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Chapter 2: Problem 23

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)?

Short Answer

Expert verified
(a) The event according to Bob is later than 100 ns. (b) Anna's arm length is 8.1 meters.

Step by step solution

01

Determining time according to Bob

First, note that the given time difference of 27 ns is due to time dilation which occurs because Bob is in a different inertial reference frame moving relative to Anna. By using the formula for time dilation which is \(t_B = \gamma t_A\) where \(t_B\) represents the time measured by Bob, \(t_A\) represents the time measured by Anna and \(\gamma = 1/ \sqrt{1 - v^2/c^2}\) is the Lorentz factor. Here, \(v = 0.6c\) is the relative velocity between Anna and Bob's frames. Since Bob is in the relatively moving frame, the time he measures is dilated or stretched with respect to what Anna measures. Therefore, the event according to Bob is later than 100 ns by 27 ns.
02

Calculating the Lorentz factor

The Lorentz factor \(\gamma\) can be calculated as \(\gamma = 1/ \sqrt{1 - (0.6)^2} = 1.25\).
03

Calculating Anna's arm length

The time delay of 27 ns that Bob observes is due to the time it takes for the light signal to travel from Anna's flashbulb (at her hand) to her center of mass. The speed of light \(c\) is about \(3 \times 10^8 m/s\). Therefore, her arm length which represents the distance the light signal travels can be calculated using the equation \(distance = speed \times time\) i.e. \(distance = c \times time\). Here, the time delay is converted from nanoseconds to seconds before the calculation. Therefore, Anna's arm length \(= c \times 27 \times 10^{-9} s = 8.1 m\).

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Key Concepts

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

Special Relativity
Special relativity, formulated by Albert Einstein in 1905, revolutionized the way we understand time and space. At its core, it's a theory that addresses the relationship between space and time under the absence of gravity and when relative velocities approach the speed of light, denoted as 'c'.

One of its most intriguing aspects is the notion that time is not absolute but relative - it can pass at different rates for observers who are in motion relative to each other. This leads to phenomena such as time dilation, where moving clocks tick more slowly compared to those at rest. The theory relies on two fundamental postulates: the laws of physics are the same in all inertial reference frames, and the speed of light in a vacuum is constant regardless of the motion of the light source or observer.

For example, in the exercise provided, Anna's perception of time while moving at a significant fraction of the speed of light (0.6c) is different from Bob’s, who is in a different frame of reference. This is due to the effects predicted by special relativity.
Lorentz Factor
The Lorentz factor is a crucial component of special relativity, featured in the time dilation and length contraction formulas. It is represented by the symbol \(\gamma\) and calculated using the equation \(\gamma = 1/ \sqrt{1 - v^2/c^2}\), where 'v' is the relative velocity of two inertial reference frames and 'c' is the speed of light in a vacuum.

It quantifies by how much time, length, and relativistic mass change for an object while it is moving at a velocity 'v'. The closer this velocity comes to the speed of light, the larger the Lorentz factor becomes, leading to more pronounced relativistic effects. For example, when Anna travels at 0.6 times the speed of light, her Lorentz factor is calculated to be 1.25, indicating a significant time dilation effect as observed by Bob.
Inertial Reference Frames
Inertial Reference Frames are key to understanding many concepts in special relativity including the exercise in question. An inertial reference frame is essentially a perspective from which an observer measures physical phenomena without the interference of any forces causing acceleration.

When observing such phenomena as motion, time, and space, it's crucial that the frame of reference is inertial, meaning it's either at rest or moving at a constant velocity. This allows for consistent laws of physics, as described by the first postulate of special relativity. For instance, Anna and Bob are in different inertial reference frames, with Bob being stationary and Anna moving at a relative speed. Due to this relative motion, the time measured for the flashbulb event is different for Bob than for Anna, exemplifying the relativity of simultaneity. This is a perfect demonstration of how crucial inertial frames are to the understanding of relativistic effects.

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Most popular questions from this chapter

You are gliding over Earth's surface at a high speed. carrying your high- precision clock. At points \(X\) and \(Y\) on the ground are similar clocks, synchronized in the ground frame of reference. As you pass over clock X, it and your clock both read 0 . (a) According to you, do clocks \(X\) and \(Y\) advance slower or faster than yours? (b) When you pass over clock \(Y\), does it read the same time, an earlier time or a later time than yours? (Make sure your answer aguces with what ground observers should see.) (c) Reconcile any seeming contradictions berween your answers to parts \((a)\) and \((b)\).

A kaon (denoted \(\left.\mathrm{K}^{\bullet}\right)\) is an unstable particle of mass \(8.87 \times 10^{-28} \mathrm{~kg}\). One of the means by which it decays is by spontaneous creation of two pions. a \(\pi^{+}\) and a \(\pi^{-}\). The decay process may be represented as $$ \mathrm{K}^{0} \rightarrow \pi^{+}+\pi^{-} $$ Both the \(\pi^{+}\) and \(\pi^{-}\) have mass \(2.49 \times 10^{-28} \mathrm{~kg}\). Suppose that a kaon moving in the \(+x\) direction decays by this process, with the \(\pi^{+}\) moving off at speed \(0.9 \mathrm{c}\) and the \(\pi^{-}\) at \(0.8 c\). (a) What was the speed of the kaon before decay? (b) In what directions do the pions move af ter the decay?

In a television picture tube, a beam of electrons is sent from the back to the front (screen) by an electron gun. When an electron strikes the screen, it causes a phosphor to glow briefly. To produce an image across the entire screen. the beam is electrically deflected up and down and left and right. The beam may sweep from left to right at a speed greater than \(c\). Why is this not a violation of the claim that no information may travel faster than the speed of light?

A \(3.000\) u ob ject moving to the right through a laboratory at \(0.8 c\) collides with a \(4.000\) u ob ject moving to the left through the laboratory at \(0.6 c\). Afterward, there are two objects. one of which is a \(6.000\) u mass at rest. (a) What are the mass and speed of the other object? (b) Determine the change in kinetic energy in this collision.

Equation \((2-30)\) is an approximation correct only if the gravitational time- dilation effect is small. In this exercise, it is also assumed to be small. but we still allow for a nonuniform gravitational field. We stan with (2-29), based on the Doppler effect in the acceleraing frame. Consider two elevations, the lower at \(r_{1}\) and the upper at \(r_{1}+d r\). Equation \((2-29)\) becomes $$ \frac{f\left(r_{1}+d r\right)}{f\left(r_{i}\right)}=\left(1-\frac{g\left(r_{1}\right) d r}{c^{2}}\right) $$ Similarly, if we consider elevations \(r_{1}+d r\) and \(r_{1}+\) \(2 d r\). we have $$ \frac{f\left(r_{1}+2 d r\right)}{f\left(r_{1}+d r\right)}=\left(1-\frac{g\left(r_{1}+d r\right) d r}{c^{2}}\right) $$ We continue the process, incrementing \(r\) by \(d r\), until we reach \(r_{2}\) $$ \frac{f\left(r_{2}\right)}{f\left(r_{2}-d r\right)}=\left(1-\frac{g\left(r_{2}-d r\right) d r}{c^{2}}\right) $$ Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply \(f\left(r_{2}\right) / f\left(r_{1}\right)\). (b) Assuming that the term \(g d r / c^{2}\) in each individual equation is very small. so that products of such terms can be ignored, argue that the right side of the product is $$ 1-\frac{1}{c^{2}} \int g(r) d r $$ (c) Deduce \(g(r)\) from Newton's universal law of gravitation, then argue that equation \((2-31)\) follows from the result, just as \((2 \cdot 30)\) does from \((2-29)\).
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