/*! 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 4 Which quantity is invariant-that... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 35: Problem 4

Which quantity is invariant-that is, has the same value-in all reference frames? a) time interval, \(\Delta t\) d) space-time interval, b) space interval, \(\Delta x\) \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\) c) velocity, \(v\)

Short Answer

Expert verified
Answer: b) space-time interval

Step by step solution

01

Definition of Lorentz Transformation

Lorentz transformation relates the space and time coordinates of an event as observed from two different inertial reference frames. One frame is moving at constant velocity, v, with respect to the other. The formula for Lorentz transformation of space and time coordinates is given by: \(x' = \gamma (x - vt)\) \(t' = \gamma (t - \frac{v}{c^2}x)\) where \(x'\) and \(t'\) are the space and time coordinates in the moving frame, \(x\) and \(t\) are the coordinates in the stationary frame, and \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) is the Lorentz factor.
02

Time Interval

In different reference frames, the time interval between two events can have different values due to time dilation. It is not invariant under Lorentz transformation. \(\Delta t' = \gamma \Delta t\)
03

Space Interval

Similarly, space interval (distance) between two events can have different values in different reference frames due to length contraction. It is not invariant in different reference frames. \(\Delta x' = \gamma \Delta x\)
04

Velocity

The velocity of an object is different in different reference frames due to Lorentz transformation, it is not invariant. Transforming velocities from one frame to another require the Velocity-Addition Formula, a consequence of Lorentz transformation.
05

Space-time Interval

The space-time interval is given by the formula: \(s^2 = c^2 \Delta t^2 - \Delta x^2\) The space-time interval between two events is an invariant quantity under Lorentz transformation, which means it has the same value in all inertial reference frames. \(s'^2 = c^2 \Delta t'^2 - \Delta x'^2 = s^2\) In conclusion, the space-time interval (option d) is the quantity that is invariant or having the same value in all reference frames.

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!

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 square of area \(100 \mathrm{~m}^{2}\) that is at rest in the reference frame is moving with a speed \((\sqrt{3} / 2) c\). Which of the following statements is incorrect? a) \(\beta=\sqrt{3} / 2\) b) \(\gamma=2\) c) To an observer at rest, it looks like another square with an area less than \(100 \mathrm{~m}^{2}\) d) The length along the moving direction is contracted by a factor of \(\frac{1}{2}\)

Consider motion in one spatial dimension. For any velocity \(v,\) define parameter \(\theta\) via the relation \(v=c \tanh \theta\) where \(c\) is the vacuum speed of light. This quantity is variously called the velocity parameter or the rapidity corresponding to velocity \(v\). a) Prove that for two velocities, which add according to the Lorentzian rule, the corresponding velocity parameters simply add algebraically, that is, like Galilean velocities. b) Consider two reference frames in motion at speed \(v\) in the \(x\) -direction relative to one another, with axes parallel and origins coinciding when clocks at the origin in both frames read zero. Write the Lorentz transformation between the two coordinate systems entirely in terms of the velocity parameter corresponding to \(v\), and the coordinates.

A particle of rest mass \(m_{0}\) travels at a speed \(v=0.20 c\) How fast must the particle travel in order for its momentum to increase to twice its original momentum? a) \(0.40 c\) c) \(0.38 c\) e) \(0.99 c\) b) \(0.10 c\) d) \(0.42 c\)

Consider two clocks carried by observers in a reference frame moving at speed \(v\) in the positive \(x\) -direction relative to ours. Assume that the two reference frames have parallel axes, and that their origins coincide when clocks at that point in both frames read zero. Suppose the clocks are separated by distance \(l\) in the \(x^{\prime}-\) direction in their own reference frame; for instance, \(x^{\prime}=0\) for one clock and \(x^{\prime}=I\) for the other, with \(y^{\prime}=z^{\prime}=0\) for both. Determine the readings \(t^{\prime}\) on both clocks as functions of the time coordinate \(t\) in our reference frame.

Show that \(E^{2}-p^{2} c^{2}=E^{2}-p^{2} c^{2},\) that is, that \(E^{2}-p^{2} c^{2}\) is a Lorentz invariant. Hint: Look at derivation showing that the space-time interval is a Lorentz invariant.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Measurements

Read Explanation

Quantum Physics

Read Explanation

Engineering Physics

Read Explanation

Thermodynamics

Read Explanation

Torque and Rotational Motion

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.