Kern- und Teilchenphysik - Blatt 1

<< Graphics`Graphics`

2. Aufgabe

DisplayTogether[
        ParametricPlot3D[{-0.488 t, -0.018 t, 2.109  t, 
            RGBColor[1, 0, 0]}, {t, 0, 1}, PlotPoints -> 75], 
        ParametricPlot3D[{-0.255 t, -0.050 t, 0.486  t, 
            RGBColor[0, 1, 0]}, {t, 0, 1}, PlotPoints -> 75], 
        ParametricPlot3D[{-0.743 t, -0.068 t, 2.595 t, 
            RGBColor[0, 0, 1]}, {t, 0, 1}, PlotPoints -> 75], 
        ImageSize -> 150];

pA=rot, pB=grün, pX0=blau

pX0 = {-0.488, -0.018, 2.109} + {-0.255, -0.050, 
          0.486}

pA2 = {-0.488, -0.018, 2.109} . {-0.488, -0.018, 
          2.109}

pB2 = {-0.255, -0.050, 0.486} . {-0.255, -0.050, 
          0.486}

pApB = {-0.488, -0.018, 2.109} . {-0.255, -0.050, 
          0.486}

4.Aufgabe

a)

mK = 497.648 10^-3; mp = 938.27203 10^-3; 
    mpi = 139.57018 10^-3; mlambda = 1115.683 10^-3;

mK^2

Zerfall1 = 
      2 mpi^2 + 
        2 (Sqrt[pA2 + mpi^2] Sqrt[pB2 + mpi^2] - pApB)

Abs[1 - Zerfall1/mK^2]

mlambda^2

Zerfall2 = 
      mp^2 + mpi^2 + 
        2 (Sqrt[pA2 + mp^2] Sqrt[pB2 + mpi^2] - pApB)

Abs[1 - Zerfall2/mlambda^2]

b)

mKp = 493.677 10^-3; mmu = 105.658369 10^-3; 
    mpi = 139.57018 10^-3; EKpi = 200;

EminPI = (mpi^2 - 
            mmu^2)/(2 (EKpi - 
                Sqrt[EKpi^2 - mpi^2] Cos[π]))

EmaxPI = (mpi^2 - 
            mmu^2)/(2 (EKpi - 
                Sqrt[EKpi^2 - mpi^2] Cos[0]))

EminK = (mKp^2 - 
            mmu^2)/(2 (EKpi - 
                Sqrt[EKpi^2 - mKp^2] Cos[π] ))

EmaxK = (mKp^2 - 
            mmu^2)/(2 (EKpi - 
                Sqrt[EKpi^2 - mKp^2] Cos[0]))