Question about relativistic energy

AI Thread Summary
In the decay of a pion at rest into a muon and an antineutrino, the conservation of energy and momentum is crucial. The muon's kinetic energy is calculated to be approximately 2.095 MeV, derived from its mass and the relativistic energy equation. The antineutrino, having no rest mass, carries an energy of -2.095 MeV, indicating that it balances the energy of the muon in the system. The calculations utilize Einstein's energy-momentum relationship and the principles of relativistic physics. This analysis highlights the importance of energy conservation in particle decay processes.
andrew410
Messages
59
Reaction score
0
A pion at rest (m_pi = 273m_e) decays to a muon (mass = 207m_e and an antineutrino (mass = 0). Find the kinetic energy of the muon and the energy of the antineutrino in electron volts.

How am I supposed to start this problem? ANy help would be great...thx!
 
Physics news on Phys.org
If the muonic antineutrino has no rest mass,then his energy & rel.momentum vector in modulus are linked through

E_{\bar{\nu}_{\mu}}=pc

The key point is that both energy (seen as the time component of the energy-momentum 4-vector) and relativistic momentum (seen as the space components of the energy-momentum 4-vector) are conserved.

Daniel.

P.S.Einstein's formula is E^{2}=m^{2}c^{4}+\left|\vec{p}\right|^{2}c^{2}.
 


To solve this problem, we can use the equation for relativistic energy: E = mc^2 / √(1-v^2/c^2). First, we need to find the velocity of the muon after the pion decays. Since the pion is at rest, its initial velocity is 0. We can use conservation of momentum to find the velocity of the muon:

0 = m_pi * v_pi + m_mu * v_mu

Since the pion is at rest, v_pi = 0. Solving for v_mu, we get:

v_mu = -m_pi / m_mu * v_pi = 0

Therefore, the velocity of the muon is also 0 after the pion decays. Now, we can plug in the values for mass and velocity into the equation for relativistic energy:

E_mu = m_mu * c^2 / √(1-0^2/c^2) = m_mu * c^2

Substituting in the given masses, we get:

E_mu = (207m_e) * (c^2) = (207 * 9.109 * 10^-31 kg) * (2.998 * 10^8 m/s)^2 = 3.358 * 10^-12 J = 2.095 MeV

To find the energy of the antineutrino, we can use the fact that the total energy of the system must be conserved. Since the pion was at rest, its initial energy is 0. Therefore, the total energy after the decay must also be 0. We can set up an equation to solve for the energy of the antineutrino:

E_pi + E_mu + E_antineutrino = 0

Substituting in the values for the masses and the energy of the muon that we just calculated, we get:

0 + 2.095 MeV + E_antineutrino = 0

Solving for E_antineutrino, we get:

E_antineutrino = -2.095 MeV

Since we are looking for the energy in electron volts, we need to convert the units:

E_antineutrino = (-2.095 MeV) * (1.602 * 10^-13 J/MeV) = -3.353 eV

Note that the negative sign indicates that the antineutrino
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top