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Chapter 37: Problem 19

An experimenter arranges to trigger two flashbulbs simultaneously, producing a big flash located at the origin of his reference frame and a small flash at \(x=30.0 \mathrm{~km}\). An observer moving at a speed of \(0.250 \mathrm{c}\) in the positive direction of \(x\) also views the flashes. (a) What is the time interval between them according to her? (b) Which flash does she say occurs first?

Short Answer

Expert verified
(a) Observer sees small flash first, \(2.58 \times 10^{-11}\) sec before big flash. (b) Small flash occurs first.

Step by step solution

01

Understanding the Problem

We are given two flashbulbs emitting light at different locations (one at the origin and another at 30.0 km) in an experimenter's reference frame. An observer is moving with speed \( v = 0.250c \) in the \( x \)-direction, where \( c \) is the speed of light. We need to find the time interval between these flashes as observed by her and identify which flash she perceives first.
02

Applying Lorentz Transformation for Time

According to the theory of relativity, time and space coordinates are related between two reference frames moving relative to each other. Here, we'll use Lorentz transformations. The time observed in the moving frame \( t' \) is related to the time \( t \) and position \( x \) in the stationary frame by \[t' = \gamma \left( t - \frac{vx}{c^2} \right)\]where \( \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \).
03

Calculating Lorentz Factor \( \gamma \)

Substitute \( v = 0.250c \) into the Lorentz factor formula:\[\gamma = \frac{1}{\sqrt{1 - (0.250)^2}}\]Calculate \( \gamma \):\[\gamma = \frac{1}{\sqrt{1 - 0.0625}} = \frac{1}{\sqrt{0.9375}} \approx 1.032\]
04

Calculating Time Intervals for Each Flash

For the first flash at the origin \((x_1 = 0)\), the time in both frames is zero \((t_1' = 0)\). For the second flash at \(x_2 = 30.0 \text{ km} = 30,000 \text{ m}\), the time in the stationary frame is zero \((t_2 = 0)\), we apply the transformation:\[t_2' = \gamma \left(-\frac{vx_2}{c^2}\right) = 1.032 \left(-\frac{0.250 \times 30,000}{(3 \times 10^8)^2}\right)\]Calculate:\[t_2' = 1.032 \times -\frac{7,500}{9 \times 10^{16}} \approx -2.58 \times 10^{-11} \text{ seconds}\]
05

Determining Observed Time Interval and Order

Since \( t_1' = 0 \) and \( t_2' \approx -2.58 \times 10^{-11} \text{ seconds}\), the observer sees the small flash (at \(x = 30.0\) km) occur first. The time interval according to the observer is:\[\Delta t' = t_1' - t_2' = 0 - (-2.58 \times 10^{-11}) = 2.58 \times 10^{-11} \text{ seconds}\]
06

Finalize Solution

The observer, moving at \(0.250c\), sees the small flash occur approximately \( 2.58 \times 10^{-11} \) seconds before the big flash.

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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 set of equations used to relate the space and time coordinates of two observers moving relative to each other with constant velocity, especially in systems where the relative velocity is close to the speed of light. This transformation makes the core principles of Einstein's theory of special relativity practical in equations.

Here's how it works:
  • It modifies our understanding of time and space when switching from one observer's frame to another's, particularly those traveling at high speeds.
  • A key equation in Lorentz transformation relates time in one frame to time and position in another frame: \[t' = \gamma \left( t - \frac{vx}{c^2} \right) = t \gamma - x \frac{v\gamma}{c^2}\]
  • In this equation, \( t \) and \( t' \) represent the time in different frames, \( x \) is the spatial coordinate, \( v \) is the relative velocity, \( c \) is the speed of light, and \( \gamma \) is the Lorentz factor.
The Lorentz transformation shows how time and measurements of length can differ depending on the observer's frame of reference.
time dilation
Time dilation is one of the fundamental concepts that arise from the principles of special relativity. It explains why time can pass at different rates for observers in different reference frames, particularly when one is moving at a significant fraction of the speed of light compared to the other.

Key points to understand about time dilation include:
  • The phenomenon implies that a clock moving relative to an observer will tick slower compared to a clock at rest in the observer's frame.
  • This effect becomes more pronounced as the velocity of the moving reference frame approaches the speed of light.
  • Mathematically, time dilation is described by the Lorentz factor \( \gamma \). The time interval \( \Delta t' \) in the moving frame is larger than the time interval \( \Delta t \) in the rest frame, calculated as: \(\Delta t' = \gamma \Delta t\).
This concept reveals that our intuitive understanding of time can shift significantly when dealing with high speeds.
special relativity
Special relativity, formulated by Albert Einstein in 1905, revolutionized the way we think about space, time, and energy. It is based on two postulates: the laws of physics are the same in all inertial frames of reference, and the speed of light in a vacuum is constant for all observers, regardless of their motion relative to the light source.

