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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?

Short Answer

Expert verified
The volume in frame \( S^{\prime} \) is \( V' = a^3 \sqrt{1 - \frac{u^2}{c^2}} \).

Step by step solution

01

Understand the Problem

In this problem, we have a cube of metal with side length \( a \) in the frame \( S \). The frame \( S^{\prime} \) moves relative to \( S \) along the x-axis with speed \( u \). We want to find the volume of the cube as observed in \( S^{\prime} \). We will use the concept of length contraction from special relativity.
02

Recall Length Contraction

In special relativity, the length of an object moving with velocity \( u \) relative to an observer is contracted along the direction of motion. The formula for length contraction is \( L' = L \sqrt{1 - \frac{u^2}{c^2}} \), where \( L' \) is the contracted length, \( L \) is the proper length, and \( c \) is the speed of light.
03

Apply Length Contraction to the Cube

Since the cube's edge is parallel to the x-axis and the frame \( S^{\prime} \) moves parallel to this axis, only this dimension is affected by length contraction. The side length \( a \) along the x-axis becomes \( a' = a \sqrt{1 - \frac{u^2}{c^2}} \) in frame \( S^{\prime} \).
04

Calculate the Volume in Frame S'

In frame \( S^{\prime} \), the lengths perpendicular to the motion (in the y and z directions) remain unchanged. So, the volume \( V' \) in frame \( S^{\prime} \) is given by the product of the contracted length and the unchanged lengths: \( V' = a' \cdot a \cdot a = (a \sqrt{1 - \frac{u^2}{c^2}}) a^2 = a^3 \sqrt{1 - \frac{u^2}{c^2}} \).
05

Conclusion

The volume of the metal cube as measured by an observer in frame \( S^{\prime} \) is \( V' = a^3 \sqrt{1 - \frac{u^2}{c^2}} \). This shows that due to relativistic effects, the volume appears reduced in the direction of motion.

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Key Concepts

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

Length Contraction
When an object moves at a velocity close to the speed of light, its length along the direction of motion appears shortened to an observer in another frame. This phenomenon is known as length contraction and is a core concept of Einstein's theory of special relativity. The formula for length contraction is:\[ L' = L \sqrt{1 - \frac{u^2}{c^2}} \]where:
  • \( L' \) is the observed, contracted length.
  • \( L \) is the proper length, or the original length measured in the object's own rest frame.
  • \( u \) is the relative velocity between the observer and the object.
  • \( c \) is the speed of light.
This contraction occurs only in the direction of motion. For objects moving at everyday speeds, this effect is negligible. However, at speeds approaching the speed of light, the contraction becomes significant, altering perceptions of time and space for observers.
Relativistic Effects
Relativistic effects are changes in physical measurements observed when an object moves at velocities close to the speed of light. Special relativity introduces concepts such as time dilation and length contraction to explain these effects. - **Time Dilation**: Time appears to move slower for moving objects compared to those at rest, as observed from another frame. - **Length Contraction**: Objects in motion appear shortened in the direction of travel. As velocities increase, these relativistic effects become more pronounced, challenging classical Newtonian physics. They predict phenomena that deviate from non-relativistic expectations, such as a contracted length or dilated time, altering how distances and durations are perceived between different frames of reference. Understanding these effects is essential when dealing with objects approaching the speed of light.
Volume Calculation
To calculate volume in the context of relativistic motion, it's crucial to consider the dimensions affected by length contraction. In our example of a cube, only the dimension parallel to the motion—along the x-axis—is contracted. Here's how the volume is calculated:- The side parallel to motion contracts: \( a' = a \sqrt{1 - \frac{u^2}{c^2}} \)- Other sides remain the same: length \( a \)The resulting volume in the moving frame \( S' \) is given by:\[ V' = a' \cdot a \cdot a = a^3 \sqrt{1 - \frac{u^2}{c^2}} \]This formula shows that the volume is reduced by a factor equivalent to the length contraction. Understanding the volume change helps grasp how space is perceived differently in moving frames, crucial for fields like astrophysics and high-energy physics.
Inertial Frames
Inertial frames are reference frames in which an object remains at rest or moves at a constant velocity unless acted upon by an external force. They are fundamental in understanding relativity because they allow for the laws of physics to be consistent.Key points about inertial frames:
  • They move at constant velocity, without acceleration.
  • The laws of physics are the same in all inertial frames.
  • Observations between inertial frames can lead to relativistic effects, like time dilation and length contraction.
In our problem, frame \( S \) is at rest, while frame \( S' \) moves at a constant velocity \( u \) along the x-axis. This setup allows for the application of special relativity, as we compare measurements like length and volume between these two inertial frames. Understanding inertial frames is essential to solving many problems in relativity, providing a foundational perspective on how motion and forces operate within different frames of reference.

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Most popular questions from this chapter

How fast must a rocket travel relative to the earth so that time in the rocket "slows down" to half its rate as measured by earth-based observers? Do present-day jet planes approach such speeds?

Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is 0.650 \(\mathrm{c}\) , and the speed of each partcle relative to the other is 0.950 \(\mathrm{c}\) . What is the speed of the second particle, as measured in the laboratory?

A \(\psi\) ( psi) particle has mass \(5.52 \times 10^{-2} \mathrm{kg}\) . Compute the rest energy of the \(\psi\) particle in MeV.

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 \(f_{0} ?(\text {Hint} \text { . In this case, successive wave crests }\) 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 \(f_{0}=345 \mathrm{MHz}\) 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, \(f-f_{0} ?(\mathrm{c})\) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) 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)?

An observer in frame \(S^{\prime}\) is moving to the right \((+x-\text { direction })\) at speed \(u=0.600 \mathrm{c}\) away from a stationary observer in frame \(S\) . The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in S measure for the particle if \((a) v^{\prime}=0.400 c ;(b) v^{\prime}=0.900 c\) (c) \(v^{\prime}=0.990 c ?\)

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