Problem Set 1

Problem 1

Part a)

Conservation of energy:

eq1 = m0 == E1 + E2

Conservation of momentum:

eq2 = 0 == p1 + p2

Mass constraints:

eq3 = (E1^2) - (p1^2) == (m1^2)

eq4 = (E2^2) - (p2^2) == (m2^2)

Find energies and momenta:

solve1 = Solve[{eq1, eq2, eq3, eq4}, {E1, E2, p1, p2}]

Part b)

The maximum laboratory momentum of a daughter particle in a two body decay will occur when the center of mass momentum of the daughter is parallel to the direction of motion of the parent.  The minimum will occur when the the cm momentum is antiparallel.

There are two ways of solving this art of the problem.  You can either solve using the energy and momentum equations as above or you can take the cm values of the energy and momentum and transform them to the lab.  Both are done below.

First solve the energy and momentum conservation equations:

eq1a = (Sqrt[(p^2) + (m0^2)]) == E1 + E2
eq2a = p == p1 + p2

solve2 = 
      Simplify[PowerExpand[
          Solve[{eq1a, eq2a, eq3, eq4}, {E1, E2, p1, p2}]]]

Part b1) Solve for pi -> mu nu

pion = {m0 ->  .13957, m1 -> 0, m2 ->  .10566, 
        p -> 100. }

solve2 /. pion

Maximum neutrino energy = 42.7 GeV,
Minimum neutrino energy = ~0.

Part b2) Solve for kaon-> mu nu

kaon = {m0 ->  .49468, m1 -> 0, m2 ->  .10566, 
        p -> 100. }

solve2 /. kaon

Maximum neutrino energy = 95.4 GeV,
Minimum neutrino energy = ~0.

Part b (again)

This time transform the center of mass energy into the lab.

gamma = ((Sqrt[(p^2) + (m0^2)])/m0)

beta = Simplify[(Sqrt[1 - (1/(gamma^2))])]

Maximum is when momentum is going forward in lab:

Elabmax = 
      gamma (E1 + beta p1) /. 
        solve1[[2]]

Minimum is when momentum is going backward in lab:

Elabmin = 
      gamma (E1 - beta p1) /. 
        solve1[[2]]

Get pion numbers:

Elabmax /. pion
Elabmin /. pion

Get kaon numbers:

Elabmax /. kaon
Elabmin /. kaon

Part c)

Let z = Cos[theta_cm]

pparalcm = z p1
pperp = (Sqrt[1 - (z^2)]) p1

Transform center of mass momentum  and energy into lab energy:

Elab = gamma (E1 + beta pparalcm)

Elab1 = ((Elab /. solve1[[1]]) /. pion)

Plot as a function of center of mass angle:

Plot[Elab1 /. z -> Cos[theta], {theta, 0, Pi}]

Plot as a function of Cos[theta_cm]:

Plot[Elab1, {z, -1, 1}]

Problem 2

Part a:

The angle in the Lab is very small.  Therefore calculate the sine.

sinthetalab = Simplify[(pperp/(Sqrt[(Elab^2) - (m1^2)]))]

Get the actual angle:

thetalab[z_] = 
      ArcSin[sinthetalab /. 
            solve1[[2]] /. 
          pion]

Part b:

The probability for a given theta in the lab = Probcm[z] dz/dtheta_lab *theta_lab dphi_lab
	~ Probcm[z] 2 Pi theta_lab/Abs[dtheta_lab/dz]

prob[z_] = 
      Simplify[2 π Abs[
            thetalab[z]/(∂_z) thetalab[z]]]

plt2a = 
      ParametricPlot[{thetalab[z], prob[z]}, {z, -1, 1}]

The maximum phase space occurs at theta_cm = Pi/2.  This would occur for a massless particle in the lab at an angle of 1/gamma.  Plot that angle.

Show[plt2a, 
      Graphics[Line[{{(1/gamma), 0}, {(1/gamma), 7}} /. pion]]]

Part C:

The Prob_Elab = Prob_cm[z]/Abs[dE_lab/dz].

probE[z_] = Simplify[PowerExpand[(1/(∂_z)) Elab]]

This is a constant.  That means that all energies between E_min and E_max are equally probable.