# Particle/High energy physics question

1. Sep 6, 2008

### BJAY

1. The problem statement, all variables and given/known data
Consider the particle decay:
A0 → p + (pi)−
(a) In the rest frame of the A0 particle, calculate the momentum of the p and of the (pi)−

(b) In the rest frame of the A0 the p and (pi)− emerge back to back. Suppose now that
the A0 is not at rest but is travelling such that it has a total energy of 2000 MeV
as observed by a stationary observer. If a large number of A0 decays was observed,
calculate the minimum and maximum proton momentum that this observer would
expect to see in his/her frame of reference.

2. Relevant equations

Particle Masses (MeV/c2):
(Not sure what its called) A0 = 1116
(pi minus) pi− = 140
(proton) p = 938
(K minus) K− = 494

this may be of use: E^2 = p^2*c^2 + m^2*c^4

3. The attempt at a solution
for part a) i put into the above equation E=1116 MeV/c^2(mass of parent particle) and m^2*c^4 = 140 + 938 MeV/c^2. Rearranged equation so its in terms of p and the ceclulated that. Im not 100% on this though so any help would be good. Part b) not very sure at all, i dont really understand the question to be honest. sorry the quesiton might be hard to understand, only my first post. thanks for any help!

Last edited: Sep 6, 2008
2. Sep 6, 2008

### malawi_glenn

in the rest frame of A0 all anlges are equal probable for the protons momentum vector. So if the A0 moves in the lab frame where the observaber is, you'll find out that the proton momentum in lab frame have a minima and a maxima.

3. Sep 8, 2008

### BJAY

Im dont know if i understand properly what your saying, because it doesnt make me see anyway that i could possibly calculate a maxima and minima

Heres my working for part a:
E = mc^2 where m =1116 MeV/c^2 (parent particle)
the c squareds cancel and gives E = 1116^2 MeV = 124 546 Mev. This is the total energy of the daughter particles.
To find the amount of energy that is in mass of daughter particles i substituted into m^2*C^4 term in the equation:
(140^2 MeV/c^4 + 938^2 MeV/c^4) * c^4 = 19600 MeV + 879 844 Mev
therefore P = sqrt(124546 - 19600 -879866) Mev/c

I thought for part b i would use same method but add 2000Mev to E, the total energy of daughter particles. This would again just give me total momentum for both daughter particles. What does the angle of proton have to do with its momentum energy?

4. Sep 8, 2008

### malawi_glenn

If the proton emission in the rest frame of the A0 is isotropic, i.e all angels are equal probable, then when the A0 is seen to be decaying in the lab-frame where it is moving, then you'll get a momentum distribution. It is quite simple to see that. If the rest frame the magnitude of momenta of protons are equal, but direction differs. If you add a velocity component to the A0, classicaly by galileo transformation, the velocities and hence the momenta of the emitted protons will be distributed. First try to get that clear to you. Then after that, you must use proper transformation, since this problem is relativistic, i.e a Lorentz transformation.