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Chapter 37: Problem 28

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; (c) 100% greater than 1?

Short Answer

Expert verified
(a) \(v \approx 0.099c\), (b) \(v \approx 0.416c\), (c) \(v \approx 0.866c\).

Step by step solution

01

Understand gamma

The Lorentz factor, \(\gamma\), is given by the formula: \[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \] Here, \(v\) is the velocity we want to find and \(c\) is the speed of light.
02

Set up the equation for (a)

For part (a), we need \(\gamma = 1.01\). So the equation becomes: \[ 1.01 = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \] We solve this to find \(v\).
03

Solve equation for (a)

Square both sides to clear the square root:\[ (1.01)^2 = \frac{1}{1 - \left(\frac{v}{c}\right)^2} \] Rearrange to solve for \(\left(\frac{v}{c}\right)^2\):\[ \left(\frac{v}{c}\right)^2 = 1 - \frac{1}{(1.01)^2} \]Compute the right side and then take the square root to find \(\frac{v}{c}\).
04

Compute (a) result

Calculate \(1 - \frac{1}{1.0201} \approx 0.0098\). Then take the square root to find that \(\frac{v}{c} \approx 0.099\).
05

Set up and solve equation for (b)

For part (b), set \(\gamma = 1.10\): \[ 1.10 = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \]Squaring both sides gives:\[ (1.10)^2 = \frac{1}{1 - \left(\frac{v}{c}\right)^2} \]Solving for \(\left(\frac{v}{c}\right)^2\):\[ \left(\frac{v}{c}\right)^2 = 1 - \frac{1}{(1.10)^2} \]Compute the right side and then take the square root.
06

Compute (b) result

Calculate \(1 - \frac{1}{1.21} \approx 0.1735\). Taking the square root gives \(\frac{v}{c} \approx 0.416\).
07

Set up and solve equation for (c)

For part (c), set \(\gamma = 2.00\):\[ 2.00 = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \]Square both sides:\[ (2.00)^2 = \frac{1}{1 - \left(\frac{v}{c}\right)^2} \]Solving for \(\left(\frac{v}{c}\right)^2\):\[ \left(\frac{v}{c}\right)^2 = 1 - \frac{1}{(2.00)^2} = 1 - 0.25\]Now compute the square root.
08

Compute (c) result

Calculate \(1 - 0.25 = 0.75\). Taking the square root gives \(\frac{v}{c} \approx 0.866\).

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Key Concepts

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

Special Relativity
Special Relativity, a theory formulated by Albert Einstein in 1905, revolutionized our understanding of how objects move through space and time.
It introduces a new framework where the laws of physics remain constant for observers moving at constant speeds relative to one another.
A key insight is that the speed of light is constant and acts as a cosmic speed limit that nothing can surpass. This leads to intriguing consequences like time dilation and length contraction, concepts which become significant at high speeds, approaching the speed of light.
The Lorentz factor, \( \gamma \), is central to these phenomena and adjusts the equations governing time and space, making relativistic effects measurable.
Relativistic Velocity
In the realm of special relativity, velocities do not add up as they do in classical physics. Relativistic velocity takes into account the finite speed of light.
When objects move at speeds close to that of light, their apparent velocity changes are linked with time dilation and length contraction.
The formula for relativistic addition of velocities ensures that the resulting speed never exceeds the speed of light, maintaining Einstein's postulate.
Understanding relativistic velocity is crucial in high-energy physics and astronomy where objects frequently move at speeds significant compared to the speed of light.
Lorentz Transformation
The Lorentz Transformation equations connect space and time coordinates of one inertial frame to another, handling high-speed motion effectively.
They mathematically formalize the concepts introduced by special relativity, adjusting measurements between observers in different inertial frames.
These transformations ensure that the speed of light remains constant and implement corrections for time dilation and length contraction.
The transformations have far-reaching implications, demonstrating that measurements of time and length are relative to the motion of the observer, shaking the foundation of classical mechanics.
Speed of Light
The speed of light, denoted as \( \text{c} \) in physics, is approximately 299,792,458 meters per second.
It serves as a fundamental constant in nature and a cornerstone of Einstein's theory of special relativity.
This speed limit is not just for light, but for all information and matter, playing a crucial role in the relativistic equations and concepts.
By establishing an upper boundary for speed, it reshapes how we perceive motion and energy, leading to a deeper understanding of the universe's operations and constraints.

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Most popular questions from this chapter

An airplane has a length of 60 m when measured at rest. When the airplane is moving at 180 m/s (400 mph) in the alternate universe, how long would the plane appear to be to a stationary observer? (a) 24 m; (b) 36 m; (c) 48 m; (d) 60 m; (e) 75 m.

Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is 0.650c, and the speed of each particle relative to the other is 0.950c. What is the speed of the second particle, as measured in the laboratory?

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 km. (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth? (b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her spaceracer at a constant speed of 0.800c relative to you. At the instant the spaceracer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled 1.20 \(\times\) 10\(^8\) m past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy is 7.50 \(\times\) 10\(^5\) eV. (a) What is the ratio of the speed \(v\) of an electron having this energy to the speed of light, \(c\)? (b) What would the speed be if it were computed from the principles of classical mechanics?
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