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One way to create exotic heavy particles is to arrange a collision between two lighter particles $$a+b \rightarrow d+e+\cdots+g$$ where \(d\) is the heavy particle of interest and \(e, \cdots, g\) are other possible particles produced in the reaction. (A good example of such a process is the production of the \(\psi\) particle in the process \(\left.e^{+}+e^{-} \rightarrow \psi, \text { in which there are no other particles } e, \cdots, g .\right)\) (a) Assuming that \(m_{d}\) is much heavier that any of the other particles, show that the minimum (or threshold) energy to produce this reaction in the CM frame is \(E_{\mathrm{cm}} \approx m_{d} c^{2} .\) (b) Show that the threshold energy to produce the same reaction in the lab frame, where the particle \(b\) is initially at rest, is \(E_{\mathrm{lab}} \approx m_{d}^{2} c^{2} / 2 m_{b} .\) (c) Calculate these two energies for the process \(e^{+}+e^{-} \rightarrow \psi,\) with \(m_{e} \approx 0.5 \mathrm{MeV} / c^{2}\) and \(m_{\psi} \approx 3100 \mathrm{MeV} / c^{2} .\) Your answers should explain why particle physicists go to the trouble and expense of building colliding-beam experiments.

Short Answer

Expert verified
The CM frame needs less energy than the lab frame, showing the efficiency of colliding beams for high-energy particle production.

Step by step solution

01

Understand the Scenario and Terminology

We have two frames of reference: the CM frame (center of mass) and the lab frame. For part (a), we focus on the CM frame where both particles are in motion, whereas in part (b), the lab frame starts with particle b at rest. We aim to find the threshold energies where the heavy particle d with mass \( m_d \) is produced.
02

Analyze CM Frame Energy Conditions

In the CM frame, total energy \( E_{\text{cm}} \) is the sum of energies of both particles before the collision. The threshold condition for producing heavy particle \( d \) is that the entire energy goes into the creation of \( d \). Given \( m_d \gg m_a, m_b \), the minimum energy \( E_{\text{cm}} \) is \( E_{\text{cm}} = m_d c^2 \) since \( e, \cdots, g \) contribute negligibly.
03

Analyze Lab Frame Energy Conditions

In the lab frame, particle \( a \) has energy \( E_a \) and \( b \) is at rest. The threshold energy \( E_{\text{lab}} \) implies that all the kinetic energy transforms into \( m_d c^2 \). Using energy and momentum conservation, the energy needed is \( E_{\text{lab}} \approx \frac{m_d^2 c^2}{2 m_b} \), derived from relativistic energy-momentum relations.
04

Substitute Given Values for the Threshold Energies

With \( m_e \approx 0.5 \text{ MeV}\, c^{-2} \) and \( m_\psi \approx 3100 \text{ MeV}\, c^{-2} \), calculate the energies. For CM: \( E_{\text{cm}} \approx 3100 \text{ MeV}\). For lab frame using \( E_{\text{lab}} \approx \frac{3100^2 \text{ MeV}^2}{2 \times 0.5\text{ MeV}} \approx 9.61 \text{ GeV} \).
05

Conclusion from Calculations

The CM energy is much smaller compared to the lab frame energy, illustrating why colliding beams are used in experiments—to achieve higher energies that are practically impossible with a single pre-accelerated particle hitting a stationary target. This is due to energy concentrating more efficiently in the CM frame collision.

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Key Concepts

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

Center of Mass Frame
The Center of Mass (CM) frame is a special reference frame in particle collisions where the total momentum is zero. This means that the momentum of the system as a whole does not move in any direction. In collisions at high energies, calculations in the CM frame are simplified because the energy is evenly distributed between the colliding particles.

In the context of particle physics, the CM frame simplifies the determination of the threshold energy needed to create new particles, such as the heavy particle "d" in the given exercise. Here, since the masses of the other particles, such as "a" and "b", are much lighter than the heavy particle "d", the minimum energy in this CM frame needed to produce "d" is approximately its rest mass energy, , which is given by the famous formula:
  • \[ E_{\text{cm}} = m_d c^2 \]
This equation illustrates that all the energy involved in the collision primarily goes into creating the heavy particle, thus the simplicity in calculations.
Lab Frame
The Lab frame is another reference frame used to analyze particle collisions. In this frame, one of the particles, typically "b", is at rest. This represents a more realistic scenario in laboratory experiments where a particle at a high speed collides with a stationary target.

While the CM frame assumes both particles are moving towards each other with equal momentum, the Lab frame simplifies calculations for scenarios such as fixed-target experiments. In our exercise, the Lab frame makes it more complex to determine threshold energy because the system’s total momentum is not zero.
  • The threshold energy in the Lab frame is calculated as:
  • \[ E_{\text{lab}} \approx \frac{m_d^2 c^2}{2 m_b} \]
This formula indicates that to produce the heavy particle "d", more energy is needed compared to the CM frame, as the extra energy compensates for the initial particle at rest.
Threshold Energy
Threshold energy is the minimum energy required for a specific particle reaction to occur. In the scenario presented, it refers to the energy needed to produce a heavy particle out of the collision between lighter particles "a" and "b".

In both the CM and Lab frames, threshold energy considerations are crucial to predict whether a collision will successfully create new particles. For high-energy particle physics experiments, threshold energy helps scientists determine the feasiblity of producing new particles.
  • In the CM frame, threshold energy is straightforward, calculated as:
  • \[ E_{\text{cm}} \approx m_d c^2 \]
  • In the Lab frame, it becomes more complex:
  • \[ E_{\text{lab}} \approx \frac{m_d^2 c^2}{2 m_b} \]
Understanding these threshold energies aids in designing experiments and accelerators to ensure collisions will yield the desired particles.
Relativistic Energy
Relativistic energy takes into account the rules of Einstein's relativity and is essential when particles travel near the speed of light. Traditional Newtonian equations do not suffice at these high velocities.

Relativistic energy relates mass and energy through the equation \[ E = \gamma m c^2 \]where \(\gamma\) is the Lorentz factor, \(1 / \sqrt{1 - v^2 / c^2 }\). This equation shows how mass and energy are interchangeable and highlights that as a particle's speed approaches the speed of light, its energy increases significantly.
  • In particle collisions, relativistic energy principles ensure the accurate application of conservation laws, allowing physicists to predict the energy requirements and outcomes.

Thus, the exercise uses relativistic energy to calculate the threshold energies needed for the creation of heavy particles, necessitating the use of energy-momentum relations.
Conservation Laws
Conservation laws are fundamental principles governing physical interactions. In particle physics collisions, these laws are crucial for determining reaction outcomes and energy requirements.

The most significant conservation laws applied to particle collisions are:
  • Conservation of Energy: The total energy before the collision must equal the total energy after the collision.
  • Conservation of Momentum: The total momentum remains unchanged throughout the interaction.
  • Conservation of Charge, Baryon Number, and Lepton Number: These also ensure consistency in particle reactions.
In our specific exercise, these laws help show that the energy used in the CM frame goes directly into creating the new particles, while in the Lab frame, calculations must account for initial kinetic and potential energy differences.

By following these universal rules, physicists can accurately predict which reactions will occur and the threshold energies needed.

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Most popular questions from this chapter

A neutral pion (Problem 15.86) is traveling with speed \(v\) when it decays into two photons, which are seen to emerge at equal angles \(\theta\) on either side of the original velocity. Show that \(v=c \cos \theta\)

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