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

Suppose that observer \(S\) fires a light beam in the \(y\) direction \(\left(v_{x}=0, v_{y}=c\right) .\) Observer \(S^{\prime}\) is moving at speed \(u\) in the \(x\) direction. \((a)\) Find the components \(v_{x}^{\prime}\) and \(v_{y}^{\prime}\) of the velocity of the light beam according to \(S^{\prime}\), and \((b)\) show that \(S^{\prime}\) measures a speed of \(c\) for the light beam.

Short Answer

Expert verified
The components \(v_{x}^{\prime}\) and \(v_{y}^{\prime}\) of the velocity of the light beam according to \(S^{\prime}\) are: \(v_{x}^{\prime} = -u\) and \(v_{y}^{\prime} = \frac{c}{\gamma}\) also \(S^{\prime}\) measures a speed of \(c\) for the light beam.

Step by step solution

01

Step 1:Find \(v_{x}^{\prime}\) component

We use the relativistic velocity transformation formula which is given by: \(v_{x}^{\prime} = \frac{v_{x} - u}{1 - \frac{v_{x}u}{c^{2}}}\). In our case, \(v_{x} = 0\) and \(u\) is some velocity in the \(x\) direction. So, after substituting values we get: \(v_{x}^{\prime} = \frac{-u}{1 - 0} = -u\)
02

Find \(v_{y}^{\prime}\) component

Similarly, using the relativistic velocity transformation for \(v_{y}\) we have : \(v_{y}^{\prime} = \frac{v_{y}}{\gamma \(1 + \frac{v_{x}u}{c^{2}}\)}\) where \(\gamma\) = \(\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}\). Substituting the given values we get: \(v_{y}^{\prime}\) = \(\frac{c}{\gamma}\)
03

Show that \(S^{\prime}\) measures a speed of \(c\) for the light beam

The speed of light according to \(S^{\prime}\) is given by the magnitude of the velocity vector: \(v^{\prime}\) = \(\sqrt{(v_{x}^{\prime})^{2} + (v_{y}^{\prime})^{2}}\). Substituting the values obtained in steps 1 and 2, we get \(v^{\prime}\) = \(\sqrt{(-u)^{2} + (\frac{c}{\gamma})^{2}}\)= \(c\), thus showing that the observer \(S^{\prime}\) also measures a speed \(c\) for the light beam.

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.

Special Relativity
Special Relativity is a theory proposed by Albert Einstein that has fundamentally changed our understanding of space and time. It's essential to realize that, according to Special Relativity, the laws of physics are the same for all observers in uniform motion relative to each other. This radical idea leads to surprising consequences, such as:
  • The relativity of simultaneity: Two events that occur simultaneously in one frame may not be simultaneous in another frame of reference.
  • Time dilation: Moving clocks tick slower compared to stationary ones from the perspective of a stationary observer.
  • Length contraction: Objects moving at high speeds are measured to be shorter in the direction of motion compared to when they are at rest.
These effects only become significant at speeds close to the speed of light, making Special Relativity crucial for high-speed space travel and particle physics. Grounded in these principles, Special Relativity sets the stage for understanding how velocities transform under different frames of reference.
Speed of Light
One of the cornerstone postulates of Special Relativity is that the speed of light in a vacuum, denoted by the letter \(c\), is constant at approximately \(3 \times 10^8\) meters per second for all observers, no matter their velocity relative to the light source. This revolutionary idea contradicts classical intuitions but is supported by extensive experimental evidence.
This unchanging speed of light leads to fascinating consequences:
  • No object with mass can reach, or exceed, the speed of light.
  • Light's behavior doesn't change regardless of the observer's motion, which aids in maintaining the uniform laws of physics across different inertial frames.
  • The concept of causality is preserved, ensuring that cause-and-effect relationships remain consistent across different observers.
Understanding the constancy of light speed is crucial for grasping how observers moving at different velocities perceive the same physical events.
Lorentz Transformation
The Lorentz Transformation is a set of equations that allow us to convert spatial and temporal coordinates from one inertial frame to another moving at a constant velocity relative to the first. It serves as a mathematical tool in Special Relativity to analyze how time and space vary between different observers.
The Lorentz Transformation involves:
  • Time coordination: Showing how time intervals between events differ for observers moving relative to each other.
  • Length transformation: Describing how the length of an object appears to contract along its direction of motion.
  • Velocity transformation: Discussed in the original exercise, it focuses on how velocities change between different reference frames. For instance, using the formulas for \(v_x'\) and \(v_y'\), we can understand velocity components in 'S' transforming to those in 'S\(^{\prime}\)', ensuring consistency with the speed of light principle.
These transformations support the understanding of how universally invariant light speed is maintained across different observers, helping us solve complex problems in relativistic scenarios.

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?

To circle the Earth in low orbit a satellite must have a speed of about \(7.91 \mathrm{~km} / \mathrm{s}\). Suppose that two such satellites orbit the Earth in opposite directions. (a) What is their relative speed as they pass? Evaluate using the classical Galilean velocity transformation equation. (b) What fractional error was made because the (correct) relativistic transformation equation was not used?

Calculate the speed of a particle \((a)\) whose kinetic energy is equal to twice its rest energy and \((b)\) whose total energy is equal to twice its rest energy.

Calculate the speed parameter \(\beta\) of a particle with a momentum of \(12.5 \mathrm{MeV} / c\) if the particle is \((a)\) an electron and \((b)\) a proton.

An unstable high-energy particle enters a detector and leaves a track \(1.05 \mathrm{~mm}\) long before it decays. Its speed relative to the detector was \(0.992 c .\) What is its proper lifetime? That is, how long would it have lasted before decay had it been at rest with respect to the detector?

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