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After being produced in a collision between elementary particles, a positive pion \(\left(\pi^{+}\right)\) must travel down a 1.90 -km-long tube to reach an experimental area. A \(\pi^{+}\) particle has an average lifetime (measured in its rest frame) of \(2.60 \times 10^{-8} \mathrm{s}\) ; the \(\pi^{+}\) we are considering has this lifetime. How fast must the \(\pi^{+}\) travel if it is not to decay before it reaches the end of the tube? (Since \(u\) will be very close to \(c,\) write \(u=(1-\Delta) c\) and give your answer in terms of \(\Delta\) rather than \(u .\) (b) The \(\pi^{+}\) has a rest energy of 139.6 \(\mathrm{MeV} .\) What is the total energy of the \(\pi^{+}\) at the speed calculated in part (a)?

Short Answer

Expert verified
\(\Delta = 3.54 \times 10^{-5}\), Total Energy = 301.2 MeV.

Step by step solution

01

Relativistic Time Dilation

First, we need to determine the time it takes for the pion to travel 1.90 km in the lab frame. In the pion's rest frame, it has a lifetime of \(2.60 \times 10^{-8} \, \text{s}\). According to time dilation in special relativity, the lab frame lifetime \(t_L\) is given by \(t_L = \gamma t_0\), where \(\gamma = \frac{1}{\sqrt{1 - \left(\frac{u}{c}\right)^2}}\) and \(t_0\) is the proper lifetime. Rearranging gives us \(t_L = \frac{2.60 \times 10^{-8} \, \text{s}}{\sqrt{1 - \left(\frac{u}{c}\right)^2}}\).
02

Calculate Required Speed Expression

For the pion to travel down the 1.90 km tube without decaying, the time in the lab frame must be at least the time it takes to traverse this distance: \(t_L = \frac{1.90 \, \text{km}}{u}\). Substituting the expression from Step 1 into this equation, we get: \(\frac{1.90 \, \text{km}}{u} = \frac{2.60 \times 10^{-8} \, \text{s}}{\sqrt{1 - \left(\frac{u}{c}\right)^2}}\).
03

Express Speed in Terms of \(\Delta\)

Since \(u = (1 - \Delta)c\), substitute \(u\) in terms of \(\Delta\) into the equation from Step 2: \(\frac{1.90 \, \text{km}}{(1 - \Delta)c} = \frac{2.60 \times 10^{-8} \, \text{s}}{\sqrt{1 - (1-\Delta)^2}}\). Solve this for \(\Delta\).
04

Calculate Total Energy at the Derived Speed

The total energy \(E\) at speed \(u\) is given by \(E = \gamma m_0c^2\), where \(m_0c^2 = 139.6 \, \text{MeV}\) is the rest energy of the pion. Now \(\gamma = \frac{1}{\sqrt{1 - (1-\Delta)^2}}\) from earlier steps. Substitute and find \(E\).
05

Solve for \(\Delta\) and Energy Calculation

From the previous steps, solve the equation for \(\Delta\) to ensure the pion travels 1.90 km without decaying, and use the derived \(\Delta\) to compute \(E\). The solution involves algebraic manipulation and solving for the required parameters.

