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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 ) $$ v_{P O}=\frac{v_{P O}+v_{o^{\prime} o}}{1+\frac{v_{P O} \cdot v_{o^{\prime}} O}{c^{2}}}=\frac{v+c}{1+(v / c)}=\frac{(v+c) c}{c+v}=c $$

Short Answer

Expert verified
The light pulse moves at speed \( c \) relative to people on Earth.

Step by step solution

01

Understanding the Scenario

We have a rocket moving at speed \( u \) relative to the Earth, sending a light pulse forward. We need to determine how fast the pulse moves according to observers on Earth.
02

Speed of Light Postulate

According to the second postulate of Special Relativity, the speed of light in a vacuum is constant and is independent of the motion of the source or observer. Hence, the speed of light observed by anyone, regardless of their relative motion, should be \( c \).
03

Velocity Addition Formula Explanation

The velocity addition formula in Special Relativity is used to calculate the observed velocity \( v_{PO} \) of an object (the light pulse, here) by another observer (on Earth, here), considering the relative velocity \( u \): \[ v_{PO} = \frac{v_{PO^{\prime}} + v_{O^{\prime}O}}{1 + \frac{v_{PO^{\prime}} \cdot v_{O^{\prime}O}}{c^2}} \] where \( v_{PO^{\prime}} = c \) is the speed of light, and \( v_{O^{\prime}O} = u \).
04

Application of Velocity Addition Formula

Given \( v_{PO^{\prime}} = c \) and \( v_{O^{\prime}O} = u \), apply the formula:\[ v_{PO} = \frac{c + u}{1 + \frac{c \cdot u}{c^2}} = \frac{c + u}{1 + \frac{u}{c}} \]Now simplify: \[ v_{PO} = \frac{(c+u)c}{c+u} = c \] Thus, the speed of the light pulse as seen by people on Earth remains \( c \).
05

Conclusion

Regardless of the method used, the answer remains consistent with the principles of Special Relativity, which asserts that the speed of light in vacuum is constant at \( c \), even for observers moving relative to the light source.

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Key Concepts

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

Speed of Light
One of the most intriguing concepts in physics is the "Speed of Light." It holds a special place, especially in the realm of Special Relativity. The speed of light in a vacuum is denoted by the symbol \( c \) and is approximately \( 299,792,458 \) meters per second. This speed is unique because it is the universal speed limit; no matter can move faster than light in a vacuum.

The constancy of the speed of light was first proposed by Albert Einstein in his theory of Special Relativity. According to this theory, the speed of light is always \( c \), regardless of whether the source of light is moving towards you or away from you. This idea was a radical departure from classical physics, which assumed that speeds simply added up. Einstein's revelation has profound implications for how we understand space and time.

To summarize, you should remember:
  • The speed of light is always \( c \), irrespective of the reference frame.
  • Nothing can move faster than light in a vacuum, setting a universal speed limit.
Velocity Addition Formula
The Velocity Addition Formula is a key concept in understanding Special Relativity. It determines how velocities add together when different reference frames are in motion relative to each other. This formula helps to resolve scenarios where speeds are significant fractions of the speed of light, unlike in classical physics where speeds simply add up.

In mathematical terms, the formula is given by:\[v_{PO} = \frac{v_{PO^{\prime}} + v_{O^{\prime}O}}{1 + \frac{v_{PO^{\prime}} \cdot v_{O^{\prime}O}}{c^2}}\]
This equation accounts for relativistic effects, ensuring that no resulting velocity exceeds the speed of light \( c \).

For example, if a rocket moving at speed \( u \) relative to the Earth emits a light pulse at speed \( c \), the speed observed on Earth still turns out to be \( c \). Even though the rocket is moving, the formula yields the same speed for the light pulse as it would in a stationary setting. This maintains adherence to the principle of constant light speed.

