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Chapter 37: Problem 2

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c,\) what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

Short Answer

Expert verified
(a) The average lifetime is approximately 5.05 碌s. (b) The muon travels about 1364 meters.

Step by step solution

01

Understanding Time Dilation

In the lab frame, the moving muon experiences time dilation, so its lifetime is longer than in its rest frame. The formula for time dilation is given by: \[ t' = \frac{t_0}{\sqrt{1 - v^2/c^2}} \] where \( t_0 \) is the proper time, \( v = 0.900c \) is the speed, and \( c \) is the speed of light.
02

Calculate Time Dilation

Given \( t_0 = 2.20 \times 10^{-6} \) s and \( v = 0.900c \), we can calculate the dilated lifetime in the lab frame:\[ t' = \frac{2.20 \times 10^{-6}}{\sqrt{1 - (0.900)^2}} \]Calculate the denominator: \[ \sqrt{1 - (0.900)^2} = \sqrt{1 - 0.81} = \sqrt{0.19} \approx 0.4359 \] Thus,\[ t' = \frac{2.20 \times 10^{-6}}{0.4359} \] \[ t' \approx 5.05 \times 10^{-6} \text{ s} \]
03

Calculate Average Distance

The average distance the muon travels in the lab frame before decaying is given by: \[ d = v \cdot t' \] We already calculated \( t' \) as \( 5.05 \times 10^{-6} \text{ s} \). Now we find \( d \):\[ d = 0.900c \cdot 5.05 \times 10^{-6} \]\[ c = 3 \times 10^8 \text{ m/s} \] \[ d = 0.900 \times 3 \times 10^8 \times 5.05 \times 10^{-6} \]\[ d \approx 1.364 \times 10^3 \text{ m} \]

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Key Concepts

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

Muon Decay
Muons are fascinating particles that are similar to electrons but much heavier. They belong to a family of particles called lepton. One big difference between a muon and an electron is that muons are unstable. This means they don't last very long before they transform or "decay" into other particles. Their average lifespan is measured as the time it takes for half of a large group of identical muons to decay.
In particle physics, understanding how long a particle lives is crucial. For muons, this lifetime is about 2.20 microseconds. This lifetime is the "proper time," measured in the particle's own frame of reference.
Using this decay property, scientists can learn more about the fundamental forces in nature and the properties of the particles involved. Studying muon decay also helps to verify various principles of physics, such as time dilation from special relativity.
Relativistic Effects
When particles like muons travel at high speeds, close to the speed of light, they exhibit relativistic effects. One such effect is known as time dilation. According to Einstein's theory of relativity, time does not behave the same way for objects moving at different speeds. Time appears to 'slow down' for an object moving close to the speed of light compared to an observer who is at rest.
This means that if a muon is moving at 90% the speed of light, it experiences more 'time' before decaying than if it were at rest. For observers on Earth, they would measure the muon's lifetime as longer than its proper time. This effect is calculated using the following formula:
\[ t' = \frac{t_0}{\sqrt{1 - v^2/c^2}} \]
where \( t' \) is the dilated time, \( t_0 \) is the proper time, and \( v \) is the muon's speed. As you can see, the closer \( v \) gets to the speed of light \( c \), the greater the dilation effect.
Proper Time
Proper time is an essential concept in relativity, referring to the time interval measured by a clock that is at rest relative to whatever event is being timed. For muons, this is the time that passes according to a clock moving along with the muon.
In relativistic physics, proper time is different from coordinate time, which is the time interval measured by an observer who is not at rest with respect to the event. The proper time for a muon is 2.20 microseconds, its lifespan when it is not moving relative to the clock measuring it.
Understanding proper time is crucial for calculating how time appears to pass differently from various observers' perspectives. It ensures we correctly apply time dilation when dealing with objects moving at significant fractions of the speed of light.
Particle Physics
Particle physics is a branch of physics that studies the fundamental constituents of matter and their interactions. It explores the tiniest bits of our universe, such as quarks and leptons, building blocks that make up protons, electrons, muons, and other particles.
In the realm of particle physics, muons play an important role. Despite their brief existence, they are key in experiments and research. Their interactions help physicists probe the mysteries of the universe. By studying particles like muons and their decay patterns, scientists can test and validate theoretical predictions in quantum mechanics and the standard model of particle physics.
This field involves complex theories, but its purpose is quite a simple one: to understand the most basic elements of nature and the forces acting upon them. By doing so, it lays the groundwork for further advancements in technology and fundamental science.

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Most popular questions from this chapter

In proton-antiproton annihilation a proton and an antiproton (a negatively charged proton) collide and disappear, producing electromagnetic radiation. If cach particle has a mass of \(1.67 \times 10^{-27} \mathrm{kg}\) and they are at rest just before the annihilation, find the total energy of the radiation. Give your answers in joules and in electron volts.

A spaceship moving at constant speed \(u\) relative to us broadcasts a radio signal at constant frequency \(f_{0}\) . As the spaceship approaches us, we receive a higher frequency \(f\) ; after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0}\) and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?(\text {Hint} \text { . In this case, successive wave crests }\) move the same distance to the observer and so they have the same transit time. Thus \(f\) equals \(1 / T .\) Use the time dilation formula to relate the periods in the stationary and moving frames) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758 c relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?(\mathrm{c})\) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2\(\mu s\) . They are produced when cosmic rays bombard the upper atmosphere about 10 \(\mathrm{km}\) above the earth's surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth's surface. (a) What is the greatest distance a muon could travel during its 2.2 -\mus lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the \(2.2-\mu\) s lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of 0.999 \(\mathrm{c}\) . What is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic rays? (c) From the point of view of the muon, it still lives for only 2.2 , \(\mu\) s, so how does it make it to the ground? What is the thickness of the 10 \(\mathrm{km}\) of atmosphere through which the muon must travel, as measured by the muon? It is now clear how the muon is able to reach the ground?

Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light-years away. (A light-year is the distance that light travels in a year.) You plan to travel at constant speed in a \(1000-\mathrm{kg}\) rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in joules, and the energy needed as a percentage of U.S. yearly use (which is 1.0 \(\times 10^{19} \mathrm{J} ) .\) For comparison, arrange your results in a table showing \(v_{\text { rocket }},\) \(t_{\text { earth }},\) \(\mathbf{t}_{\text { rouket }} \boldsymbol{E}\) \((\text { in } \mathrm{J}),\) and \(E\) (as \(\%\) of U.S. use). The rocket ship's speed is (a) \(0.50 \mathrm{c} ;\) (b) 0.99 \(\mathrm{c}\) (c) \(0.9999 \mathrm{c} .\) On the basis of your results, does it seem likely that any government will invest in such high-speed space travel any time soon?

A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. (A light-year is the distance light travels in a year.) The probe travels with a speed of 0.9910 \(\mathrm{c}\) . An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella?
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