Question

Right before take-off, a passenger on a plane flying from town \(A\) to town B synchronizes his clock with the clock of his friend who is waiting for him in town \(\mathrm{B}\). The plane flies with a constant velocity of \(240 \mathrm{~m} / \mathrm{s}\). The moment the plane touches the ground, the two friends check their clocks simultaneously. The clock of the passenger on the plane shows that it took exactly \(3.00 \mathrm{~h}\) to travel from \(\mathrm{A}\) to \(\mathrm{B}\). Ignoring any effects of acceleration: a) Will the clock of the friend waiting in \(B\) show a shorter or a longer time interval? b) What is the difference between the readings of the two clocks?

Short answer

Question: In a scenario involving time dilation, a passenger on an airplane observes a time of 3 hours. Determine (a) if the time observed by a friend waiting at their destination would be shorter or longer than the passenger's observed time, and (b) the difference between the clock readings of the two individuals, given that the airplane's speed is 240 m/s. Answer: (a) The friend waiting at the destination will observe a longer time interval compared to the passenger on the airplane. (b) The difference between the readings of the two clocks is approximately 0.0051 seconds.

Step-by-step solution

Understanding the time dilation formula

Time dilation is a phenomenon that occurs in special relativity, where the observed time in a moving frame of reference (such as an airplane) is different from the time in the rest frame. The time dilation formula is given by: \[t = t_0\sqrt{1 - \frac{v^2}{c^2}}\] Where \(t\) is the time observed in the rest frame (friend on the ground), \(t_0\) is the time observed in the moving frame (passenger on the plane), \(v\) is the relative velocity between the two frames, and \(c\) is the speed of light in a vacuum.

Determine if the time observed by the friend in B is shorter or longer

In our case, the passenger on the plane observes a time \(t_0 = 3.00 \text{ h}\). Since the fraction \(\frac{v^2}{c^2}\) would be less than 1, the value inside the square root would be greater than 0 and less than 1. Thus, the time observed by the friend on the ground, t, will be longer due to the square root term. Answer for (a): The clock of the friend waiting in \(B\) will show a longer time interval.

Convert hours to seconds

To perform the calculations, we need to convert the given time in hours to seconds. We have \(t_0 = 3.00 \text{ h}\), and knowing that 1 hour equals 3600 seconds, we can convert it as follows: \[t_0 = 3.00 \text{ h} \times \frac{3600 \text{ s}}{1 \text{ h}} = 10800 \text{ s}\]

Calculate the time observed by the friend in B

Now, we can substitute the known values into the time dilation formula and calculate the time observed by the friend in B: \[t = t_0\sqrt{1 - \frac{v^2}{c^2}}\] \[t = 10800\sqrt{1 - \frac{(240)^2}{(3.00 \times 10^8)^2}}\] \[t \approx 10800.0051\text{ s}\]

Calculate the difference between the readings of the two clocks

Finally, we can calculate the difference between the readings of the two clocks by subtracting the passenger’s time (\(t_0\)) from the friend's time (t): \[\Delta{t} = t - t_0\] \[\Delta{t} = 10800.0051 - 10800\] \[\Delta{t} \approx 0.0051 \text{ s}\] Answer for (b): The difference between the readings of the two clocks is approximately \(0.0051 \text{ s}\).

Key concepts

Special Relativity

Special relativity, conceptualized by Albert Einstein in 1905, refers to the structure of spacetime and the relationship between space and time when objects move at a constant speed in a straight line, particularly at speeds close to the speed of light. According to this theory, there's no single absolute frame of reference. Every observation of time and space is relative to the observer's unique frame of reference.

The most famous equation from special relativity, E=mc^2, shows that energy (E) and mass (m) are interchangeable; they are different forms of the same thing. Special relativity has many counterintuitive implications, such as time dilation, length contraction, and the relativity of simultaneity—all of which were once considered science fiction.

When applying the concept of special relativity to everyday speeds, the effects are imperceptible, which is why our daily experiences seem to follow the rules of classical, Newtonian physics. However, as the speed of an object approaches the speed of light, these relativistic effects become significant and must be accounted for, as in the time dilation problem from the textbook exercise.

Frame of Reference

A frame of reference is essentially a viewpoint—a set of axes or a coordinate system used to measure the position, orientation, and other properties of objects. Physics problems often involve comparing two different frames of reference. One might be 'stationary' (like the ground or town B in the exercise), and another could be 'moving' (like the plane travelling from town A to town B).

In the context of the textbook exercise, there are two key frames of reference: the viewpoint of the passenger on the plane (the moving frame of reference) and the viewpoint of the friend waiting on the ground (the rest frame of reference). Events that happen simultaneously in one frame of reference do not necessarily happen at the same time in another. This is because the concept of simultaneity is relative—each frame of reference has its own 'clock', and these clocks can run at different rates, particularly when the frames are moving relative to each other at significant fractions of the speed of light.

Understanding frames of reference is crucial because it allows us to grasp why the passenger and the friend on the ground might record different times on their clocks for the same journey. This difference is not a result of the clocks' mechanics but of the laws of physics that govern time itself.

Speed of Light

The speed of light in a vacuum, commonly denoted as 'c', is a fundamental constant of nature. Its value is approximately 299,792,458 meters per second. Remarkably, the speed of light is the same for all observers, regardless of their relative motion or the motion of the source of the light. This invariance of the speed of light is a key postulate of special relativity and helps explain the phenomenon of time dilation.

Because the speed of light is so fast compared to speeds we are familiar with in everyday life, the effects of special relativity are generally not noticeable at low speeds. In the airplane scenario, even though the plane's speed is high by human standards, it's minuscule compared to the speed of light. This small ratio means that time dilation effects are incredibly small, as the textbook solution shows.

However, the speed of light being constant also means that nothing with mass can reach or exceed it. As objects with mass accelerate close to the speed of light, they gain relativistic mass, requiring more and more energy to continue to accelerate, effectively making it impossible to reach the speed of light. This ensures that causality is maintained—the sequence of cause and effect is upheld—and prevents paradoxes that could arise if information traveled faster than light.