/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 An observer in frame \(S^{\prime... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 37: Problem 15

An observer in frame \(S^{\prime}\) is moving to the right \((+x-\text { direction })\) at speed \(u=0.600 \mathrm{c}\) away from a stationary observer in frame \(S\) . The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in S measure for the particle if \((a) v^{\prime}=0.400 c ;(b) v^{\prime}=0.900 c\) (c) \(v^{\prime}=0.990 c ?\)

Short Answer

Expert verified
(a) \(v \approx 0.806c\), (b) \(v \approx 0.974c\), (c) \(v \approx 0.997c\).

Step by step solution

01

Identify the Relativistic Velocity Transformation Formula

In special relativity, to find the velocity of a particle in another frame, we use the relativistic velocity transformation formula: \[ v = \frac{v' + u}{1 + \frac{v'u}{c^2}} \] where \(v\) is the velocity as measured by the stationary observer in frame \(S\), \(v'\) is the velocity in the moving frame \(S'\), and \(u\) is the velocity of the moving frame \(S'\) relative to \(S\). \(c\) is the speed of light.
02

Solve for (a) \(v' = 0.400c\)

Substitute \(v' = 0.400c\) and \(u = 0.600c\) into the formula: \[ v = \frac{0.400c + 0.600c}{1 + \frac{0.400c \times 0.600c}{c^2}} \] Simplify the numerator: \(0.400c + 0.600c = 1.000c\). Simplify the denominator: \(1 + \frac{0.240c^2}{c^2} = 1.240\). Thus, \[ v = \frac{1.000c}{1.240} \approx 0.806c \].
03

Solve for (b) \(v' = 0.900c\)

Substitute \(v' = 0.900c\) and \(u = 0.600c\) into the formula: \[ v = \frac{0.900c + 0.600c}{1 + \frac{0.900c \times 0.600c}{c^2}} \] Simplify the numerator: \(0.900c + 0.600c = 1.500c\). Simplify the denominator: \(1 + \frac{0.540c^2}{c^2} = 1.540\). Thus, \[ v = \frac{1.500c}{1.540} \approx 0.974c \].
04

Solve for (c) \(v' = 0.990c\)

Substitute \(v' = 0.990c\) and \(u = 0.600c\) into the formula: \[ v = \frac{0.990c + 0.600c}{1 + \frac{0.990c \times 0.600c}{c^2}} \] Simplify the numerator: \(0.990c + 0.600c = 1.590c\). Simplify the denominator: \(1 + \frac{0.594c^2}{c^2} = 1.594\). Thus, \[ v = \frac{1.590c}{1.594} \approx 0.997c \].

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 91影视!

Key Concepts

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

Special Relativity
Albert Einstein introduced the theory of special relativity in 1905. It fundamentally changes how we understand time, space, and motion. At its core, special relativity states that the laws of physics are the same for all non-accelerating observers, no matter their speed relative to each other. A surprising result of this theory is that the speed of light is always constant 鈥� about 299,792,458 meters per second 鈥� no matter the observer's frame of reference.
  • The theory combines space and time into a single four-dimensional fabric called spacetime.
  • It challenges our classical notions, showing that time can slow down and lengths can contract for objects moving close to the speed of light.
  • Core principles include time dilation and length contraction, which become noticeable at high velocities.
This theory underpins modern physics, influencing fields from cosmology to quantum mechanics and even GPS technology. It's crucial to process these abstract concepts first to understand phenomena such as relativistic velocity transformation.
Frame of Reference
A frame of reference is essentially a viewpoint from which an observer measures and perceives physical quantities like position, velocity, and acceleration. In the context of special relativity, frames of reference become even more significant.
  • Inertial frames are those where an object either remains at rest or moves at a constant velocity unless acted upon by an external force.
  • Different observers in varying frames can measure the same event differently, particularly when dealing with high speeds close to the speed of light.
  • The concept is crucial for understanding relativistic effects because the measurements can change based on the observer's movement relative to what they are observing.
In the original exercise, the stationary observer holds the frame \(S\), and the moving observer holds the frame \(S'\). To find how a moving observer's perceived velocity compares to a stationary observer's, we use the relativistic velocity transformation formula, showing how these frames interact at high speeds.
Speed of Light
The speed of light, denoted by \(c\), is one of the most critical constants in physics. It represents the ultimate speed limit in the universe and is fundamental in the theory of special relativity. In the relativistic velocity transformation exercise, the speed of light plays a vital role.
  • The constancy of light's speed leads to many peculiar effects described by special relativity, such as time dilation and the relativity of simultaneity.
  • No information or physical object can travel faster than \(c\), providing a universal speed limit.
  • The speed of light ensures that the velocity equations, like the relativistic velocity transformation, do not result in speeds surpassing \(c\).
Understanding why \(c\) is constant requires abandoning the notion of absolute space and time, embracing instead that measurements of time and distance can change based on the observer's motion. This is pivotal for solving exercises involving high velocities, where seemingly strange outcomes result from these relativistic effects.

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

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)?

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.190 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 \(\mathrm{ms}\) . (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth expressed as a fraction of the speed of light \(c\) ?

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?

As you have seen, relativistic calculations usually involve the quantity \(\gamma .\) When \(\gamma\) is appreciably greater than \(1,\) we must use relativistic formulas instead of Newtonian ones. For what speed \(v\) (in terms of \(c )\) is the value of \(\gamma(\mathrm{a}) 1.0 \%\) greater than \(1 ;\) (b) 10\(\%\) greater than 1 ; (c) 100\(\%\) greater than 1\(?\)

(a) How fast must you be spproaching a red traffic light \((\lambda=675 \mathrm{nm})\) for it to appear yellow \((\lambda=575 \mathrm{nm}) ?\) Express your answer in terms of the speed of light. (b) If you used this as a reason not to get a ticket for running a red light, how much of a fine would you get for speeding? Assume that the fine is \(\$ 1.00\) for each kilometer per hour that your speed exceds the posted limit of 90 \(\mathrm{km} / \mathrm{h}\).
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Waves Physics

Read Explanation

Engineering Physics

Read Explanation

Electrostatics

Read Explanation

Electricity and Magnetism

Read Explanation

Electricity

Read Explanation
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.