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Chapter 37: Problem 10

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 ft 11 ft = 0.3048 m2-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?

Short Answer

Expert verified
The meter stick is moving at approximately \(2.85 \times 10^8\) m/s relative to you.

Step by step solution

01

Identify the Concept

We are dealing with a problem that involves length contraction, a concept in special relativity. When an object moves at high speeds relative to an observer, its length appears contracted along the direction of motion.
02

Set Up the Length Contraction Formula

The length contraction formula is given by \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where \( L \) is the observed contracted length, \( L_0 \) is the proper length, \( v \) is the velocity of the object, and \( c \) is the speed of light. Given \( L_0 = 1 \text{ m} \) and \( L = 0.3048 \text{ m} \).
03

Rearrange the Formula to Solve for Speed

Rearrange the formula to solve for velocity \( v \). First, square both sides of the equation to get:\[ \left(\frac{L}{L_0}\right)^2 = 1 - \frac{v^2}{c^2} \]Then solve for \( \frac{v^2}{c^2} \):\[ \frac{v^2}{c^2} = 1 - \left(\frac{L}{L_0}\right)^2 \]
04

Substitute the Known Values

Substitute \( L = 0.3048 \text{ m} \), \( L_0 = 1 \text{ m} \), and \( c = 3 \times 10^8 \text{ m/s} \) into the equation:\[ \frac{v^2}{c^2} = 1 - \left(\frac{0.3048}{1}\right)^2 \] Calculate \[ \frac{v^2}{c^2} = 1 - 0.0927 = 0.9073 \]
05

Solve for v

Finally, solve for \( v \):\[ v = c \sqrt{0.9073} \]Plug in the known value of \( c \):\[ v = 3 \times 10^8 \text{ m/s} \times \sqrt{0.9073} \]\[ v \approx 2.85 \times 10^8 \text{ m/s} \]

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Key Concepts

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

Understanding Length Contraction
Length contraction is a fascinating concept from Einstein's theory of special relativity. It comes into play when objects move at a significant fraction of the speed of light relative to an observer. In simple terms, when you observe a fast-moving object, it appears shorter along the direction of motion, compared to when it is at rest.
This isn't an illusion but a real effect that Einstein's theories predict. The effect becomes noticeable only at speeds close to the speed of light.The formula for length contraction is \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]- **L**: contracted length (what the observer measures)- **L_0**: proper length (the object's length at rest)- **v**: velocity of the object - **c**: speed of light When you apply this formula, it allows you to calculate how much the length contracts based on its speed. This concept helps us understand that measurements of space and time are not absolute but depend on the observer's relative motion.
Calculating Velocity in Special Relativity
In special relativity, calculating the velocity of a moving object involves more than just dividing distance by time. This is due to relativistic effects, including length contraction and time dilation. For the given exercise, we're interested in finding the velocity of a moving meter stick observed to be shorter than its rest length. Using the length contraction formula, we rearrange it to solve for velocity.Start by squaring both sides. This leads to: \[ \left(\frac{L}{L_0}\right)^2 = 1 - \frac{v^2}{c^2} \]Next, isolate \( \frac{v^2}{c^2} \) by moving terms around: \[ \frac{v^2}{c^2} = 1 - \left(\frac{L}{L_0}\right)^2 \]From this equation, you can find the square of velocity as a ratio involving the speed of light. Finally, take the square root and multiply by the speed of light to solve for \( v \). This process highlights how dramatically different relativistic velocity calculations are from classical ones.
The Role of the Speed of Light
The speed of light, denoted by \( c \), is a fundamental constant in physics and plays a crucial role in special relativity. It's about \( 3 \times 10^8 \text{ m/s} \), and it serves as a universal speed limit. No object with mass can reach or exceed this speed.In the realm of special relativity, \( c \) not only limits speed but also affects time and space. For instance, as objects approach the speed of light, time dilation and length contraction become significant.This constant appears in the mathematical expressions for relativity, including the length contraction formula:\[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]Here, \( c \) ensures that velocity \( v \) is compared to a scale that defines maximum possible speed. It also emphasizes Einstein鈥檚 insight that space and time are intertwined in the fabric of the universe. Understanding how \( c \) interacts with other variables helps illustrate why it's impossible to observe objects at or above this speed. It serves as a reminder of the unique and counterintuitive aspects of our universe.

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Most popular questions from this chapter

Many of the stars in the sky are actually \(binary\space stars\), 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 \(spectroscopic\space binary\space stars. \textbf{Figure P37.68}\) 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\(^{14}\) Hz. In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between 4.567710 \(\times\) 10\(^{14}\) Hz and 4.568910 \(\times\) 10\(^{14}\) 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\(^{30}\) kg. Compare the value of \(R\) to the distance from the earth to the sun, 1.50 \(\times\) 10\(^{11}\) 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) At what speed is the momentum of a particle twice as great as the result obtained from the nonrelativistic expression \(mv\)? Express your answer in terms of the speed of light. (b) A force is applied to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light.

Everyday Time Dilation. Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 m/s and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 h. By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (\(Hint\): Since \(u \ll c\), you can simplify \(\sqrt{1 - u^2/c^2}\) by a binomial expansion.)

A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.600c. A scientist on Coruscant measures the length of the moving spacecraft to be 74.0 m. The spacecraft later lands on Coruscant, and the same scientist measures the length of the now stationary spacecraft. What value does she get?

As you have seen, relativistic calculations usually involve the quantity \(\gamma\). When \(\gamma\) is appreciably greater than 1, we must use relativistic formulas instead of Newtonian ones. For what speed \(v\) (in terms of \(c\)) is the value of \(\gamma\) (a) 1.0% greater than 1; (b) 10% greater than 1; (c) 100% greater than 1?
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