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Chapter 37: Problem 72

The French physicist Armand Fizeau was the first to measure the speed of light accurately. He also found experimentally that the speed, relative to the lab frame, of light traveling in a tank of water that is itself moving at a speed \(V\) relative to the lab frame is $$v=\frac{c}{n}+k V$$ where \(n=1.333\) is the index of refraction of water. Fizeau called \(k\) the draging coefficient and obtained an experimental value of \(k=0.44 .\) What value of \(k\) do you calculate from relativistic transformations?

Short Answer

Expert verified
The theoretical value of the dragging coefficient \(k\) is approximately 0.435.

Step by step solution

01

Understand the Problem

We need to find the theoretical value of the dragging coefficient \(k\) using relativistic transformations. The formula given is \(v=\frac{c}{n}+kV\), where \(n=1.333\) and Fizeau's experimental \(k=0.44\). We will use the relativistic velocity addition to find \(k\).
02

Apply Relativistic Velocity Addition

Relativistic velocity addition states that if an object moves at speed \(u\) and the system itself moves at speed \(V\), then the speed \(v\) in another frame is given by: \[ v = \frac{u + V}{1 + \frac{uV}{c^2}} \]Here, \(u = \frac{c}{n}\), which is the speed of light in water.
03

Substitute Known Values

Substitute \(u = \frac{c}{n} = \frac{c}{1.333}\) into the relativistic equation:\[ v = \frac{\frac{c}{1.333} + V}{1 + \frac{\left(\frac{c}{1.333}\right)V}{c^2}} \]
04

Simplify the Equation

Simplify the denominator where:\[ \frac{\left(\frac{c}{1.333}\right)V}{c^2} = \frac{V}{c \cdot 1.333} \]This gives us:\[ v = \frac{\frac{c}{1.333} + V}{1 + \frac{V}{c \cdot 1.333}} \]
05

Approximate for Small Velocities

In practical situations, \( V \) is much smaller than \( c \), allowing us to use a first-order approximation for small values. Thus, the denominator simplifies to approximately 1:\[ v \approx \frac{c}{1.333} + V \cdot \left(1 - \frac{V}{c \cdot 1.333} \right) \]
06

Identify the Dragging Coefficient

In the approximation: \[ v \approx \frac{c}{n} + \left(1 - \frac{1}{n^2}\right)V \]By comparing this expression to \(v = \frac{c}{n} + kV\), we identify that the theoretical value of \(k\) is: \[ k = 1 - \frac{1}{n^2} \] Substitute \(n = 1.333\): \[ k = 1 - \frac{1}{(1.333)^2} \]
07

Calculate the Theoretical Value of k

Calculate the expression: \[ k = 1 - \frac{1}{(1.333)^2} = 1 - \frac{1}{1.778889} \approx 0.435 \] Hence, the theoretical dragging coefficient \(k\) is approximately 0.435.

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Key Concepts

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

Speed of Light
The speed of light is a fundamental constant in physics, denoted by the symbol \(c\). It is approximately \(299,792,458\) meters per second in a vacuum. Light does not always travel at this speed, as its velocity changes depending on the medium through which it passes. This notion is crucial for phenomena like refraction or light moving through different substances such as air, water, or glass.

Light's speed in any medium other than a vacuum is slower, and this adjustment in speed is due to the medium's "index of refraction."

One interesting problem concerning the speed of light occurs when light moves through a medium that itself is moving, such as light passing through a tank of water in motion. In such scenarios, relativistic effects need to be considered, specifically how velocities add up in relativistic contexts.

In classical mechanics, simple addition suffices for velocity calculations. However, relativistic circumstances, such as near-light speed or involving light itself, necessitate using the relativistic velocity addition formula. This accounts for the reduction of combined speeds when considering the limit set by the speed of light.

  • Symbol for Speed of Light: \(c\)
  • Speed in Vacuum: \(299,792,458\) m/s
  • Speed varies by medium: Determined by the medium's refractive index
Index of Refraction
The index of refraction is a critical factor that describes how much light bends, or refracts, when it enters a different medium. It is represented by the symbol \(n\) and is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. Mathematically, it is expressed as \( n = \frac{c}{v} \), where \(v\) is the speed of light in the medium.

The index of refraction is always greater than or equal to 1. For instance, in water, \(n\) is approximately \(1.333\). This means that light travels 1.333 times slower in water than it does in a vacuum.

Understanding the index of refraction helps explain phenomena like why a straw appears bent when placed in a glass of water. It also plays a significant role in things like making corrective lenses or developing fiber-optic technologies.

  • Symbol: \(n\)
  • For water: \(n = 1.333\)
  • Role: Determines light's speed and how much it bends in a medium
Dragging Coefficient
The dragging coefficient in the context of Fizeau's experiment refers to the factor \(k\) used to describe how the speed of light in a moving medium, like water, is affected by the medium's velocity. It is a reflection of how light "drags" in a moving fluid.

In Fizeau's work, the observational formula was \(v = \frac{c}{n} + kV\), where \(v\) is the light speed in the moving medium, \(c\) is the speed of light in vacuum, and \(V\) is the velocity of the water relative to the observer.

The dragging coefficient \(k\) captures the degree to which the water's motion affects the speed of light. Fizeau's experimental value for this coefficient was \(0.44\), while the theoretically derived value using relativistic considerations is approximately \(0.435\).

Such measurements show the nuances of light's interaction with moving media and emphasize how even non-intuitive physical phenomena can be scientifically explained and quantified.

  • Represents: The effect of medium's motion on light speed
  • Fizeau's Experimental Value: \(k = 0.44\)
  • Theoretical Value: \(k \approx 0.435\)

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Most popular questions from this chapter

The starships of the Solar Federation are marked with the symbol of the federation, a circle, while starships of the Denebian Empire are marked with the empire's symbol, an ellipse whose major axis is 1.40 times longer than its minor axis \((a=1.40 b\) in Fig. P37.51). How fast, relative to an observer, does an empire ship have to travel for its marking to be confused with the marking of a federation ship?

A nuclear bomb containing 12.0 \(\mathrm{kg}\) of plutonium explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in \(10^{4} .\) (a) How much energy is released in the explosion? (b) If the explosion takes place in 4.00\(\mu \mathrm{s}\) , what is the average power developed by the bomb? (c) What mass of water could the released energy lift to a height of 1.00 \(\mathrm{km} ?\)

A proton (rest mass \(1.67 \times 10^{-27} \mathrm{kg} )\) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the speed of the proton?

An extraterrestrial spaceship is moving away from the earth after an unpleasant encounter with its inhabitants. As it departs, the spaceship fires a missile toward the earth. An observer on earth measures that the spaceship is moving away with a speed of 0.600\(c .\) An observer in the spaceship measures that the missile is moving away from him at a speed of 0.800\(c .\) As measured by an observer on earth, how fast is the missile approaching the earth?

In high-energy physics, new particles can be created by collisions of fast- moving projectile par- ticles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive kaon \(\left(\mathrm{K}^{+}\right)\) $$p+p \rightarrow p+p+\mathrm{K}^{-}+\mathrm{K}^{+}$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 \(\mathrm{MeV}\) , and the rest energy of each proton is 938.3 \(\mathrm{MeV}\) . (Hint: It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)
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