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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 m/s and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 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 clocks differ by approximately \(1.736 \times 10^{-12}\) hours. The plane's clock displays the shorter time.

Step by step solution

01

Understand the Problem

We need to find how much the time measurements differ between two clocks using the concept of time dilation. The clock on the airplane should run slower because of its motion relative to the clock in New York.
02

Write the Time Dilation Formula

The time dilation formula in the context of special relativity is given by: \( t' = \frac{t}{\sqrt{1 - \frac{u^2}{c^2}}} \), where \( t' \) is the dilated time on the plane, \( t \) is the stationary time in New York (4.00 h), \( u \) is the speed of the plane (250 m/s), and \( c \) is the speed of light (approximately \(3 \times 10^8 \) m/s).
03

Simplify the Formula Using Binomial Expansion

Since \( u \ll c \), we can use the binomial expansion to approximate \( \sqrt{1 - \frac{u^2}{c^2}} \approx 1 - \frac{u^2}{2c^2} \). This allows us to simplify the time dilation equation to \( t' \approx t \left( 1 + \frac{u^2}{2c^2} \right) \).
04

Calculate the Time Difference

Substitute the provided values into the simplified formula: \( t' = 4.00 \text{ h} \left( 1 + \frac{(250)^2}{2 \times (3 \times 10^8)^2} \right) \). First, calculate \( \frac{(250)^2}{2 \times (3 \times 10^8)^2} \), which is a very small number. After computations, \( t' \approx 4.00 \text{ h} + 1.736 \times 10^{-12} \text{ h}\).
05

Calculate the Difference in Time Readings

The difference in readings is \( t' - t = 1.736 \times 10^{-12} \text{ h} \). This difference is due to time dilation, making the clock on the plane show less time.
06

Determine Which Clock Shows the Shorter Time

Since the plane clock experiences time dilation, it shows the shorter elapsed time. Hence, the plane's clock will display slightly less than 4.00 h compared to the stationary clock.

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Key Concepts

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

Special Relativity
Albert Einstein’s theory of special relativity revolutionized our understanding of space and time. In this theory, time is not absolute but is relative and can vary for different observers depending on their relative motion. A critical prediction of this theory is time dilation, which means time appears to move slower for an object moving at high speeds relative to a stationary observer.

In the context of our everyday time dilation problem, the key idea is that the atomic clock on the airplane, which is moving at a relatively high speed, will experience less time passing compared to the clock on the ground in New York. This happens because the airplane clock is in motion, leading to time dilation. The formula to calculate the effect of time dilation comes from special relativity: \[ t' = \frac{t}{\sqrt{1 - \frac{u^2}{c^2}}} \]where:
  • \( t' \) is the time observed on the moving clock
  • \( t \) is the time observed on the stationary clock
  • \( u \) is the velocity of the moving object
  • \( c \) is the speed of light
The outcome of this principle is that the faster you move through space, the slower you move through time.
Binomial Expansion
The binomial expansion is a useful mathematical tool that allows us to simplify expressions that involve powers of sums. It is particularly helpful when dealing with expressions like \[ \sqrt{1 - \frac{u^2}{c^2}} \]when \( u \ll c \). In such cases, we can approximate this expression using binomial expansion.

For small values of \( x \) where \( 1 - x \approx 1 + (-x) \), the binomial expansion approximation \[ \sqrt{1 - x} \approx 1 - \frac{x}{2} \]is applied. For the time dilation problem, this means:\[ \sqrt{1 - \frac{u^2}{c^2}} \approx 1 - \frac{u^2}{2c^2} \]which simplifies the calculation by turning a more complex algebraic expression into a simple linear term. This approximation is invaluable in making quick and effective predictions without complicated calculations, particularly when the speed \( u \) is much less than the speed of light, \( c \).
Atomic Clocks
Atomic clocks are highly precise timekeeping devices that use the vibrations of atoms, typically cesium or rubidium, to measure time. They are far more accurate than traditional mechanical or quartz clocks because they rely on natural atomic oscillations which occur at very stable frequencies.

In our problem, atomic clocks are used to demonstrate how even small relativistic effects can be detected due to their extreme precision. When one atomic clock is onboard an airliner and another remains stationary in New York, even the negligible relativistic effects caused by speeds well below the speed of light result in noticeable differences over time.
  • Atomic clocks can measure incredibly tiny differences in time, to the tune of billionths of a second.
  • They ensure that any discrepancy due to speed and relativity can be detected, as demonstrated by our time dilation problem.
This ability to observe relativity on a practical scale helps confirm the theory's predictions and highlights the vital role of atomic clocks in scientific research and technology, such as in GPS systems where time dilation must be accounted for to provide accurate locations.

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Most popular questions from this chapter

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).

A rocket ship flies past the earth at 91.0% of the speed of light. Inside, an astronaut who is undergoing a physical examination is having his height measured while he is lying down parallel to the direction in which the ship is moving. (a) If his height is measured to be 2.00 m by his doctor inside the ship, what height would a person watching this from the earth measure? (b) If the earth-based person had measured 2.00 m, what would the doctor in the spaceship have measured for the astronaut's height? Is this a reasonable height? (c) Suppose the astronaut in part (a) gets up after the examination and stands with his body perpendicular to the direction of motion. What would the doctor in the rocket and the observer on earth measure for his height now?

A spaceship moving at constant speed u relative to us broadcasts a radio signal at constant frequency \(f_0\) . As the spaceship approaches us, we receive a higher frequency \(f\) ; after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_0\) , and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_0\)? (\(Hint\): In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals 1 /T. Use the time dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_0\) = 345 MHz as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758\(c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f - f_0\)? (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift (\(f - f_0\)) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

A nuclear bomb containing 12.0 kg of plutonium explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in 10\(^4\). (a) How much energy is released in the explosion? (b) If the explosion takes place in 4.00 \(\mu\)s, what is the average power developed by the bomb? (c) What mass of water could the released energy lift to a height of 1.00 km ?

The French physicist Armand Fizeau was the first to measure the speed of light accurately. He also found experimentally that the speed, relative to the lab frame, of light traveling in a tank of water that is itself moving at a speed \(V\) relative to the lab frame is $$v = {c \over n} + kV$$ where \(n\) = 1.333 is the index of refraction of water. Fizeau called \(k\) the dragging coefficient and obtained an experimental value of \(k\) = 0.44. What value of \(k\) do you calculate from relativistic transformations?

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