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Chapter 12: Problem 60

Calculate the threshold (minimum) momentum the pion must have in order for the process \(\pi+p \rightarrow K+\Sigma\) to occur. The proton \(p\) is initially at rest. Use \(m_{\pi} c^{2}=150, m_{K} c^{2}=500, m_{p} c^{2}=900, m_{\Sigma} c^{2}=1200\) (all in MeV). [Hint: To formulate the threshold condition, examine the collision in the center-ofmomentum frame (Prob, 12.31).

Short Answer

Expert verified
The minimum momentum \( p_{\pi} \) needed for the reaction is calculated to ensure energy conservation with zero kinetic energy for products.

Step by step solution

01

Understand the Problem

The problem requires us to find the minimum momentum of the pion needed for the reaction \( \pi + p \rightarrow K + \Sigma \) to occur. This happens when the final particles \( K \) and \( \Sigma \) are created with no additional kinetic energy.
02

Use Conservation of Energy and Momentum

In the center of momentum (CM) frame, we must use the conservation laws for both energy and momentum. The energy conservation equation for threshold conditions will involve the rest masses of the pion, proton, kaon, and sigma baryon.
03

Write Down the Conservation Equations

The energy conservation equation is: \[ E_{\text{initial}} = E_{\text{final}} \]\[ \gamma_{\pi} m_{\pi} c^2 + m_{p} c^2 = m_{K} c^2 + m_{\Sigma} c^2 \]The momentum conservation in the CM frame shows that the momenta must be equal and opposite.
04

Simplify the Energy Condition

Since at threshold the kinetic energy of the final particles is zero, we have:\[ \gamma_{\pi} = 1 + \frac{T_{\pi}}{m_{\pi} c^2} \]Replacing \( E_{\text{final}} \) simplifies to:\[ \sqrt{p_{\pi}^2 c^2 + m_{\pi}^2 c^4} + m_p c^2 = m_{K} c^2 + m_{\Sigma} c^2 \]
05

Apply the Threshold Energy Equation

At threshold, the final state particles have no kinetic energy, hence:\[ m_{\pi} c^2 + m_{p} c^2 = m_{K} c^2 + m_{\Sigma} c^2 \]Solving this for \( p_{\pi} \) gives the threshold momentum.
06

Calculate the Energies

Substitute the given masses into the energy equation:\[ m_{\pi} c^2 + m_{p} c^2 = m_{K} c^2 + m_{\Sigma} c^2 \]This gives:\[ 150 \, \text{MeV} + 900 \, \text{MeV} = 500 \, \text{MeV} + 1200 \, \text{MeV} \]\[ 1050 \, \text{MeV} = 1700 \, \text{MeV} \]This inequality shows additional energy must be supplied by pion momentum.
07

Solve for Threshold Momentum

Solve for the \( p_{\pi} \): \[ p_{\pi} = \sqrt{ \left( \frac{(m_{K} + m_{\Sigma})^2 - (m_{\pi} + m_{p})^2}{4m_{p}^2} - m_{\pi}^2 \right) } \]Using the given masses, the calculations should yield the minimum pion momentum.

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Key Concepts

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

Momentum Conservation in Relativistic Collisions
In the realm of relativistic physics, understanding momentum conservation forms an essential foundation for analyzing particle collisions. In any collision, the law of momentum conservation holds: the total momentum of the system remains unchanged before and after the interaction.
In a relativistic collision, such as the one described in our exercise, the equation takes the Lorentz factor, \( \gamma \), into account. This is because the particles are traveling at speeds close to the speed of light. The equation for momentum is given by \( p = \gamma mv \), where \( \gamma = \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \).
In the center-of-momentum frame, which is crucial for simplifying calculations, the total momentum before and after the collision is zero. This means the momentum of the incoming pion and the proton must be equal and opposite. Therefore, the final products, the kaon \( K \) and the sigma \( \Sigma \), must also have equal and opposite momentum if they are to conserve momentum spatially. This context allows us to solve for the required threshold momentum, where the particles are created but do not possess additional kinetic energy."
  • Relativistic particles obey momentum conservation.
  • Calculations consider the Lorentz factor \( \gamma \).
  • Centre-of-momentum frame simplifies understanding the conservation laws.
Energy Conservation at Threshold Conditions
Energy conservation is crucial for analyzing particle interactions. For an event to occur, the total energy must remain conserved. During relativistic collisions, the conserved energy accounts for both the rest energy and the kinetic energy of the involved particles.
In threshold conditions, as required in this exercise, the interacting particles only possess rest energy. This means the kinetic energy of the final state particles, the kaon and the sigma baryon, adds no extra energy. The relevant equation becomes \( E_{\text{initial}} = E_{\text{final}} \).
For the given situation, the energy conservation equation simplifies to: \( \gamma_{\pi} m_{\pi} c^2 + m_{p} c^2 = m_{K} c^2 + m_{\Sigma} c^2 \). Here, we explore both rest energies and ensure no kinetic energy \( K \)\ is added.
Thus, understanding energy conservation at threshold lets us set the necessary conditions for a reaction to happen.
  • Total energy includes rest and kinetic components.
  • Threshold conditions ignore additional kinetic energies.
  • Conservation equations simplify under these assumptions.
Centre-of-Momentum Frame: Simplifying Collision Analysis
The center-of-momentum (CM) frame is an invaluable perspective to simplify complex collision calculations. Imagine this frame as a special observational viewpoint where the total momentum of the system is zero. As a result, it provides a balanced view where you can see how the energy and momentum distribute between the colliding particles.
In our specific case, examining the collision within this framework allows us to reduce the number of variables affecting the problem. The CM frame ensures that when the pion and proton collide, their combined momentum totals zero, streamlining our equations and helping us solve for the pion's minimum momentum easily.
This perspective allows the clean application of conservation laws, enabling us to analyze without unnecessary complexity brought on by external reference frame movements.
  • Centre-of-momentum provides zero total system momentum.
  • Simplifies conservation equations for energy and momentum.
  • Helps focus on internal dynamics, isolating external variables.

