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Chapter 41: Problem 35

As a spacecraft moving at \(0.92 \mathrm{c}\) travels past an observer on Earth, the Earthbound observer and the occupants of the craft each start identical alarm clocks that are set to ring after \(6.0 \mathrm{~h}\) have passed. According to the Earthling, what does the Earth clock read when the spacecraft clock rings?

Short Answer

Expert verified
According to the Earth observer, the Earth clock reads 15.3 hours when the spacecraft clock rings.

Step by step solution

01

Understanding Time Dilation

In the context of special relativity, time dilation occurs for clocks moving at a significant fraction of the speed of light compared to a stationary observer. The formula for time dilation is \( t = \frac{t_0}{\sqrt{1-v^2/c^2}} \), where \( t \) is the dilated time (Earth time in this case), \( t_0 \) is the proper time (6 hours on the spacecraft), \( v \) is the velocity (0.92c), and \( c \) is the speed of light.
02

Substitute Given Values

Set \( t_0 = 6 \) hours and \( v = 0.92c \). The speed of light \( c \) is just a symbol here because it will cancel out. Substitute these values into the time dilation formula: \[ t = \frac{6}{\sqrt{1-(0.92)^2}} \]
03

Calculate the Denominator

Calculate \( 1-(0.92)^2 \). First, \( (0.92)^2 = 0.8464 \). Thus, \( 1 - 0.8464 = 0.1536 \).
04

Compute the Square Root

Determine the square root of 0.1536: \[ \sqrt{0.1536} \approx 0.3919 \]
05

Final Calculation for Dilated Time

Substitute the value from the previous step back into the formula: \[ t = \frac{6}{0.3919} \approx 15.3 \text{ hours} \] This means that, according to the Earth observer, 15.3 hours elapse on Earth when the spacecraft's clock rings.

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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 foundational concept in physics, introduced by Albert Einstein in 1905. It describes how the laws of physics remain the same for all non-accelerating observers and it introduces the intriguing phenomenon where time and space are interrelated. One of the main tenets of special relativity is that the speed of light is constant in all inertial frames of reference, regardless of the motion of the light source or observer.
Special relativity provides the framework for understanding how time dilation occurs. This starts to happen at velocities approaching the speed of light. When objects move at these high speeds, time may pass differently than it would in a stationary frame of reference. This is not just a theoretical idea but has been confirmed by experiments over the years. For example, particles moving at high speeds within accelerators have shown measurable effects of time dilation.
Another essential aspect to remember is that special relativity requires us to rethink our classical notions of absolute time and space. Time isn't a uniform flow, but can be experienced differently based on speed, as illustrated by the classic twin paradox thought experiment.
Time Dilation Formula
The time dilation formula is a mathematical expression that quantifies time dilation effects as per the principles of special relativity. The formula is given by \[ t = \frac{t_0}{\sqrt{1 - v^2/c^2}} \]where:
  • \( t \) represents the time experienced by a stationary observer, often called the "dilated time," which is longer than the proper time.
  • \( t_0 \) is the "proper time," and it represents the time experienced by an observer moving along with the event (e.g., in a spacecraft).
  • \( v \) is the velocity of the moving object relative to the stationary observer, often measured as a fraction of the speed of light (\( c \)).
  • \( c \) is the speed of light in a vacuum, a constant with approximately 299,792,458 meters per second.
The formula shows how as velocity \( v \) gets closer to the speed of light \( c \), the term \( \sqrt{1 - v^2/c^2} \) becomes smaller, which results in a larger value for \( t \). This means time runs slower for objects moving near the speed of light, relative to a stationary observer.
Proper Time
Proper time is a key concept in understanding time dilation. It refers to the time measured by an observer who is at rest relative to the event being observed. This means that, to the person actually experiencing the event (like someone on a speeding spacecraft), the passage of time feels normal and unaltered. In our exercise, the 6 hours experienced by people inside the moving spacecraft is the proper time.
For the space travelers, life goes on as usual. Their clocks, watches, and heartbeats tick at what seems like the normal pace. However, in the context of special relativity, proper time can be different from the time measured by an observer who's in a different inertial frame (such as someone on Earth).
Proper time is denoted with \( t_0 \) in the time dilation formula. It's an intrinsic time measurement unaffected by relative speeds or gravitational fields. Only the observer who is co-located with the event being timed can measure proper time directly.
Speed of Light
The speed of light, often denoted by \( c \), is a fundamental constant in physics that plays a crucial role in the theory of relativity. Its value is approximately 299,792,458 meters per second, making it the fastest thing in the universe. This speed acts as a universal speed limit, as nothing with mass can accelerate to reach or exceed \( c \).
Light's constant speed is central to the differences encountered in time and space measurements between observers. It is why time dilation occurs, and why we need the time dilation formula to account for differences in elapsed time between moving and stationary observers.
The constancy of the speed of light leads to many counterintuitive effects explored in special relativity, such as the aforementioned time dilation and length contraction. These effects are only noticeable when dealing with speeds approaching that of light, but they have profound implications for our understanding of the universe.

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Most popular questions from this chapter

A certain strain of bacteria doubles in number each 20 days. Two of these bacteria are placed on a spaceship and sent away from the Earth for 1000 Earth-days. During this time, the speed of the ship is \(0.9950 \mathrm{c}\). How many bacteria are aboard when the ship lands on the Earth?

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 certain light source sends out \(2 \times 10^{15}\) pulses each second. As a spaceship travels parallel to the Earth's surface with a speed of \(0.90 \mathrm{c}\), it uses this source to send pulses to the Earth. The pulses are sent perpendicular to the path of the ship. How many pulses are recorded on Earth each second?

The Sun radiates energy equally in all directions. At the position of the Earth \(\left(r=1.50 \times 10^{11} \mathrm{~m}\right)\), the irradiance of the Sun's radiation is \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). How much mass does the Sun lose per day because of the radiation? The area of a spherical shell centered on the Sun and passing through the Earth is $$ \text { Area }=4 \pi r^{2}=4 \pi\left(1.50 \times 10^{11} \mathrm{~m}\right)^{2}=2.83 \times 10^{23} \mathrm{~m}^{2} $$ Through each square meter of this area, the Sun radiates an energy per second of \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). Therefore, the Sun's total radiation per second is $$ \text { Energy } / \mathrm{s}=(\text { area })\left(1400 \mathrm{~W} / \mathrm{m}^{2}\right)=3.96 \times 10^{26} \mathrm{~W} $$ The energy radiated in one day \((86400 \mathrm{~s})\) is Energy/day \(=\left(3.96 \times 10^{26} \mathrm{~W}\right)(86400 \mathrm{~s} /\) day \() \times 3.42 \times 10^{31} \mathrm{~J} /\) day Because mass and energy are related through \(\Delta \mathrm{E}_{0}=\Delta m \mathrm{c}^{2}\), the mass loss per day is $$ \Delta m=\frac{\Delta E_{0}}{c^{2}}=\frac{3.42 \times 10^{31} \mathrm{~J}}{\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}}=3.8 \times 10^{14} \mathrm{~kg} $$ For comparison, the Sun's mass is \(2 \times 10^{30} \mathrm{~kg}\).

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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