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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

Calculate the magnitude of the force required to give a 0.145-kg baseball an acceleration \(a =\) 1.00 m/s\(^2\) in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) 10.0 m/s; (b) 0.900c; (c) 0.990c. (d) Repeat parts (a), (b), and (c) if the force and acceleration are perpendicular to the velocity.

Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth at a speed of 0.520c. If the radiation has a frequency of 8.64 \(\times\) 10\(^{14}\) Hz in the rest frame of the star, what is the frequency measured by an observer on earth?

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An observer in frame \(S'\) is moving to the right (+\(x\)-direction) at speed \(u\) = 0.600c away from a stationary observer in frame S. The observer in \(S'\) measures the speed \(v'\) of a particle moving to the right away from her. What speed \(v'\) does the observer in S measure for the particle if (a) \(v'\) = 0.400c; (b) \(v'\) = 0.900c; (c) \(v'\) = 0.990c?

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