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Chapter 37: Problem 23

(a) How fast must you be approaching a red traffic (\(\lambda=\) 675 nm) for it to appear yellow (\(\lambda=\) 575 nm)? Express your answer in terms of the speed of light. (b) If you used this as a reason not to get a ticket for running a red light, how much of a fine would you get for speeding? Assume that the fine is $1.00 for each kilometer per hour that your speed exceeds the posted limit of 90 km/h.

Short Answer

Expert verified
(a) Speed: 0.075c; (b) Fine: $80,999,910.

Step by step solution

01

Understanding the Problem

To solve this problem, we need to use the Doppler effect principle for light. The observed wavelength changes due to the relative motion between the observer and the source. We will calculate this speed using the formula for the Doppler effect in light.
02

Applying the Doppler Effect Formula for Light

For light, the relativistic Doppler effect is given by the formula: \( \frac{\lambda'}{\lambda} = \sqrt{\frac{1+\beta}{1-\beta}} \), where \(\lambda'\) is the observed wavelength, \(\lambda\) is the original wavelength, and \(\beta = \frac{v}{c}\) is the velocity as a fraction of the speed of light \(c\).
03

Plugging in Given Values

We are given \(\lambda = 675 \text{ nm}\) and \(\lambda' = 575 \text{ nm}\). We plug these into the formula: \( \frac{575}{675} = \sqrt{\frac{1+\beta}{1-\beta}} \).
04

Solving for \(\beta\)

Square both sides to eliminate the square root: \(\left( \frac{575}{675} \right)^2 = \frac{1+\beta}{1-\beta} \). Simplify and solve for \(\beta\) using algebraic manipulation. You'll get: \(\beta \approx 0.075 \).
05

Converting \(\beta\) to Velocity

The velocity \(v\) in terms of the speed of light \(c\) is \(v = \beta c\). So, \(v = 0.075c\).
06

Calculating Actual Speed in km/h

Since \(c = 3 \times 10^5 \text{ km/s}\), the speed \(v = 0.075 \times 3 \times 10^5 = 22,500 \text{ km/s}\). To convert this into km/h, multiply by 3600 (seconds per hour): \(v = 22,500 \times 3600 = 81,000,000 \text{ km/h}\).
07

Determining the Speed Over Limit

The speed limit is 90 km/h. Your speed is \(81,000,000 \text{ km/h}\). So, you are exceeding the speed limit by \(81,000,000 - 90 = 80,999,910 \text{ km/h}\).
08

Calculating the Fine

The fine is \\(1.00 for each km/h over the speed limit. Thus, the fine will be \\)80,999,910.

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Key Concepts

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

Relativistic Doppler Effect
The relativistic Doppler effect is a specific phenomenon that applies to light and other electromagnetic waves. It occurs when the source and observer move relative to each other at speeds close to the speed of light. In these conditions, the familiar Doppler effect we experience with sound waves needs to be adjusted for the effects of special relativity.
Unlike sound waves, where the medium (air, water) influences wave speed, light waves move through the vacuum of space at a constant speed. The formula used to describe the relativistic Doppler effect for light is:
  • \(\frac{\lambda'}{\lambda} = \sqrt{\frac{1+\beta}{1-\beta}}\)
where \(\lambda'\) is the observed wavelength, \(\lambda\) is the source wavelength, and \(\beta = \frac{v}{c}\) is the velocity as a fraction of the speed of light \(c\).
This formula accounts for the relativistic effects due to high velocity. When calculating such changes for fast-moving objects, this formula ensures observations accurately reflect the phenomenon's relativistic nature.
Wavelength Change
Wavelength change in the context of the Doppler effect describes how the color of light alters when the source or observer moves at significant speed. For example, as explained in the exercise, a red traffic light with a wavelength of 675 nm might appear yellow (575 nm) to someone moving toward it.
To solve problems involving wavelength change due to relative motion, we use the ratio \(\frac{\lambda'}{\lambda}\). This division gives us the change factor, influenced by the velocity of the motion, which can be plugged into the relativistic Doppler effect formula. The alteration in wavelength is why distant galaxies often appear redshifted, or shifted toward longer wavelengths, if they move away from us at high speed.
This effect is crucial in astrophysics to measure the universe's expansion and observe celestial objects' motion.
Speed of Light
The speed of light, denoted as \(c\), is a fundamental constant of nature. It is approximately \(3 \times 10^8 \text{ m/s}\) in a vacuum. Light's consistent speed forms the backbone of Einstein's theory of relativity, influencing both time and space's behavior.
In problems like the original exercise, the speed of light assists in calculating speed as a fraction of \(c\). For instance, using \(\beta = \frac{v}{c}\) helps determine how close a velocity is to light speed. When we say a speed is a percentage of \(c\), such as 7.5% in the exercise, it contextualizes within the maximum possible speed limit set by the universe.
The unwavering pace of light unifies electromagnetic wave observations, ensuring that any calculations related to the relativistic Doppler effect remain consistent and grounded in reality. This constancy allows light's properties to be predictable and reliable across numerous scientific endeavors.

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Most popular questions from this chapter

A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600c. The pursuit ship is traveling at a speed of 0.800c relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the velocity of the cruiser relative to the pursuit ship be directed toward or away from the pursuit ship? (b) What is the speed of the cruiser relative to the pursuit ship?

An imperial spaceship, moving at high speed relative to the planet Arrakis, fires a rocket toward the planet with a speed of 0.920c relative to the spaceship. An observer on Arrakis measures that the rocket is approaching with a speed of 0.360c. What is the speed of the spaceship relative to Arrakis? Is the spaceship moving toward or away from Arrakis?

(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 \(S'\).

Two protons (each with rest mass \(M\) = 1.67 \(\times\) 10\(^{-27}\) kg) are initially moving with equal speeds in opposite directions. The protons continue to exist after a collision that also produces an \(\eta_0\) particle (see Chapter 44). The rest mass of the \(\eta_0\) is m = 9.75 \(\times\) 10\(^{-28}\) kg. (a) If the two protons and the \(\eta_0\) are all at rest after the collision, find the initial speed of the protons, expressed as a fraction of the speed of light. (b) What is the kinetic energy of each proton? Express your answer in MeV. (c) What is the rest energy of the \(\eta_0\), expressed in MeV? (d) Discuss the relationship between the answers to parts (b) and (c). 37.39 .

Inside a spaceship flying past the earth at three-fourths the speed of light, a pendulum is swinging. (a) If each swing takes 1.80 s as measured by an astronaut performing an experiment inside the spaceship, how long will the swing take as measured by a person at mission control (on earth) who is watching the experiment? (b) If each swing takes 1.80 s as measured by a person at mission control, how long will it take as measured by the astronaut in the spaceship?
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