Question

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

Muon decay mode: \(\beta \approx 0.271\); Electron decay mode: \(\beta \approx 0.99917\).

Step-by-step solution

Understanding Conservation of Energy and Momentum

In the decay process, the total energy and momentum must remain conserved. Since the pion decays at rest, the initial momentum is zero, and the energy is simply the rest energy of the pion: \(E_{\pi} = m_{\pi}c^2\). In the final state, the energy is the sum of the energies of the muon and the neutrino: \(E_{\mu} + E_u = E_{\pi}\). The final momentum should also be zero as the neutrino carries away the momentum opposite to that of the muon.

Applying the Conservation of Momentum

Since the initial momentum is zero and the final particles have oppositely directed momenta, we can write the conservation of momentum as: \(p_{\mu} = -p_{u}\). As \(p = EV/c^2\) and for the neutrino, \(E_u = p_u c\), we have \(p_{\mu} = E_u/c\).

Solving for Muon Energy and Velocity

Using conservation of energy: \(E_{\mu} + E_u = m_{\pi}c^2\). Substitute \(E_u = p_{\mu}c = \sqrt{E_{\mu}^2 - m_{\mu}^2c^4}\): \(E_{\mu} = m_{\pi}c^2 - \sqrt{E_{\mu}^2 - m_{\mu}^2c^4}\). Solve this equation for \(E_{\mu}\).

Finding Muon Speed

With \(E_{\mu}\) and knowing \(E_{\mu} = \gamma m_{\mu}c^2\), solve for \(\gamma\). Use \(\beta = \sqrt{1 - 1/\gamma^2}\). To find \(\gamma\), use: \(E_{\mu} = \gamma m_{\mu}c^2 = \frac{m_{\pi}^2 - m_{\mu}^2}{2m_{\pi}}c^2\).\ Hence, \(\beta = \frac{m_{\pi}^2 - m_{\mu}^2}{m_{\pi}^2 + m_{\mu}^2}\).

Numerical Evaluation for \(\pi^+ \rightarrow \mu^+ + \nu\)

Use the given masses: \(m_{\pi} = 140\ \text{MeV/c}^2\) and \(m_{\mu} = 106\ \text{MeV/c}^2\). Calculate \(\beta = \frac{(140)^2 - (106)^2}{(140)^2 + (106)^2}\).

Numerical Evaluation for \(\pi^+ \rightarrow e^+ + \nu\)

Repeat steps 1 through 5 for electron decay with \(m_{e} = 0.5\ \text{MeV}/c^2\), instead of \(m_{\mu}\), to get the \(\beta\) value specific to this decay mode.

Key concepts

Conservation of Energy

In the context of pion decay, the conservation of energy is a fundamental principle. This principle states that the total energy before the decay must equal the total energy after the decay.

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

The conservation of momentum is another key principle that applies to pion decay. Momentum, like energy, must be conserved in every process. This means that the total momentum before decay must match the total momentum after decay.

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

Muon decay is central to our understanding of the processes that govern subatomic particle transformations. In the decay of a positive pion, a muon is one of the resulting particles. This specific decay is written as: \(\pi^{+} \rightarrow \mu^{+} + u\).

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

Neutrinos are fascinating particles that intrigue physicists, particularly because of their incredibly small mass and elusive nature. In the case of pion decay \(\pi^{+} \rightarrow \mu^{+} + u\), the neutrino involved is assumed to be massless for simplification in calculations.

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

Relativistic energy is an essential concept when dealing with particles moving close to the speed of light, such as in pion decay. Energy is not only from mass but also from the motion and velocity of particles in these high-speed scenarios.

The relativistic energy of a particle is provided by \(E = \gamma m c^2 \), where \(\gamma\) is the Lorentz factor given by

• \(\gamma = \frac{1}{\sqrt{1 - \beta^2}}\)

.\(\beta\) is the ratio of the particle's speed to the speed of light (v/c).

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.