/*! 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 62 A spaceship moving at constant s... [FREE SOLUTION] | 91Ó°ÊÓ

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A spaceship moving at constant speed u relative to us broadcasts a radio signal at constant frequency \(f_0\) . As the spaceship approaches us, we receive a higher frequency \(f\) ; after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_0\) , and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_0\)? (\(Hint\): In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals 1 /T. Use the time dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_0\) = 345 MHz as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758\(c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f - f_0\)? (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift (\(f - f_0\)) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

Short Answer

Expert verified
(a) Frequency when passing is \( \frac{f_0}{\gamma} \), lower than \( f_0 \). (b) 897 MHz approaching, 133 MHz receding. (c) 220 MHz passing, shift less than in (b).

Step by step solution

01

Understanding Time Dilation

Time dilation is a concept from the theory of relativity which states that time between two events appears longer from a stationary frame than from a moving frame. The time dilation formula is \( T = \gamma \cdot T_0 \), where \( \gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \) is the Lorentz factor, \( u \) is the speed of the spaceship, and \( c \) is the speed of light. Here, \( T_0 \) is the period of the signal in the spaceship's frame.
02

Frequency Relationship and Derivation

In the spaceship's frame, the period of the signal is \( T_0 = \frac{1}{f_0} \). When the spaceship passes us, the frequency \( f \) we observe is given by \( f = \frac{1}{T} \). Using time dilation, \( T = \gamma \cdot T_0 = \gamma \frac{1}{f_0} \), so \( f = \frac{1}{T} = \frac{f_0}{\gamma} \). Since \( \gamma > 1 \), \( f < f_0 \); thus, upon passing, the frequency we receive is lower.
03

Calculation for Approaching Spaceship

As the spaceship approaches, the received frequency is altered by the Doppler effect. The relativistic Doppler formula is \( f = f_0 \sqrt{\frac{1+\frac{u}{c}}{1-\frac{u}{c}}} \). For \( f_0 = 345 \) MHz and \( u = 0.758c \), calculate \( f \): \[ f = 345 \times \sqrt{\frac{1+0.758}{1-0.758}} \approx 897 \text{ MHz} \]. The shift is \( f - f_0 = 552 \text{ MHz} \).
04

Calculation for Receding Spaceship

When moving away, use the same formula with reversed speed: \( f = f_0 \sqrt{\frac{1-\frac{u}{c}}{1+\frac{u}{c}}} \). Calculate \( f \): \[ f = 345 \times \sqrt{\frac{1-0.758}{1+0.758}} \approx 133 \text{ MHz} \]. The shift is \( f - f_0 = -212 \text{ MHz} \).
05

Instantaneous Frequency as Ship Passes

Applying the situation from part (a), when the spaceship passes directly by, we use \( f = \frac{f_0}{\gamma} \) with \( \gamma = \frac{1}{\sqrt{1 - (0.758)^2}} \approx 1.565 \). So \( f \approx \frac{345}{1.565} \approx 220 \text{ MHz} \). The shift is \( 220 \text{ MHz} - 345 \text{ MHz} = -125 \text{ MHz} \).
06

Comparing Frequency Shifts

When the spaceship passes, the frequency shift is a decrease of 125 MHz, reflecting time dilation effects. Previously, shifts due to the Doppler effect were larger because they included relative motion changes, showing dramatic rises or falls depending on approach or departure.

