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

The Lorentz transformation equations have \(x\) and \(t\) and \(x^{\prime}\) and \(t^{\prime}\). Why no \(v\) and \(v^{\prime} ?\)

Short Answer

Expert verified
The Lorentz transformations are designed to convert spacetime coordinates not velocities. The varied velocities v in the Lorentz equations represent the constant relative velocity between the two reference frames, not a quantity whose transformation is being considered. To transform velocities, one would need another set of formulas known as 'velocity transformations'.

Step by step solution

01

Understanding Lorentz Transformation Equations

The Lorentz Transformation Equations are given by: \[x' = \gamma (x - vt) \] and \[ t' = \gamma (t - vx/c^2) \] where \( x', t' \) are space and time coordinates in the moving frame, \( x, t \) are in the stationary frame, \(\gamma \) is the Lorentz factor \( \gamma = 1/ \sqrt{1 - v^2/c^2 } \), \( v \) is the relative velocity between the frames, and \( c \) is the speed of light.
02

The role of velocities in Lorentz Transformation

The relative velocity v between the frames is taken into account in the transformation equations. Therefore, \( v \) is in the transformation equations, but it's not transforming from one frame to another. It is the relative velocity between the two inertial frames. The variable \( v \) in the Lorentz transformations describes the constant, relative velocity between the two reference frames.
03

Reasoning for lack of transformed velocities

The Lorentz transformations are concerned with transforming coordinates of an event in spacetime, not the velocities of a moving object. This is because velocities can change, while the spacetime coordinates of an event, a specific point in space at a specific moment in time, won't. Furthermore, to transform velocities, you'd need to use velocity addition formulas, which are separate from the position and time transformation equations.

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.

Relative Velocity
When studying the Lorentz Transformation, we encounter the concept of "relative velocity." This refers to the speed at which two inertial frames are moving in relation to one another. In Lorentz transformations, this relative velocity is symbolized by \( v \) and plays a crucial role in transforming spacetime coordinates between these two reference frames.
This means that relative velocity helps us understand how different observers, each in their own reference frame, perceive the timing and positioning of an event.

In practice:
  • Imagine two spaceships traveling in space.
  • If one spaceship is moving at a velocity \( v \) relative to the other, the Lorentz equations help transform coordinates measured in one frame to those in the other.
It's important to note that the Lorentz transformation only considers one constant relative velocity for simplicity, assuming linear, uniform motion. That means \( v \) is consistent and does not change over time.
This constant velocity aspect is essential when calculating time dilation and length contraction effects that are consistent with special relativity.
Inertial Frames
The concept of "inertial frames" is fundamental in discussing the Lorentz transformations. An inertial frame is a coordinate system or a point of view where an observer moves at a constant velocity, neither speeding up nor slowing down. These frames are crucial because the laws of physics, especially relativity, are designed to be simple and uniform across all inertial frames.

Key aspects of inertial frames include:
  • They always move at constant velocities relative to each other without acceleration.
  • Inertial frames make analyzing relative motion straightforward.
  • They are pivotal in deriving and applying Lorentz transformations, as transformations occur between such frames.
In contexts where two inertial frames observe an event, their observations can be transformed using the Lorentz equations. This ensures that each observer can translate their specific data, like space and time coordinates, into an equivalent form in another frame.
Furthermore, the consistency across all inertial frames is a key postulate of Einstein's theory of relativity; it asserts identical physical laws apply in all these frames, making studies such as time dilation universally valid when using the Lorentz transformations.
Spacetime Coordinates
"Spacetime coordinates" is a term that merges spatial and temporal measures, illustrating how space and time are interconnected in physics. In the realm of Lorentz transformation, spacetime coordinates represent specific points described by both space \( x \) and time \( t \) within a particular reference frame.
When an event happens in one inertial frame, it possesses coordinates \( (x, t) \). Lorentz transformations provide the tools to convert these coordinates into another frame's system, denoted \( (x', t') \).

Key elements include:
  • The coordinates \( x \) and \( x' \) describe spatial positions or locations within different frames.
  • Time coordinates \( t \) and \( t' \) indicate when an event occurs relative to observers in each frame.
  • Transformation between these coordinates helps analyze how events are perceived differently in motion-bound observers' frames.
This concept is pivotal since transformations allow for a unified understanding of events across various observers and positions, explaining phenomena like time dilation and why time might pass differently for moving observers.

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

In the twin paradox situation, a fellow student objects to the argument that Anna's acceleration is the root of the asymmetry. "All motion is relative! Anna is accelerating relative to Bob. but Bob is accelerating relative to Anna." Answer this objection.

It is possible to define new types in \(\mathrm{C}++\). For example, the definition struct point \\{ float \(\mathrm{x}\) float y; \\} defines a new type named point. A value of type point contains two values of type float. What mathematical operation corresponds to the construction of this data type? Why?

You fire a light signal at \(60^{\circ}\) norh of west. (a) Find the velocity components of this signal according to an observer moving eastward relative to you at half the speed of light. From them, determine the magnitude and direction of the light signal's velocity according to this other observer. (b) Find the components according to a different observer, moving westward relative to you at half the speed of light.

A famous experiment detected 527 muons per hour at the top of Mt. Washington, New Hampshire, elevation \(1910 \mathrm{~m}\). At sea level. the same equipment detected 395 muons per hour. A discriminator selected for muons whose speed was between \(0.9950 c\) and \(0.9954 c\). Given that the mean lifetime \(\tau\) of a muons in a frame in which it is at rest is \(2.2 \mu \mathrm{s}\) and that in this frame the number of moons decays exponentially with time accounting \(\operatorname{to} N=N_{0} e^{-d r},\) show that the results obtained in the experiment are sensible.

Determine the momentum of an electron moving (a) at speed \(2.4 \times 10^{4} \mathrm{~m} / \mathrm{s}\) (about three times escape velocity) and (b) at speed \(2.4 \times 10^{8} \mathrm{~m} / \mathrm{s} .\) (c) In each case, by how much is the classical formula in error?

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