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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 \( S' \) is \( a^3 \sqrt{1 - \frac{u^2}{c^2}} \).

Step by step solution

01

Understanding Special Relativity

Due to the effects of special relativity, when an object moves with respect to an observer, its length contracts along the direction of motion. This phenomenon is known as "length contraction."
02

Calculate Length Contraction Factor

The length contraction factor is given by the Lorentz factor \[ \gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}\]where \( u \) is the speed of the moving frame \( S' \) relative to \( S \) and \( c \) is the speed of light.
03

Apply Length Contraction to Cube's Edge

The edge of the cube parallel to the \( x \)-axis contracts in the frame \( S' \) by the factor \( \frac{1}{\gamma} \). Thus, the new length \( a' \) is \[a' = \frac{a}{\gamma}\]
04

Compute Volume in Moving Frame

In the frame \( S' \), only the length along the \( x \)-axis changes. Therefore, the volume of the cube in \( S' \) is given by \[ V' = a' a a = \frac{a^3}{\gamma}\]
05

Substitute Length Contraction Factor

Substitute the expression for \( a' \) into the volume equation: \[ V' = a \times a \times \frac{a}{\gamma} = \frac{a^3}{\gamma}\]Substituting \( \gamma \) gives\[ V' = a^3 \times \sqrt{1 - \frac{u^2}{c^2}}\]

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Key Concepts

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

Length Contraction
In the world of special relativity, things behave a bit differently than we're used to in classical mechanics. One key effect is called length contraction. This happens when an object moves at a high speed relative to an observer. The fascinating part is that the object appears shorter along the direction of motion. This effect isn't something you'd notice in everyday life, since it only becomes significant at speeds close to the speed of light.

For a cube resting in one frame, when viewed from another moving frame, the edge of the cube parallel to the motion will appear contracted. This is essential to calculating the new volume of shapes like cubes in moving frames. The contraction can be calculated using a special factor called the Lorentz factor, which helps us understand how much shorter the object appears. We'll explore that in the next section.
Lorentz Factor
The Lorentz factor is a crucial concept in special relativity. It's named after the physicist Hendrik Lorentz, who contributed to the theory's foundation. This factor tells us how time and space transform for objects moving at significant fractions of the speed of light.

Mathematically, the Lorentz factor, often symbolized as \( \gamma \), is given by:
  • \( \gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \)
Here, \( u \) is the velocity of the moving frame, and \( c \) is the speed of light. As you can see, the formula consists of a square root involving the object's velocity squared divided by the speed of light squared. This makes \( \gamma \) always greater than or equal to 1.

As speed \( u \) approaches the speed of light, \( \gamma \) grows larger, indicating stronger relativistic effects. For length contraction, the contracted length \( a' \) can be found by dividing the original length \( a \) by \( \gamma \). This calculation helps us when we need to consider effects like volume change, as seen with our metal cube example.
Relativistic Volume
In special relativity, when you talk about volume, you're dealing with how space itself appears to change due to motion. For our cube sitting in a moving frame, only the side parallel to the direction of motion is contracted, thanks to length contraction. This results in a change in the measured volume.

The original volume of the cube is \( a^3 \), which holds true in its rest frame. In the new, moving frame, only one of the cube’s dimensions is affected by the speed of motion. So, we apply the contraction to that specific length, while the other dimensions remain unchanged. The volume in this moving frame \( V' \) then becomes:
  • \( V' = \frac{a^3}{\gamma} \)
This new contracted volume depends on the Lorentz factor \( \gamma \). As the speed increases, \( \gamma \) increases, resulting in a smaller volume.

In essence, the volume appears less than what it is at rest in the moving observer's perspective. Importantly, this understanding helps in realizing how space and time are interconnected in the vast cosmos, influenced largely by the object's speed relative to the observer.

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Most popular questions from this chapter

Tell It to the Judge. (a) How fast must you be approaching a red traffic light \((\lambda=675 \mathrm{nm})\) for it to appear yellow \((\lambda=575 \mathrm{nm})\) ? 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 \(\mathrm{km} / \mathrm{h}\) .

After being produced in a collision between elementary particles, a positive pion \(\left(\pi^{+}\right)\) must travel down a 1.90 -km-long tube to reach an experimental area. A \(\pi^{+}\) particle has an average lifetime (measured in its rest frame) of \(2.60 \times 10^{-8} \mathrm{s}\) ; the \(\pi^{+}\) we are considering has this lifetime. How fast must the \(\pi^{+}\) travel if it is not to decay before it reaches the end of the tube? (Since \(u\) will be very close to \(c,\) write \(u=(1-\Delta) c\) and give your answer in terms of \(\Delta\) rather than \(u .\) (b) The \(\pi^{+}\) has a rest energy of 139.6 \(\mathrm{MeV} .\) What is the total energy of the \(\pi^{+}\) at the speed calculated in part (a)?

Compute the kinetic energy of a proton (mass \(1.67 \times\) \(10^{-27} \mathrm{kg}\) ) using both the nonrelativistic and relativistic expressions, and compute the ratio of the two results (relativistic divided by nonrelativistic for speeds of (a) \(8.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) and (b) \(2.85 \times\) \(10^{8} \mathrm{m} / \mathrm{s} .\)

Two events observed in a frame of reference \(S\) have positions and times given by \(\left(x_{1}, t_{1}\right)\) and \(\left(x_{2}, t_{2}\right),\) respectively. (a) Frame \(S^{\prime}\) moves along the \(x\) -axis just fast enough that the two events occur at the same position in \(S^{\prime} .\) Show that in \(S^{\prime},\) the time interval \(\Delta t^{\prime}\) between the two events is given by $$\Delta t^{\prime}=\sqrt{(\Delta t)^{2}-\left(\frac{\Delta x}{c}\right)^{2}}$$ where \(\Delta x=x_{2}-x_{1}\) and \(\Delta t=t_{2}-t_{1}\) . Hence show that if \(\Delta x>c \Delta t,\) there is \(n o\) frame \(S^{\prime}\) in which the two events occur at the same point. The interval \(\Delta t^{\prime}\) is sometimes called the proper time interval for the events. Is this term appropriate? (b) Show that if \(\Delta x>c \Delta t,\) there is a different frame of reference \(S^{\prime}\) in which the two events occur simultaneously. Find the distance between the two events in \(S^{\prime} ;\) express your answer in terms of \(\Delta x, \Delta t,\) and \(c\) This distance is sometimes called a proper length. Is this term appropriate? (c) Two events are observed in a frame of reference \(S^{\prime}\) to occur simultaneously at points separated by a distance of 2.50 \(\mathrm{m} .\) In a second frame \(S\) moving relative to \(S^{\prime}\) along the line joining the two points in \(S^{\prime},\) the two events appear to be separated by 5.00 \(\mathrm{m} .\) What is the time interval between the events as measured in \(S ?[\)Hint : Apply the result obtained in part (b).]

In high-energy physics, new particles can be created by collisions of fast- moving projectile par- ticles 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 kaon \(\left(\mathrm{K}^{+}\right)\) $$p+p \rightarrow p+p+\mathrm{K}^{-}+\mathrm{K}^{+}$$ (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 \(\mathrm{MeV}\) . (Hint: It is useful here to work in the frame in which the total momentum is zero. But note that 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 kinetic 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.)

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