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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 fundamental aspect of Albert Einstein's theory of special relativity. It implies that time is not an absolute entity but rather can vary for observers in different frames of reference. Think of it like a stretchy rubber band: time can stretch and contract depending on how fast you are moving.
In the case of the positive pion, its lifetime, measured in its own rest frame, is just a few billionths of a second (\(2.60 \times 10^{-8} \text{s}\)). However, to an observer in the lab frame (where the pion is moving), this lifetime appears longer due to time dilation. This phenomenon allows the pion to travel farther than it could if time ticked at the same rate in every frame.
The key formula to denote this is\[ t_L = \gamma t_0 \]where \(t_L\) is the dilated time or lifetime in the lab frame, \(t_0\) is the proper time or lifetime in the pion's rest frame, and \(\gamma\) is the gamma factor which we will discuss later. This formula illustrates how velocity affects the perception of time.
Relativistic Speed
In special relativity, speeds close to the speed of light present unique challenges and interesting phenomena. When dealing with particles like the positive pion, these speeds are termed 'relativistic speeds'. Such speeds involve significant fractions of the speed of light \(c\).
To solve problems like our pion traveling through the tube, we sometimes express their speed in terms of a small difference from the speed of light, noted as \((1 - \Delta)c\). Here, \(\Delta\) represents a tiny fraction showing just how short of light speed the particle travels. This approach is practical when the speed \(u\) is so close to \(c\) that distinguishing it by small differences is more insightful.
Relativistic speeds demonstrate the limit of how ordinary physics change at velocities approaching light speed. This affects not just the perception of time, but also mass and energy.
Particle Physics
Particle physics studies the fundamental particles of the universe, like the pion in our example. These particles often exist only for short periods before decaying into other particles. Particles such as pions are produced by high-energy collisions, such as those seen in particle accelerators.
Pions belong to a category known as mesons, which are made of one quark and one antiquark. They are key in studying nuclear forces because they mediate the strong force that holds nuclei together.
  • Quarks: Fundamental constituents of matter
  • Antiquarks: Antimatter counterparts to quarks
  • Mesons: Particles, including pions, that participate in the strong force
Understanding pions and other elementary particles helps physicists probe the foundations of matter and the forces governing the universe.
Rest Energy
Rest energy in particle physics is the energy an object possesses from simply existing with mass, even when it is not moving. This concept arises from Einstein's famous equation:\[ E= m_0c^2 \]where \(E\) is the rest energy, \(m_0\) is the rest mass, and \(c\) is the speed of light.
For the pion in our problem, its rest energy is crucial to calculating the total energy it has while traveling at relativistic speeds. The rest energy of the pion is \(139.6 \, \text{MeV}\).This rest energy can also be transformed into other forms of energy when the pion interacts with other particles, or upon decaying.
  • Reflects mass-energy equivalence
  • Crucial for energy conservation calculations
Rest energy informs the total energy of the particle when motion is considered.
Gamma Factor
The gamma factor, denoted \(\gamma\), is a central component in understanding how time, length, and relativistic momentum transform at high velocities. It quantifies the strength of relativistic effects seen as speeds approach that of light.
The gamma factor is defined as:\[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{u}{c}\right)^2}} \]where \(u\) is the object's speed and \(c\) is the speed of light. As \(u\) nears \(c\), \(\gamma\) grows significantly above 1, illustrating increased effects of time dilation and relativistic mass.
The gamma factor not only affects time dilation but also plays a vital role in calculating total energy at relativistic speeds, as it amplifies the mass-energy equivalence effect, meaning more energy is needed as speed increases.

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Most popular questions from this chapter

The starships of the Solar Federation are marked with the symbol of the federation, a circle, while starships of the Denebian Empire are marked with the empire's symbol, an ellipse whose major axis is 1.40 times longer than its minor axis \((a=1.40 b\) in Fig. P37.51). How fast, relative to an observer, does an empire ship have to travel for its marking to be confused with the marking of a federation ship?

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Determining the Masses of Stars. Many of the stars in the sky are actually binary stars, in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they empler. Stars for which this is the case are called spectroscopic binary stars. Figure \(\mathrm{P3} 7.75\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m,\) orbiting their center of mass in a circle of radius \(R\) . The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of \(4.568110 \times\) \(10^{14}\) Hz. In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between \(4.567710 \times 10^{14} \mathrm{Hz}\) and \(4.568910 \times 10^{14} \mathrm{Hz}\) . Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. Hint: The speeds involved are much less than \(c,\) so you may use the approximate result \(\Delta f / f=u / c\) given in Section 37.6 . (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, \(1.99 \times 10^{30}\) kg. Compare the value of \(R\) to the distance from the earth to the sun, \(1.50 \times 10^{11}\) m. (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

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