Problem Set 9 - Problem 4

There are two ways of solving this part of the problem.  You can either use conservation of energy and momentum in the laboratory frame and solve as in the previous section, or you can transform the results of the previous section into the laboratory.

Conservation of Energy and Momentum

Conservation of energy:

eq1  =  Elambda  ==  Ep  +  Epi

Conservatoin of momentum:

eq2  =  plambda  ==  pp  +  ppi

Constituent equations:

Elambda  =  Sqrt[plambda^2  + mlambda^2]Ep  =  Sqrt[pp^2  +  mp^2]Epi  =  Sqrt[ppi^2  +  mpi^2]

Solve:

solve1  =  Simplify[Solve[Evaluate[{eq1, eq2}], {ppi, pp}]]

The two solution correspond to the pion going parallel to the lambda's motion in the center of mass ans anti-parallel to the motion.

Input the masses and lambda momentum:

rule = {plambda -> 15. ,  mpi ->  .1396,  mp ->  .9383,  
        mlambda -> 1.1156}

solve2  =  solve1 /. rule

Obviously, the first solution is the anti-parallel decay and the second is the parallel decay.  In either case, the proton takes most of the momentum in the lab.

Transformation

First find cm energy as in the previous part of this problem:

Epicm  =  
      (mlambda^2  +  mpi^2  -  mp^2)/(2 mlambda) /. rule

Calculate the center of mass momentum:

ppicm  =  Sqrt[Epicm^2 - mpi^2] /. rule

Calculate the beta and gamma of the center of mass:

gamma  =  Elambda/mlambda /. rule

beta  =  Sqrt[1 - 1/gamma^2]

Calculate the momentum of the pion when its momentum is parallel to beta:

ppilab1  =  gamma*(ppicm  +  beta Epicm)

Calculate the momentum of the pion when its momentum is anti-parallel to beta:

ppilab2  =  gamma*(-ppicm  +  beta Epicm)