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

Derive Eqs. \(20-17\) for the inverse Lorentz transformation by algebraically inverting the equations for the Lorentz transformation, Eqs. \(20-14\).

Short Answer

Expert verified
The inverse Lorentz transformations are \( x = \frac{x' + vt'}{\sqrt{1 - v^2/c^2}} \) and \( t = \frac{t' + vx'/c^{2}}{\sqrt{1 - v^2/c^2}} \).

Step by step solution

01

Set up the Lorentz transformation equations

First, recall the Lorentz transformations from Eqs. 20-14: \( x' = \gamma (x - vt) \) and \( t' = \gamma (t - vx/c^2) \) where \( \gamma = 1/\sqrt{1 - v^2/c^2} \) is the Lorentz factor.
02

Solve for x in the first equation

From the first transformation equation, rearrange to solve for x: \( x = \frac{x'}{\gamma} + vt \). Here we have substituted \(\gamma\) back in the equation.
03

Substitute t' in the equation for x

Now, we substitute the expression for t derived from the second Lorentz transformation into this equation: \( x = \frac{x'}{\gamma} + v\left(\frac{t'}{\gamma} + \frac{vx}{c^2}\right) \). We end up with the equation for x in terms of the primed variables and v.
04

Solve the equation for x

Manipulate this equation to isolate x on one side, obtaining \( x = \frac{x' + vt'}{\sqrt{1 - v^2/c^2}} \). This is the inverse Lorentz transformation for x.
05

Repeat the process for t

Repeat the same process for the equation for t: \( t = \frac{t' + vx'/c^{2}}{\sqrt{1 - v^2/c^2}} \). Thus, we have the inverse Lorentz transformations.

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.

Understanding the Lorentz Factor
At the heart of understanding special relativity is the Lorentz factor, denoted as \( \text{\(\gamma\)} \). It plays a crucial role in the Lorentz transformation, which describes how the measurements of space and time by two observers are related. Essentially, it quantifies how much time dilation and length contraction occur due to relative motion.

The Lorentz factor is mathematically defined as \( \text{\(\gamma = 1/\sqrt{1 - v^2/c^2}\)} \), where \( \text{\(v\)} \) is the relative velocity between two observers, and \( \text{\(c\)} \) is the speed of light. It's important to note that as the velocity \( \text{\(v\)} \) approaches the speed of light, \( \text{\(\gamma\)} \) increases significantly, leading to more pronounced relativistic effects such as significant time dilation and length contraction.
Special Relativity: The Fabric of Space and Time
Developed by Albert Einstein, special relativity revolutionized our understanding of space, time, and how they interlink. It challenged conventional views, reflecting that the laws of physics are the same for all non-accelerating observers and that the speed of light is constant in a vacuum for all observers, regardless of the motion of the light source.

Two fundamental implications arising from special relativity are time dilation and length contraction, both underpinned by the concept of relative motion. In essence, special relativity tells us that time will appear to pass slower and lengths will appear to shorten for an object in motion, as observed from a different inertial frame. These phenomena are numerically captured by the Lorentz transformations.
Time Dilation: The Stretching of Time
Time dilation is an intriguing consequence of special relativity, causing a clock moving relative to an observer to tick more slowly compared to the observer's own clock. This is not an illusion but a real physical effect; as relative velocity increases, the effect becomes more noticeable.

To understand time dilation, let's consider an observer in a stationary frame of reference and another observer moving at high speed. The Lorentz transformation shows that the time interval \( \text{\(t'\)} \) in the moving frame will be longer than the interval \( \text{\(t\)} \) in the stationary frame, as given by the formula \( \text{\(t' = \gamma(t - vx/c^2)\)} \). As a result, if we're observing something moving close to the speed of light, it will seem to age more slowly.
Length Contraction: Compressing Space
Complementary to time dilation is the phenomenon of length contraction, which states that objects physically shorten in the direction of their motion relative to an observer. Just as with time dilation, the effect becomes significant only at velocities approaching the speed of light.

The Lorentz transformation equations describe the relation between the lengths measured in different inertial frames. The length measured in the moving frame \( \text{\(l'\)} \) is shorter than that in the stationary frame \( \text{\(l\)} \), according to the relationship \( \text{\(l' = l/\gamma\)} \). This contraction occurs only in the direction of motion and illustrates how motion can warp our perceived dimensions of space.

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

Quite apart from effects due to the Earth's rotational and orbital motions, a laboratory frame is not strictly an inertial frame because a particle placed at rest there will not, in general, remain at rest; it will fall under gravity. Often, however, events happen so quickly that we can ignore free fall and treat the frame as inertial. Consider, for example, a \(1.0-\mathrm{MeV}\) electron (for which \(v=0.941 c\) ) projected horizontally into a laboratory test chamber and moving through a distance of \(20 \mathrm{~cm}\). (a) How long would it take, and \((b)\) how far would the electron fall during this interval? What can you conclude about the suitability of the laboratory as an inertial frame in this case?

Observer \(S\) assigns to an event the coordinates \(x=100 \mathrm{~km}\), \(t=200 \mu \mathrm{s}\). Find the coordinates of this event in frame \(S^{\prime}\), which moves in the direction of increasing \(x\) with speed \(0.950 c\). Assume that \(x=x^{\prime}\) at \(t=t^{\prime}=0\).

Consider the following, all moving in free space: a \(2.0\) -eV photon, a \(0.40-\mathrm{MeV}\) electron, and a \(10-\mathrm{MeV}\) proton. \((a)\) Which is moving the fastest? (b) The slowest? ( \(c\) ) Which has the greatest momentum? \((d)\) The least? (Note: A photon is a light particle of zero mass.)

An alpha particle with kinetic energy \(7.70 \mathrm{MeV}\) strikes a \({ }^{14} \mathrm{~N}\) nucleus at rest. An \({ }^{17} \mathrm{O}\) nucleus and a proton are produced, the proton emitted at \(90^{\circ}\) to the direction of the incident alpha particle and carrying kinetic energy \(4.44 \mathrm{MeV}\). The rest energies of the various particles are: alpha particle, \(3730.4 \mathrm{MeV}\); \({ }^{14} \mathrm{~N}, 13,051 \mathrm{MeV} ;\) proton, \(939.29 \mathrm{MeV} ;{ }^{17} \mathrm{O}, 15,843 \mathrm{MeV}\) (a) Find the kinetic energy of the \({ }^{17} \mathrm{O}\) nucleus. (b) At what angle with respect to the direction of the incident alpha particle does the \({ }^{17} \mathrm{O}\) nucleus move?

A particle has a momentum equal to \(m c .\) Calculate its speed.

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