were valid?

Short answer

The fraction of muons that will survive to reach the ground can be calculated using the decay formula and incorporating the effects of time dilation due to the muons' near-light velocity. A different fraction would result under classical mechanics, where there is no time dilation.

Step-by-step solution

Understand the decay formula

The given decay formula \( N = N_0 e^{-\nu / \tau} \) represents exponential decay. \( N_0 \) is the initial quantity (or number) of muons; \( N \) is the quantity after some time \( \nu \); \( \tau \) is the mean lifetime, which measures the time taken for the quantity to reduce by a certain fraction (approximately 63.2%). Here, \( \nu \) corresponds to the time taken for the muons to travel from their initial height to the ground.

Apply the relativistic time dilation formula

Because the muons are moving at a speed close to that of light, we must take into account time dilation. This effect means that a moving observer (in our case the muons) measures a longer time interval than a stationary observer. The formula for time dilation is \( t = t_0 / \sqrt{1-v^2/c^2} \), where \( t \) is the dilated time, \( t_0 \) is the proper time, \( v \) is the velocity of the moving observer, and \( c \) is the speed of light. Using this formula, we can calculate the dilated time that the muons take to reach the ground.

Calculate the fraction of surviving muons

Substitute the dilated time from step 2 into the decay formula in place of \( \nu \) to get the number of surviving muons \( N \). The fraction of surviving muons can be calculated as \( N/N_0 \).

Classical mechanics scenario

In classical mechanics, there is no time dilation. Therefore, \( t \) in the decay formula is the same as \( t_0 \), the real time taken for the muons to reach the ground. Substitute \( t_0 \) into the decay formula to get the number of muons \( N' \) that would reach the ground according to classical mechanics. The fraction of muons that would survive is then \( N' / N_0 \).

Key concepts

Exponential Decay

Understanding exponential decay is crucial when studying the behavior of muons, particularly how their numbers decrease over time. This concept is represented by the formula \( N = N_0 e^{-u / \tau} \), where \( N_0 \) is the initial number of muons, \( N \) is the number of muons remaining after a time \( u \), and \( \tau \) is the mean lifetime, in this case, 2.2 microseconds.
Exponential decay implies that over each equal time period, the quantity of muons decreases by the same factor, allowing us to predict how many will remain after a given time. It's like a continuous process of halving or reducing by a constant percentage, which naturally results in a curve when plotted.
In the context of this exercise, this formula helps us calculate the fraction of muons that reach the ground from a set height before they decay away.

Relativistic Time Dilation

Time dilation is a fundamental concept in Einstein's theory of relativity. It's especially important when objects move at speeds close to the speed of light, such as muons descending towards Earth at 0.93c. Because of time dilation, time appears to run slower for the muons from our vantage point compared to how they experience it.
The formula for calculating time dilation is \( t = t_0 / \sqrt{1-v^2/c^2} \), where \( t_0 \) is the proper time (time experienced by an observer moving with the muons), \( v \) is the velocity of the muons, and \( c \) is the speed of light.
This phenomenon allows more muons to reach the Earth than would be expected if they experienced only their proper lifetime. Consequently, in this problem, the time it takes for muons to reach the ground must be perceived from their perspective, accounting for the slower passage of time due to their high speed.

Classical Mechanics

Classical mechanics provides a simpler view where time dilation does not occur. In this approach, the decay of muons is calculated purely based on the real time they experience as if they were at rest.
Under classical mechanics, time \( t \) in the decay formula \( N = N_0 e^{-t / \tau} \) is not adjusted for relative motion. Thus, it would consider only the real time it takes to travel to the ground, assuming a straightforward application of gravity and speed.
This provides a contrasting result to relativistic calculations, showing fewer muons surviving the journey to the ground because the decay is not slowed by relativistic effects. It serves to highlight the impact and necessity of considering relativistic effects at high velocities.

Mean Lifetime

Mean lifetime, denoted by \( \tau \), is a critical component of the exponential decay formula, representing the average time a particle, like a muon, exists before decaying. For muons, this is approximately 2.2 microseconds.
In the context of decay, the mean lifetime helps us understand how quickly or slowly a group of muons will reduce to a small fraction of its original number. It's a statistical measure—individual muons may last longer or shorter than this duration, but overall, it's significantly useful for predictions.
When calculating how many muons will survive to reach a starting point, mean lifetime becomes integral. It signifies how deeply muons are affected by decay both in classical mechanics and within the realm of relativistic effects.