# Decaying particle

1. Mar 26, 2009

### 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.

Last edited by a moderator: Mar 15, 2017
2. Mar 15, 2017

### 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.