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Calculate the speed of a particle \((a)\) whose kinetic energy is equal to twice its rest energy and \((b)\) whose total energy is equal to twice its rest energy.

Short Answer

Expert verified
The velocity of the particle when its kinetic energy is twice its rest energy is \(v=\sqrt{8}/3 c\). The velocity of the particle when its total energy is twice its rest energy is \(v=\sqrt{3}/2 c\).

Step by step solution

01

Define the known and find equations

The problem involves principles of relativity and energy. Let's denote the rest energy as \(E_0=m_0c^2\), where \(m_0\) is the rest mass of the particle and \(c\) is the speed of light. Kinetic energy is \(KE=E_{total}-E_0=((\gamma-1)m_0)c^2\), where \(\gamma\) is the Lorentz factor and it equals \(1/\sqrt{1-(v/c)^2}\). Total energy is \(E_{total}=\gamma m_0 c^2\). The term 'twice' in the problem indicates we need to set kinetic energy equal to \(2E_0\) and total energy equal to \(2E_0\) to solve for \(v\).
02

Solve part (a)

To find the velocity of the particle when kinetic energy is twice the rest energy, we set the equation as \(KE=2E_0\), which gives us \((\gamma-1)m_0c^2=2m_0c^2\). Solving for \(\gamma\) gives \(\gamma=3\). Substituting it into the formula for \(\gamma\), \(1/\sqrt{1-(v/c)^2}\)=3, we can solve for \(v\), which gives us \(v=\sqrt{8}/3 c\).
03

Solve part (b)

To find the velocity when the total energy is twice rest energy, we set \(E_{total}=2E_0\), which gives us \(\gamma m_0c^2=2m_0c^2\). Solving this yields \(\gamma=2\). Substitute \(\gamma=2\) into the formula for \(\gamma\), \(1/\sqrt{1-(v/c)^2}\)=2, solving for \(v\) gives us \(v=\sqrt{3}/2 c\).

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Key Concepts

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

Lorentz factor
In relativistic physics, the Lorentz factor \( \gamma \) plays a crucial role in understanding how time, length, and energy transform when an object approaches the speed of light. It is defined by the formula \( \gamma = \frac{1}{\sqrt{1-(v/c)^2}} \), where \( v \) is the object's speed, and \( c \) is the speed of light.

As the speed \( v \) of an object gets closer to the speed of light \( c \), the Lorentz factor \( \gamma \) increases significantly. This implies that at very high speeds, even small increases in speed require large amounts of energy, impacting how we calculate kinetic and total energy.
  • When \( v = 0 \), \( \gamma = 1 \)
  • As \( v \to c \), \( \gamma \to \infty \)
These transformations explain why classical mechanics doesn't hold at high velocities and highlight the need for relativistic physics.
Kinetic energy
In the context of relativity, kinetic energy \( KE \) is the energy a particle has due to its motion, calculated as \( KE = (\gamma - 1)m_0c^2 \). Here \( \gamma \) is the Lorentz factor, \( m_0 \) is the rest mass, and \( c \) is the speed of light.

This altered equation from classical mechanics reflects how energy grows with velocity, especially when nearing the speed of light.
  • Kinetic energy increases rapidly as velocity approaches \( c \)
  • When \( KE = 2E_0 \), it indicates high relativistic speeds
Understanding kinetic energy in this framework helps us solve problems involving high-speed particles, like in particle physics experiments.
Rest energy
Rest energy \( E_0 \) represents the intrinsic energy a particle possesses due to its mass when at rest. It is defined as \( E_0 = m_0c^2 \), a famous result of Einstein's theory of relativity.

This concept highlights that energy and mass are interchangeable, a core principle of relativistic physics. Rest energy accounts for the energy potential of matter irrespective of its motion, contributing to the total energy of an object.
  • A key component in equations encompassing total and kinetic energy
  • Forms the benchmark for measuring energy changes at relativistic speeds
This relationship underscores the profound nature of mass-energy equivalence in the universe.
Total energy
Total energy \( E_{total} \) in relativistic physics quantifies the complete energy of a particle, comprising both rest energy and kinetic energy. It is calculated using \( E_{total} = \gamma m_0c^2 \).

At high speeds, the Lorentz factor \( \gamma \) significantly affects total energy, indicating large increases in energy requirements as objects approach light speed.
  • Combines rest and kinetic energy
  • Vital for understanding high-energy physics data
  • Dictates energy transformation dynamics in space
Grasping total energy is essential for delving into phenomena like particle accelerators and cosmic events, where particles exhibit extreme velocities.

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Most popular questions from this chapter

Galaxy A is reported to be receding from us with a speed of \(0.347 c\). Galaxy B, located in precisely the opposite direction, is also found to be receding from us at this same speed. What recessional speed would an observer on galaxy A find \((a)\) for our galaxy and \((b)\) for galaxy \(\mathrm{B}\) ?

A \(100-\mathrm{MeV}\) electron, for which \(v=0.999987 c\), moves along the axis of an evacuated tube that has a length of \(2.86 \mathrm{~m}\) as measured by a laboratory observer \(S\) with respect to whom the tube is at rest. An observer \(S^{\prime}\) moving with the electron, however, would see this tube moving past with speed \(v\). What length would this observer measure for the tube?

Find the momentum of a particle of mass \(m\) in order that its total energy be three times its rest energy.

Quite apart from effects due to the Earth's rotational and orbital motions, a laboratory frame is not strictly an inertial frame because a particle placed at rest there will not, in general, remain at rest; it will fall under gravity. Often, however, events happen so quickly that we can ignore free fall and treat the frame as inertial. Consider, for example, a \(1.0-\mathrm{MeV}\) electron (for which \(v=0.941 c\) ) projected horizontally into a laboratory test chamber and moving through a distance of \(20 \mathrm{~cm}\). (a) How long would it take, and \((b)\) how far would the electron fall during this interval? What can you conclude about the suitability of the laboratory as an inertial frame in this case?

An unstable high-energy particle enters a detector and leaves a track \(1.05 \mathrm{~mm}\) long before it decays. Its speed relative to the detector was \(0.992 c .\) What is its proper lifetime? That is, how long would it have lasted before decay had it been at rest with respect to the detector?

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