/*! 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 34 A particle moves with speed \(v^... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle moves with speed \(v^{\prime}\) in the \(x^{\prime} y^{\prime}\) plane of the rocket frame and in a direction that makes an angle \(\phi^{\prime}\) with the \(x^{\prime}\) axis. Find the angle \(\phi\) that the velocity vector of this particle makes with the \(x\) axis of the laboratory frame. (Hint: Transform space and time displacements rather than velocities.)

Short Answer

Expert verified
\[ \phi = \tan^{-1} \left( \frac{ v^{\prime} \sin \phi^{\prime} \sqrt{1 - \frac{u^2}{c^2}}}{v^{\prime} \cos \phi^{\prime} + u} \right) \]

Step by step solution

01

Understand the Problem

Determine the angle \(\phi\) that the velocity vector of a particle in the rocket frame makes with the \(x\)-axis of the laboratory frame. Use transformations between the rocket frame and the laboratory frame.
02

Define Coordinates and Angles

The particle's speed in the rocket frame is denoted as \(v^{\prime}\), and it makes an angle \(\phi^{\prime}\) with the \(x^{\prime}\)-axis in this frame.
03

Transform Space and Time Displacements

Use the Lorentz transformation equations to relate the space and time coordinates from the rocket frame to the laboratory frame.The transformations are:\[ x = \gamma \left( x^{\prime} + ut^{\prime} \right) \]\[ t = \gamma \left( t^{\prime} + \frac{u x^{\prime}}{c^2} \right) \] where \(u\) is the velocity of the rocket relative to the laboratory frame and \(\gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \) is the Lorentz factor.
04

Relate Velocities Between Frames

The velocity components of the particle in the rocket frame are: \( v_{x^{\prime}} = v^{\prime} \cos \phi^{\prime} \) and \( v_{y^{\prime}} = v^{\prime} \sin \phi^{\prime} \). Using the Lorentz velocity transformation equations, calculate the velocity components in the laboratory frame: \[ v_{x} = \frac{v_{x^{\prime}} + u}{1 + \frac{uv_{x^{\prime}}}{c^2}} \] \[ v_{y} = \frac{v_{y^{\prime}} \sqrt{1 - \frac{u^2}{c^2}}}{1 + \frac{uv_{x^{\prime}}}{c^2}} \]
05

Find the Angle \(\phi\) in the Laboratory Frame

Determine the angle \(\phi\) the velocity vector makes with the \(x\)-axis in the laboratory frame using the transformed velocity components: \[ \tan \phi = \frac{v_y}{v_x} = \frac{ v^{\prime} \sin \phi^{\prime} \sqrt{1 - \frac{u^2}{c^2}}}{v^{\prime} \cos \phi^{\prime} + u} \]Thus, \[ \phi = \tan^{-1} \left( \frac{ v^{\prime} \sin \phi^{\prime} \sqrt{1 - \frac{u^2}{c^2}}}{v^{\prime} \cos \phi^{\prime} + u} \right) \]

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Key Concepts

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

Velocity Transformation
In special relativity, velocities are not transformed in a straightforward way between different reference frames due to the effects of time dilation and length contraction. To find the velocity of an object in one frame given its velocity in another, we use the Lorentz velocity transformation equations. These equations account for the fact that time and space are interconnected and change depending on the relative motion of observers.

In our scenario, the particle's velocity components in the rocket frame, denoted as \(v_{x'}\) and \(v_{y'}\), are given in terms of the particle's speed \(v'\) and the angle \(\fph'\) it makes with the \(x'\)-axis. The equations for transforming these components to the laboratory frame are:
  • \( v_x = \frac{v_{x'} + u}{1 + \frac{u v_{x'}}{c^2}} \)
  • \( v_y = \frac{v_{y'} \text{ sqrt(1 - \frac{u^2}{c^2})}}{1 + \frac{u v_{x'}}{c^2}} \)
Here, \(u\) is the relative velocity between the two frames, and \(c\) is the speed of light. Using these, we can determine the new velocity components in the laboratory frame, which are essential for finding the angle \(\fph\) in this frame.
Special Relativity
Special relativity, formulated by Albert Einstein, revolutionized our understanding of space, time, and motion. It is based on two key principles: the laws of physics are the same for all observers, and the speed of light in a vacuum is constant for all observers, regardless of their relative motion. This leads to several non-intuitive effects such as time dilation, length contraction, and the relativity of simultaneity.

