What is the momentum of the muon in a pi-meson decay?

  • Thread starter Thread starter Abdul.119
  • Start date Start date
  • Tags Tags
    Momentum Muon
AI Thread Summary
The discussion focuses on calculating the momentum of a muon produced from the decay of a pi-meson at rest, with specific energy values provided for both particles. The user initially encounters issues with negative values under the square root when attempting to solve for momentum using the energy-momentum relationship equations. Clarifications are made regarding the importance of units in calculations, as well as the source of the equations used. Ultimately, the user resolves their confusion and successfully finds the solution. The conversation highlights the significance of understanding conservation laws and proper equation application in particle physics.
Abdul.119
Messages
73
Reaction score
2

Homework Statement


pi-meson's rest energy is 139.57MeV, it decays into a muon and a neutrino. The muon has a rest energy of 105.45MeV, and a proper lifetime of 2.197*10^-6. The neutrino's mass can be treated as massless in the process.

1) Assuming the pi-meson decays at rest, what is the momentum of the muon?

Homework Equations


E_muon + E_neutrino = E_pi-meson
E_muon = c^2* ((m_pi-meson)^2 - (m_muon)^2)) / (2m_pi-meson)
E_muon = sqrt((P_muon^2 c^2) + (m_muon c^2)^2)

The Attempt at a Solution


Ok I used used the second equation and it gives about 9.866*10^18, then, use this value in the 3rd equation and solving for the momentum, I get a negative value under the square root, which doesn't make sense. In the 2nd equation, the values I plugged in for m_pi-meson and m_muon are just 139.57 and 105.47, and used 3*10^8 for c. Where did I go wrong?
 
Last edited:
Physics news on Phys.org
Abdul.119 said:
E_muon = c^2* ((m_pi-meson)^2 - (m_muon)^2)) / (2m_pi-meson)
E_muon = sqrt((P_muon^2 c^2) + (m_muon c^2)^2)
Where do those equations come from?
Abdul.119 said:
Ok I used used the second equation and it gives about 9.866*10^18
9.866*1018 what? Kilometers, apples, ...? Units are important.
 
mfb said:
Where do those equations come from?
9.866*1018 what? Kilometers, apples, ...? Units are important.
The first equation is based on the conservation laws, you can see from here https://teachers.web.cern.ch/teache...h/mbitu/energy_and_momentum_conservation1.htm I used equation (2.5)
For the second equation, I used the fact that E^2 = p^2 c^2 + m^2 c^4 , squared both sides to get E, and tried to solve for the momentum from here
 
Oh never mind, I solved it. Thanks for the help.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top