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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?

Short Answer

Expert verified
Distances along the y and y'-axes do not differ in both frames because there is no length contraction along these axes. When the origins of \(S\) and \(S'\) cross, no sawing occurs since the posts along y and y'-axis remain the same length \(L_{0}\) on both frames.

Step by step solution

01

Understanding the Setup

Recognize that we are dealing with two identical posts of length \(L_{0}\) in different frames of reference, \(S\) and \(S'\). The post in \(S\) is along the +y-axis, while the direction of relative motion is the x or x'-axis.
02

Identifying symmetry

Understand that because physical laws are the same in both directions along any given axis, contrasts along the y and y'-axes should be identical whether we consider motion to be in positive or negative direction.
03

Applying Length Contraction

The Length Contraction formula is given by \(L = L_{0}/\gamma\), where \(L_{0}\) is the proper length and \(\gamma\) is the Lorentz factor, which is greater than 1 for any relative speed. Note that the length contraction formula only applies along the direction of motion. In this case, the direction of motion is the x-axis, not the y-axis. Therefore, length contraction does not occur along the y-axis.
04

Analyze Outcome

When the origins of \(S\) and \(S'\) cross, no sawing will occur since both posts in their respective frames will remain unchanged along the y-axis. This indicates that on both frames, the y and y'-lengths remain \(L_{0}\), resulting in no differences.

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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 within Einstein's theory of relativity, capturing how objects physically contract along the direction of their motion relative to an observer. This contraction occurs only along the direction of travel and not along perpendicular directions. For example, consider a spaceship traveling at speeds close to the speed of light, along the x-axis. From an observer's point of view at rest relative to the spaceship, the ship appears shorter in the direction of motion (x-axis) compared to its proper length when at rest. The contraction is described by the formula \( L = \frac{L_0}{\gamma} \), where \( L_0 \) is the proper length (the length of the object in its rest frame), and \( \gamma \) is the Lorentz factor. This factor is given by \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( v \) is the relative velocity and \( c \) is the speed of light.
  • Contraction is noticeable only at extremely high speeds, close to the speed of light.
  • Only the length along the direction of movement changes; other dimensions remain unaffected.
Lorentz Transformation
The Lorentz Transformation provides the mathematical framework to understand how measurements of space and time by two observers are related to each other when they are in relative motion. Imagine two observers, one stationary and one moving at a constant velocity relative to one another. The Lorentz Transformation helps translate measurements from one observer's frame to the other's.The transformation equations relate coordinates \((x, y, z, t)\) in one frame, say frame \( S \), to coordinates \((x', y', z', t')\) in another, say frame \( S' \), moving with respect to a constant velocity \( v \), as follows:\[x' = \gamma (x - vt) \t' = \gamma \left(t - \frac{vx}{c^2}\right)\]The coordinates \( y \) and \( z \) remain unchanged since motion occurs along the x-axis.
  • Frames of reference play a key role in understanding measurements, as no measurement is absolute—it all depends on the observer's state of motion.
  • The transformation ensures the constancy of the speed of light in all frames of reference.
Frames of Reference
Frames of reference are crucial in the study of physics, particularly in understanding motion and relativity. A frame of reference is essentially a set of coordinates that an observer uses to measure the position, orientation, and other physical quantities of objects. It is akin to the vantage point from which an observer perceives the universe. Every observer has a unique frame of reference depending on their motion. For instance, if you are sitting in a car moving at a steady speed, you might consider the car as stationary while everything outside moves. However, an observer standing by the roadside sees you moving with the car.
  • Inertial Frame: Here, the observer moves at a constant velocity—no accelerations involved.
  • Non-inertial Frame: The observer experiences acceleration, like sitting in a car that’s speeding up.
Understanding frames of reference helps explain and predict how different observers might describe the same event differently, all using valid perspectives.

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Most popular questions from this chapter

A friend says, "It makes no sense that Anna could turn on lights in her hands simultaneously in her frame but that they don't turn on simultaneously in another - she decides to do it, and it's done" By considering the tractable, if somewhat unrealistic, situation of Anna's thought being communicated to her hands by light signals, answer this objection.

Does the asymmetric aging of an Earth-bound observer and his twin who travels away and back demand "relativistic speed"? (Overlook the fact that each has a limited life span.)

In the collision shown, energy is conserved. because both objects have the same speed and mass after as before the collision. Since the collision merely reverses the velocities, the final (total) momentum is opposite the initial. Thus, momentum can be conserved ooly if it is zero. (a) Using the relativistically correct expression for momentum, show that the total momentum is zero that momentum is conserved. (Masses are in arbitrary units.) (b) Using the relativistic velocity transformation, find the four velocities in a frame moving to the right at \(0.6 c\) (c) Verify that momentum is conserved in che new frame.

(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.

Anna is on a railroad flatcar moving at \(0.6 c\) relative to Bob. (Their clocks read 0 as Anna's center of mass passes Bob's.) Anna's arm is outstretched in the direction the flatcar moves, and in her hand is a flashbulb. According to the wristwatch on Anna's hand. the flashbulb goes off at \(100 \mathrm{~ns}\). The time of this event according to Bob differs by 27 ns. (a) Is it earlier or later than 100 ns? (b) How long is Anna's arm (i.e., from her hand to her center of mass)?

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