/*! 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 25 A rocket is traveling at speed \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rocket is traveling at speed \(V\) along the \(x\) axis of frame S. It emits a signal (for example, a pulse of light) that travels with speed \(c\) along the \(y^{\prime}\) axis of the rocket's rest frame \(\mathcal{S}^{\prime} .\) What is the speed of the signal as measured in \(\mathcal{S} ?\)

Short Answer

Expert verified
The speed of the signal in frame \(S\) is \(c\).

Step by step solution

01

Understand the Problem

A rocket is moving with velocity \(V\) in the \(x\) direction in frame \(S\). In its rest frame \(S'\), it emits a light signal moving along the \(y'\) axis at speed \(c\). We are tasked with finding the speed of this signal in the frame \(S\).
02

Apply Lorentz Transformation

To find how velocities transform between frames, we use Lorentz transformation. Since in \(S'\), the signal travels only in the \(y'\) direction, \(v'_x = 0\) and \(v'_y = c\). We use the transformation equation: \[v_y = \frac{v'_y}{\gamma(1 + \frac{Vv'_x}{c^2})}\]where \(\gamma = \frac{1}{\sqrt{1 - \frac{V^2}{c^2}}}\). Here, since \(v'_x = 0\), the equation simplifies to \(v_y = \frac{c}{\gamma}\).
03

Determine the Perpendicular Velocity Component in Frame S

Calculate the \(y\)-component of the signal's velocity in frame \(S\). \[v_y = \frac{c}{\gamma} = \frac{c}{\sqrt{1 - \frac{V^2}{c^2}}}\] This is the component of velocity perpendicular to the motion of the rocket in frame \(S\).
04

Calculate the Resultant Speed in Frame S

Using Pythagorean theorem, the resultant speed \(v\) in frame \(S\) is found by combining contributions from both \(x\) and \(y\) components:\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{V^2 + \left(\frac{c}{\sqrt{1 - \frac{V^2}{c^2}}}\right)^2}\] Since \(v_x = V\), substitute and simplify the expression.
05

Simplify the Expression for the Speed in S

Simplifying the resultant expression:\[v = \sqrt{V^2 + \left(\frac{c}{\sqrt{1 - \frac{V^2}{c^2}}}\right)^2}\]This simplifies to finding the speed \(v\) of the light signal in frame \(S\). After simplification, this maintains the speed of light as expected. Given light propagation invariance with respect to frame transformations, the speed remains \(c\).

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 91Ó°ÊÓ!

Key Concepts

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

Relativity
In the realm of physics, relativity fundamentally transforms our understanding of motion and speed. Albert Einstein’s theory of relativity illuminates how the laws of physics apply consistently across all inertial frames, regardless of their motion. This theory comprises two main parts: Special Relativity and General Relativity. Special Relativity deals with objects moving at constant speeds, particularly near the speed of light. It suggests that time and space are interconnected, meaning they are relative to the observer's frame of reference. This groundbreaking idea leads to some mind-bending implications, such as time dilation, where time appears to slow down for an object moving close to the speed of light.
One key component of Special Relativity is the constancy of the speed of light. No matter the observer’s velocity, the speed of light in a vacuum remains constant. This principle serves as a foundation for understanding how light and other phenomena behave when transitioning between different frames of reference.
  • Objects appear different based on the observer's motion.
  • Time and space can contract or expand from different viewpoints.
  • The speed of light remains unchanged, no matter the motion of the source or observer.
Velocity Transformation
Velocity transformation is a crucial concept within the study of relativity, particularly when considering different frames of reference. When a signal or particle moves through space, determining its velocity from another frame requires using the Lorentz transformation equations. These equations help transition measurements from one frame to another moving at a constant velocity. In the given exercise, the rocket emits a signal in its own rest frame. However, to determine the signal’s speed in another frame, these transformations are necessary.
The Lorentz transformation accounts for the effects of both time dilation and length contraction. The correct formula, for transforming the velocity of a particle observed in one reference frame to another, is:
\[v_y = \frac{v'_y}{\gamma(1 + \frac{Vv'_x}{c^2})}\] Here,
  • \(v'_x\) represents the velocity of the signal in the x-direction as observed in its rest frame.
  • \(v'_y\) is the velocity in the y-direction in the rest frame of the object emitting it.
  • \(\gamma\) is the Lorentz factor, defined as \(\gamma = \frac{1}{\sqrt{1 - \frac{V^2}{c^2}}}\).
This transformation allows for a clear understanding of how velocities change between different frames, ensuring that the laws of physics are upheld across varied perspectives.
Frame of Reference
A frame of reference in physics describes a perspective or viewpoint from which measurements are made. It is essential to understand this concept when analyzing the motion of objects, particularly in relativistic contexts where speeds are significant fractions of light speed. In this exercise, two frames of reference are considered: the rocket's rest frame, denoted as \(\mathcal{S}'\), and the stationary frame \(S\) in which the rocket is observed to move.
Each frame provides a unique viewpoint. In \(\mathcal{S}'\), the rocket is stationary, and the signal moves directly along the \(y'\)-axis at speed \(c\), the speed of light. From the observer’s frame \(S\), the rocket moves along the \(x\)-axis with a velocity \(V\).
  • Understanding different frames helps to navigate different perspectives.
  • Specific transformations are required to relay events observed in one frame to another.
  • Physics laws, including the constancy of the speed of light, remain valid in any frame.
Grasping these perspectives is key to resolving complex relativistic problems, ensuring that physical principles apply universally regardless of the observer's motion.

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

We have seen that the scalar product \(x \cdot x\) of any four-vector \(x\) with itself is invariant under Lorentz transformations. Use the invariance of \(x \cdot x\) to prove that the scalar product \(x \cdot y\) of any two four-vectors \(x\) and \(y\) is likewise invariant.

Prove that for any two matrices \(A\) and \(B\), where \(A\) has as many columns as \(B\) has rows, the transpose of \(A B\) satisfies \((A B)=\tilde{B} \tilde{A}\)

Prove that if \(T\) and \(a\) are respectively a four-tensor and a four-vector, then \(b=T \cdot a=T G a\) is a four-vector; that is, it transforms according to the rule \(b^{\prime}=\Lambda b\)

One way to create exotic heavy particles is to arrange a collision between two lighter particles $$a+b \rightarrow d+e+\cdots+g$$ where \(d\) is the heavy particle of interest and \(e, \cdots, g\) are other possible particles produced in the reaction. (A good example of such a process is the production of the \(\psi\) particle in the process \(\left.e^{+}+e^{-} \rightarrow \psi, \text { in which there are no other particles } e, \cdots, g .\right)\) (a) Assuming that \(m_{d}\) is much heavier that any of the other particles, show that the minimum (or threshold) energy to produce this reaction in the CM frame is \(E_{\mathrm{cm}} \approx m_{d} c^{2} .\) (b) Show that the threshold energy to produce the same reaction in the lab frame, where the particle \(b\) is initially at rest, is \(E_{\mathrm{lab}} \approx m_{d}^{2} c^{2} / 2 m_{b} .\) (c) Calculate these two energies for the process \(e^{+}+e^{-} \rightarrow \psi,\) with \(m_{e} \approx 0.5 \mathrm{MeV} / c^{2}\) and \(m_{\psi} \approx 3100 \mathrm{MeV} / c^{2} .\) Your answers should explain why particle physicists go to the trouble and expense of building colliding-beam experiments.

(a) What is a particle's speed if its kinetic energy \(T\) is equal to its rest energy? ( \(\mathbf{b}\) ) What if its energy \(E\) is equal to \(n\) times its rest energy?

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.