/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Everyday Time Dilation. Two atom... [FREE SOLUTION] | 91Ó°ÊÓ

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Everyday Time Dilation. Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 \(\mathrm{m} / \mathrm{s}\) and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 \(\mathrm{h}\) . By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (Hint: since \(u \ll c,\) you can simplify \(\sqrt{1-u^{2} / c^{2}}\) by a binomial expansion.)

Short Answer

Expert verified
The moving clock shows a shorter time by approximately \( 1.39 \times 10^{-12} \) seconds.

Step by step solution

01

Identify the Relevant Equation

The phenomenon described involves time dilation due to relative velocity, which can be calculated using the formula \( \Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{u^2}{c^2}}} \), where \( \Delta t \) is dilated time, \( \Delta t_0 \) is the proper time, \( u \) is the velocity of the object, and \( c \) is the speed of light.
02

Simplify the Time Dilation Formula

Since \( u \ll c \), we can simplify the expression \( \sqrt{1 - \frac{u^2}{c^2}} \) using the binomial expansion: \( \sqrt{1 - \frac{u^2}{c^2}} \approx 1 - \frac{1}{2}\frac{u^2}{c^2} \). Substituting back, the dilated time becomes \( \Delta t \approx \Delta t_0 \left(1 + \frac{1}{2}\frac{u^2}{c^2}\right) \).
03

Calculate the Change in Time

Given that \( \Delta t_0 = 4 \) hours \( = 4 \times 3600 \) seconds and \( u = 250 \) m/s, substitute these values to find the difference in time. Calculate \( \Delta t_0 \times \frac{1}{2}\frac{u^2}{c^2} \) where \( c = 3 \times 10^8 \) m/s.
04

Perform the Calculations

First, calculate \( \frac{u^2}{c^2} = \left( \frac{250}{3 \times 10^8} \right)^2 \). Then, calculate \( \Delta t_0 \times \frac{1}{2} \left( \frac{u^2}{c^2} \right) \). This will give you the time difference between the two clocks.
05

Determine the Time Difference

Perform the calculation: \( \Delta t \approx 4 \times 3600 \times 0.5 \times \frac{(250)^2}{(3 \times 10^8)^2} \), which results in \( \Delta t \approx 1.39 \times 10^{-12} \) seconds.
06

Compare Clocks and Conclude

The clock on the plane (moving clock) experiences time dilation and runs slower, so it shows a shorter elapsed time by approximately \( 1.39 \times 10^{-12} \) seconds.

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Key Concepts

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

Relative Velocity
Understanding relative velocity is crucial when examining concepts in special relativity. It refers to the velocity of an object as observed from a particular frame of reference. In the context of the exercise, we observe the velocity of the airliner compared to the stationary clock in New York. This relative velocity is given as 250 m/s, representing the speed at which the airplane travels relative to the Earth.

For special relativity equations, like the time dilation formula, knowing the relative velocity ( u in this case) allows us to determine the time variation experienced by the traveling clock due to its motion. Because this velocity is significantly smaller than the speed of light ( c ), it allows for certain simplifications, such as using binomial expansion.
Binomial Expansion
Binomial expansion is a mathematical method used to simplify calculations when dealing with expressions raised to a power. In this exercise, it is applied to the time dilation formula: \( \sqrt{1 - \frac{u^2}{c^2}} \).

Since u is much smaller than c , we can use the approximation:
  • \( \sqrt{1 - \frac{u^2}{c^2}} \approx 1 - \frac{1}{2} \frac{u^2}{c^2} \)
This simplification means we can replace the square root with this simpler expression, making calculations more manageable. It's like saying, "for very small values, complex expressions can be approximated by simpler forms," allowing us to compute the time difference without overly complicated math.
Atomic Clocks
Atomic clocks are extremely precise timekeeping devices, and they play a critical role in experiments involving time dilation, such as the one described in the exercise. They work by measuring the vibrations of atoms, usually cesium or rubidium, providing remarkably stable and exact time measurements.

These clocks are used to test predictions of special relativity by comparing the elapsed time on a clock that moves (on the plane) with one that remains stationary. Because they can measure time differences down to tiny fractions of a second, atomic clocks are perfect for experimentally verifying relativistic effects. In this exercise, the atomic clock on the stationary platform serves as the reference for "proper time," while the traveling clock measures "dilated time."
This setup elegantly demonstrates the principles of special relativity, revealing how motion affects time.
Special Relativity
Special relativity, formulated by Albert Einstein, fundamentally changed our understanding of physics. It posits that the laws of physics are the same for all non-accelerating observers and introduced the concept that the speed of light is constant regardless of the observer's motion. A key consequence is time dilation: when an object moves relative to an observer, its clock ticks more slowly according to that observer.

In the exercise, the time dilation effect is observed by comparing two synchronized atomic clocks—one stationary and the other moving with the airline's speed. When the plane returns, the clock on board shows less time elapsed than the one on the ground. This confirms that relative motion affects the perceived passage of time, showcasing the fascinating time-bending aspects of special relativity. This concept challenges our everyday perceptions, illustrating how the universe operates under vastly different rules at high speeds.

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Most popular questions from this chapter

A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600\(c .\) The pursuit ship is traveling at a speed of 0.800\(c\) relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the velocity of the cruiser relative to the pursuit ship be directed toward or away from the pursuit ship? (b) What is the speed of the cruiser relative to the pursuit ship?

A proton (rest mass \(1.67 \times 10^{-27} \mathrm{kg} )\) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the speed of the proton?

After being produced in a collision between elementary particles, a positive pion \(\left(\pi^{+}\right)\) must travel down a 1.90 -km-long tube to reach an experimental area. A \(\pi^{+}\) particle has an average lifetime (measured in its rest frame) of \(2.60 \times 10^{-8} \mathrm{s}\) ; the \(\pi^{+}\) we are considering has this lifetime. How fast must the \(\pi^{+}\) travel if it is not to decay before it reaches the end of the tube? (Since \(u\) will be very close to \(c,\) write \(u=(1-\Delta) c\) and give your answer in terms of \(\Delta\) rather than \(u .\) (b) The \(\pi^{+}\) has a rest energy of 139.6 \(\mathrm{MeV} .\) What is the total energy of the \(\pi^{+}\) at the speed calculated in part (a)?

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800\(c .\) 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).

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