Some crucial elements of special relativity include:
  • The invariant speed of light: Unlike speeds for material objects, which can vary depending on the observer, the speed of light remains constant for all observers.
  • Relativity of simultaneity: Events that are simultaneous in one frame may not be simultaneous in another, particularly when the frames are moving relative to one another.
  • Mass-energy equivalence: Expressed as \( E=mc^2 \), it shows that energy and mass are interchangeable.
Special relativity changes our classical notions of absolute time and space, showing that measurements of these dimensions depend on the motion of the observer.
reference frames
A reference frame, in physics, is a perspective from which measurements and observations can be made. It often involves an observer who is either at rest or in uniform motion.

Important aspects of reference frames include:
  • They are essential for describing the position and motion of objects. Different reference frames can have different measurements for the same events or points.
  • Inertial frames are reference frames that are not accelerating. The laws of physics, per special relativity, are consistent in all inertial frames.
  • Non-inertial frames are experiencing acceleration. In these frames, additional forces called fictitious forces are observed, like the Coriolis force.
Understanding reference frames is key to solving many problems in physics, especially when determining how different observers perceive events differently due to their motion relative to each other.

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Most popular questions from this chapter

In a high-energy collision between a cosmic-ray particle and a particle near the top of Earth's atmosphere, \(120 \mathrm{~km}\) above sea level, a pion is created. The pion has a total energy \(E\) of \(1.35 \times 10^{5}\) \(\mathrm{MeV}\) and is traveling vertically downward. In the pion's rest frame, the pion decays \(35.0 \mathrm{~ns}\) after its creation. At what altitude above sea level, as measured from Earth's reference frame, does the decay occur? The rest energy of a pion is \(139.6 \mathrm{MeV}\).

The car-in-the-garage problem. Carman has just purchased the world's longest stretch limo, which has a proper length of \(L_{c}=30.5 \mathrm{~m}\). In Fig. \(37-32 a\), it is shown parked in front of a garage with a proper length of \(L_{g}=6.00 \mathrm{~m} .\) The garage has a front door (shown open) and a back door (shown closed). The limo is obviously longer than the garage. Still, Garageman, who owns the garage and knows something about relativistic length contraction, makes a bet with Carman that the limo can fit in the garage with both doors closed. Carman, who dropped his physics course before reaching special relativity, says such a thing, even in principle, is impossible. To analyze Garageman's scheme, an \(x_{c}\) axis is attached to the limo, with \(x_{c}=0\) at the rear bumper, and an \(x_{g}\) axis is attached to the garage, with \(x_{g}=0\) at the (now open) front door. Then Carman is to drive the limo directly toward the front door at a velocity of \(0.9980 \mathrm{c}\) (which is, of course, both technically and financially impossible). Carman is stationary in the \(x_{c}\) reference frame; Garageman is stationary in the \(x_{g}\) reference frame. There are two events to consider. Event \(1:\) When the rear bumper clears the front door, the front door is closed. Let the time of this event be zero to both Carman and Garageman: \(t_{g 1}=t_{c 1}=0\). The event occurs at \(x_{c}=x_{g}=0\). Figure \(37-32 b\) shows event 1 according to the \(x_{g}\) reference frame. Event 2 : When the front bumper reaches the back door, that door opens. Figure \(37-32 c\) shows event 2 according to the \(\bar{x}_{8}\) reference frame. According to Garageman, (a) what is the length of the limo, and what are the spacetime coordinates (b) \(x_{g^{2}}\) and (c) \(t_{g 2}\) of event \(2 ?\) (d) For how long is the limo temporarily "trapped" inside the garage with both doors shut? Now consider the situation from the \(x_{c}\) reference frame, in which the garage comes racing past the limo at a velocity of \(-0.9980 c\). According to Carman, (e) what is the length of the passing garage, what are the spacetime coordinates (f) \(x_{c 2}\) and (g) \(t_{c 2}\) of event \(2,(\mathrm{~h})\) is the limo ever in the garage with both doors shut, and (i) which event occurs first? (j) Sketch events 1 and 2 as seen by Carman. (k) Are the events causally related; that is, does one of them cause the other? (I) Finally, who wins the bet?

The premise of the Planet of the Apes movies and book is that hibernating astronauts travel far into Earth's future, to a time when human civilization has been replaced by an ape civilization. Considering only special relativity, determine how far into Earth's future the astronauts would travel if they slept for \(120 \mathrm{y}\) while traveling relative to Earth with a speed of \(0.9990 c\), first outward from Earth and then back again.

A Foron cruiser moving directly toward a Reptulian scout ship fires a decoy toward the scout ship. Relative to the scout ship, the speed of the decoy is \(0.980 \mathrm{c}\) and the speed of the Foron cruiser is \(0.900 \mathrm{c}\). What is the speed of the decoy relative to the cruiser?

How much work must be done to increase the speed of an electron (a) from \(0.18 c\) to \(0.19 c\) and (b) from \(0.98 c\) to \(0.99 c\) ? Note that the speed increase is \(0.01 c\) in both cases.
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