Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A particle \(a\) traveling along the positive \(x\) axis of frame \(\mathcal{S}\) with speed 0.5c decays into two identical particles, \(a \rightarrow b+b,\) both of which continue to travel on the \(x\) axis. (a) Given that \(m_{a}=2.5 m_{b},\) find the speed of either \(b\) particle in the rest frame of particle \(a .\) (b) By making the necessary transformation on the result of part (a), find the velocities of the two \(b\) particles in the original frame S.

Short Answer

Expert verified
The velocities of the two \(b\) particles in frame S are \(0.846c\) and \(-0.143c\).

Step by step solution

01

Analyze the Decay Process in Rest Frame of Particle a

In the rest frame of particle \(a\), the conservation of momentum must hold. Since particle \(a\) is at rest, the two \(b\) particles must have equal and opposite velocities to conserve momentum.Let the velocity of one \(b\) particle be \(v'_b\) in the rest frame of \(a\). Since the particles are identical and the total momentum is zero, \(v'_b = -v'_b\). We need to solve for \(v'_b\) using conservation of energy.
02

Apply Conservation of Energy in Rest Frame of Particle a

In the rest frame of particle \(a\), energy is conserved:\[ m_a c^2 = 2 \times \gamma_b m_b c^2 \]where \(\gamma_b\) is the Lorentz factor for particle \(b\):\[ \gamma_b = \frac{1}{\sqrt{1 - \left(\frac{v'_b}{c}\right)^2}} \]Given that \(m_a = 2.5 m_b\), we can substitute and solve:\[ 2.5 m_b c^2 = 2 \times \gamma_b m_b c^2 \]\[ \frac{2.5}{2} = \gamma_b \]\[ \gamma_b = 1.25 \]
03

Calculate the Velocity of b Particles in Rest Frame of a

Use \(\gamma_b = 1.25\) to find \(v'_b\):\[ \gamma_b = \frac{1}{\sqrt{1 - \left(\frac{v'_b}{c}\right)^2}} \]Solve for \(v'_b\):\[ 1.25 = \frac{1}{\sqrt{1 - \left(\frac{v'_b}{c}\right)^2}} \]\[ \sqrt{1 - \left(\frac{v'_b}{c}\right)^2} = \frac{1}{1.25} \]\[ 1 - \left(\frac{v'_b}{c}\right)^2 = \left(\frac{4}{5}\right)^2 \]\[ \left(\frac{v'_b}{c}\right)^2 = 1 - \frac{16}{25} \]\[ \left(\frac{v'_b}{c}\right)^2 = \frac{9}{25} \]\[ v'_b = 0.6c \]
04

Perform Lorentz Transformation to Find Velocities in Frame S

In frame \(S\), particle \(a\) has velocity \(u = 0.5c\) and the velocities of the \(b\) particles need to be calculated. Using velocity addition formula:\[ v_{b1} = \frac{v'_b + u}{1 + \frac{v'_b u}{c^2}} \]\[ v_{b2} = \frac{-v'_b + u}{1 - \frac{v'_b u}{c^2}} \]Substitute \(v'_b = 0.6c\):\[ v_{b1} = \frac{0.6c + 0.5c}{1 + \frac{0.6c \times 0.5c}{c^2}} = \frac{1.1c}{1.3} = 0.846c \]\[ v_{b2} = \frac{-0.6c + 0.5c}{1 - \frac{0.6c \times 0.5c}{c^2}} = \frac{-0.1c}{0.7} = -0.143c \]
05

Conclusion

The velocity of the \(b\) particle in the rest frame of \(a\) is \(0.6c\). In frame \(S\), the velocities of the \(b\) particles are \(0.846c\) and \(-0.143c\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Lorentz Transformation
Understanding the Lorentz Transformation is essential in the realm of relativistic physics, where objects move at speeds close to the speed of light. This transformation allows us to convert the coordinates of an event from one inertial frame to another.

When transitioning between frames, the Lorentz transformation helps calculate how measurements of space and time vary. It accounts for the effects of time dilation and length contraction. The transformations are:
  • For time: \( t' = \gamma (t - \frac{vx}{c^2}) \)
  • For position: \( x' = \gamma (x - vt) \)
Where \( \gamma \) is the Lorentz factor, \( v \) is the relative speed between the frames, \( t \) is time in the stationary frame, and \( x \) is the position.

