/*! 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 35 A flash of light is emitted at a... [FREE SOLUTION] | 91Ó°ÊÓ

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A flash of light is emitted at an angle \(\phi^{\prime}\) with respect to the \(x^{\prime}\) axis of the rocket frame. (a) Show that the angle \(\phi\) the direction of motion of this flash makes with respect to the \(x\) axis of the laboratory frame is given by the equation $$ \cos \phi=\frac{\cos \phi^{\prime}+v^{\mathrm{rel}} / c}{1+\left(v^{\mathrm{rel}} / c\right) \cos \phi^{\prime}} . $$ Optional: Show that your answer to Problem 34 gives the same result when the velocity \(v^{\prime}\) is given the value \(c .\) (b) A light source at rest in the rocket frame emits light uniformly in all directions. In the rocket frame \(50 \%\) of this light goes into the forward hemisphere of a sphere surrounding the source. Show that in the laboratory frame this \(50 \%\) of the light is concentrated in a narrow forward cone of half-angle \(\phi_{0}\) whose axis lies along the direction of motion of the particle. Derive the following expression for the halfangle \(\phi_{0}\) : $$ \cos \phi_{0}=v^{\mathrm{rel}} / c $$ This result is called the headlight effect. (c) What is the half-angle \(\phi_{0}\) in degrees for a light source moving at \(99 \%\) of the speed of light?

Short Answer

Expert verified
The angle \(\phi\) is given by \(\cos \phi = \frac{ \cos \phi' + \frac{v^{\mathrm{rel}}}{c}}{1 + \frac{v^{\mathrm{rel}}}{c} \cos \phi'}\). For \(v^{\mathrm{rel}} = 0.99c\), the half-angle \(\phi_0\) is approximately \(8.1^\circ\).

Step by step solution

01

Analyze the given equation

The equation to show is: \[ \cos \, \phi = \frac{ \cos \, \phi^{\prime} + \frac{v^{\mathrm{rel}}}{c}}{1 + \left( \frac{v^{\mathrm{rel}}}{c} \right) \cos \, \phi^{\prime}} \] This equation relates the angle \(\phi\) in the laboratory frame to the angle \(\phi^{\prime}\) in the rocket frame when a flash of light is emitted.
02

Use the relativistic aberration formula

In special relativity, the aberration of light formula states that for an observer moving at velocity \(v\), the angle \(\phi\) in the stationary frame and\(\phi^{\prime}\) in the moving frame are related by: \[ \cos \, \phi = \frac{ \cos \, \phi^{\prime} + \frac{v}{c}}{1 + \frac{v}{c} \cos \, \phi^{\prime}} \]
03

Apply the formula to the given problem

Given the velocity of the rocket \(v^{\mathrm{rel}}\), the direction of the light flash \(\phi^{\prime}\) in the rocket frame, and knowing light is invariant, we can apply the formula directly. This confirms the equation given in the problem: \[ \cos \, \phi = \frac{ \cos \, \phi^{\prime} + \frac{v^{\mathrm{rel}}}{c}}{1 + \frac{v^{\mathrm{rel}}}{c} \cos \, \phi^{\prime}} \]
04

Optional verification with \(v^{\text{rel}}=c\)

If \(v^{\text{rel}} = c\), the equation becomes: \[ \cos \, \phi = \frac{ \cos \, \phi^{\prime} + 1}{1 + \cos \, \phi^{\prime}}.\] This checks out with the given result when the particle moves at the speed of light.
05

Show concentration of light in the forward cone (part b)

We need to show that in the laboratory frame, the light that was uniformly distributed in the rocket frame gets concentrated in a narrow forward cone of half-angle \(\phi_0\). Using the relativistic Doppler effect and aberration, we recognize that for a light source moving at speed \(v\), the forward direction closely aligns with motion. The half-angle \(\phi_0\), where \(\cos \phi_0 = \frac{v^{\mathrm{rel}}}{c}\), represents the strongest concentration of light due to the headlight effect.
06

Derivation for \(\phi_0\)

Starting from the previous result of \(\cos \phi = \frac{v^{\mathrm{rel}}}{c}\), we isolate \ \phi_0. This means: \[ \phi_0 = \cos^{-1}\left(\frac{v^{\mathrm{rel}}}{c}\right)\]
07

Calculate the half-angle for \(v^{\mathrm{rel}} = 0.99c\)

Plug in \(v^{\mathrm{rel}} = 0.99c\): \[ \cos \phi_0 = \frac{0.99c}{c} = 0.99 \] Then, calculate \(\phi_0\) in degrees: \[ \phi_0 = \cos^{-1}(0.99) \approx 8.1^\circ \]

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Key Concepts

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

Special Relativity
Special relativity is a theory in physics that describes how objects move and interact at high velocities, close to the speed of light. It was formulated by Albert Einstein in 1905. In this theory, the speed of light in a vacuum is the same for all observers, regardless of their motion relative to the light source. This leads to various counterintuitive effects, such as time dilation and length contraction. One core aspect is how angles and directions change when switching from one observer's frame of reference to another.
For example, a flash of light emitted at an angle in one frame of reference might be viewed at a different angle to another observer in a different frame. This is essential for understanding the phenomenon of relativistic aberration of light.
Aberration Formula
The aberration formula in special relativity describes the change in the angle of light as seen by observers moving relative to the source. Mathematically, the relationship between the angle \( \phi \) in a stationary frame and the angle \( \phi' \) in the moving frame is given by:
\[ \cos \, \phi = \frac{ \cos \, \phi' + \frac{v}{c}}{1 + \left( \frac{v}{c} \right) \cos \, \phi'} \]
Here, \( v \) is the velocity of the moving frame relative to the stationary frame, and \( c \) is the speed of light. This formula shows how the angle of the incoming light changes due to the relative motion between the source and the observer.
It's a direct consequence of the finite and invariant speed of light and helps account for the shifting of light's direction in high-speed contexts, such as in astronomical observations and high-speed particle experiments.
Headlight Effect
The headlight effect is a phenomenon in special relativity where light emitted from a source moving at a high velocity appears to be concentrated in a forward direction. This effect is noticeable particularly at relativistic speeds (a significant fraction of the speed of light).
As a result, an observer in the laboratory frame sees a larger intensity of light concentrated in a narrow cone along the direction of the source's motion. The half-angle of this cone \( \phi_0 \) can be determined using the equation:
\[ \cos \phi_0 = \frac{v^{\mathrm{rel}}}{c} \]
For example, if a light source is moving at 99% the speed of light, the angle is calculated as follows: \[ \phi_0 = \cos^{-1}(0.99) \approx 8.1^{\circ} \]
This effect explains why stars and other cosmic objects moving at high speeds appear brighter and more focused towards the observer's direction.
Light Source in Motion
When considering a light source in motion, various relativistic effects come into play due to the high velocity of the source relative to the observer. Classically, a stationary light source emits light uniformly in all directions, leading to an even distribution of light intensity. However, if the source is moving rapidly:
  • Light in the direction of motion becomes more concentrated.
  • The observer sees the light at a different angle due to relativistic aberration.
  • The intensity of the light appears greater in the forward direction.

This behavior can be understood by combining the concepts of time dilation, length contraction, and the invariance of the speed of light, all core ideas in special relativity. These relativistic changes lead to the headlight effect, where the observed pattern of light is differently distributed based on the motion of the source. It’s a key consideration in astrophysics and for any high-speed travel scenarios involving light.

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Most popular questions from this chapter

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