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Chapter 35: Problem 24

What is the value of \(\gamma\) for a particle moving at a speed of \(0.8 c ?\)

Short Answer

Expert verified
Answer: The Lorentz factor for a particle moving at a speed of 0.8c is approximately 1.67.

Step by step solution

01

Understand the Lorentz Factor Formula

The Lorentz factor, denoted by \(\gamma\), is a fundamental concept in special relativity. It relates the time, length, and mass of an object moving at relativistic speeds to those measured in its rest frame. The formula to calculate the Lorentz factor is: $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ Where: - \(v\) is the speed of the particle - \(c\) is the speed of light
02

Insert the Given Values

We are given that the particle is moving at a speed of \(0.8c\). To find the value of \(\gamma\), we'll substitute the given values into the formula: $$\gamma = \frac{1}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}}$$
03

Simplify the Expression

Now, let's simplify the expression by squaring the speed of the particle, dividing by the square of the speed of light, and subtracting the result from 1: $$\gamma = \frac{1}{\sqrt{1 - \frac{(0.64c^2)}{c^2}}}$$
04

Cancel out the Speed of Light

The \(c^2\) in the numerator and denominator will cancel each other out, which leaves us with: $$\gamma = \frac{1}{\sqrt{1 - 0.64}}$$
05

Solve for Gamma

Finally, we can calculate the value of \(\gamma\) by finding the square root: $$\gamma = \frac{1}{\sqrt{0.36}}$$ $$\gamma = \frac{1}{0.6}$$ $$\gamma \approx 1.67$$ Therefore, the value of \(\gamma\) for a particle moving at a speed of \(0.8c\) is approximately 1.67.

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Most popular questions from this chapter

More significant than the kinematic features of the special theory of relativity are the dynamical processes that it describes that Newtonian dynamics does not. Suppose a hypothetical particle with rest mass \(1.000 \mathrm{GeV} / c^{2}\) and \(\mathrm{ki}-\) netic energy \(1.000 \mathrm{GeV}\) collides with an identical particle at rest. Amazingly, the two particles fuse to form a single new particle. Total energy and momentum are both conserved in the collision. a) Find the momentum and speed of the first particle. b) Find the rest mass and speed of the new particle.

At rest, a rocket has an overall length of \(L .\) A garage at rest (built for the rocket by the lowest bidder) is only \(L / 2\) in length. Luckily, the garage has both a front door and a back door, so that when the rocket flies at a speed of \(v=0.866 c\), the rocket fits entirely into the garage. However, according to the rocket pilot, the rocket has length \(L\) and the garage has length \(L / 4\). How does the rocket pilot observe that the rocket does not fit into the garage?

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?

Sam sees two events as simultaneous: (i) Event \(A\) occurs at the point (0,0,0) at the instant 0: 00: 00 universal time; (ii) Event \(B\) occurs at the point \((500, \mathrm{~m}, 0,0)\) at the same moment. Tim, moving past Sam with a velocity of \(0.999 c \hat{x}\), also observes the two events. a) Which event occurred first in Tim's reference frame? b) How long after the first event does the second event happen in Tim's reference frame?

An astronaut on a spaceship traveling at a speed of \(0.50 c\) is holding a meter stick parallel to the direction of motion. a) What is the length of the meter stick as measured by another astronaut on the spaceship? b) If an observer on Earth could observe the meter stick, what would be the length of the meter stick as measured by that observer?
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