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Chapter 26: Problem 24

The Tevatron is a particle accelerator at Fermilab that accelerates protons and antiprotons to high energies in an underground ring. Scientists observe the results of collisions between the particles. The protons are accelerated until they have speeds only \(100 \mathrm{m} / \mathrm{s}\) slower than the speed of light. The circumference of the ring is \(6.3 \mathrm{km} .\) What is the circumference according to an observer moving with the protons? [Hint: Let \(v=c-u\) where $v \text { is the proton speed and } u=100 \mathrm{m} / \mathrm{s} .$]

Short Answer

Expert verified
Based on the step-by-step solution provided, calculate the circumference of the Tevatron particle accelerator according to an observer moving with the protons. Recall the given information: the stationary circumference \(L_0 = 6.3 \, km = 6300 \, m\), and the difference between the proton's speed and the speed of light is \(100 \, m/s\).

Step by step solution

01

Write down the known quantities

We are given: 1. The stationary circumference \(L_0 = 6.3 \, km = 6300 \, m\) 2. The difference between the proton's speed \(v\) and the speed of light \(c\) is \(100 \, m/s\)
02

Calculate the proton's speed

We are given the hint to let \(v = c - u\) where \(u = 100 \, m/s\) and \(c\) is the speed of light. Therefore, we can write the protons' speed as \(v = c - 100 \, m/s\). Since the speed of light \(c\) is approximately \(3 \times 10^8 \, m/s\), the proton's speed \(v\) is: $$ v = (3 \times 10^8 \, m/s) - 100 \, m/s $$
03

Determine the Lorentz factor

The Lorentz factor, \(\gamma\), is a crucial quantity in special relativity, particularly for dealing with length contraction and time dilation. It's given by the formula: $$ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $$ Substitute the protons' speed \(v\) found in Step 2 into the above equation to find the Lorentz factor \(\gamma\).
04

Find the moving observer's circumference using length contraction

The length contraction formula is given by: $$ L = \frac{L_0}{\gamma} $$ Where \(L\) is the contracted length (in this case, the moving observer's circumference), \(L_0\) is the stationary length (the given \(6.3 \, km\)), and \(\gamma\) is the Lorentz factor found in Step 3. Substitute the values of \(L_0\) and \(\gamma\) into the formula to find the moving observer's circumference, \(L\). Following these steps will give you the circumference of the Tevatron particle accelerator according to an observer moving with the protons.

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Most popular questions from this chapter

Muons are created by cosmic-ray collisions at an elevation \(h\) (as measured in Earth's frame of reference) above Earth's surface and travel downward with a constant speed of \(0.990 c .\) During a time interval of \(1.5 \mu \mathrm{s}\) in the rest frame of the muons, half of the muons present at the beginning of the interval decay. If one fourth of the original muons reach Earth before decaying, about how big is the height \(h ?\)
Refer to Example \(26.2 .\) One million muons are moving toward the ground at speed \(0.9950 c\) from an altitude of \(4500 \mathrm{m} .\) In the frame of reference of an observer on the ground, what are (a) the distance traveled by the muons; (b) the time of flight of the muons; (c) the time interval during which half of the muons decay; and (d) the number of muons that survive to reach sea level. [Hint: The answers to (a) to (c) are not the same as the corresponding quantities in the muons' reference frame. Is the answer to (d) the same?]
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