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

p
+ D --> n + π^{0}

The rest masses of the particles involved are m_{p}c^{2} ≈ m_{n}c^{2}
≈ m_{D}c^{2}/2 ≈ 939 MeV and m_{π0}c^{2} = 135
MeV. Find the total energy of the emitted π^{0}.

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 = 3m_{p}c^{2}= E_{n}+ E_{π0}. (1)

Momentum conservation: p_{n}= p_{π0}. (2)

E_{n}^{2}= p_{n}^{2}c^{2}+ m_{p}^{2}c^{4}_{, }E_{π0}^{2}= p_{n}^{2}c^{2}+ m_{π0}^{2}c^{4},

E_{n}^{2}- E_{π0}^{2}= (E_{n}+ E_{π0})(E_{n}- E_{π0}) = m_{p}^{2}c^{4}_{ }- m_{π0}^{2}c^{4}. (3)

(E_{n}- E_{π0}) = (m_{p}^{2}c^{4}_{ }- m_{π0}^{2}c^{4})/(3m_{p}c^{2}). (4) = (3)/(1)

2E_{π0}= 3m_{p}c^{2}- (m_{p}^{2}c^{4}_{ }- m_{π0}^{2}c^{4})/(3m_{p}c^{2}). (1) - (4)

E_{π0}= (8m_{p}^{2}c^{4}_{ }+ m_{π0}^{2}c^{4})/(3m_{p}c^{2}) = 1255 MeV.

A K^{0} meson decays in flight into two pions,
K^{0} --> π^{+} + π^{-}. Let the mass of each pion
be m_{π} and the mass of the K^{0} be m_{K} > 2m_{π}.

(a) Find the speed of the K^{0} 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 K^{0}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 K^{0}meson we have for the forward emitted pion,

E/c = m_{K}c/2, p_{x}= (m_{K}^{2}c^{2}/4 – m_{π}^{2}c^{2})^{½}= (m_{K}c/2)(1 – 4m_{π}^{2}/m_{K}^{2})^{½}.

For the backward emitted pion we have

E/c = m_{K}c/2, p_{x}= -(m_{K}^{2}c^{2}/4 – m_{π}^{2}c^{2})^{½}= -(m_{K}c/2)(1 – 4m_{π}^{2}/m_{K}^{2})^{½}.

A Lorentz transformation to the lab frame yields:

Forward emitted pion: E_{f}= γ[(m_{K}c/2) + β(m_{K}c/2)(1 – 4m_{π}^{2}/m_{K}^{2})^{½}]

Backward emitted pion: E_{b}= γ [(m_{K}c/2) - β(m_{K}c/2)(1 – 4m_{π}^{2}/m_{K}^{2})^{½}]

E_{f}/E_{b}= α = [1 + β(1 – 4m_{π}^{2}/m_{K}^{2})^{½}]/[1 - β(1 – 4 m_{π}^{2}/m_{K}^{2})^{½}]

β = v/c = (1 – 4m_{π}^{2}/m_{K}^{2})^{-½}(α – 1)/(α + 1)

(b) We want p_{x}of the backward emitted pion to be positive in the lab frame

p_{x}= γ [β(m_{K}c/2) - (m_{K}c/2)(1 – 4m_{π}^{2}/m_{K}^{2})^{½}] = 0.

β = (1 – 4m_{π}^{2}/ m_{K}^{2})^{½}, α = (1 + β^{2})/(1 – β^{2}) = m_{K}^{2}/(2m_{π}^{2}) – 1

For α > m_{K}^{2}/(2m_{π}^{2}) – 1 there will there be no π-meson flying into the backward hemisphere?

A Higgs boson (M = 178 GeV/c^{2})
at rest decays into a pair of leptons (φ --> τ^{+}τ^{-}).
The τ^{-} has a mass of m = 1.78 GeV/c^{2}. 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.

2E_{τ}= Mc^{2}. E_{τ}= (m^{2}c^{4}+ p_{τ}^{2}c^{2})^{½}.

p_{τ}= (M^{2}/4 - m^{2})^{½}c, E_{τ}= Mc^{2}/2,

β = p_{τ}c/ E_{τ}= (1 – 4m^{2}/M^{2})^{½}= (1 – 4*10^{-4})^{½}= 0.9998… .

A Φ particle (m_{Φ }= 1.020GeV/c^{2}) has a momentum of
3GeV/c along the z-axis in the laboratory frame.

It decays Φ --> K^{+}K^{-} into two charged kaons (m_{K}^{+
}= m_{K}^{- }= 0.494GeV/c^{2} ).

(a) Calculate the momentum **P**_{K} of each kaon in the laboratory
frame if the decay axis coincides with the z-axis.

