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Chapter 2: Problem 8

High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of the track depends on the speed of the particle and its lifetime. A particle moving at \(0.993 c\) leaves a track \(1.15 \mathrm{mm}\) long. What is the proper lifetime of the particle?

Short Answer

Expert verified
The proper lifetime of the particle is approximately \( 5.13 \times 10^{-13} \text{ seconds} \).

Step by step solution

01

Understand the given data

The problem gives the speed of a particle, which is moving at 0.993 c (where c is the speed of light), and the length of the track it leaves, which is 1.15 mm.
02

Relate track length to the distance traveled

The length of the track (1.15 mm) is the distance traveled by the particle in the laboratory frame. Use the formula: \[ d = v t \ \text{where } d \text{ is the distance, } v \text{ is the velocity, and } t \text{ is the time} \]
03

Solve for the time in the laboratory frame

Since \( v = 0.993 c \) and \( d = 1.15 \times 10^{-3} \text{ meters} \), we relate the speed and distance to find the time in the laboratory frame. \[ t' = \frac{d}{v} = \frac{1.15 \times 10^{-3} \text{ m}}{0.993 c} \] \( c = 3 \times 10^8 \text{ m/s} \), so \[ t' = \frac{1.15 \times 10^{-3}}{(0.993 \times 3 \times 10^8)} \text{ seconds} \] \[ t' \approx 3.868 \times 10^{-12} \text{ seconds} \]
04

Apply time dilation concept

The proper lifetime (\( t_0 \)) is the time experienced by the particle itself (in its rest frame). According to special relativity, \[ t' = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}} \]
05

Solve for the proper lifetime

Rearrange the time dilation formula to solve for \( t_0 \): \[ t_0 = t' \sqrt{1 - \frac{v^2}{c^2}} \] Substituting \( t' \approx 3.868 \times 10^{-12} \text{ seconds} \) and \( v = 0.993 c \): \[ t_0 = 3.868 \times 10^{-12} \sqrt{1 - (0.993)^2} \] Calculate the value: \[ t_0 \approx 5.13 \times 10^{-13} \text{ seconds} \]

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Key Concepts

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

Special Relativity
Special relativity is a theory in physics proposed by Albert Einstein. It revolutionized our understanding of space, time, mass, and energy. The theory introduces two key postulates:

  • The laws of physics are the same in all inertial frames of reference
  • The speed of light in a vacuum is constant and is independent of the motion of the source or observer

The implications of special relativity can be quite surprising. It tells us that time and space are not absolute. They are relative to the observer's motion. If an object moves very fast (close to the speed of light), we see some interesting effects like time dilation and length contraction.

Understanding special relativity helps us solve problems related to high-speed particles, like the one in our exercise. It allows us to predict how particles behave when they approach light speed.
Time Dilation
When we talk about time dilation, we refer to one of the most fascinating consequences of special relativity. Time dilation means that time moves slower for an object in motion compared to an object at rest.

In our particle exercise, we used the time dilation formula:

t' = \frac{t_0}{\text{鈭殅(1 - \frac{v^2}{c^2})}
Here,
  • \( t' \) represents the time measured in the laboratory frame (where the particle is moving)
  • \( t_0 \) is the proper lifetime of the particle (its time in the rest frame)
  • \( v \) is the speed of the particle
  • \( c \) is the speed of light

This formula tells us that the proper time (or lifetime) of the particle is shorter than the time we observe in the laboratory frame. As the speed of the particle approaches the speed of light, the observed time stretches out significantly.
Particle Speed
In our exercise, the particle speed is vital for understanding the result. The particle moves at 99.3% of the speed of light (\(0.993c\)). This high speed brings relativistic effects into play.

To find how long the particle lives in the laboratory, we started with the track length it leaves, which is 1.15 mm. We used the formula:

d = v t
Given the distance \(d\) and speed \(v\), we can solve for time \(t\).

Speed plays a crucial role because only at speeds close to the speed of light do the effects of special relativity become prominent. For everyday speeds, these relativistic effects are negligible. However, when dealing with particles in accelerators or cosmic rays, these effects are significant.

This understanding helps scientists and engineers design better experiments and technologies that account for relativistic effects.

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Most popular questions from this chapter

By carrying the binomial expansion one term farther, find the next term after \(\frac{1}{2} m v^{2}\) in the classical approximation of the relativistic kinetic energy. For what value of the speed does this term throw the classical value off by \(0.1 \% ?\)

A pion has a rest energy of 135 MeV. It decays into two gamma-ray photons, bursts of electromagnetic radiation that travel at the speed of light. A pion moving through the laboratory at \(v=0.98 c\) decays into two gamma-ray photons of equal energies, making equal angles \(\theta\) with the original direction of motion. Find the angle \(\theta\) and the energies of the two gamma ray photons.

Two twins make a round-trip journey from Earth to a star that is 12 light- years distant. Alice travels at a speed of \(0.6 c .\) Bob departs 10 years after Alice and travels at a speed of \(0.8 c .\) (a) Show that the two twins arrive back on Earth at the same time. \((b)\) Which twin is the younger when they return?

Suppose we want to send an astronaut on a round trip to visit a star that is 200 light- years distant and at rest with respect to Earth. The life support systems on the spacecraft enable the astronaut to survive at most 20 years. (a) At what speed must the astronaut travel to make the round trip in 20 years of spacecraft time? \((b)\) How much time passes on Earth during the round trip?

A physics professor claims in court that the reason he went through the red light \((\lambda=650 \mathrm{nm})\) was that, due to his motion, the red color was Doppler shifted to green \((\lambda=550 \mathrm{nm}) .\) How fast was he going?
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