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The clock on a spaceship shows that a robot on board took \(10.0 \mathrm{~s}\) to do some job. The ship flies passed a station, and someone watching the robot also notes how long it took to do the job using her own wristwatch. If she computes \(y\) for the ship to be \(1.08\), how long will she say the robot took to do the job? [Hint: Time dilation means time slows and durations appear longer.]

Short Answer

Expert verified
The observer measures the robot taking 10.8 seconds.

Step by step solution

01

Understand Time Dilation Formula

The time dilation effect in relativity is given by the formula \( t' = \gamma t \), where \( t' \) is the time measured by an observer watching from a stationary point, \( t \) is the proper time measured by an observer moving with the object (robot), and \( \gamma \) is the Lorentz factor.
02

Identify Given Values

From the problem, the proper time \( t = 10.0 \) seconds and the Lorentz factor \( \gamma = 1.08 \). We're asked to find \( t' \), the time as measured by the stationary observer.
03

Apply the Lorentz Factor Formula

Use the formula \( t' = \gamma t \). Substitute the given values: \( t' = 1.08 \times 10.0 \).
04

Calculate the Dilated Time

Perform the multiplication: \( t' = 10.8 \). Thus, the stationary observer measures the time taken by the robot as 10.8 seconds.

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Key Concepts

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

Special Relativity
Special relativity is a theory formulated by Albert Einstein. It describes the physics of objects moving close to the speed of light. The theory revolutionized how we think about space and time. It tells us that they are not absolute but relative and interconnected in a space-time fabric. In special relativity, the speed of light is constant in all frames of reference.
One key idea is that the laws of physics are the same for all observers, no matter their constant velocity. It's like being in a train moving smoothly; you can't tell you're moving without looking outside. This concept leads to fascinating consequences, like time dilation and length contraction.
Special relativity challenges our everyday experiences of motion and time. Instead of time being a separate entity, it merges with space. Observers moving relative to one another may have different measurements of time and distance, yet both are correct in their own frames.
Lorentz Factor
The Lorentz factor, denoted as \(\gamma\), is a crucial element in the equations of special relativity. It's defined as: \[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \(v\) is the relative velocity between two frames and \(c\) is the speed of light.
It's interesting how this factor affects time and space. As an object's speed approaches the speed of light, \(\gamma\) increases significantly. This results in more pronounced effects of time dilation and length contraction.
  • When \(v\) is very small compared to \(c\), \(\gamma\) is close to 1, meaning minimal relativistic effects.
  • As \(v\) grows close to \(c\), \(\gamma\) can become very large, affecting time perception notably.
For the problem mentioned, the Lorentz factor is given as \(1.08\), indicating that the robot's time appears slightly dilated to the observer at the station.
Proper Time
Proper time \(t\) refers to the time interval measured by an observer moving with the object being observed. Think of it as carrying a clock while you travel and noting the time passed according to your clock. It is the shortest time interval possible between two events.
  • In the given problem, the robot's proper time is measured to be \(10.0\) seconds by the clock on the spaceship.
  • Proper time does not change despite the observer's frame of reference.
Proper time provides a baseline to compare how time is perceived differently in other reference frames. The beauty of special relativity is how it highlights the differences in time perception due to relative motion.
Time Perception in Physics
Time perception in physics, especially in the context of special relativity, can differ drastically from our daily experiences. Relativity shows us that time isn't constant and can warp depending on how fast you are moving compared to the speed of light.
Imagine two twins: one stays on Earth while the other travels through space near the speed of light. Upon reuniting, the traveling twin is younger due to time dilation. This thought experiment known as the "Twin Paradox" illustrates how time perception is subjective in different inertial frames.
  • From the perspective of the robot on the spaceship, its job took '\(10.0\)' seconds.
  • The astronaut observer notes it as '\(10.8\)' seconds due to her high-speed relative to the spaceship.
These discrepancies are not errors but insights to the flexible nature of time in physics. Understanding these differences helps scientists develop technologies like GPS, sensitive to relativistic effects.

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Most popular questions from this chapter

A proton has a mass of \(1.6726 \times 10^{-27} \mathrm{~kg}\) and is traveling at \(\mathrm{a}\) speed where \(y=2.294157\); that's \(90 \%\) the speed of light. Determine its total energy. Four significant figures will do. Discuss your answer as it compares with classical energy.

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As a rocket ship sweeps past the Earth with speed \(u\), it sends out a pulse of light ahead of it. How fast does the light pulse move according to people on the Earth? Method 1 With speed \(c\) (by the second postulate of Special Relativity). Method 2 Here \(v_{O^{\prime} O}=u\) and \(v_{P O^{\prime}}=c\). According to the velocity addition formula, the observed speed will be (since \(u=\mathrm{c}\) in this case ) $$ v_{P O}=\frac{v_{P O}+v_{o^{\prime} o}}{1+\frac{v_{P O} \cdot v_{o^{\prime}} O}{c^{2}}}=\frac{v+c}{1+(v / c)}=\frac{(v+c) c}{c+v}=c $$

A \(2.0\) -kg object is lifted from the floor to a tabletop \(30 \mathrm{~cm}\) above the floor. By how much did the mass of the system consisting of the Earth and the object increase because of this increased \(\mathrm{PE}_{\mathrm{G}}\) ? We use \(\Delta \mathrm{E}_{\mathrm{O}}=(\Delta m) \mathrm{c}^{2}\), with \(\Delta \mathrm{E}_{\mathrm{O}}=m g h .\) Therefore, $$ \Delta m=\frac{\Delta E_{0}}{c^{2}}=\frac{m g h}{c^{2}}=\frac{(2.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.30 \mathrm{~m})}{\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}}=6.5 \times 10^{-17} \mathrm{Kg} $$

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