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Chapter 2: Problem 21

According to an observer on Earth, a spacecraft whizzing by at \(0.6 c\) is \(35 \mathrm{~m}\) long. What is the length of the spacecraft according to passengers on board?

Short Answer

Expert verified
The length of the spacecraft according to the passengers on board (i.e., the proper length) is \(43.75m\).

Step by step solution

01

Identifying given values

The speed of the spacecraft according to an observer on earth \(v = 0.6c\) and its observed length \(L = 35m\)
02

Analyzing the concept of length contraction

According to the concept of length contraction in special relativity, the length of an object in motion (from the viewpoint of an observer at rest) is always shorter than its proper length, (i.e., its length in its own rest frame). This is given by the equation \(L=L_0\sqrt{1- v^2/c^2}\), where \(L_0\) is the proper length, \(L\) is the contracted length, \(v\) is the velocity of the object and \(c\) is the speed of light.
03

Calculating the proper length

We need to find the proper length \(L_0\), so we rearrange the equation to \(L_0= L/\sqrt{1- v^2/c^2}\), and substitute the given values into the equation to obtain: \(L_0= 35m/\sqrt{1- (0.6c)^2/c^2}= 35m/\sqrt{1- 0.36}= 35m/\sqrt{0.64} = 43.75m\).

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Key Concepts

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

Special Relativity
Imagine a universe where high-speed travel could make the world around us change shape and time stretch or squeeze. This is not the stuff of science fiction, but very much a reality, explained by special relativity, a groundbreaking theory proposed by Albert Einstein in 1905. At its core, special relativity is about how space and time are not absolute concepts, but rather, they are interlinked in a four-dimensional space-time.

When objects move at velocities close to the speed of light (denoted as 'c'), their length, the flow of time, and even mass are perceived differently by observers in motion relative to those objects and those at rest. For example, if you were standing still and watching a spacecraft zip past you at a significant fraction of the speed of light, it would appear shorter than if you were riding inside it. This phenomenon is known as length contraction, one of the mind-bending implications of relativity that becomes significant at high velocities.
Proper Length
To delve deeper into the weirdness of special relativity, let's talk about proper length. This is the length of an object as measured by an observer who is at rest relative to the object. Hence, proper length is also called 'rest length', as it is the length of the object when it's not whizzing past at relativistic speeds.

On a day-to-day basis, measuring lengths doesn't require contemplation about relative velocities. This is because common speeds are much too slow to have any noticeable relativistic effects. However, when we start considering objects like our aforementioned spacecraft traveling at 0.6 times the speed of light, these effects can't be ignored. In our exercise, the proper length is the 'real' length of the spacecraft as measured by passengers on board who are moving with it and not subject to length contraction.
Lorentz Transformation
So, how do we switch between the viewpoints of different observers in special relativity? Enter the Lorentz transformation, a set of equations that connect the space and time coordinates of one frame of reference to another that is in uniform motion relative to the first. These transformations are named after the Dutch physicist Hendrik Lorentz, who actually formulated them before Einstein's special theory of relativity.

The Lorentz transformation takes into account that the speed of light is the same in all inertial frames of reference. They are also how the variables in our exercise change: from the length observed by someone at rest on Earth to the proper length experienced by passengers in the spacecraft. Mathematically, these transformations beautifully encapsulate the way distances contract and clocks dilate, radically altering our classical, intuitive notions of space and time. Without the Lorentz transformation, there would be no way to accurately describe the relativistic effects observed in the exercise or any similar scenario where relativistic speeds are involved.

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Most popular questions from this chapter

If it is fundamental to nature that a given mass has a critical radius at which something extraordinary happens (i.e., a black hole forms), we might guess that this radius should depend only on the mass and fundamental constants of nature. Assuming that \(r_{\text {critical }}\) depends only' on \(M, G\), and \(c\), show that dimensional analysis gives the equation for the Schwarzschild radius to within a multiplicative constant.

A muon has a mean lifetime of \(2.2 \mu \mathrm{s}\) in its rest frame. Suppose muons are traveling at \(0.92 c\) relative to Earth. What is the mean distance a muon will travel as measured by an observer on Earth?

Both classically and relativistically, the force on an object is what causes a time rate of change of its momentum: \(F=d p / d t\). (a) Using the relativistically cotrect expression for momentum, show that $$ F=\gamma_{u}^{3} m \frac{d u}{d l} $$ (b) Under what condition does the classical equation \(F=m a\) hold? \(?\) (c) Assuming a constant force and that the speed is zero at \(t=0\), separate \(t\) and \(u\), then integrate to show that $$ u=\frac{1}{\sqrt{1+(F t / m c)^{2}}} \frac{F}{m} t $$ (d) Plot \(u\) versus \(t\). What happens to the velocity of an object when a constant force is applied for an indefinite length of time?

If an object actually occupies less space physically when moving, it cannot depend on the direction we define as positive. As we know, an object aligned with the direction of relative motion is contracted whether it is fixed in frame \(S\) and viewed from \(S^{\prime}\), or the other way around. Use this idea to argue that distances along the \(y\) - and \(y^{\prime}\) -axes cannot differ at all. Consider a post of length \(L_{0}\) fixed in frame \(S\), jutting up from the origin along the \(+y\) -axis, with a saw at the top poised to slice off anything extending any higher in the passing frame \(S\). Also consider an identical post fixed in frame \(S\). What happens when the origins cross?

(a) Determine the Lorentz transformation matrix giving position and time in frame \(S^{\prime}\) from those in frame \(S\) for the case \(v=0.5 c\) ( (b) If frame \(S^{\prime \prime}\) moves at \(0.5 c\) relative to frame \(S^{\prime}\), the Lorentz transformation matrix is the same as the previous one. Find the product of the two matrices, which gives \(x^{\prime \prime}\) and \(t^{\prime \prime}\) from \(x\) and \(t\). (c) To what single speed does the transformation correspond? Explain this result.
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