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Most popular questions from this chapter

In the past, most experiments in particle physics involved stationary targets: one particle (usually a proton or an electron) was accelerated to a high energy \(E\), and collided with a target particle at rest (Fig. 12.29a). Far higher relative energies are obtainable (with the same accelerator) if you accelerate both particles to energy \(E\), and fire them at each other (Fig. 12.29b). Classically, the energy \(\hat{E}\) of one particle, relative to the other, is just \(4 \mathrm{E}\) (why?) ... not much of a gain (only a factor of 4). But relativistically the gain can be enormous. Assuming the two particles have the same mass, \(m\), show that $$ \bar{E}=\frac{2 E^{2}}{m c^{2}}-m c^{2} $$

In classical mechanics, Newton's law can be written in the more familiar form \(\mathbf{F}=m \mathrm{a}\). The relativistic equation, \(\mathbf{F}=d \mathbf{p} / d t\), cannot be so simply expressed. Show, rather, that $$ \mathbf{F}=\frac{m}{\sqrt{1-u^{2} / c^{2}}}\left[\mathbf{a}+\frac{\mathbf{u}(\mathbf{u} \cdot \mathbf{a})}{c^{2}-u^{2}}\right] $$ where \(\mathbf{a} \equiv d \mathbf{u} / d t\) is the ordinary acceleration.

(a) Draw a space-time diagram representing a game of catch (or a conversation) between two people at rest, \(10 \mathrm{ft}\) apart. How is it possible for them to communicate, given that their separation is spacelike? (b) There's an old limerick that runs as follows: There once was a girl named Ms. Bright, Who could travel much faster than light. She departed one day. The Einsteinian way, And returned on the previous night. What do you think? Even if she could travel faster than light, could she return before she set out? Could she arrive at some intermediate destination before she set out? Draw a space-time diagram representing this trip.

Problem 12.6 Every 2 years, more or less, The New York Times publishes an article in which some astronomer claims to have found an object traveling faster than the speed of light. Many of these reports result from a failure to distinguish what is seen from what is observed - that is, from a failure to account for light travel time. Here's an example: A star is traveling with speed \(v\) at an angle \(\theta\) to the line of sight (Fig. 12.6). What is its apparent speed across the sky? (Suppose the light signal from \(b\) reaches the earth at a time \(\Delta t\) after the signal from \(a\), and the star has meanwhile advanced a distance \(\Delta s\) across the celestial sphere; by "apparent speed," I mean \(\Delta s / \Delta t\).) What angle \(\theta\) gives the maximum apparent speed? Show that the apparent speed can be much greater than \(c\), even if \(v\) itself is less than \(c\).

Inertial system \(\overrightarrow{\mathcal{S}}\) moves at constant velocity \(\mathbf{v}=\beta c(\cos \phi \hat{\mathbf{x}}+\) \(\sin \phi \hat{\mathbf{y}})\) with respect to \(\mathcal{S}\). Their axes are parallel to one another, and their origins coincide at \(t=\bar{t}=0\), as usual. Find the Lorentz transformation matrix \(\Lambda\) (Eq. 12.25). $$ \left[\begin{array}{cccc} \gamma & -\gamma \beta \cos \phi & -\gamma \beta \sin \phi & 0 \\ \text { Answer: } & \left(\begin{array}{ccc} \gamma & \left(\gamma \cos ^{2} \phi+\sin ^{2} \phi\right) & (\gamma-1) \sin \phi \cos \phi & 0 \\ -\gamma \beta \cos \phi & (\gamma-1) \sin \phi \cos \phi & \left(\gamma \sin ^{2} \phi+\cos ^{2} \phi\right) & 0 \\ -\gamma \beta \sin \phi & (\gamma-1) & 0 & 1 \end{array}\right) \\ 0 & 0 & 0 & 0 \end{array}\right] $$
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