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Chapter 2: Problem 41

A muon has a mean lifetime of \(2.2 \mu \mathrm{s}\) in its rest frame. Suppose muons are traveling at \(0.92 c\) relative to Earth. What is the mean distance a muon will travel as measured by an observer on Earth?

Short Answer

Expert verified
The mean distance a muon will travel as measured by an observer on Earth can be calculated using the calculation of dilated time and the given velocity of the muons.

Step by step solution

01

Understanding Time Dilation

We need to calculate the mean lifetime of the muon as measured by an observer on Earth, which is different from that in its rest frame, due to the effects of time dilation. According to the theory of special relativity, time dilation is given by \(t' = \gamma t = \frac{t}{\sqrt{1 - (v^2 / c^2)}}\), where \(t\) is the proper time (rest frame), \(t'\) is the dilated time (moving frame), \(v\) is the speed of moving frame relative to rest frame, and \(c\) is the speed of light.
02

Calculation of Dilated Time

Substitute \(t = 2.2 \mu \mathrm{s}\), \(v = 0.92 c\), and \(c\) (the speed of light) into the equation for time dilation to get the dilated time. This gives us \(t' = \frac{2.2 \mu \mathrm{s}}{\sqrt{1 - 0.92^2}}\)
03

Calculation of Distance

The mean distance a muon will travel as measured by an observer on Earth can be calculated using the formula \(d = vt'\). Substituting \(v = 0.92 c\) and \(t'\) calculated in the previous step, we get the mean distance.

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Key Concepts

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

Special Relativity
Special Relativity is a fundamental theory in physics proposed by Albert Einstein in 1905. It revolutionized the way we understand space, time, and how they are interwoven. According to this theory:
  • The laws of physics are the same for all observers, regardless of their relative motion.
  • The speed of light in vacuum, denoted as \(c\), is constant and always observed to be the same, no matter the relative speed of the observer and the light source.
An important consequence of these principles is the concept of time dilation. When objects move at speeds close to the speed of light, time experienced by the moving object differs from that of a stationary observer. This prediction leads to several interesting effects, such as length contraction and relativistic mass increase, but in this context, we're focusing on time dilation.
Proper Time
Proper time is a key concept in understanding time dilation. It refers to the time measured by a clock that is at rest relative to the event being observed. In other words, it is the time interval between two events as measured in the instantaneous rest frame of the object where the events occur.
For example, the mean lifetime of a muon in its rest frame is considered as proper time. A muon is a subatomic particle that naturally decays after a short period. When it is stationary, its lifetime is exactly 2.2 microseconds (\(\mu \text{s}\)). This is the proper time because it is measured by an observer moving along with the muon.
Proper time is different from dilated time, which is observed by someone who sees the muon moving at a high speed.
Dilated Time
Dilated time is the time interval as perceived by an observer watching an object moving relative to them at such a high speed that relativistic effects come into play. According to special relativity, when an object moves at a velocity approaching the speed of light, time appears to slow down for that object from the perspective of a stationary observer. This phenomenon is described by the time dilation formula: \[t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}}\]
  • \(t\) represents the proper time.
  • \(t'\) is the dilated time observed by someone for whom the object is in motion.
  • \(v\) is the velocity of the moving object relative to the observer.
  • \(c\) is the speed of light.
For our muon traveling at 0.92 times the speed of light, the dilated time is longer than its proper time of 2.2 microseconds. This is due to the relativistic effects as the speed approaches that of light, making the muon live longer from the Earth's observer's point of view. Using the dilated time, we can then calculate how far the muon will travel before decaying, offering real-world insight into the practical applications of special relativity.

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Most popular questions from this chapter

A famous experiment detected 527 muons per hour at the top of Mt. Washington, New Hampshire, elevation \(1910 \mathrm{~m}\). At sea level. the same equipment detected 395 muons per hour. A discriminator selected for muons whose speed was between \(0.9950 c\) and \(0.9954 c\). Given that the mean lifetime \(\tau\) of a muons in a frame in which it is at rest is \(2.2 \mu \mathrm{s}\) and that in this frame the number of moons decays exponentially with time accounting \(\operatorname{to} N=N_{0} e^{-d r},\) show that the results obtained in the experiment are sensible.

From a standstill, you begin jogging at \(5 \mathrm{~m} / \mathrm{s}\) directly toward the galaxy Centaurus A. which is on the horizon \(2 \times 10^{23} \mathrm{~m}\) away, (a) There is a clock in Centaurus A. According to you, how will readings on this clock differ before and after you begin jogging? (Remember: You change frames.) (b) The planet Neptune is between Earth and Centaurus A. \(4.5 \times 10^{9} \mathrm{~m}\) from Earth. How much would readings on a clock there differ? (c) What would be the time differences if you had instead begun jogging in the opposite direction? (d) What do these results tell you about the observations of a traveling twin who accelerates toward his Earth-bound twin? How do these observations depend on the distance between the twins?

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?

You are strapped into a rear-facing seat at the middle of a long bus accelerating from rest at about \(10 \mathrm{~m} / \mathrm{s}^{2}\) (a rather violent acceleration for a bus). As the back of the bus passes a warning sign alongside the street, a red light of precisely 650 nm wavelength on the sign tums on. Do you see this precise 650 nm wavelength? Does your friend sitting at the front of the bus see the wavelength you see? How could the same observations be produced with the bus and sign stationary?

You are gliding over Earth's surface at a high speed. carrying your high- precision clock. At points \(X\) and \(Y\) on the ground are similar clocks, synchronized in the ground frame of reference. As you pass over clock X, it and your clock both read 0 . (a) According to you, do clocks \(X\) and \(Y\) advance slower or faster than yours? (b) When you pass over clock \(Y\), does it read the same time, an earlier time or a later time than yours? (Make sure your answer aguces with what ground observers should see.) (c) Reconcile any seeming contradictions berween your answers to parts \((a)\) and \((b)\).
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