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Chapter 37: Problem 26

Show that when the source of electromagnetic waves moves away from us at 0.600 \(\mathrm{c}\) , the frequency we measure is half the value measured in the rest frame of the source.

Short Answer

Expert verified
The frequency measured is half the value when the source moves at 0.600c.

Step by step solution

01

Identify the Formula for Doppler Effect

The Doppler effect for electromagnetic waves, like light, can be calculated using the relativistic Doppler effect formula: \( f' = f \sqrt{\frac{1 - \beta}{1 + \beta}} \), where \( f' \) is the observed frequency, \( f \) is the emitted frequency (rest frequency), and \( \beta = \frac{v}{c} \), with \( v \) being the velocity of the source relative to the observer and \( c \) being the speed of light.
02

Substitute Known Values

In the problem, the source moves away from us with \( v = 0.600c \). Therefore, \( \beta = 0.600 \). Substitute \( \beta = 0.600 \) into the equation: \( f' = f \sqrt{\frac{1 - 0.600}{1 + 0.600}} \).
03

Simplify the Expression Inside the Square Root

Calculate the values inside the square root: \( 1 - 0.600 = 0.400 \) and \( 1 + 0.600 = 1.600 \). Therefore, the expression becomes \( f' = f \sqrt{\frac{0.400}{1.600}} \).
04

Calculate the Square Root

Simplify the fraction: \( \frac{0.400}{1.600} = \frac{1}{4} = 0.25 \). Therefore, \( f' = f \sqrt{0.25} \). Calculate the square root: \( \sqrt{0.25} = 0.5 \).
05

Conclude the Calculation

Substitute the square root back into the equation: \( f' = f \times 0.5 \). This shows that the observed frequency \( f' \) is half of the rest frequency \( f \).

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Key Concepts

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

Electromagnetic Waves
Electromagnetic waves are a type of wave that travel through space at the speed of light. These waves include visible light, radio waves, X-rays, and more. They do not need a medium to travel through, which means they can propagate through the vacuum of space. Understanding electromagnetic waves is essential as they are fundamental to how we communicate and perceive the universe.

The behavior of electromagnetic waves depends on their frequency and wavelength. Higher frequency waves tend to have more energy. When an electromagnetic wave is emitted from a source, it may undergo changes depending on the observer's motion relative to the source. This is where the Doppler effect comes into play, particularly affecting the frequency of light waves as perceived by an observer.
Observed Frequency
Observed frequency refers to the frequency that a stationary or moving observer measures a wave at, differing from the frequency emitted by the source. According to the Doppler effect for electromagnetic waves, if a source of electromagnetic waves is moving away from an observer, the observed frequency will be lower than the emitted frequency.

This phenomenon can be explained with the relativistic Doppler effect formula:
  • \( f' = f \sqrt{\frac{1 - \beta}{1 + \beta}} \)
  • Here, \( f' \) is the observed frequency, and \( f \) is the emitted frequency.
When the source is moving away from the observer, as in the exercise, the observed frequency decreases due to the increase in the wavelength of the waves reaching the observer.
Speed of Light
The speed of light, denoted by \( c \), is a universal constant that is approximately equal to \( 299,792,458 \) meters per second. It represents the maximum speed at which energy, matter, or information can travel in the universe.
  • Light's speed is always the same, regardless of the motion of the source or observer.
  • This constant speed is vital in the equations of the relativistic Doppler effect.
Light speed plays a crucial role in modern physics and is central to the theory of relativity, which affects everything from how we calculate distances in space to how we understand time itself. In the context of the Doppler effect, the speed of light helps determine how the observed frequency is altered due to relative motion.
Velocity of Source
Velocity of the source can significantly impact how electromagnetic waves are observed. In the context of the Doppler effect, the velocity of the source relative to the observer influences the measured frequency of the wave. Velocity is described as the speed of an object in a specific direction. In physics, it is usually denoted as \( v \).
  • In the given exercise, the source moves away from the observer at \( 0.600c \).
  • This velocity is used to calculate \( \beta \), a dimensionless number that represents the normalized speed of the source relative to the speed of light: \( \beta = \frac{v}{c} \).
As seen in the original solution, substituting a value of 0.600 for \( \beta \) in the equation reveals how it directly impacts the observed frequency, confirming the broader principles of the Doppler effect.

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Most popular questions from this chapter

A spaceraft ties away from the earth with a speed of \(4.80 \times 10^{6} \mathrm{m} / \mathrm{s}\) relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days ( 1 year) later, as measured by the clock that remained on earth. What is the difference in the elapsed times on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on earth, shows the shortest elapsed time?

Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light-years away. (A light-year is the distance that light travels in a year.) You plan to travel at constant speed in a \(1000-\mathrm{kg}\) rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in joules, and the energy needed as a percentage of U.S. yearly use (which is 1.0 \(\times 10^{19} \mathrm{J} ) .\) For comparison, arrange your results in a table showing \(v_{\text { rocket }},\) \(t_{\text { earth }},\) \(\mathbf{t}_{\text { rouket }} \boldsymbol{E}\) \((\text { in } \mathrm{J}),\) and \(E\) (as \(\%\) of U.S. use). The rocket ship's speed is (a) \(0.50 \mathrm{c} ;\) (b) 0.99 \(\mathrm{c}\) (c) \(0.9999 \mathrm{c} .\) On the basis of your results, does it seem likely that any government will invest in such high-speed space travel any time soon?

In high-energy physics, new particles can be created by collisions of fast- moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive \(\mathrm{kaon}\left(\mathrm{K}^{+}\right) :\) $$p+p \rightarrow p+p+\mathbf{K}^{-}+\mathbf{K}^{+}$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 \(\mathrm{MeV}\) , and the rest energy of each proton is 938.3 MeV. (Hint: It is useful here to work in the frame in which the total momentum is zero. See Problem 8.100 , but note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kimetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

Many of the stars in the sky are actually binary stars, in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called spectruscopic binary stars. Figure 37.30 (next page) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m,\) orbiting their center of mass in a circle of radius \(R\) . The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of \(4.568110 \times 10^{14} \mathrm{Hz}\) . In the light received from the stars by a tele- scope on the earth, hydrogen light is observed to vary in frequency between \(4.567710 \times 10^{14} \mathrm{Hz}\) and \(4.568910 \times 10^{14} \mathrm{Hz}\) . Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (Hint: The speeds involved are much less than \(c,\) so you may use the approximate result \(\Delta f|f=u| c\) given in Section \(37.6 . )\) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, \(1.99 \times 10^{30} \mathrm{kg}\) . Compare the value of \(R\) to the distance from the earth to the sun, \(1.50 \times 10^{11} \mathrm{m}\) . (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

A meter stick moves past you at great speed. Its motion relative to you is parallel to its long axis. If you measure the length of the moving meter stick to be \(1.00 \mathrm{ft}(1 \mathrm{ft}=0.3048 \mathrm{m})-\) for example, by comparing it to a 1 -foot ruler that is at rest relative to you \(-\) at what speed is the meter stick moving relative to you?
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