/*! 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 12 Is the relativistic Doppler effe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Is the relativistic Doppler effect consistent with the classical Doppler effect in the respect that \(\lambda_{\text {obs }}\) is larger for motion away?

Short Answer

Expert verified
In both the classical and relativistic Doppler effects, the observed wavelength \(\lambda_{obs}\) is larger for motion away from the observer, as shown by the relations \(\lambda_{obs} = \frac{c + v}{c - v} \cdot \lambda_{src}\) (classical) and \(\lambda_{obs} = \gamma (1 + \beta) \cdot \lambda_{src}\) (relativistic). This consistency holds because, in both cases, the relative velocity terms in the respective formulas contribute to an increase in the observed wavelength for motion away.

Step by step solution

01

Classical Doppler Effect

In the classical Doppler effect, the change in the observed frequency (and wavelength) is due to the relative motion of the source and the observer. The classical Doppler effect for light can be given by the following formula: \(f_{obs} = \frac{c \pm v_{obs}}{c \pm v_{src}} \cdot f_{src}\) Here, \(f_{src}\) is the emitted frequency, \(f_{obs}\) is the observed frequency, \(v_{obs}\) is the velocity of the observer, \(v_{src}\) is the velocity of the source, and \(c\) is the speed of light. The choice of signs (positive or negative) depends on the relative motion between the source and the observer (away or towards). Now, we need to find the relation between observed and emitted wavelength to figure out if the observed wavelength is larger for motion away. We know the relation between frequency and wavelength: \(f = \frac{c}{\lambda}\) Applying this relation to both the emitted and observed frequencies and solving for \(\lambda_{obs}\): \(\lambda_{obs} = \frac{c \pm v_{obs}}{c \pm v_{src}} \cdot \lambda_{src}\)
02

Relativistic Doppler Effect

The relativistic Doppler effect takes into account the time dilation experienced due to relative motion. The relation for the relativistic Doppler effect is: \(f_{obs} = \frac{1}{\gamma (1 \pm \beta)} \cdot f_{src}\) Here, \(\beta = \frac{v}{c}\) (v is the relative velocity) and \(\gamma = \frac{1}{\sqrt{1 - \beta^2}}\) is the Lorentz factor. The choice of signs depends on the relative motion between the source and the observer (away or towards). Similarly, we can find the relation between observed and emitted wavelength for the relativistic Doppler effect: \(\lambda_{obs} = \gamma (1 \pm \beta) \cdot \lambda_{src}\)
03

Comparing Classical and Relativistic Doppler Effects

Now, let's compare both the classical and relativistic formulas to see if the observed wavelength is larger for motion away: - Classical Doppler effect when motion is away: \(\lambda_{obs} = \frac{c + v}{c - v} \cdot \lambda_{src}\) Since \(\frac{c+v}{c-v}>1\) (as \(v>0\)), \(\lambda_{obs} > \lambda_{src}\). - Relativistic Doppler effect when motion is away: \(\lambda_{obs} = \gamma (1 + \beta) \cdot \lambda_{src}\) Since \(\gamma \geq 1\) and \(\beta > 0\), \(\lambda_{obs} > \lambda_{src}\). Therefore, both the classical and relativistic Doppler effects are consistent with the behavior that the observed wavelength is larger for motion away from the observer.

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 91Ó°ÊÓ!