In our problem, special relativity tells us that space and time coordinates in one frame are related to those in another by the Lorentz transformations. These transformations allow us to convert between different reference frames and understand how objects move and interact at high velocities, close to the speed of light.
Reference Frames
A reference frame is essentially a perspective from which an observer measures and records the physical phenomena around them. In classical mechanics, we usually deal with inertial frames, which are either stationary or moving at constant velocity. In special relativity, reference frames moving at high velocities relative to each other are common subjects of study.

In our exercise, we have two primary frames: the rocket frame (denoted with primes, e.g., \(x'\), \(y'\), \(t'\)) and the laboratory frame (denoted without primes). Transforming quantities such as position, time, and velocity between these frames requires us to account for relativistic effects, which are described by the Lorentz transformations.

Understanding how to switch between these frames is crucial for analyzing the motion of particles and applying the principles of special relativity correctly.
Relativistic Mechanics
Relativistic mechanics extends the principles of classical mechanics to scenarios where objects are moving at significant fractions of the speed of light. It incorporates the corrections predicted by special relativity, ensuring that the laws of physics remain consistent at all speeds.

One key aspect of relativistic mechanics is how velocities combine when shifting between frames of reference. Simple addition is insufficient; instead, we use the Lorentz velocity transformation equations. These equations ensure that the resulting velocities do not exceed the speed of light and reflect the nature of space and time in relativistic contexts.

In the given problem, we've applied these principles to transform the velocity components of a particle from one frame to another and then determine the angle of its motion. Understanding these transformations is essential for anyone studying objects moving at high velocities relative to different observers.

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Most popular questions from this chapter

You wish to make \(a\) round trip from Earth in a spaceship, traveling at constant speed in a straight line for 6 months on your watch and then returning at the same constant speed. You wish, further, to find Earth to be 1000 years older on your return. (a) What is the value of your constant speed with respect to Earth? (b) How much do you age during the trip? (c) Does it matter whether or not you travel in a straight line? For example, could you travel in a huge circle that loops back to Earth?

An unpowered spaceship whose rest length is 350 meters has a speed \(0.82 c\) with respect to Earth. A micrometeorite, also with speed of \(0.82 c\) with respect to Earth, passes the spaceship on an antiparallel track that is moving in the opposite direction. How long does it take the micrometeorite to pass the spaceship as measured on the ship?

Review Problem 40 , in which we concluded that a limo of proper length \(30 \mathrm{~m}\) can fit into a garage of proper length \(6 \mathrm{~m}\) with room to spare. This result is possible because the speeding limo is observed by Garageman to be Lorentz -contracted. Carman protests that in the rest frame of the limo (in which the limo is its full proper length) it is the garage that is Lorentz-contracted. As a result, he claims, there is no possibility whatever that the limo can fit into the garage. What could be the possible basis for resolving this paradox? (Hint: Think about the space and time locations of two events: event A, front garage door closes and event \(\mathrm{B}\), rear garage door opens.)

One kilogram of hydrogen combines chemically with 8 kilograms of oxygen to form water; about \(10^{8} \mathrm{~J}\) of energy is released. Ten metric tons \(\left(10^{4} \mathrm{~kg}\right)\) of hydrogen combines with oxygen to produce water. (a) Does the resulting water have a greater or less mass than the original hydrogen plus oxygen? (b) What is the numerical magnitude of this difference in mass? (c) A smaller amount of hydrogen and oxygen is weighed, then combined to form water, which is weighed again. A very good chemical balance is able to detect a fractional change in mass of 1 part in \(10^{8}\). By what factor is this sensitivity more than enough -or insufficient - to detect the fractional change in mass in this reaction?

A gamma ray (an energetic photon) falls on a nucleus of initial mass \(m\), initially at rest. The energy \(E_{\mathrm{p}}\) of the incoming gamma ray matches the energy separation between the lowest energy of the nucleus and its first excited state, so the incident photon is absorbed. We want to know the mass \(m^{*}\) of the excited nucleus. (see Fig. \(38-13 .\) ) (a) Show that the conservation of energy and momentum equations are, in an obvious notation: and $$ \begin{array}{c} E_{\mathrm{p}}+m c^{2}=E_{m^{*}} \\ \frac{E_{\mathrm{p}}}{c}=p_{m^{*}}=\frac{\left(E_{m^{*}}^{2}-m^{* 2} c^{4}\right)^{1 / 2}}{c} . \end{array} $$ (b) Combine the two conservation equations to find an expression for \(m^{*}\) as a function of \(E_{\mathrm{p}}, m\), and \(c\). (c) Show that for very small values of \(E_{\mathrm{p}}\) the limiting result is \(m^{*}=m .\) Explain why this limiting result is reasonable.

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