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Chapter 35: Problem 28

If a muon is moving at \(90.0 \%\) of the speed of light, how does its measured lifetime compare to when it is in the rest frame of a laboratory, where its lifetime is \(2.2 \cdot 10^{-6}\) s?

Short Answer

Expert verified
Answer: When moving at 90% the speed of light, the measured lifetime of a muon is approximately 5.046 microseconds, which is more than 2 times longer than its lifetime in the rest frame (2.2 microseconds).

Step by step solution

01

Identify the time dilation formula

The time dilation formula, as per special relativity, is given by: \(T = T_{0} \cdot \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\), where T is the dilated time (measured time), \(T_{0}\) is the proper time (time in the rest frame), v is the relative velocity of the moving frame, and c is the speed of light.
02

Determine the moving frame's relative velocity

The muon is moving at 90% of the speed of light. To find its relative velocity v, we multiply the speed of light (c) by 0.9 (90% in decimal form): \(v = 0.9c\).
03

Substitute values in the time dilation formula

We have the proper time \(T_{0} = 2.2 \cdot 10^{-6}\) s, and the relative velocity \(v = 0.9c\). Substitute these values into the equation: \(T = (2.2 \cdot 10^{-6}) \cdot \frac{1}{\sqrt{1-\frac{(0.9c)^2}{c^2}}}\).
04

Simplify the equation

Now we need to simplify the expression in the equation: \(T = (2.2 \cdot 10^{-6}) \cdot \frac{1}{\sqrt{1-\frac{(0.81c^2)}{c^2}}}\).
05

Cancel the terms and calculate the time dilation factor

The \(c^2\) in the numerator and the denominator cancel each other out: \(T = (2.2 \cdot 10^{-6}) \cdot \frac{1}{\sqrt{1-0.81}}\). Calculate the time dilation factor: \(\frac{1}{\sqrt{1-0.81}} = \frac{1}{\sqrt{0.19}} \approx 2.294\).
06

Determine the measured lifetime of the moving muon

Multiply the proper time by the time dilation factor: \(T \approx (2.2 \cdot 10^{-6}) \cdot 2.294 \approx 5.046 \cdot 10^{-6}\) s. So, the measured lifetime of the muon when it is moving at 90% the speed of light is approximately 5.046 microseconds.
07

Compare the measured lifetime to the lifetime in the rest frame

The lifetime in the rest frame of the laboratory is 2.2 microseconds, while the measured lifetime when it is moving at 90% of the speed of light is approximately 5.046 microseconds. This means that the muon's lifetime appears to be more than 2 times longer when moving at a high speed compared to its lifetime in the rest frame.

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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, first proposed by Albert Einstein in 1905. It revolutionized the way we understand space, time, and motion. At its core, special relativity reveals how the laws of physics are the same for all observers, regardless of their relative velocity. One of the crucial postulates is that the speed of light is constant for all observers, no matter how fast they move relative to each other. This leads to some fascinating outcomes, including time dilation.

Time dilation is the concept that time can pass at different rates for different observers, depending on their relative velocity. When an object moves close to the speed of light, time appears to pass more slowly for it compared to a stationary observer. This effect has been confirmed by many experiments, making it a key prediction of special relativity. Understanding special relativity allows us to explore peculiar phenomena such as how fast-moving particles behave and is critical for interpreting results in high-energy physics and cosmology. By diving deep into special relativity, we gain insight into the relationship between time, space, and motion.
Muon Lifetime
Muons are elementary particles similar to electrons, but they are around 200 times heavier, and they have a short lifespan. In the context of special relativity, muons provide compelling evidence for time dilation through their observed lifetimes. Normally, muons have a lifetime of about 2.2 microseconds when at rest. However, when they move at speeds close to that of light, such as 90% of the speed of light, their lifetimes as observed from a stationary reference frame鈥攍ike a lab鈥攊ncrease significantly.

