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Chapter 1: Problem 37

An observer in reference frame \(S\) sees two events as simultaneous. Event \(A\) occurs at the point \((50.0 \mathrm{~m}, 0,0)\) at the instant 9:00:00 Universal time, 15 January 2001 . Event \(B\) occurs at the point \((150 \mathrm{~m}, 0,0)\) at the same moment. A second observer, moving past with a velocity of \(0.800 c \hat{1}\), also observes the two events. In her reference frame \(\mathrm{S}^{\prime}\), which event occurred first and what time elapsed between the events?

Short Answer

Expert verified
In reference frame \( S' \), event B occurred first followed by event A. The time elapsed between the two events is \( 133.33 \, \mathrm{ns} \).

Step by step solution

01

Determine the observers' relative velocity

The relative velocity between the two observers is given as 0.800 times the speed of light, \(c\). Therefore, \( v = 0.800c \)
02

Calculate the Lorentz factor

The Lorentz factor, \( \gamma \), is defined as \( \gamma = \frac{1}{\sqrt{1 - (v/c)^2}} \). Substituting \( v = 0.800c \) into the expression, we find that \( \gamma = \frac{1}{\sqrt{1 - (0.800)^2}} = 1.6667 \).
03

Apply Lorentz transformation for time

The Lorentz transformation for time, \( t' \), is given by: \( t' = \gamma(t - \frac{vx}{c^2}) \), where \( x \) is the event's position in the \( S \) frame. Since both the events A and B were observed simultaneously in frame S, \( t = 0 \). Therefore, for event A, \( t'_A = \gamma(\frac{vx_A}{c^2}) = 1.6667(-\frac{0.800c(50.0 \, \mathrm{m})}{c^2}) = -66.67 \, \mathrm{ns} \) , and similarly for event B, \( t'_B = \gamma(\frac{vx_B}{c^2}) = 1.6667(-\frac{0.800c(150 \, \mathrm{m})}{c^2}) = -200 \, \mathrm{ns} \) .
04

Determine which event occurred first

Since \(t'_A > t'_B\), we conclude that in reference frame \(S'\), event B occurred before event A.
05

Calculate time elapsed between events

The time elapsed between the events is given by the difference \( \Delta t' = t'_A - t'_B = -66.67 \, \mathrm{ns} - (-200 \, \mathrm{ns}) = 133.33 \, \mathrm{ns} \) .

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Key Concepts

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

Lorentz transformation
The Lorentz transformation is a crucial concept in Einstein's theory of relativity. It mathematically describes how the measurements of time and space by two observers moving relative to each other are related. This transformation accounts for the differences in their view of events based on their relative velocity. Through the Lorentz transformation equations:
  • Time and space coordinates are adjusted to reflect the differences in observations.
  • Time dilation and length contraction are key effects derived from this transformation.
In the exercise, the Lorentz transformation revealed that even though two events were simultaneous in one reference frame, they were not in another, highlighting how different observers might view the order of events differently.
reference frames
A reference frame is essentially a point of view from which measurements are taken. When dealing with relativity, it's important to specify the reference frame, as measurements such as time and distance can vary between frames. There are two types, typically:
  • Inertial reference frames, where an observer is either at rest or moving at a constant velocity.
  • Non-inertial frames, where the observer accelerates or moves with varying speeds.
In our exercise, frame \( S \) is an inertial frame where the observer sees two events as simultaneous. Frame \( S' \), moving at a significant fraction of the speed of light, allows us to understand how different reference frames lead to different observations of time for the same events.
relativity of simultaneity
Simultaneity is not absolute in Einstein's relativity. An event that appears simultaneous to one observer may not appear so to another if they are in different states of motion. This phenomenon is known as the relativity of simultaneity.The exercise demonstrated this when two events seemed simultaneous in frame \( S \) but were non-simultaneous in frame \( S' \).
  • This results from the finite speed of light, which causes different observers to "receive" information about events at different times depending on their relative motion.
  • Such discrepancies are only noticeable at high velocities, particularly those approaching the speed of light.
observer motion
The motion of an observer affects how events are perceived and measured. Observers traveling at different velocities, particularly close to the speed of light, will perceive time and space differently from stationary observers, due to relativistic effects.
  • Velocity changes how observers measure the timing of events, leading to phenomena like time dilation.
  • In our exercise, the observer in frame \( S' \) was moving past the original events at \( 0.800c \), showing how higher speeds can alter the perception of the order and timing of events.
This concept emphasizes the need to consider the relative motion of observers when analyzing scenarios in relativistic contexts, as it can dramatically shift the interpretations of what happens and when.

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Most popular questions from this chapter

Suppose our Sun is about to explode. In an effort to escape, we depart in a spaceship at \(v=0.80 c\) and head toward the star Tau Ceti, 12 lightyears away. When we reach the midpoint of our journey from the Earth, we see our Sun explode and, unfortunately, at the same instant we see Tau Ceti explode as well. (a) In the spaceship's frame of reference, should we conclude that the two explosions occurred simultaneously? If not, which occurred first? (b) In a frame of reference in which the Sun and Tau Ceti are at rest, did they explode simultaneously? If not, which exploded first?

In 1962, when Scott Carpenter orbited Earth 22 times, the press stated that for each orbit he aged 2 millionths of a second less than if he had remained on Earth. (a) Assuming that he was \(160 \mathrm{~km}\) above Earth in an eastbound circular orbit, determine the time difference between someone on Earth and the orbiting astronaut for the 22 orbits. (b) Did the press report accurate information? Explain.

A rod of length \(L_{0}\) moves with a speed \(v\) along the horizontal direction. The rod makes an angle of \(\theta_{0}\) with respect to the \(x\) -axis. (a) Show that the length of the rod as measured by a stationary observer is given by \(L=L_{0}\left[1-\left(v^{2} / c^{2}\right) \cos ^{2} \theta_{0}\right]^{1 / 2}\). (b) Show that the angle that the rod makes with the x-axis is given by the expression tan \(\theta=\gamma \tan \theta_{0}\). These results show that the rod is both contracted and rotated. (Take the lower end of the rod to be at the origin of the primed coordinate system.)

The identical twins Speedo and Goslo join a migration from Earth to Planet \(\mathrm{X}\). It is \(20.0\) ly away in a reference frame in which both planets are at rest. The twins, of the same age, depart at the same time on different spaceships. Speedo's ship travels steadily at \(0.950 c\), and Goslo's at \(0.750 c\). Calculate the age difference between the twins after Goslo's spaceship reaches Planet \(\mathrm{X}\). Which twin is the older?

A clock on a moving spacecraft runs \(1 \mathrm{~s}\) slower per day relative to an identical clock on Earth. What is the relative speed of the spacecraft? (Hint: For \(v / c<1\), note that \(\gamma \approx 1+v^{2} / 2 c^{2}\).)
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