In practical problems like our exercise, using the Lorentz transformation simplifies finding the velocities in different frames, ensuring consistent results across varying perspectives.
Conservation of Momentum
In physics, momentum conservation is a fundamental principle that states the total momentum of a closed system remains constant over time, if it is not influenced by external forces. In the context of our exercise, momentum conservation is crucial when dealing with particle decays.

When particle \( a \) decays into two identical particles \( b \), the system's momentum before and after the decay must remain unchanged. In the rest frame of particle \( a \), which initially has zero momentum, the sum of the momenta of the two particles \( b \) must also be zero.

This leads to the condition:
\[ m_b v'_b + m_b (-v'_b) = 0 \]
This condition emphasizes that both particles travel at equal speeds but opposite directions, ensuring no net change in momentum.
Conservation of Energy
Within any closed physical system, the energy conservation principle maintains that the total energy remains constant over time. This principle is vital in analyzing high-speed particle interactions in relativistic mechanics.

In the given exercise, during the decay of particle \( a \) into two particles \( b \), the initial energy of particle \( a \) is entirely converted into the energies of the two \( b \) particles. This can be expressed as:
  • Initial energy: \( E_a = m_a c^2 \)
  • Final energy: \( 2 \times \gamma_b m_b c^2 \)
Equating these gives: \[ m_a c^2 = 2 \gamma_b m_b c^2 \] Substituting the given mass relationship, \( m_a = 2.5 m_b \), lets us solve for \( \gamma_b \): \[ \gamma_b = \frac{1.25}{1} \] This equation ensures that we respect the energy conservation rule, allowing us to find missing variables like velocity in the decay process.
Velocity Addition Formula
Velocity addition in Special Relativity differs from classical physics, due to the involvement of the speed of light. The relativistic velocity addition formula is an essential tool when dealing with moving reference frames.

In our scenario, to find particle \( b \)'s velocity in the original stationary frame \( S \), where particle \( a \) was initially traveling, we use the formula:
  • Velocity of particle \( b \) moving in direction of particle \( a \):\[ v_{b1} = \frac{v'_b + u}{1 + \frac{v'_b u}{c^2}} \]
  • Velocity of particle \( b \) moving opposite the direction of particle \( a \):\[ v_{b2} = \frac{-v'_b + u}{1 - \frac{v'_b u}{c^2}} \]
In this formula, \( v'_b \) is the velocity in the rest frame of particle \( a \), \( u \) is the velocity of the frame, and \( c \) is the speed of light.

By substituting known values, we ensure all velocities are calculated accurately, adhering to relativistic principles. The formula confirms that as velocities approach the speed of light, they never surpass it, honoring Einstein's principle of relativity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the tale of the physicist who is ticketed for running a red light and argues that because he was approaching the intersection, the red light was Doppler shifted and appeared green. How fast would he have to have been going? \(\left(\lambda_{\text {red }} \approx 650 \mathrm{nm} \text { and } \lambda_{\text {green }} \approx 530 \mathrm{nm} .\right)\)

A particle of unknown mass \(M\) decays into two particles of known masses \(m_{a}=0.5 \mathrm{GeV} / c^{2}\) and \(m_{b}=1.0 \mathrm{GeV} / c^{2},\) whose momenta are measured to be \(\mathbf{p}_{a}=2.0 \mathrm{GeV} / \mathrm{c}\) along the \(x_{2}\) axis and \(\mathbf{p}_{b}=1.5 \mathrm{GeV} / c\) along the \(x_{1}\) axis. \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV} .\right)\) Find the unknown mass \(M\) and its speed.

We have seen that the scalar product \(x \cdot x\) of any four-vector \(x\) with itself is invariant under Lorentz transformations. Use the invariance of \(x \cdot x\) to prove that the scalar product \(x \cdot y\) of any two four-vectors \(x\) and \(y\) is likewise invariant.

A particle of mass 12 MeV/c \(^{2}\) has a kinetic energy of 1 \(\mathrm{MeV}\). What are its momentum (in MeV/c) and its speed (in units of \(c\) )?

Two particles \(a\) and \(b\) with masses \(m_{a}=0\) and \(m_{b}>0\) approach one another. Prove that they have a CM frame (that is, a frame in which their total three-momentum is zero). [Hint: As you should explain, this is equivalent to showing that the sum of two four-vectors, one of which is forward light-like and one forward time-like, is itself forward time-like.]

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free