This change is a direct result of time dilation as predicted by special relativity. For example, as demonstrated in the solution to the exercise, a muon moving at 90% the speed of light will have a measured lifetime of about 5.046 microseconds in the laboratory rest frame. This is because time effectively 'stretches' for the muon when it moves at such high speeds. The study of muons and their behavior in motion not only supports the theory of special relativity but also helps physicists probe fundamental questions about the universe and the nature of particles.
Speed of Light
The speed of light, denoted by the symbol \(c\), is one of the most important constants in the universe. It is approximately equal to \(299,792,458\) meters per second. This constant represents the ultimate speed limit in the universe, according to the theory of special relativity. No information or matter can travel faster than the speed of light. This limitation has profound implications for our understanding of space, time, and the universe at large.

The constancy of the speed of light ensures that it remains the same for all observers, regardless of their motion or the motion of the light source. This invariance is a cornerstone of special relativity and leads to phenomena like time dilation and length contraction, allowing us to better comprehend the universe at both very high speeds and in strong gravitational fields. Light's speed defines the maximum rate at which information can be transmitted, affecting fields as varied as telecommunications, astrophysics, and space exploration. Understanding the speed of light helps us grasp not only practical applications but also the philosophical implications of our place in the space-time continuum.

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Most popular questions from this chapter

Show that momentum and energy transform from one inertial frame to another as \(p_{x}^{\prime}=\gamma\left(p_{x}-v E / c^{2}\right) ; p_{y}^{\prime}=p_{y}\) \(p_{z}^{\prime}=p_{p} ; E^{\prime}=\gamma\left(E-v p_{x}\right) .\) Hint: Look at the derivation for the space-time Lorentz transformation.

An astronaut in a spaceship flying toward Earth's Equator at half the speed of light observes Earth to be an oblong solid, wider and taller than it appears deep, rotating around its long axis. A second astronaut flying toward Earth's North Pole at half the speed of light observes Earth to be a similar shape but rotating about its short axis. Why does this not present a contradiction?

In some proton accelerators, proton beams are directed toward each other for head-on collisions. Suppose that in such an accelerator, protons move with a speed relative to the lab of \(0.9972 c\). a) Calculate the speed of approach of one proton with respect to another one with which it is about to collide head on. Express your answer as a multiple of \(c\), using six significant digits. b) What is the kinetic energy of each proton beam (in units of \(\mathrm{MeV}\) ) in the laboratory reference frame? c) What is the kinetic energy of one of the colliding protons (in units of \(\mathrm{MeV}\) ) in the rest frame of the other proton?

Although it deals with inertial reference frames, the special theory of relativity describes accelerating objects without difficulty. Of course, uniform acceleration no longer means \(d v / d t=g,\) where \(g\) is a constant, since that would have \(v\) exceeding \(c\) in a finite time. Rather, it means that the acceleration experienced by the moving body is constant: In each increment of the body's own proper time \(d \tau,\) the body acquires velocity increment \(d v=g d \tau\) as measured in the inertial frame in which the body is momentarily at rest. (As it accelerates, the body encounters a sequence of such frames, each moving with respect to the others.) Given this interpretation: a) Write a differential equation for the velocity \(v\) of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the "ground frame"). You can simplify your equation, remembering that squares and higher powers of differentials can be neglected. b) Solve this equation for \(v(t),\) where both \(v\) and \(t\) are measured in the ground frame. c) Verify that your solution behaves appropriately for small and large values of \(t\). d) Calculate the position of the body \(x(t),\) as measured in the ground frame. For convenience, assume that the body is at rest at ground-frame time \(t=0,\) at ground-frame position \(x=c^{2} / g\) e) Identify the trajectory of the body on a space-time diagram (Minkowski diagram, for Hermann Minkowski) with coordinates \(x\) and \(c t,\) as measured in the ground frame. f) For \(g=9.81 \mathrm{~m} / \mathrm{s}^{2},\) calculate how much time it takes the body to accelerate from rest to \(70.7 \%\) of \(c,\) measured in the ground frame, and how much ground-frame distance the body covers in this time.

Robert, standing at the rear end of a railroad car of length \(100 . \mathrm{m},\) shoots an arrow toward the front end of the car. He measures the velocity of the arrow as \(0.300 c\). Jenny, who was standing on the platform, saw all of this as the train passed her with a velocity of \(0.750 c .\) Determine the following as observed by Jenny: a) the length of the car b) the velocity of the arrow c) the time taken by arrow to cover the length of the car d) the distance covered by the arrow
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