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Inside a spaceship flying past the earth at three-fourths the speed of light, a pendulum is swinging. (a) If each swing takes 1.50 s as measured by an astronaut performing an experiment inside the spaceship, how long will the swing take as measured by a person at mission control on earth who is watching the experiment? (b) If each swing takes 1.50 s as measured by a person at mission control on earth, how long will it take as measured by the astronaut in the spaceship?

Short Answer

Expert verified
(a) 2.26 s as measured by a person at mission control; (b) 0.996 s as measured inside the spaceship.

Step by step solution

01

Understand time dilation

Time dilation is a phenomenon described by the theory of relativity, where time appears to pass at different rates in different reference frames. For an observer watching the spaceship from Earth while it's moving at 3/4 the speed of light, time inside the spaceship will appear slower.
02

Identify relevant formula

To calculate time dilation, use the Lorentz transformation formula: \[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \( t \) is the proper time (time measured by the moving observer), \( t' \) is the dilated time, \( v \) is the velocity of the spaceship, and \( c \) is the speed of light.
03

Solve part (a)

In part (a), the time observed inside the spaceship (proper time \( t \)) is 1.50 s. The spaceship's speed is three-fourths the speed of light, so \( v = \frac{3}{4}c \). Substitute these values into the formula:\[ t' = \frac{1.50}{\sqrt{1 - \left(\frac{3}{4}\right)^2}} \]This simplifies to:\[ t' = \frac{1.50}{\sqrt{1 - \frac{9}{16}}} = \frac{1.50}{\sqrt{\frac{7}{16}}} = \frac{1.50}{\frac{\sqrt{7}}{4}} = \frac{1.50 \times 4}{\sqrt{7}}\]Calculate \( t' \):\[ t' \approx \frac{6.0}{\sqrt{7}} \approx 2.26 \, \text{s}\]So, the observer on Earth will measure approximately 2.26 s for each swing.
04

Solve part (b)

In part (b), if the mission controller on Earth measures the swing time as 1.50 s, this is the dilated time \( t' \). We need to find the proper time \( t \) measured by the astronaut: \[ t = t' \times \sqrt{1 - \frac{v^2}{c^2}} \]Substitute \( t' = 1.50 \) s and \( v = \frac{3}{4}c \): \[ t = 1.50 \times \sqrt{1 - \frac{9}{16}} \]This simplifies to: \[ t = 1.50 \times \sqrt{\frac{7}{16}} = 1.50 \times \frac{\sqrt{7}}{4} \]Calculate \( t \): \[ t \approx 1.50 \times \frac{\sqrt{7}}{4} \approx 0.996 \, \text{s} \]So, the astronaut will measure approximately 0.996 s for each swing.

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Key Concepts

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

Theory of Relativity
The theory of relativity, proposed by Albert Einstein, revolutionized our understanding of space and time. It consists of the special and general theories of relativity. Special relativity, in particular, introduces the idea that the laws of physics are the same for all non-accelerating observers. It shows how space and time are interconnected, forming what we call "spacetime."
This theory proposes that the speed of light is constant for all observers, regardless of their relative speed. In practical terms, this means that time and space are not absolute. They are malleable and change depending on the observer's velocity relative to the speed of light.
One fascinating consequence of this is time dilation, where time is observed to pass at different rates in different inertial frames. When an object moves at speeds near the speed of light, time will appear to slow down relative to a stationary observer. This is a key concept when dealing with time experienced on fast-moving spaceships versus stationary observers.
Lorentz Transformation Formula
The Lorentz transformation formula is an essential mathematical tool used in the theory of relativity. It provides the means to convert physical measurements from one inertial frame of reference to another.
The formula is particularly important for understanding time dilation and length contraction. It mathematically describes how, at relativistic speeds, time and space are transformed between two observers moving at a constant velocity relative to each other.
For time dilation, the relevant formula is:\[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]Here, \( t \) is the proper time measured in the moving reference frame, \( t' \) is the dilated time observed in the stationary reference frame, \( v \) is the relative velocity, and \( c \) is the speed of light.
This equation shows that as an object's velocity \( v \) approaches the speed of light \( c \), time \( t' \) for that object in the stationary frame will appear to increase. This is because of the term \( \sqrt{1 - \frac{v^2}{c^2}} \), which becomes smaller, thus increasing the time \( t' \).
Proper Time
Proper time is a crucial concept in understanding time dilation. It represents the time measured by an observer who is at rest relative to the event being timed. In simpler terms, it's the time interval measured by someone who is experiencing the events without moving relative to them.
In the context of our pendulum problem, the proper time is the 1.50 seconds measured by the astronaut inside the spaceship. This is because the astronaut is at rest relative to the pendulum he/she is observing. To them, the pendulum swings naturally, without any perceived differences.
Proper time differs from dilated time, which is the time interval measured by an observer in a different inertial frame moving relative to the proper time observer. Understanding the distinction is vital when applying the Lorentz transformation to calculate how time behaves at relativistic speeds.
Reference Frames
Reference frames are essential for understanding and applying the concepts of relativity. A reference frame is simply a perspective or a point of view from which measurements like time and space are made.
There are two types of reference frames: inertial, where objects move at a constant velocity, and non-inertial, where objects are accelerating. Inertial frames are the ones often discussed in special relativity because they apply constant velocities which are crucial for using the Lorentz transformations.
In our problem scenario, the spaceship and Earth act as two distinct inertial reference frames. The astronaut inside the spaceship has a different frame compared to someone on Earth. Because the spaceship is moving at three-fourths the speed of light relative to Earth, each observer measures time (and space) differently. The key takeaway is that no single reference frame is more correct than the other—they're just different perspectives of the same events.

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Most popular questions from this chapter

Tell It to the Judge. (a) How fast must you be approaching a red traffic light \((\lambda=675 \mathrm{nm})\) for it to appear yellow \((\lambda=575 \mathrm{nm})\) ? Express your answer in terms of the speed of light. (b) If you used this as a reason not to get a ticket for running a red light, how much of a fine would you get for speeding? Assume that the fine is \(\$ 1.00\) for each kilometer per hour that your speed exceeds the posted limit of 90 \(\mathrm{km} / \mathrm{h}\) .

When a particle meets its antiparticle, they annihilate each other and their mass is converted to light energy. The United States uses approximately \(1.0 \times 10^{20} \mathrm{J}\) of energy per year. (a) If all this energy came from a futuristic antimatter reactor, how much mass of matter and antimatter fuel would be consumed yearly? (b) If this fuel had the density of iron \(\left(7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) and were stacked in bricks to form a cubical pile, how high would it be? (Before you get your hopes up, antimatter reactors are a long way in the future-if they ever will be feasible.

A person is standing at rest on level ground. How fast would she have to run to (a) double her total energy and (b) increase her total energy by a factor of 10?

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540\(c\) relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 \(\mathrm{km} .\) (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 \(\mathrm{km}\) to the surface of the earth? (b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2\(\mu \mathrm{s}\) . They are produced when cosmic rays bombard the upper atmosphere about 10 \(\mathrm{km}\) above the earth's surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth's surface. (a) What is the greatest distance a muon could travel during its 2.2 -\mus lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the \(2.2-\mu \mathrm{s}\) lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of \(0.999 c,\) what is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic rays? (c) From the point of view of the muon, it still lives for only \(2.2 \mu s,\) so how does it make it to the ground? What is the thickness of the 10 \(\mathrm{km}\) of atmosphere through which the muon must travel, as measured by the muon? Is it now clear how the muon is able to reach the ground?

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