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Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when the front of Mavis's ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate x and t as measured by Stanley for the event of turning on the light. (b) Use the time dilation formula, Eq. (37.6), to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Compare to the value of \(t\) you calculated in part (a). (c) Multiply the time interval by Mavis's speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on. Compare to the value of \(x\) you calculated in part (a).

Short Answer

Expert verified
(a) \(x = 5.00c\), \(t = 6.25 \) s; (b) \(t = 6.25\) s; (c) \(x = 5.00c\).

Step by step solution

01

Understand the Lorentz Transformation

The Lorentz transformation equations relate the time and position of events as measured by two observers in relative motion. For an observer in a stationary frame observing an event in a moving frame, the transformations are \( t' = \gamma (t - \frac{vx}{c^2}) \) and \( x' = \gamma (x - vt) \). Here, \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor.
02

Calculate x and t using Lorentz Transformation

Given Mavis reads 5.00 s on her timer, we need to find the corresponding time \( t \) and position \( x \) in Stanley's frame. The relative speed \( v = 0.800c \), thus \( \gamma = \frac{1}{\sqrt{1 - (0.800)^2}} = \frac{1}{0.6} = 1.25 \). Using \( t' = 5.00 \) s, \( t = \gamma (t' + \frac{vx'}{c^2}) \), where \( x' = 0 \) since Mavis starts from rest relative to her own frame. Therefore, \( t = 1.25 \times 5.00 = 6.25 \) s and \( x = vt = 0.800c \times 6.25 = 5.00c \) s in Stanley's frame.
03

Calculate Time Interval using Time Dilation

The time dilation formula is \( t = \gamma t' \). Given \( t' = 5.00 \) s, \( \gamma = 1.25 \), then \( t = 1.25 \times 5.00 = 6.25 \) s. This confirms the time calculated using the Lorentz transformations.
04

Calculate Distance using Speed and Time Interval

The distance Mavis travels in Stanley's frame is given by the product of speed and time interval. From Step 3, \( t = 6.25 \) s and \( v = 0.800c \). Therefore, \( x = vt = 0.800c \times 6.25 = 5.00c \) s, which matches the distance calculated via the Lorentz transformation.

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Key Concepts

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

Time Dilation
Time dilation is a fascinating consequence of Einstein's theory of relativity. Imagine you are moving at a significant fraction of the speed of light, just like Mavis in her spaceship. As you zoom through space, the passage of time for you feels completely normal. However, for an observer at rest, like Stanley on the ground, time appears to move differently for you.

This phenomenon is known as time dilation. It means that a moving clock ticks slower relative to a stationary observer's clock. The formula governing this is \( t = \gamma t' \), where \( t \) is the time measured by the stationary observer, \( t' \) is the time for the moving observer (like Mavis), and \( \gamma \) is the Lorentz factor. The Lorentz factor \( \gamma \) depends on the speed \( v \) of the moving object and is given by \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \).

In Mavis's case, her timer reads 5 seconds, but because she is moving at 0.800 times the speed of light, Stanley measures that time to be 6.25 seconds. This "stretching" of time is vital in understanding how events occur differently in varying frames of reference.
Relativity
The principle of relativity is at the heart of Einstein's groundbreaking work. This principle asserts that the laws of physics are the same for all observers, regardless of their relative motion. It means whether you're on a speeding train or standing still, the basic rules governing nature don't change.

Einstein took this idea and expanded it into the theory of special relativity. Within this framework, the speed of light \( c \) remains constant for all observers, contrary to what might be intuitively expected. As a consequence, space and time become intertwined into what is known as spacetime. Events that seem simultaneous in one frame may differ in another.

For Stanley watching Mavis, time and distance calculations adjust according to their relative velocities. Hence, the Lorentz transformations come into play, enabling us to convert measurements from one observer's frame to another's. This ensures that the fundamental laws of physics retain their form, reinforcing the universality of nature's laws.
Special Relativity
Special relativity is a theory that revolutionized how we view space and time. Developed by Albert Einstein in 1905, this theory focuses specifically on observers moving at constant speeds relative to one another, without the complications of gravity, which comes into play in general relativity.

A critical insight of special relativity is that the speed of light is the ultimate speed limit of the universe. This limit leads to surprising effects, such as time dilation and length contraction, where moving objects are measured to be shorter in the direction of motion by stationary observers.

In the scenario with Mavis and Stanley, special relativity predicts how they perceive each other's time and distance. The Lorentz transformations give us the tools to calculate how Mavis's time and her traveled distance, as measured by Stanley, differ from her own readings. This fundamental theory not only changes our understanding of movement and speed but also influences modern technology, from GPS systems to particle accelerators, proving its immense practicality and accuracy in the real world.

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Most popular questions from this chapter

What is the speed of a particle whose kinetic energy is equal to (a) its rest energy and (b) five times its rest energy?

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If a muon is traveling at 0.999c, what are its momentum and kinetic energy? (The mass of such a muon at rest in the laboratory is 207 times the electron mass.)

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In high-energy physics, new particles can be created by collisions of fast- moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon (\(K^-\)) and a positive kaon (\(K^+\)): $$p + p \rightarrow p + p + K^- + K^+$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV. (\(Hint\): It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-totalmomentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

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