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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?

Short Answer

Expert verified
The meter stick is moving at approximately \( 2.8563 \times 10^8 \text{ m/s} \).

Step by step solution

01

Understand Length Contraction in Special Relativity

According to the theory of special relativity, the observed length of an object moving relative to an observer is shorter than when it is at rest. This phenomenon is known as length contraction and is given by the formula: \( L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \), where \( L \) is the length of the object as measured by the observer, \( L_0 \) is the length of the object in its rest frame (proper length), \( v \) is the velocity of the object, and \( c \) is the speed of light.
02

Identify Given Values

From the problem, you have the proper length, \( L_0 = 1 \text{ m} \), and the contracted length measured by you, \( L = 1 \text{ ft} = 0.3048 \text{ m} \). We need to find the speed \( v \) of the meter stick relative to the observer.
03

Rearrange the Length Contraction Formula

Rearrange the length contraction formula to solve for \( v \): \( v = c \sqrt{1 - \left( \frac{L}{L_0} \right)^2} \).
04

Substitute Known Values Into the Equation

Substitute the known values into the equation: \( L = 0.3048 \text{ m} \), \( L_0 = 1 \text{ m} \), and \( c \approx 3 \times 10^8 \text{ m/s} \). The equation becomes \( v = 3 \times 10^8 \sqrt{1 - (0.3048)^2 } \).
05

Calculate the Result

Perform the calculation: First, calculate \( \left( \frac{L}{L_0} \right)^2 = (0.3048)^2 \approx 0.0929 \). Then find \( 1 - 0.0929 = 0.9071 \). Finally, calculate \( \sqrt{0.9071} \approx 0.9521 \). Therefore, \( v \approx 3 \times 10^8 \times 0.9521 \approx 2.8563 \times 10^8 \text{ m/s} \).
06

Conclusion: Express the Result

The speed of the meter stick relative to the observer is approximately \( 2.8563 \times 10^8 \text{ m/s} \).

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Key Concepts

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

Length Contraction
Length contraction is a fascinating concept in the realm of special relativity. It describes how the length of an object moving at high speeds becomes shorter when observed from a stationary point of view. The formula to understand this phenomenon is:
  • \( L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \)
Here:
  • \( L \) is the contracted length observed.
  • \( L_0 \) is the actual length when at rest.
  • \( v \) is the speed of the object.
  • \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \).

As the speed of the object approaches the speed of light, the length contraction becomes more pronounced. It's important to note that this contraction is only noticeable at speeds close to the speed of light. For everyday speeds, like a moving car or plane, length contraction is not apparent. This concept helps us understand the behavior of particles in accelerators and is crucial in astrophysics.
Lorentz Transformation
The Lorentz Transformation equations are the mathematical tools that allow us to change the coordinates of events in spacetime from one inertial frame to another. They form the backbone of Einstein’s theory of special relativity and ensure that the speed of light is constant in all frames of reference.
The transformations are useful to calculate how time and distance change for objects moving at significant fractions of the speed of light.
  • It connects time and space coordinates so that the speed of light remains the same for everyone, regardless of their motion.
  • While the transformation primarily deals with time dilation and length contraction, it's vital in demonstrating that space and time are interwoven into a single continuum known as spacetime.

For instance, if two observers are moving relative to each other, the Lorentz Transformation helps us understand how one observer perceives the time and space measurements taken by the other. This is key when working with high-speed objects that require relativity to explain their behavior accurately.
Speed of Light
The speed of light, denoted as \( c \), is one of the most fundamental constants in physics. Sitting at roughly \( 3 \times 10^8 \text{ meters/second} \), it's pivotal in theories of relativity. Not only is it the speed at which light travels, but it is also the maximum speed at which information and matter can travel in the universe.
  • The constancy of the speed of light, regardless of the motion of the light source or observer, is a cornerstone of Einstein’s theory of special relativity.
  • This constancy leads to phenomena like time dilation and length contraction, which seem counterintuitive but have been confirmed by numerous experiments.

It is also involved in the famous equation \( E = mc^2 \), showing the equivalence of energy and mass. The speed of light's constant value means that as objects move faster, their behavior diverges from classical physics, requiring relativistic mechanics to describe them accurately.
Velocity Calculation
Calculating the velocity of an object simply means determining how fast and in what direction it is moving. In the context of the exercise with the meter stick, we're interested in finding its velocity relative to an observer using length contraction results.
To find the velocity, we used the rearranged length contraction formula.
  • \( v = c \sqrt{1 - \left( \frac{L}{L_0} \right)^2} \)
This gives us insight into how fast the object must be moving to appear contracted to a particular length when viewed by the observer. Substituting the known values into the equation, we get its high-speed movement measured in meters per second.
Understanding the velocity in such problems is essential because it ties together the observable effects predicted by special relativity, like time dilation and length contraction, allowing physicists to make precise predictions about high-speed objects.

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Most popular questions from this chapter

A cube of metal with sides of length \(a\) sits at rest in a frame \(S\) with one edge parallel to the \(x\) -axis. Therefore, in \(S\) the cube has volume \(a^{3} .\) Frame \(S^{\prime}\) moves along the \(x\) -axis with a speed \(u\) . As measured by an observer in frame \(S^{\prime},\) what is the volume of the metal cube?

A \(0.100-\mu g\) speck of dust is accelerated from rest to a speed of 0.900\(c\) by a constant \(1.00 \times 10^{6} \mathrm{N}\) force. (a) If the nonrelativistic form of Newton's second law \((\Sigma F=m a)\) is used, how far does-the object travel to reach its final speed? (b) Using the correct relativistic treatment of Section 37.8 , how far does the object travel to reach its final speed? (c) Which distance is greater? Why?

(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 \(m c^{2}-)(c)\) How do your answers in parts \((a)\) and \((b)\) compare?

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.)

Calculate the magnitude of the force required to give a \(0.145-\mathrm{kg}\) baseball an acceleration \(a=1.00 \mathrm{m} / \mathrm{s}^{2}\) in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) \(10.0 \mathrm{m} / \mathrm{s} ;\) (b) 0.900 \(\mathrm{c}\) (c) \(0.990 c .(\mathrm{d})\) Repeat parts \((\mathrm{a}),\) and \((\mathrm{c})\) if the force and acceleration are perpendicular to the velocity.

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