Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 km. (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth? (b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

Short Answer

Expert verified
(a) 1.508 × 10^-4 s; (b) 4307 m; (c) 1.574 × 10^-3 s, results agree.

Step by step solution

01

Calculate time as measured by the scientist

To find the time taken for the particle to travel 45.0 km at speed 0.99540c, use the formula \( t = \frac{d}{v} \), where \( d = 45000 \text{ m} \) and \( v = 0.99540c \). First calculate the velocity: \[ v = 0.99540 \times 3 \times 10^8 \text{ m/s} = 2.9862 \times 10^8 \text{ m/s} \]Next, use the formula to calculate time:\[ t = \frac{45000}{2.9862 \times 10^8} = 1.508 \times 10^{-4} \text{ s} \]
02

Calculate contracted distance in particle's frame

Using the length-contraction formula \( L = L_0 \sqrt{1 - v^2/c^2} \), where \( L_0 = 45000 \text{ m} \) and \( v = 0.99540c \), find \( L \):\[ L = 45000 \times \sqrt{1 - (0.99540)^2} \]Calculate the contraction factor:\[ \sqrt{1 - (0.99540)^2} = \sqrt{1 - 0.99082} = \sqrt{0.00918} = 0.0958 \]So, contracted distance:\[ L = 45000 \times 0.0958 = 4307 \text{ m} \]
03

Calculate time in particle's frame using time dilation

Using the time dilation formula \( t' = \frac{t}{\sqrt{1 - v^2/c^2}} \), substitute \( t = 1.508 \times 10^{-4} \text{ s} \) and solve for \( t' \):\[ t' = \frac{1.508 \times 10^{-4}}{0.0958} = 1.574 \times 10^{-3} \text{ s} \]
04

Calculate time in particle's frame using distance

For the particle's frame, use \( t' = \frac{L}{v} \) with \( L = 4307 \text{ m} \) and \( v = 0.99540c = 2.9862 \times 10^8 \text{ m/s} \):\[ t' = \frac{4307}{2.9862 \times 10^8} = 1.574 \times 10^{-3} \text{ s} \]
05

