Chapter 15: Problem 92
A positive pion decays at rest into a muon and neutrino, \(\pi^{+} \rightarrow \mu^{+}+\nu .\) The masses involved are \(m_{\pi}=140 \mathrm{MeV} / c^{2}, m_{\mu}=106 \mathrm{MeV} / c^{2},\) and \(m_{\nu}=0 .\) (There is now convincing evidence that \(m_{\nu}\) is not exactly zero, but it is small enough that you can take it to be zero for this problem.) Show that the speed of the outgoing muon has \(\beta=\left(m_{\pi}^{2}-m_{\mu}^{2}\right) /\left(m_{\pi}^{2}+m_{\mu}^{2}\right) .\) Evaluate this numerically. Do the same for the much rarer decay mode \(\pi^{+} \rightarrow \mathrm{e}^{+}+\nu,\left(m_{\mathrm{e}}=0.5 \mathrm{MeV} / c^{2}\right)\)
Short Answer
Step by step solution
Understanding Conservation of Energy and Momentum
Applying the Conservation of Momentum
Solving for Muon Energy and Velocity
Finding Muon Speed
Numerical Evaluation for \(\pi^+ \rightarrow \mu^+ + \nu\)
Numerical Evaluation for \(\pi^+ \rightarrow e^+ + \nu\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservation of Energy
Before the pion decays, it is at rest, meaning its entire energy is in the form of rest energy, which can be expressed as \[ E_{\pi} = m_{\pi}c^2 \]. Here, \(m_{\pi}\) is the rest mass of the pion and \(c\) is the speed of light.
After the decay, the pion transforms into a muon and a neutrino. Each of these particles will have their energy. The muon has its rest energy, \(m_{\mu}c^2\), and kinetic energy, making up its total energy, \(E_{\mu}\). The neutrino, on the other hand, is considered massless for this exercise, so its energy is mainly kinetic, given by \(E_u = p_uc\).
Hence, the conservation of energy equation for this decay process is:
\[ E_{\pi} = E_{\mu} + E_u \]
This equation ensures that the total energy remains constant, explaining how energy transitions from one form (rest) to another through kinetic components.
Conservation of Momentum
Initially, the pion is at rest, so the total initial momentum is zero. When the pion decays into a muon and a neutrino, their momenta must be equal and opposite to satisfy the momentum conservation law. This is expressed as:
- \(p_{\mu} + p_{u} = 0\)
- Therefore, \(p_{\mu} = -p_{u}\)
Here, \(p_{\mu}\) and \(p_{u}\) are the momenta of the muon and the neutrino respectively. Since the neutrino is considered massless, its energy \(E_u\) relates directly to its momentum via \(E_u = p_uc\), which can be coupled with conserved energy results to find expressions for all involved variables.
This relationship reflects how the decay products' speeds and directions are constrained to maintain the momentum balance established by the initial conditions.
Muon Decay
The muon, noted as \(\mu^{+}\), is a charged lepton similar to an electron but with greater mass. It carries away a portion of the pion's initial energy. The mass of the muon is given as 106 MeV/c². This particular decay is significant as it highlights the transient nature of some particles; muons eventually decay further, underlining the cycles within particle physics.
The muon's journey from being part of a pion to becoming a free particle is governed by the rules of conservation, ensuring the integrity of physics laws.
Neutrino Mass
Interestingly, recent advancements have shown that neutrinos indeed possess a small but non-zero mass. However, concerning the mass of the pion (140 MeV/c²) and muon (106 MeV/c²), the neutrino mass is negligible and hence omitted here for ease.
This assumption simplifies solving related equations, keeping focus on other variables like muon speed. Even though this mass was considered zero in past calculations, the evolving scope of physics continually refines and corrects such models.
Relativistic Energy
The relativistic energy of a particle is provided by \(E = \gamma m c^2 \), where \(\gamma\) is the Lorentz factor given by
- \(\gamma = \frac{1}{\sqrt{1 - \beta^2}}\)
In pion decay, calculating the relativistic energy of the muon gives insight into its motion post-decay. Understanding this concept gives us the equation:
- \(E_{\mu} = \gamma m_{\mu} c^2\)
The term \(\gamma\) accounts for relativistic effects, adjusting energy calculations based on velocity, ensuring the calculated speeds adhere to relativistic constraints.