The Velocity Addition Formula is vital when dealing with high-speed phenomena, such as space travel or particle physics, where classical assumptions no longer hold true.
Constant Speed of Light
The principle of the "Constant Speed of Light" is one of the cornerstones of modern physics. It asserts that, in a vacuum, the speed of light \( c \) is the same for all observers, regardless of the motion of the light source or the observer. This idea forms the second postulate of Einstein's theory of Special Relativity.

What makes it so fascinating is that it defies our everyday experience, where speeds seem to simply add up. Imagine standing in a moving train and throwing a ball forward. The speed of the ball would seem faster to someone on the ground. However, for light, this is not the case. Whether an observer is stationary, moving towards, or away from the light source, the speed remains \( c \).

This constant nature of light speed has various implications:
  • It leads to the realization that time and space are interconnected, giving rise to space-time.
  • It implies that time can dilate and lengths can contract depending on the relative motion between observers.

These revelations have reshaped our understanding of the universe, opening windows to new areas of physics and technology, like GPS systems and high-energy particle phenomena. Einstein's insight into the nature of light velocity is thus pivotal in modern scientific thought.

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Most popular questions from this chapter

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.]

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.

Two cells that subdivide on Earth every \(10.0 \mathrm{~s}\) start from the Earth on a journey to the Sun \(\left(1.50 \times 10^{11} \mathrm{~m}\right.\) away) in \(\mathrm{a}\) spacecraft moving at \(0.850 \mathrm{c}\). How many cells will exist when the spacecraft crashes into the Sun? According to Earth observers, with respect to whom the cells are moving, the time taken for the trip to the Sun is the distance traveled \((x)\) over the speed \((v)\) $$ \Delta t_{\mathrm{M}}=\frac{x}{v}=\frac{1.50 \times 10^{11} \mathrm{~m}}{(0.850)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}=588 \mathrm{~s} $$ Because spacecraft clocks are moving with respect to the planet, they appear from Earth to run more slowly. The time these clocks read is $$ \Delta t_{\mathrm{s}}=\Delta f_{\mathrm{M}} / \gamma=\Delta t_{\mathrm{M}} \sqrt{1-(v / \mathrm{c})^{2}} $$ and so $$ \Delta l_{\mathrm{s}}=310 \mathrm{~s} $$ The cells divide according to the spacecraft clock, a clock that is at rest relative to them. They therefore undergo 31 divisions in this time, since they divide each \(10.0 \mathrm{~s}\). Therefore, the total number of cells present on crashing is $$ (2)^{31}=2.1 \times 10^{9} \text { cells } $$

Two twins are \(25.0\) years old when one of them sets out on a journey through space at nearly constant speed. The twin in the spaceship measures time with an accurate watch. When he returns to Earth, he claims to be \(31.0\) years old, while the twin left on Earth knows that she is \(43.0\) years old. What was the speed of the spaceship? The spaceship clock as seen by the space-twin reads the trip time to be \(\Delta t_{\mathrm{S}}\), which is \(6.0\) years long. The Earth-bound twin sees her brother age \(6.0\) years, but her clocks tell her that a time \(\Delta t_{\mathrm{M}}=18.0\) years has actually passed. Hence, \(\Delta t_{\mathrm{M}}=y \Delta t_{\mathrm{S}}\) becomes \(\Delta t_{\mathrm{S}}=\Delta t_{\mathrm{M}} \sqrt{1-(v / \mathrm{c})^{2}}\) and so $$ 6=18 \sqrt{1-(v / \mathrm{c})^{2}} $$ from which \((\mathrm{u} / \mathrm{c})^{2}=1-0.111\) or \(v=0.943 \mathrm{c}=2.83 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

Starting with the definition of linear momentum and the relation between mass and energy, prove that \(\mathrm{E}^{2}=p^{2} c^{2}+m^{2} \mathrm{c}^{4}\). Use this relation to show that the translational KE of a particle of mass \(m\) is \(\sqrt{m^{2} \mathrm{c}^{4}+p^{2} \mathrm{c}^{2}}-m \mathrm{c}^{2}\)

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