Question

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

The clocks differ by approximately \(1.736 \times 10^{-12}\) hours. The plane's clock displays the shorter time.

Step-by-step solution

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.

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).

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 \( t' \approx t \left( 1 + \frac{u^2}{2c^2} \right) \).

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, \( t' \approx 4.00 \text{ h} + 1.736 \times 10^{-12} \text{ h}\).

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.

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.

Key concepts

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.