Problem Set 1

Problem 3

Part A

Minimum energy = m1 (momentum is conserved by m2 and m3 having equal and opposite momenta).  Maximum energy occurs when m2 and m3 are moving opposite m1 and have the minimum effective mass between them (ie. they are moving with the same velocity).  Then they act as a single particle with effective mass of m2+m3

eq1 = m0 == (Sqrt[(p^2) + (m1^2)]) + (Sqrt[(p^2) + (((m2 + m3))^2)])

solve1 = Simplify[Solve[eq1, p]]

Get the maximum energy:

Em = 
      PowerExpand[
        Simplify[
          (Sqrt[(p^2) + (m1^2)]) /. 
            solve1[[1]]]]

Choose eta' decay:

particles = {m0 -> 958. , m1 -> 140. , m2 -> 140. , 
        m3 -> 547. }

Maximum energy:

Emax = Em /. particles

Emin = m1 /. particles

Part B:

Given an E1 between Emin and Emax, determine E2 as a function of Cos[theta]:

Clear[E1]

eq1 = m0 == E1 + E2 + E3

eq2 = eq1 /. E3 -> (Sqrt[(p3^2) + (m3^2)])

eq3 = eq2 /. p3 -> (Sqrt[(p1^2) + (p2^2) + 2 p1 p2 costh])

eq4 = eq3 /. p1 -> (Sqrt[(E1^2) - (m1^2)])

eq5 = Simplify[eq4 /. p2 -> (Sqrt[(E2^2) - (m2^2)])]

Now solve for E2 in terms of E1 and Cos[theta]:

solve2 = Solve[eq5, E2]

The two solutions correspond to the uper and lower branches of the curve if Cos[theta]=+/- 1.

Plot upper branch:

plot1 = 
      Plot[Evaluate[
          E2 /. solve2[[1]] /. 
              particles /. costh -> 1], {E1, Emin, Emax}, 
        PlotRange -> {{100, 300}, {100, 300}}]

Plot lower branch:

plot2 = 
      Plot[Evaluate[
          E2 /. solve2[[2]] /. 
              particles /. costh -> 1], {E1, Emin, Emax}, 
        PlotRange -> {{100, 300}, {100, 300}}]

Plot both and energy conservation triangle:

dalitzplot = 
      Show[plot1, plot2, 
        Graphics[
          Line[{{m1, m2}, {m0 - m3 - m2, m2}, {m1, m0 - m3 - m1}, {m1, 
                  m2}} /. particles]], AspectRatio -> 1, 
        AxesLabel -> {"Pi- Energy", "Pi+ Energy"}]