What Are the Threshold Kinetic Energies in Particle Decay Reactions?

nv125
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Electrodynamics and Relativity. Please Help!

Homework Statement



Consider the decay of a particle.

(a) Find the threshold kinetic energies of the incident particles in each of
the following reactions:

(i) ∏+ + p →∏++ ∏° + p
(ii) p + n →p + n + ∏° + ∏°

The incident particle is the first particle listed in the reactions; the target particle is at rest. The masses of the ∏+ and ∏° ions are 139.567MeV/c2 and 134.963 MeV/c2 respectively while the masses of the neutron and proton are respectively 939.566 MeV/c2 and 938.272 MeV/c2.




Homework Equations



The teacher suggested that I will need to use the following equations to get the solution
(Here P is the momentum and Px is the momentum of the incident particle ∏+in the X direction, assuming it hits the target particle p(proton) horizontally, so no P(momentum) in y or z direction)

ƩPiu=0

PuPu=m2c2= E2/C2-P2

Also said that using this notations will be useful to solve the problem

For ∏+, the incident particle:
Pu=(E/C, Px, 0, 0)
Pu=(E/C, -Px, 0, 0) where Px is the momentum of the incident particle, and no need to worry about y and z direction so just put zeros.


The Attempt at a Solution


This is what I have done so far to solve for part i, but I have no clue how to continue.
Since we are solving for the threshold kinetic energy, so assuming that ∏+, ∏° and p are at rest after the reaction.

(E/C, Px, 0, 0) +(mpC2, 0, 0, 0) - (m+C2,0, 0, 0 ) -(m∏°C2,0, 0, 0 ) - (mpC2,0, 0, 0 ) = 0

mp, m+, m∏° are just the rest mass.

PLEASE PLEASE help me to do this problem! thank you so much!
 
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nv125 said:
Since we are solving for the threshold kinetic energy, so assuming that ∏+, ∏° and p are at rest after the reaction.

(E/C, Px, 0, 0) +(mpC2, 0, 0, 0) - (m+C2,0, 0, 0 ) -(m∏°C2,0, 0, 0 ) - (mpC2,0, 0, 0 ) = 0

As you see, trying to have all 3 final state particles at rest is impossible due to conservation of momentum. You need to allow for some final momenta and then think about what being at threshold implies.
 
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