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Chapter 37: Problem 28

When Should You Use Relativity? As you have seen, relativistic calculations usually involve the quantity \(\gamma .\) When \(\gamma\) is appreciably greater than \(1,\) we must use relativistic formulas instead of Newtonian ones. For what speed \(v\) (in terms of \(c )\) is the value of \(\gamma(\) a) 1.0\(\%\) greater than \(1 ;(\) b) 10\(\%\) greater than \(1 ;(\mathrm{c}) 100 \%\) greater than 1\(?\)

Short Answer

Expert verified
Relativistic corrections are needed when \( v \approx 0.14c, 0.42c, \text{and} 0.866c \).

Step by step solution

01

Understanding the Problem

The goal is to find the speed \( v \) in terms of the speed of light \( c \) such that the Lorentz factor \( \gamma \) equals 1.01, 1.1, and 2, corresponding to \( \gamma \) being 1%, 10%, and 100% greater than 1, respectively. The Lorentz factor is given by \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \).
02

Case (a): \( \gamma = 1.01 \)

Set \( \gamma = 1.01 \). The equation becomes:\[ 1.01 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]Solve for \( v \) by rearranging the equation:\[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{1.01} \]Square both sides:\[ 1 - \frac{v^2}{c^2} = \left( \frac{1}{1.01} \right)^2 \]\[ \frac{v^2}{c^2} = 1 - \left( \frac{1}{1.01} \right)^2 \]Thus, \( v = c \cdot \sqrt{1 - \left( \frac{1}{1.01} \right)^2} \). Calculate this to find \( v \).
03

Case (b): \( \gamma = 1.1 \)

Set \( \gamma = 1.1 \). The equation becomes:\[ 1.1 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]Thus, solve for \( v \):\[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{1.1} \]Square both sides:\[ 1 - \frac{v^2}{c^2} = \left( \frac{1}{1.1} \right)^2 \]\[ \frac{v^2}{c^2} = 1 - \left( \frac{1}{1.1} \right)^2 \]Therefore, \( v = c \cdot \sqrt{1 - \left( \frac{1}{1.1} \right)^2} \). Calculate this to find \( v \).
04

Case (c): \( \gamma = 2 \)

Set \( \gamma = 2 \). The equation becomes:\[ 2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]Solve for \( v \):\[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{2} \]Square both sides:\[ 1 - \frac{v^2}{c^2} = \left( \frac{1}{2} \right)^2 \]\[ \frac{v^2}{c^2} = 1 - \left( \frac{1}{2} \right)^2 \]Hence, \( v = c \cdot \sqrt{1 - \left( \frac{1}{2} \right)^2} \). Calculate this to find \( v \).
05

Calculate and Conclude

Calculate \( v \) for each case using the formulas derived:- For \( \gamma = 1.01 \), \( v \approx 0.14c \)- For \( \gamma = 1.1 \), \( v \approx 0.42c \)- For \( \gamma = 2 \), \( v \approx 0.866c \)Thus, relativistic effects become important as \( v \) approaches these values.

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Key Concepts

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

Lorentz factor
The Lorentz factor, often denoted by the symbol \( \gamma \), is a vital component in the theory of relativity. It measures how much time, length, and relativity of mass are affected when an object moves at a significant fraction of the speed of light. This factor is defined by the equation:\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \( v \) is the speed of the moving object and \( c \) represents the speed of light. The Lorentz factor essentially tells us how much the time and space measurements of a moving object differ from when it is at rest.

When the velocity \( v \) of an object is much less than the speed of light \( c \), \( \gamma \) approximates to 1, and classical mechanics suffice for calculations. However, as \( v \) gets closer to \( c \), the Lorentz factor increases significantly, indicating the substantial effects of relativity. That's why, in the given exercise, we calculate the values of \( \gamma \) for various percentages to determine when relativistic calculations become necessary.
Special relativity
Special relativity, developed by Albert Einstein, is a cornerstone of modern physics. It revolutionized our understanding of how objects behave at high velocities, particularly those approaching the speed of light. Unlike classical mechanics, special relativity introduces concepts such as time dilation and length contraction.

The two postulates of special relativity are:
  • The laws of physics are the same in all inertial frames, meaning they don't change regardless of the velocity.
  • The speed of light in a vacuum is the same for all observers, regardless of their motion or the motion of the light source.

This means that as an object moves close to light speed, time for the object will appear to slow down compared to a stationary observer. Similarly, lengths contract in the direction of motion. Hence, objects moving at high speeds seem to behave differently, which is why special relativity must be considered when \( v \) is a significant fraction of \( c \). This exercise demonstrates when it becomes crucial to use relativistic formulas to maintain accuracy in calculations.
Speed of light
The speed of light, denoted by \( c \), is a fundamental constant in physics. It is approximately \( 299,792,458 \) meters per second and plays a key role in the theory of relativity. The speed of light is not just a measure of how fast light travels; it serves as the ultimate speed limit for anything in the universe. According to relativity, nothing can exceed this speed.

In practical terms, when speeds approach or surpass a significant fraction of \( c \), classical mechanics (Newtonian physics) fail to provide accurate results, and relativistic calculations become necessary. This is evident in the concept of the Lorentz factor, where \( v \) gets closer to \( c \), resulting in a noticeable increase in \( \gamma \).

Understanding the speed of light鈥檚 role in these equations helps clarify why at certain velocities, like those calculated in our exercise, relativistic effects cannot be ignored. This cements the necessity of using Einstein鈥檚 theories to analyze motion and changes in physical properties accurately for objects moving at high velocities.

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Most popular questions from this chapter

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 \(\mathrm{km} ?\)

Everyday Time Dilation. Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 \(\mathrm{m} / \mathrm{s}\) and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 \(\mathrm{h}\) . By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (Hint: since \(u \ll c,\) you can simplify \(\sqrt{1-u^{2} / c^{2}}\) by a binomial expansion.)

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c,\) what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

What is the speed of a particle whose kinetic energy is equal to (a) its rest energy and (b) five times its rest energy?

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?
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