Key Concepts

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

Classical Doppler Effect
The Classical Doppler Effect describes the change in frequency and wavelength of a wave in relation to an observer who is moving relative to the wave source. When the source and observer are moving towards each other, the observed frequency increases resulting in a shorter observed wavelength. Conversely, if the source and observer are moving away from each other, the observed frequency decreases, leading to a longer observed wavelength.
In mathematical terms, this change is captured by:
  • \(f_{\text{obs}} = \frac{c \pm v_{\text{obs}}}{c \pm v_{\text{src}}} \cdot f_{\text{src}}\)
  • \(\lambda_{\text{obs}} = \frac{c \pm v_{\text{obs}}}{c \pm v_{\text{src}}} \cdot \lambda_{\text{src}}\)
Here, \(v_{\text{obs}}\) is the velocity of the observer, \(v_{\text{src}}\) is the velocity of the source, \(c\) is the speed of light, \(f_{\text{src}}\) is the emitted frequency, and \(\lambda_{\text{src}}\) is the emitted wavelength. The choice of sign depends on whether the source and observer are moving towards or away from each other.
Relativistic Doppler Effect
The Relativistic Doppler Effect incorporates the element of time dilation, a fundamental concept in Einstein's theory of relativity, into the analysis of wave frequency shifts due to motion. This effect is most noticeable at velocities approaching the speed of light. In this realm, the factor \(\gamma\), known as the Lorentz factor, becomes significant and is defined as:
  • \(\gamma = \frac{1}{\sqrt{1 - \beta^2}}\)
  • \(\beta = \frac{v}{c}\)
These variables point out that as \(v\) approaches \(c\), \(\gamma\) significantly modifies the relationship between observed and emitted quantities.
The formula for the relativistic Doppler effect is:
  • \(f_{\text{obs}} = \frac{1}{\gamma (1 \pm \beta)} \cdot f_{\text{src}}\)
  • \(\lambda_{\text{obs}} = \gamma (1 \pm \beta) \cdot \lambda_{\text{src}}\)
The positive sign is used when the object is moving away, resulting in an increase in the observed wavelength \(\lambda_{\text{obs}}\), thus demonstrating consistency with the classical approach.
Frequency and Wavelength Relationship
Understanding the relationship between frequency and wavelength is crucial in physics, as it forms the foundation for concepts like the Doppler Effect. Frequency \(f\) and wavelength \(\lambda\) are inversely proportional and related by the speed of light \(c\) for electromagnetic waves:
  • \(f = \frac{c}{\lambda}\)
This means that an increase in the observed wavelength corresponds to a decrease in frequency, and vice versa. This fundamental relation aids in interpreting both classical and relativistic Doppler effects.
In both effects, regardless of classical or relativistic conditions, when an observer moves away from a wave source, the wavelength appears longer (redshifted), and the frequency appears lower (blueshifted). Thus, observing action in frequency terms provides a reciprocal view of changes observed in wavelength terms, enhancing comprehension of wave behavior through motion dynamics.

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

An observer standing by the railroad tracks sees two bolts of lightning strike the ends of a 500 -m-long train simultaneously at the instant the middle of the train passes him at \(50 \mathrm{m} / \mathrm{s}\). Use the Lorentz transformation to find the time between the lightning strikes as measured by a passenger seated in the middle of the train.

All galaxies farther away than about \(50 \times 10^{6}\) ly exhibit a red shift in their emitted light that is proportional to distance, with those farther and farther away having progressively greater red shifts. What does this imply, assuming that the only source of red shift is relative motion?

(a) Calculate the speed of a 1.00 - \(\mu\) g particle of dust that has the same momentum as a proton moving at \(0.999 c\) (b) What does the small speed tell us about the mass of a proton compared to even a tiny amount of macroscopic matter?

A spacecraft starts from being at rest at the origin and accelerates at a constant rate \(g\), as seen from Earth, taken to be an inertial frame, until it reaches a speed of \(c / 2\). (a) Show that the increment of proper time is related to the elapsed time in Earth's frame by: \(d \tau=\sqrt{1-v^{2} / c^{2}} d t\) (b) Find an expression for the elapsed time to reach speed C/2 as seen in Earth's frame. (c) Use the relationship in (a) to obtain a similar expression for the elapsed proper time to reach \(c / 2\) as seen in the spacecraft, and determine the ratio of the time seen from Earth with that on the spacecraft to reach the final speed.

A man standing still at a train station watches two boys throwing a baseball in a moving train. Suppose the train is moving east with a constant speed of \(20 \mathrm{m} / \mathrm{s}\) and one of the boys throws the ball with a speed of \(5 \mathrm{m} / \mathrm{s}\) with respect to himself toward the other boy, who is 5 m west from him. What is the velocity of the ball as observed by the man on the station?

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.