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Chapter 26: Problem 21

A futuristic train moving in a straight line with a uniform speed of \(0.80 c\) passes a series of communications towers. The spacing between the towers, according to an observer on the ground, is \(3.0 \mathrm{km} .\) A passenger on the train uses an accurate stopwatch to see how often a tower passes him. (a) What is the time interval the passenger measures between the passing of one tower and the next? (b) What is the time interval an observer on the ground measures for the train to pass from one tower to the next?

Short Answer

Expert verified
Answer: The time interval measured by the train passenger is approximately \(1.25 \times 10^{-5} \mathrm{s}\), and the time interval measured by the observer on the ground is approximately \(2.08 \times 10^{-5} \mathrm{s}\).

Step by step solution

01

Identifying given information

We have the following information from the exercise: 1. The train's velocity is \(v = 0.80c\), where \(c\) is the speed of light. 2. The distance between communication towers is \(3.0 \mathrm{km}\) in the ground frame. 3. We need to calculate the time interval for both the passenger on the train and the observer on the ground.
02

Calculate the proper time interval for the train passenger

For the train passenger, the communication towers are moving towards him at a speed of \(0.80c\). So, the time interval he measures for the towers to pass can be calculated by dividing the distance between the towers by the relative speed of the towers. The proper time interval, denoted as \(\Delta t'\), is given by: \(\Delta t' = \frac{d}{v} = \frac{3.0 \mathrm{km}}{0.80c}\)
03

Convert the distance to meters

In the above equation, we need to convert the distance from kilometers to meters. To do this, simply multiply the distance by 1000. \(d = 3.0 \mathrm{km} \times 1000 = 3.0 \times 10^3 \mathrm{m}\) Now, plug in the converted distance into the equation from Step 2: \(\Delta t' = \frac{3.0 \times 10^3 \mathrm{m}}{0.80c}\)
04

Calculate the time interval for the train passenger

Using the equation from Step 3, we can calculate the time interval for the train passenger. \(\Delta t' = \frac{3.0 \times 10^3 \mathrm{m}}{0.80(3.0 \times 10^8 \mathrm{m/s})} = 1.25 \times 10^{-5} \mathrm{s}\) So, the time interval measured by the train passenger between the passing of one tower and the next is approximately \(1.25 \times 10^{-5} \mathrm{s}\).
05

Calculate the time dilation factor

Now, we need to calculate the time dilation factor to get the time interval for the observer on the ground. The time dilation factor, denoted as \(\gamma\), is given by: \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) Plug in the given train velocity: \(\gamma = \frac{1}{\sqrt{1 - \frac{(0.80c)^2}{c^2}}}\) Dividing by the square of the speed of light, we get: \(\gamma = \frac{1}{\sqrt{1 - 0.80^2}}\)
06

Calculate the time dilation factor

Using the equation from Step 5, we can calculate the time dilation factor: \(\gamma = \frac{1}{\sqrt{1 - 0.80^2}} = 1.667\)
07

Calculate time interval for the observer on the ground

Finally, we can find the time interval for the observer on the ground by multiplying the time dilation factor with the time interval measured by the train passenger: \(\Delta t = \gamma \Delta t' = 1.667 \times 1.25 \times 10^{-5} \mathrm{s} = 2.08 \times 10^{-5} \mathrm{s}\) So, the time interval measured by the observer on the ground for the train to pass from one tower to the next is approximately \(2.08 \times 10^{-5} \mathrm{s}\).

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