(b) Calculate the momentum **P**_{K} 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 m_{Φ}, the 4-vector momentum of the Φ is (E/c,p) = (m_{Φ}c, 0).

Each component of the 4-vector is conserved. After the decay we therefore have for each particle:

E_{k}^{2}= ¼m_{Φ}^{2}c^{4}= m_{k}^{2}c^{4}+ p_{k}^{2}c^{2}, p_{k}^{2}= c^{2}(¼m_{Φ}^{2}- m_{k}^{2}).

E_{k}= 0.51 GeV, p_{k}= 0.127 GeV/c.

The laboratory frame moves with respect to the rest frame in the negative z-direction with speed v.

E_{Φ}' = γE_{Φ}, E_{Φ}'^{2}= γ^{2}E_{Φ}^{2}, m_{Φ}^{2}c^{4}+ p_{Φ}^{2}c^{2}= γ^{2}E_{Φ}^{2}, γ^{2}= [1.02^{2}+ 3^{2}]/1.02^{2}, 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 = E_{k}, p_{z}= p_{k}, p_{x}= p_{y}= 0. p_{z}' = 1.89 GeV/c.

Particle 2: E = E_{k}, p_{z}= -p_{k}, p_{x}= p_{y}= 0. p_{z}' = 1.11 GeV/c.(b) Particle 1: E = E

_{k}, p_{z}= 0, p_{x}= p_{k}, p_{y}= 0. p_{z}' = 1.5 GeV/c, p_{x}' = 0.127 GeV/c.

Particle 2: E = E_{k}, p_{z}= 0, p_{x}= -p_{k}, p_{y}= 0. p_{z}' = 1.5 GeV/c, p_{x}' = -0.127 GeV/c.

For both particles: |**p**'| = (p_{x}'^{2}+p_{z}'^{2})^{½}= 1.505 GeV/c.

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: Mc^{2}= 4mc^{2}= (m^{2}c^{4}+ p^{2}c^{2})^{½}+ (4m^{2}c^{4}+ p^{2}c^{2})^{½}.

4mc^{2}- (m^{2}c^{4}+ p^{2}c^{2})^{½}= (4m^{2}c^{4}+ p^{2}c^{2})^{½}. Solve for pc!

pc = 1.281 mc^{2}.

E = (m^{2}c^{4}+ 1.64 m^{2}c^{4})^{½}= 1.625 mc^{2}.

E' = (4m^{2}c^{4}+ 1.64 m^{2}c^{4})^{½}= 2.375 mc^{2}.

(c) p = γmv. E = γmc^{2}. v = pc^{2}/E

v/c = 1.281 mc^{2}/1.625 mc^{2}= 0.788.

v'/c = 1.281 mc^{2}/2.375 mc^{2}= 0.539.

Velocity addition formula:

Let v'' be the speed of m in the rest frame of 2m. v'' = (v + v')/(1 + vv'/c^{2}) = 0.93c.

An unstable particle decays in its flight into
three charged pions (mass 140 MeV/c^{2}). The tracks recorded are shown
below, the event being coplanar. The kinetic energies and the emission angles
are T_{1} = 190 MeV, T_{2}
= 321 MeV, T_{3}
= 58 MeV, θ_{1}
= 12.25^{o},
θ_{2}
= 22.4^{o}.
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 p^{2}c^{2}= E^{2}- m^{2}c^{4}. - We use energy and momentum conservation to find the mass of the decaying particle.
- Details of the calculation:

E_{1}= T_{1 }+ m_{π}c^{2}= 330 MeV, p_{1}c = 299 MeV

E_{2}= T_{2 }+ m_{π}c^{2}= 461 MeV, p_{2}c = 439 MeV

E_{3}= T_{3 }+ m_{π}c^{2}= 198 MeV, p_{3}c = 140 MeV

Energy conservation: E = E_{1}+ E_{2}+ E_{3}= 989 MeVE is the total energy of the decaying particle.

Let the positive x-direction be the direction of motion of pion 2.

p_{1x}c = 299 MeV*cos(22.4^{o}), p_{1y}c = -299 MeV*sin(22.4^{o}).

p_{2x}c = 439 MeV, p_{2y}c = 0.

P_{3x}c = 140 MeV*cos(12.25^{o}), p_{3y}c = 140 MeV*sin(12.25^{o}).

p_{x}c = 852.25 MeV, p_{y}c = -84.23 MeV, are the momentum components of the decaying particle (times c).

pc = 856.4 MeV is the momentum of the decaying particle.

mc^{2}= (989^{2}– 856.4^{2})^{½}= 494.7 MeV.

The decaying particle's mass is 494.7 MeV/c^{2}. It is a kaon.

tanθ = p_{y}/p_{x}, 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.