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A muon is created 55.0 km above the surface of the earth (as measured in the earth's frame). The average lifetime of a muon, measured in its own rest frame, is 2.20 \(\mu\)s, and the muon we are considering has this lifetime. In the frame of the muon, the earth is moving toward the muon with a speed of 0.9860\(c\). (a) In the muon's frame, what is its initial height above the surface of the earth? (b) In the muon's frame, how much closer does the earth get during the lifetime of the muon? What fraction is this of the muon's original height, as measured in the muon's frame? (c) In the earth's frame, what is the lifetime of the muon? In the earth's frame, how far does the muon travel during its lifetime? What fraction is this of the muon's original height in the earth's frame?

Short Answer

Expert verified
(a) 9.14 km; (b) 7.12%; (c) 13.24 μs and 3.92 km, 7.13%

Step by step solution

01

Calculate the initial height in the muon's frame

To find the initial height from the muon's frame, we need to use the Lorentz contraction formula. The formula for Lorentz contraction is: \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where \( L_0 \) is the original length (55.0 km), \( v = 0.9860c \), and \( c \) is the speed of light.Plug in the values:\[ L = 55.0 \text{ km} \times \sqrt{1 - (0.9860)^2} \approx 55.0 \text{ km} \times 0.1662 \approx 9.14 \text{ km} \]Thus, the initial height is 9.14 km.
02

Calculate distance traveled by Earth in muon's frame

In the muon's frame, we consider the distance the Earth moves towards the muon during the muon's lifetime. This can be calculated using the formula:\[ d = v \times \Delta t \]where \( v = 0.9860c \) and \( \Delta t = 2.20 \text{ } \mu s = 2.20 \times 10^{-6} \text{ s} \).Calculate \( d \) as follows:\[ d = 0.9860c \times 2.20 \times 10^{-6} s \]Since \( c = 3 \times 10^8 \) m/s, substitute this value to get:\[ d \approx 0.9860 \times 3 \times 10^8 \times 2.20 \times 10^{-6} \approx 651 \text{ meters} \]
03

Calculate the fraction of the initial height

To find the fraction of the initial height that the Earth moves during the muon's lifetime, use the formula:\[ \text{Fraction} = \frac{d}{L} \]where \( d = 651 \text{ meters} \) and \( L = 9.14 \times 10^3 \text{ meters} \).Calculate:\[ \text{Fraction} = \frac{651}{9140} \approx 0.0712 \]So, the Earth moves approximately 7.12% of the muon’s initial height.
04

Determine lifetime of muon in Earth's frame

In the Earth's frame, the lifetime of the muon changes due to time dilation.The time dilation formula is:\[ \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \( \Delta t = 2.20 \) μs and \( v = 0.9860c \).Calculate \( \Delta t' \) as follows:\[ \Delta t' = \frac{2.20 \times 10^{-6}}{\sqrt{1 - (0.9860)^2}} \approx \frac{2.20 \times 10^{-6}}{0.1662} \approx 13.24 \times 10^{-6} \text{ s} \]The lifetime of the muon in the Earth’s frame is approximately 13.24 μs.
05

Calculate distance traveled by muon in Earth's frame

In the Earth's frame, use the same formula \( d = v \times \Delta t' \) to calculate the distance traveled by the muon.\[ d = 0.9860c \times 13.24 \times 10^{-6} s \]\[ d \approx 0.9860 \times 3 \times 10^8 \times 13.24 \times 10^{-6} \approx 3.92 \text{ km} \]
06

Find fraction of original height in Earth's frame

Calculate the fraction of the original height the muon travels in Earth's frame using:\[ \text{Fraction} = \frac{Distance \text{ traveled by the muon}}{\text{Initial height (Earth's frame)}} \]where Distance traveled is \( 3.92 \text{ km} \) and Initial height is \( 55.0 \text{ km} \).Calculate:\[ \text{Fraction} = \frac{3.92}{55.0} \approx 0.0713 \]The muon travels about 7.13% of its original height in the Earth's frame.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Time Dilation
In the realm of special relativity, time dilation is an intriguing consequence of traveling at high speeds, close to the speed of light. This effect means that time seems to pass differently for observers moving relative to one another.
For an observer moving at a faster speed, less time has passed compared to an observer who is stationary. This is mathematically depicted by the time dilation formula: \[ \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \] where \( \Delta t \) is the proper time (rest frame time), \( \Delta t' \) is the dilated time (moving frame time), \( v \) is the velocity of the moving object, and \( c \) is the speed of light.

In the context of a muon moving towards Earth, its lifetime extends from 2.20 microseconds in its own rest frame to about 13.24 microseconds when measured in Earth's frame. This longer time allows the muon to travel a greater distance than it would otherwise, and it's a prime example of how time dilation affects subatomic particles, letting them traverse much further than expected.
Muon Lifetime
Muons are unstable subatomic particles, and they have an average lifetime when at rest. In their rest frame, muons typically live for 2.20 microseconds before decaying. Now, when these particles are moving at speeds close to that of light, which often happens in cosmic ray interactions in Earth's atmosphere, their lifetimes appear quite different to observers on Earth.
  • In the rest frame of the muon, its lifetime remains 2.20 microseconds.
  • Due to relativistic effects, particularly time dilation, the observed lifetime on Earth extends to 13.24 microseconds.
  • This increased lifespan allows more muons to reach the Earth's surface before decaying, explaining their higher than expected incidence rate detected by ground experiments.
    This phenomenon illustrates the profound impact of relativistic effects on our understanding of subatomic particles, showing that what appears stationary to one observer can be drastically different for another.
Special Relativity
Special Relativity, a groundbreaking theory developed by Albert Einstein, reshapes how we perceive time and space. This theory asserts two key postulates:
  • The laws of physics are the same for all observers in uniform motion relative to one another.
  • The speed of light in a vacuum is constant and will always measure the same for any observer, regardless of the observer's motion.
    These postulates lead to essential conclusions in relativity, like time dilation and Lorentz contraction, which are critical for explaining the behavior of rapidly moving particles such as muons. When contemplating muons, special relativity helps explain why these particles reach the Earth's surface more frequently than expected.
Special relativity is fundamentally entwined with how we analyze high-speed systems, and without it, we'd struggle to fully comprehend phenomena at the cosmic and subatomic levels.
Relativistic Effects
Relativistic effects refer to the phenomena that arise as objects approach the speed of light. These include both time dilation and Lorentz contraction, underscoring the departure from classical mechanics intuitions.

For muons moving at relativistic speeds, these effects ensure their longevity and altered path measurements:
  • **Time dilation** causes their lifetime to lengthen from our viewpoint.
  • **Lorentz contraction** describes the observed reduction in distance traveled by light-speed approaching objects.
    Using Lorentz contraction, we find the muon's initial height relative to Earth shrinks from 55.0 km to about 9.14 km in the muon's frame, showcasing relativity's counterintuitive nature. These effects illustrate the fluid nature of reality at high velocities, reshaping our understanding of space and time for both cosmic travelers and everyday physics problems.

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Most popular questions from this chapter

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