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Key Concepts

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

Time Dilation
Time dilation is a fascinating concept from Einstein's theory of relativity that describes how time can slow down depending on the speed of an object relative to an observer. It's crucial to understand that when an object moves close to the speed of light, time for that object moves slower compared to a stationary observer. This can be perceived through the time dilation formula:
  • \( T = \gamma \cdot T_0 \)
  • Here, \( T \) is the time observed in the stationary frame, \( T_0 \) is the proper time (time observed in the moving frame), and \( \gamma \) is the Lorentz factor.
  • The Lorentz factor \( \gamma \) is given by \( \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \), where \( u \) is the velocity of the moving object and \( c \) is the speed of light.
Time dilation can be observed using light signals, as in the case of a spaceship sending radio signals. To a stationary observer, the period of the received signal seems longer due to time dilation, providing a basis for understanding related phenomena such as the Doppler effect.
Lorentz Factor
The Lorentz factor \( \gamma \) is an essential component of both time dilation and other relativistic effects like length contraction and relativistic mass. It serves as a mathematical tool that quantifies deviations in time and distance measurements between moving and stationary frames.
  • The formula for the Lorentz factor is \( \gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \).
  • As \( u \) approaches the speed of light \( c \), \( \gamma \) increases significantly, illustrating greater relativistic effects.
  • In many physics problems, including those involving the Doppler effect, \( \gamma \) helps compute how much time or distance is altered due to high velocities.
Understanding the Lorentz factor is crucial when analyzing how motion affects perception of signals, as it directly impacts equations used in Doppler shift calculations where time dilation plays a role.
Doppler Shift Calculation
The Doppler shift refers to the change in frequency (or wavelength) of a wave in relation to an observer moving relative to the wave source. In the realm of special relativity, the Doppler effect must be modified to account for high velocities approaching the speed of light, hence the term "relativistic Doppler effect."
  • For a source approaching the observer, the frequency \( f \) is given by \( f = f_0 \sqrt{\frac{1+\frac{u}{c}}{1-\frac{u}{c}}} \).
  • For a source moving away, it is \( f = f_0 \sqrt{\frac{1-\frac{u}{c}}{1+\frac{u}{c}}} \).
  • These formulas illustrate how perceived frequency increases as a source approaches and decreases as it moves away, compared to the original frequency \( f_0 \) emitted by the source.
The relativistic Doppler effect is different from the classical one because it includes adjustments for time dilation, ensuring that calculations remain true even at very high speeds.
Special Relativity
Special relativity is a theory proposed by Albert Einstein that revolutionized our understanding of space, time, and gravity. One of its fundamental postulates is that the laws of physics are the same in all inertial frames, and the speed of light is constant in a vacuum for all observers, regardless of their motion.
  • Special relativity combines space and time into a single four-dimensional continuum known as spacetime.
  • Time dilation and length contraction are two direct consequences of this theory, explaining why moving objects appear "shortened" in the direction of motion and "time-stretched."
  • The theory applies when dealing with objects moving close to the speed of light, requiring adjustments to classical mechanics.
Special relativity provides the framework for understanding phenomena such as the Doppler effect in a relativistic context, where the changes in observed frequencies or wavelengths become significant because of the high speeds involved. Through its principles, a wealth of modern physics including quantum mechanics and general relativity have been built.

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Most popular questions from this chapter

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of 99.95\(\%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at 75.00\(\%\) the speed of light in the laboratory reference frame, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving and (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts (a) and (b), find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth at a speed of 0.520c. If the radiation has a frequency of 8.64 \(\times\) 10\(^{14}\) Hz in the rest frame of the star, what is the frequency measured by an observer on earth?

(a) How much work must be done on a particle with mass \(m\) to accelerate it (a) from rest to a speed of 0.090\(c\) and (b) from a speed of 0.900\(c\) to a speed of 0.990\(c\) ? (Express the answers in terms of \(mc^2\).) (c) How do your answers in parts (a) and (b) compare?

A source of electromagnetic radiation is moving in a radial direction relative to you. The frequency you measure is 1.25 times the frequency measured in the rest frame of the source. What is the speed of the source relative to you? Is the source moving toward you or away from you?

(a) Consider the Galilean transformation along the \(x\)-direction : \(x' = x - vt\) and \(t' = t\). In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $${\partial^2E(x, t)\over \partial x^2} - {1 \over c^2} {\partial^2E(x, t)\over \partial t^2} = 0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S'\) is found to be $$(1 - {v^2 \over c^2} ) {\partial^2E(x', t') \over \partial{x'^2}} + {2v \over c^2} {\partial^2E(x', t') \over \partial{x'} \partial{t'}} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'^2}} = 0$$ This has a different form than the wave equation in \(S\). Hence the Galilean transformation \(violates\) the first relativity postulate that all physical laws have the same form in all inertial reference frames. (\(Hint\): Express the derivatives \(\partial/\partial{x}\) and \(\partial/\partial{t}\) in terms of \(\partial/\partial{x'}\) and \(\partial/\partial{t'}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S'\) the wave equation has the same form as in frame \(S\): $${\partial^2E(x', t') \over \partial{x'}^2} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'}^2} = 0$$ Explain why this shows that the speed of light in vacuum is c in both frames \(S\) and \(S'\).

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