Conserving Momentum & Energy in Particle A Decay

[Alpha&Omega]

1. Consider the decay of particle A with rest mass M_{0A} into two particles, labelled particle 1 and particle 2. The energy of particle 1 is denoted by E_1 and the rest mass by m_{01}, similarly for particle 2 the energy is E_2 and the rest mass m_{02}.


i). in the rest frame of particle A write down expressions describing the conservation of momentum and energy in this decay process.

Since this is the rest frame of particle A, the particle is stationary. This implies it has no momentum so it's total energy is just E_{initial}=\gamma M_{0A} c^2.

The question is unclear about whether the particles produced have a velocity or not. I decided to say they did (since I could always say it was 0) and called them v_1 for particle 1 and v_2 for particle 2.

The total energy of particle 1 is therefore E_1=\gamma m_{01} c^2 and the total energy of particle 2 is therefore E_2=\gamma m_{02} c^2.

Since energy has to be conserved, E=E_1+E_2 or \gamma M_{0A}c^2=\gamma m_{01} c^2+\gamma m_{02}c^2 \ \Rightarrow \ M_{0A}=m_{01}+m_{02}.

This didn't look right (even though I can't see where it could be wrong) so I decided to use the equation E^2=\rho ^2 c^2 + m^2 c^4:

E remains the same since \rho=0.
E_1=\sqrt{\rho_1^2 c^2+m_{01}^2c^4}
E_2=\sqrt{\rho_2^2c^2+m_{02}^2c^4

Hence E=E_1+E_2 \Rightarrow \ M_{0A}c^2=\sqrt{\rho_1^2 c^2+m_{01}^2c^4}+\sqrt{\rho_2^2c^2+m_{02}^2c^4.

This looked a bit better except I can't get the second part of the question from this!

Any help would be appreciated.

 

[scottdave]

Remember that momentum is a vector, so each individual particle can have momentum, as long as the vector sum of these two momentums equals the momentum beforehand. So if the momentum beforehand is zero, a vector momentum in one direction will cancel out an equal magnitude momentum in the opposite direction. I realize that this has not answered your question, but hopefully points you on the right track.