Compare results

Both methods to find time in the particle's frame give \( t' = 1.574 \times 10^{-3} \text{ s} \). The results agree, confirming the calculations.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Time Dilation
Time dilation is a fascinating concept in physics that comes from Einstein's theory of special relativity. It tells us that time passes differently for two observers who are moving relative to each other. To understand this, imagine a scientist and a fast-moving particle.
For the scientist, who is stationary on Earth, time ticks at a normal pace. However, for the particle whizzing toward Earth at nearly the speed of light (0.99540c, where c is the speed of light), time slows down. This slowing of time means that the particle, in its own "frame of reference," experiences less time passing compared to the scientist who is observing from a stationary position.
To calculate the time experienced by the particle, we use the time dilation formula \( t' = \frac{t}{\sqrt{1 - v^2/c^2}} \). Here, \( t \) is the time measured by the observer (the scientist), and \( t' \) is the dilated time experienced by the particle. Hence, even though the scientist measures the journey to take \( 1.508 \times 10^{-4} \) seconds, the particle experiences a longer time: \( 1.574 \times 10^{-3} \) seconds.
This example shows how high velocities can significantly affect time perception, which is only observable at speeds close to the speed of light. It’s exciting to think about how flexible time becomes in such extreme conditions.
Length Contraction
Length contraction is another intriguing result of special relativity. It explains how the length of an object is perceived to be shorter when it is moving at high speeds relative to an observer. Imagine the path the particle takes from the atmosphere to the Earth's surface.
From the scientist's perspective on Earth, the distance is a standard \( 45.0 \) kilometers or \( 45000 \) meters. But for the particle moving at the speed of \( 0.99540c \), the distance appears shorter. According to the formula \( L = L_0 \sqrt{1 - v^2/c^2} \), the length \( L \) becomes \( 4307 \) meters in the particle's frame.
This occurs because, at such high speeds, the dimensions along the direction of motion contract. This effect doesn’t affect motions or objects we encounter in our daily lives but becomes significant when dealing with particles or objects moving at speeds close to that of light.
Length contraction is a key piece of understanding how interstellar travel or high-speed journeys would alter human perceptions and measurements, a captivating point for future explorations.
Special Relativity
Special relativity is a theory introduced by Albert Einstein in 1905. It reshaped our understanding of space and time. One of its core ideas is that the laws of physics are the same for all non-accelerated observers, meaning there are no "special" or "preferred" reference frames.
One of the most famous outcomes of special relativity is the equation \( E=mc^2 \), linking mass (m) and energy (E). However, the theory also leads to phenomena like time dilation and length contraction, as discussed earlier. These phenomena reveal how different observers, moving relative to one another, can perceive time and space in contrasting ways.
Special relativity becomes especially crucial when understanding cosmic rays or particles traveling at light-like speeds because it gives us the tools to calculate how time and space change in such extraordinary conditions. Its implications for high-speed travel, particle physics, and even the fabric of the universe continue to be areas of active research and wonder.
Whether it's calculating times or distances in unique scenarios, special relativity helps us comprehend the universe's intricate dance as particles flit across it, governed by these timeless laws.
Cosmic Rays
Cosmic rays are high-energy particles that travel through space and sometimes enter Earth's atmosphere. They can be electrons, protons, and other atomic nuclei. When these cosmic rays strike the Earth's atmosphere, they often produce a shower of secondary particles.
These particles collide with atmospheric atoms and create more particles, a phenomenon scientists observe at high altitudes. In the provided problem, the creation of an unstable particle from a cosmic ray in the upper atmosphere is being studied.
The particle's high speed (0.99540c) as it descends toward Earth brings special relativity into play. Time dilation and length contraction reflect the incredible speeds involved. Studying cosmic rays isn't just an exercise in theory but an exploration of nature’s powerful, energetic phenomena that might give us clues about the universe's origins.
Cosmic rays offer a fascinating avenue for both experimental and theoretical physics, allowing scientists to test theories like special relativity under extreme conditions and providing insights into high-energy processes far beyond Earth’s boundaries.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An airplane has a length of 60 m when measured at rest. When the airplane is moving at 180 m/s (400 mph) in the alternate universe, how long would the plane appear to be to a stationary observer? (a) 24 m; (b) 36 m; (c) 48 m; (d) 60 m; (e) 75 m.

A nuclear bomb containing 12.0 kg of plutonium explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in 10\(^4\). (a) How much energy is released in the explosion? (b) If the explosion takes place in 4.00 \(\mu\)s, what is the average power developed by the bomb? (c) What mass of water could the released energy lift to a height of 1.00 km ?

If a muon is traveling at 0.999c, what are its momentum and kinetic energy? (The mass of such a muon at rest in the laboratory is 207 times the electron mass.)

The positive muon (\(\mu^+\)), an unstable particle, lives on average 2.20 \(\times\) 10\(^{-6}\) 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 0.900c, what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

After being produced in a collision between elementary particles, a positive pion (\(\pi^+\)) must travel down a 1.90-km-long tube to reach an experimental area. A \(\pi^+\) particle has an average lifetime (measured in its rest frame) of 2.60 \(\times\) 10\(^{-8}\) s; the \(\pi^+\) we are considering has this lifetime. (a) How fast must the \(\pi^+\) travel if it is not to decay before it reaches the end of the tube? (Since \(u\) will be very close to \(c\), write \(u\) = (1 - \(\Delta\))c and give your answer in terms of \(\Delta\) rather than \(u\).) (b) The \(\pi^+\) has a rest energy of 139.6 MeV. What is the total energy of the \(\pi^+\) at the speed calculated in part (a)?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free