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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 0.99540\(c\) 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 \(\mathrm{km} .\) (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 \(\mathrm{km}\) 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?

Step by step solution

01

Calculate Time Measured by Earth Observer

The scientist measures the distance as 45.0 km and the speed of the particle as 0.99540\(c\). To find the time it takes to travel, use the formula: \[ t = \frac{d}{v} \]Substitute \( d = 45.0 \text{ km} = 45000 \text{ m} \) and \( v = 0.99540c \), where \( c = 3 \times 10^8 \text{ m/s} \):\[ t = \frac{45000}{0.99540 \times 3 \times 10^8} \approx 1.505 \times 10^{-4} \text{ s} \]

02

Calculate Contracted Distance (Particle's Frame)

In the particle's reference frame, the distance to the Earth is contracted. Use the length contraction formula:\[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where \( L_0 = 45.0 \text{ km} = 45000 \text{ m} \), and \( v = 0.99540c \).Calculate the Lorentz factor \( \gamma \) first:\[ \gamma = \frac{1}{\sqrt{1 - \frac{0.99540^2 c^2}{c^2}}} \approx 10.096 \]Therefore, the contracted length is:\[ L = \frac{L_0}{\gamma} = \frac{45000}{10.096} \approx 4460.5 \text{ m} \]

03

Calculate Time in Particle's Frame Using Time Dilation

In the particle's frame, we calculate the time using the time dilation formula:\[ t' = \frac{t}{\gamma} \]where \( t \approx 1.505 \times 10^{-4} \text{ s} \) and \( \gamma \approx 10.096 \). Hence:\[ t' = \frac{1.505 \times 10^{-4}}{10.096} \approx 1.490 \times 10^{-5} \text{ s} \]

04

Calculate Time in Particle's Frame Using Contracted Distance

To find the time using the contracted distance, use the formula:\[ t' = \frac{L}{v} \]where \( L \approx 4460.5 \text{ m} \) and \( v = 0.99540c \) with \( c = 3 \times 10^8 \text{ m/s} \):\[ t' = \frac{4460.5}{0.99540 \times 3 \times 10^8} \approx 1.490 \times 10^{-5} \text{ s} \]

05

Check Consistency of Results

Comparing both calculations of time \( t' \) in the particle's frame (from Steps 3 and 4), both methods yield \( t' \approx 1.490 \times 10^{-5} \text{ s} \), verifying consistency in the calculated time.

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Key Concepts

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

Time Dilation

Time dilation is a fascinating concept from Einstein's Theory of Relativity. It explains how time can pass at different rates depending on the speed of an object relative to an observer.

• In this context, time dilation means that a moving clock ticks slower compared to a clock that is at rest.
• When an object moves very fast, close to the speed of light, time appears to slow down for that object from the perspective of a stationary observer.

For the particle descending from the atmosphere, the scientist measures its journey to take about 1.505 脳 10^-4 seconds. But from the particle's own frame of reference, less time has elapsed.

• This difference arises because of the relativistic effect of time dilation.
• The actual time taken in the particle鈥檚 frame, as calculated, is approximately 1.490 脳 10^-5 seconds.

This shows the difference between the time experienced by the particle and the time observed by the scientist due to the high speed of the particle.

Lorentz Factor

The Lorentz factor, denoted as \( \gamma \), is crucial for understanding both time dilation and length contraction. It shows the amount by which time, length, and relativistic mass change for an object while it is moving.

• Mathematically, it is defined as: \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)
• Where \( v \) is the speed of the object and \( c \) is the speed of light.
• As \( v \) approaches \( c \), \( \gamma \) becomes very large.

For the particle traveling towards Earth's surface at 0.99540\( c \), the Lorentz factor is approximately 10.096. This results in a significant reduction in measured time and contracted distance, emphasizing its role in extreme speeds close to the speed of light. Understanding \( \gamma \) helps us see why the scientist and the particle have different measures for time and distance.

Speed of Light

The speed of light, denoted as \( c \), is a universal constant and one of the foundations of Einstein's theory. It is the maximum speed at which all energy, matter, and information in the universe can travel.

• Its value is roughly \( 3 \times 10^8 \) meters per second in a vacuum.
• Nothing with mass can move faster than the speed of light, but particles can travel at speeds close to it.

When an object travels at velocities approaching \( c \), relativistic effects like time dilation and length contraction become significantly noticeable. In this exercise, the particle travels at 0.99540 times the speed of light, which is extremely fast.

• This high velocity results in the particle experiencing massive changes in time perception and space measurements when viewed from different frames of reference.
• These phenomena underscore the intriguing nature of relativity and its reliance on the constancy of the speed of light across all frames of reference.

By appreciating \( c \)'s role, we get a clearer picture of how special relativity affects the universe at high speeds.