### Decay involving only massive particles

#### Problem:

A slowly moving antiproton is captured by a deuteron at rest producing a neutron and a neutral pion.

p + D --> n + π0

The rest masses of the particles involved are mpc2 ≈ mnc2 ≈ mDc2/2 ≈ 939 MeV and mπ0c2 = 135 MeV.  Find the total energy of the emitted π0.

Solution:

• Concepts:
Relativistic collisions, energy and momentum conservation
• Reasoning:
In relativistic collisions energy and momentum are always conserved.
• Details of the calculation:
We assume that the initial particles have no kinetic energy.
Energy conservation:  E = 3mpc2 = En + Eπ0.                      (1)
Momentum conservation: pn = pπ0.                                   (2)
En2 = pn2c2 + mp2c4Eπ02 = pn2c2 + mπ02c4,
En2 - Eπ02 = (En + Eπ0)(En - Eπ0) = mp2c4  - mπ02c4.              (3)
(En - Eπ0) = (mp2c4  - mπ02c4)/(3mpc2).                              (4) = (3)/(1)
2Eπ0 = 3mpc2 - (mp2c4  - mπ02c4)/(3mpc2).                          (1) - (4)
Eπ0 = (8mp2c4  + mπ02c4)/(3mpc2) = 1255 MeV.

#### Problem:

A K0 meson decays in flight into two pions,  K0 --> π+ + π-.  Let the mass of each pion be mπ and the mass of the K0 be mK > 2mπ.
(a)  Find the speed of the K0 if the greatest possible energy of a π-meson from this decay is α times larger than the smallest possible energy.
(b)  For what α will there be no π-meson flying into the backward hemisphere?

Solution:

• Concepts:
Relativistic "collisions", energy and momentum conservation, Lorentz transformation
• Reasoning:
The decay of a particle is a relativistic problem.  In relativistic "collisions" energy and momentum are always conserved.
• Details of the calculation:
(a)  Let the K0 meson move in the positive x direction. The pion with the greatest possible energy in the laboratory is emitted into the +x direction and the pion with the smallest possible energy is emitted in the -x direction.
In the rest frame of the K0 meson we have for the forward emitted pion,
E/c = mKc/2, px = (mK2c2/4 - mπ2c2)½ = (mKc/2)(1 - 4mπ2/mK2)½.
For the backward emitted pion we have
E/c = mKc/2, px = -(mK2c2/4 - mπ2c2)½ = -(mKc/2)(1 - 4mπ2/mK2)½.
A Lorentz transformation to the lab frame yields:
Forward emitted pion: Ef = γ[(mKc/2) + β(mKc/2)(1 - 4mπ2/mK2)½]
Backward emitted pion: Eb = γ [(mKc/2) - β(mKc/2)(1 - 4mπ2/mK2)½]
Ef/Eb = α = [1 + β(1 - 4mπ2/mK2)½]/[1 - β(1 - 4 mπ2/mK2)½]
β = v/c = (1 - 4mπ2/mK2)(α - 1)/(α + 1)

(b)  We want px of the backward emitted pion to be positive in the lab frame
px = γ [β(mKc/2) - (mKc/2)(1 - 4mπ2/mK2)½] = 0.
β = (1 - 4mπ2/ mK2)½, α = (1 + β2)/(1 - β2) = mK2/(2mπ2) - 1
For α > mK2/(2mπ2) - 1 there will there be no π-meson flying into the backward hemisphere?

#### Problem:

A Higgs boson (M = 178 GeV/c2) at rest decays into a pair of leptons (φ --> τ+τ-).  The τ- has a mass of m = 1.78 GeV/c2.  What is the recoil β = v/c of the τ+?

Solution:

• Concepts:
Relativistic "collisions", energy and momentum conservation
• Reasoning:
The decay of a particle is a relativistic problem.  In relativistic "collisions" energy and momentum are always conserved.
• Details of the calculation:
The tau leptons emerge from the decay in opposite directions with equal energy.
2Eτ = Mc2.  Eτ = (m2c4 + pτ2c2)½.
pτ = (M2/4 - m2)½c,  Eτ = Mc2/2,
β = pτc/ Eτ  = (1 - 4m2/M2)½ = (1 - 4*10-4)½ = 0.9998... .

#### Problem:

A Φ particle (mΦ = 1.020GeV/c2) has a momentum of 3GeV/c along the z-axis in the laboratory frame.
It decays Φ --> K+K- into two charged kaons (mK+ = mK- = 0.494GeV/c2 ).
(a)  Calculate the momentum PK of each kaon in the laboratory frame if the decay axis coincides with the z-axis.
(b)  Calculate the momentum PK of each kaon in the laboratory frame if the decay axis is perpendicular to the z-axis.

