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Use the relativistic velocity addition to reconfirm that the speed of light with respect to any inertial reference frame is \(c\). Assume one-dimensional motion along a common \(x\) -axis.

Short Answer

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Question: Use the relativistic velocity addition formula to confirm that the speed of light is the same in any inertial reference frame in a one-dimensional motion along the common x-axis. Answer: By applying the relativistic velocity addition formula and setting the velocity of the object relative to both observers as equal to the speed of light (c), we derived the equation \(c + v = v + c\). This result confirms that the speed of light remains constant in any inertial reference frame, regardless of the relative velocity (v).

Step by step solution

01

Write down the relativistic velocity addition formula

The relativistic velocity addition formula is given by: \(u = \frac{v +u'}{1 + \frac{vu'}{c^2}}\) where \(u\) is the velocity of an object relative to an observer, \(v\) is the relative velocity of two observers, and \(u'\) is the velocity of the object relative to the second observer. The speed of light is denoted by \(c\).
02

Set the speed of light as the velocity of the object relative to both observers

Since we want to show that the speed of light remains constant in any inertial reference frame, we will set the velocity of the object relative to both observers as equal to the speed of light, \(c\). Thus, \(u = c\) and \(u' = c\).
03

Substitute the values of \(u\), \(u'\) and \(c\) into the relativistic velocity addition formula

Substitute the values of \(u = c\) and \(u' = c\) into the relativistic velocity addition formula: \(c = \frac{v+c}{1 + \frac{vc}{c^2}}\) Now, we will simplify the equation to show that the speed of light remains constant.
04

Simplify the equation

First, cancel the \(c^2\) term in the denominator: \(c = \frac{v + c}{1 + \frac{v}{c}}\) Next, multiply both sides of the equation by the denominator to get rid of the fraction: \(c(1 + \frac{v}{c}) = v + c\) Distribute \(c\) through the parentheses: \(c + v = v + c\)
05

Verify the speed of light remains constant

As per the result, the equation simplifies to \(c + v = v + c\). This result implies that the speed of light remains constant in any inertial reference frame, regardless of the relative velocity \(v\). Since the speed of light is \(c\) with respect to both observers, the problem is solved and the exercise is complete.

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Key Concepts

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

Inertial Reference Frame
An **inertial reference frame** is a key concept in physics, particularly when dealing with relativity. It is essentially a viewpoint, or a coordinate system, where an observer is not experiencing any acceleration. Think of it as standing on a platform that isn’t moving or rotating. This makes the laws of physics simpler and more uniform.

Inertial frames are where both Newton’s laws and Einstein’s relativistic theories work best. For instance:
  • Newton’s first law: An object in motion remains in motion, and an object at rest stays at rest unless acted upon by an external force, is easily observed in an inertial frame.
  • In relativity, regardless of the speed of observers in different inertial frames, the speed of light is always constant, demonstrating a fundamental principle of Einstein's theory of relativity.
Understanding inertial frames helps clarify why the speed of light is constant in all situations and why relativistic effects only become noticeable under high velocities or accelerations.
Speed of Light
One of the most fascinating constants in the universe is the **speed of light**, often denoted by the symbol \(c\). In a vacuum, it measures approximately \(299,792,458\) meters per second. This constant is significant in many physics equations, including Einstein’s famous equation, \(E=mc^2\).

Notably, Einstein’s postulates in special relativity stress that:
  • The speed of light is always the same, no matter what inertial frame you’re in. It’s why when observing light, whether we’re stationary or moving at a high speed, we calculate it as \(c\).
  • This constant nature reinforces the principle that light’s speed is the universe’s speed limit. No information or object can travel faster than \(c\).
The exercise we tackled reinforces that the velocity of light remains unchanged when evaluated through different inertial frames using the relativistic velocity addition, showcasing the impressive symmetry and consistency of natural laws.
One-Dimensional Motion
The concept of **one-dimensional motion** simplifies our understanding of movement by considering only a single direction or line. Imagine motion confined to a straight track, where only forward or backward movement matters.

In the study of relativity and velocity addition, one-dimensional motion makes calculations manageable and less complex. Here's why:
  • It lets us focus on motion along a single axis, the \(x\)-axis, mainly avoiding complications with multi-dimensional dynamics.
  • For phenomena like light speed calculations, this simplification shows how consistency in speed occurs even when simplifying the scenario.
Thus, breaking motion down to one dimension is not just a mathematical convenience but a way to better visualize how principles of physics like the constancy of light’s speed operate regardless of direction or relative velocity between frames.

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Most popular questions from this chapter

You shouldn't invoke time dilation due to your relative motion with respect to the rest of the world as an excuse for being late to class. While it is true that relative to those at rest in the classroom, your time runs more slowly, the difference is likely to be negligible. Suppose over the weekend you drove from your college in the Midwest to New York City and back, a round trip of \(2200 .\) miles, driving for 20.0 hours each direction. By what amount, at most, would your watch differ from your professor's watch?

A spacecraft travels along a straight line from Earth to the Moon, a distance of \(3.84 \cdot 10^{8} \mathrm{~m}\). Its speed measured on Earth is \(0.50 c\). a) How long does the trip take, according to a clock on Earth? b) How long does the trip take, according to a clock on the spacecraft? c) Determine the distance between Earth and the Moon if it were measured by a person on the spacecraft.

A rocket ship approaching Earth at \(0.90 c\) fires a missile toward Earth with a speed of \(0.50 c,\) relative to the rocket ship. As viewed from Earth, how fast is the missile approaching Earth?

If a muon is moving at \(90.0 \%\) of the speed of light, how does its measured lifetime compare to when it is in the rest frame of a laboratory, where its lifetime is \(2.2 \cdot 10^{-6}\) s?

Suppose you are explaining the theory of relativity to a friend, and you have told him that nothing can go faster than \(300,000 \mathrm{~km} / \mathrm{s}\). He says that is obviously false: Suppose a spaceship traveling past you at \(200,000 \mathrm{~km} / \mathrm{s}\), which is perfectly possible according to what you are saying, fires a torpedo straight ahead whose speed is \(200,000 \mathrm{~km} / \mathrm{s}\) relative to the spaceship, which is also perfectly possible; then, he says, the torpedo's speed is \(400,000 \mathrm{~km} / \mathrm{s}\). How would you answer him?

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