Solution:

• Concepts:
Relativistic "collisions", energy and momentum conservation
• Reasoning:
The decay of a particle is a relativistic problem.  In relativistic "collisions" energy and momentum are always conserved.  Often the physics is best visualized in the center of momentum frame.
• Details of the calculation:
In the rest frame of the mΦ, the 4-vector momentum of the Φ is (E/c,p) = (mΦc, 0).
Each component of the 4-vector is conserved.  After the decay we therefore have for each particle:
Ek2 = ¼mΦ2c4 = mk2c4 + pk2c2,  pk2 = c2(¼mΦ2 - mk2).
Ek = 0.51 GeV,  pk = 0.127 GeV/c.
The laboratory frame moves with respect to the rest frame in the negative z-direction with speed v.
EΦ' = γEΦ,  EΦ'2 = γ2EΦ2,  mΦ2c4 + pΦ2c2 = γ2EΦ2,  γ2 = [1.022 + 32]/1.022 , γ  = 3.1.
b = 0.947, v =  0.947c.
We can now find the momentum of each kaon in the laboratory frame.

(a)  Particle 1: E = Ek, pz = pk, px = py = 0.  pz' = 1.89 GeV/c.
Particle 2: E = Ek, pz = -pk, px = py = 0.  pz' = 1.11 GeV/c.

(b)  Particle 1:  E = Ek, pz = 0, px = pk, py = 0.  pz' = 1.5 GeV/c, px' = 0.127 GeV/c.
Particle 2:  E = Ek, pz = 0, px = -pk, py = 0.  pz' = 1.5 GeV/c, px' = -0.127 GeV/c.
For both particles: |p'| = (px'2 +pz'2)½ = 1.505 GeV/c.

#### Problem:

A particle of mass M is at rest in the laboratory.  Suddenly it disintegrates into two particles, one of mass m and one of mass 2m.  Let M = 4m.  The final particles have momenta p and p', and energies E and E'.
(a)  What is the relationship between p and p'?
(b)  Find expressions for p, p', E, and E'.
(c)  Determine the speed of each final particle in the laboratory frame.
(d)  Determine the speed of m in the rest frame of 2m.

Solution:

• Concepts:
Relativistic "collisions", energy and momentum conservation
• Reasoning:
The decay of a particle is a relativistic problem.  In relativistic "collisions" energy and momentum are always conserved.
• Details of the calculation:
(a)  Momentum conservation requires that p = -p'.
(b)  Energy conservation:  Mc2 = 4mc2 = (m2c4 + p2c2)½ + (4m2c4 + p2c2)½.
4mc2 - (m2c4 + p2c2)½ = (4m2c4 + p2c2)½.  Solve for pc!
pc = 1.281 mc2.
E = (m2c4 + 1.64 m2c4 )½ = 1.625 mc2.
E' = (4m2c4 + 1.64 m2c4 )½ = 2.375 mc2.
(c) p = γmv.  E = γmc2.  v = pc2/E
v/c = 1.281 mc2/1.625 mc2 = 0.788.
v'/c =  1.281 mc2/2.375 mc2 = 0.539.
Let v'' be the speed of m in the rest frame of 2m.  v'' = (v + v')/(1 + vv'/c2) = 0.93c.

#### Problem:

An unstable particle decays in its flight into three charged pions (mass 140 MeV/c2).  The tracks recorded are shown below, the event being coplanar.  The kinetic energies and the emission angles are T1 = 190 MeV, T2 = 321 MeV, T3 = 58 MeV, θ1 = 12.25o, θ2 = 22.4o.  Estimate the mass of the primary particle.  In what direction was it moving?

Solution:

• Concepts:
Energy and momentum conservation
• Reasoning:
We find the momenta of three pions by using the formula p2c2 = E2 - m2c4.
• We use energy and momentum conservation to find the mass of the decaying particle.
• Details of the calculation:
E1 = T1 + mπc2 = 330 MeV,  p1c = 299 MeV
E2 = T2 + mπc2 = 461 MeV,  p2c = 439 MeV
E3 = T3 + mπc2 = 198 MeV,  p3c = 140 MeV
Energy conservation:  E = E1 + E2 + E3 = 989 MeVE is the total energy of the decaying particle.
Let the positive x-direction be the direction of motion of pion 2.
p1xc = 299 MeV*cos(22.4o),  p1yc = -299 MeV*sin(22.4o).
p2xc = 439 MeV,  p2yc = 0.
P3xc = 140 MeV*cos(12.25o),  p3yc = 140 MeV*sin(12.25o).
pxc = 852.25 MeV, pyc = -84.23 MeV, are the momentum components of the decaying particle (times c).
pc = 856.4 MeV is the momentum of the decaying particle.
mc2 = (9892 - 856.42)½ = 494.7 MeV.
The decaying particle's mass is 494.7 MeV/c2.  It is a kaon.
tanθ = py/px,  θ = -5.64 o.
The decaying particle's momentum was 856.4 MeV/c and it pointed in a direction making an angle of -5.64 o with the positive x-direction.