Collisions involving only massive particles
Problem:
An energetic proton collides with a proton at rest. What is the minimum
kinetic energy the incident proton must have to make the reaction p + p > p +
p + p(bar)
+ p possible?
Solution:
 Concepts:
Relativistic collisions, energy and momentum conservation, frame
transformations
For each component p_{μ} of the 4vector (p_{0}, p_{1},
p_{2}, p_{3}) we have
∑_{particles_in} p_{μ} = ∑_{particles_out} p_{μ},
or ∑_{i} (p_{i})_{μ} = ∑_{j} (p_{j})_{μ},
where i denotes the particles going into the collision and j denotes the
particles emerging from the collision.
For transformations between
reference frames we have
(P_{0},P)∙(P_{0},P)
= (P_{0}',P')∙(P_{0}',P').
Here
P_{0} = ∑_{particles} p_{0} and P
= ∑_{particles} p.
 Reasoning:
In relativistic collisions between free particles energy and momentum are
always conserved. Often the physics is best visualized in the center of
momentum frame.
 Details of the calculation:
The proton has minimum energy when the reaction products are at rest in the
CM frame.
CM frame:
In the CM frame after the collision we have for the length of the momentum
4vector
P_{0}^{2}  P^{2} = P_{0}^{2} = (∑p_{0})^{2
}= (4 mc)^{2} = 16 m^{2}c^{2}.
Lab frame:
The "length" of the momentum 4vector is 16 m^{2}c^{2}
before and after the collision.
Before the collision we have
P_{0}^{2}  P^{2} = (p_{0a} + p_{0b})^{2}
 (p_{a} + p_{b})^{2} = p_{0a}^{2}
+ p_{0b}^{2}  p_{a}^{2} + 2p_{0a}p_{0b}.
For any free particle we have p_{0}^{2}
p^{2}
= m^{2}c^{2}. We therefore have:
P_{0}^{2}  P^{2} = 16m^{2}c^{2} =
2m^{2}c^{2} + 2p_{0a}p_{0b}, p_{0a}p_{0b}
= 7 m^{2}c^{2}. p_{0a}mc = 7 m^{2}c^{2}.
E_{a}/c = 7 mc. mc^{2} = 938 MeV, E_{a} = 6.6 GeV.
The proton must have E_{a}  mc^{2} = 6 mc^{2} of
kinetic energy to make the reaction possible.
Problem:
A particle of mass 2 m and kinetic energy 8 mc^{2} collides with a
particle of mass m at rest in the laboratory. A particle of mass 3 m_{
}and a particle of mass M emerge from the collision. What is the
maximum possible value of M?
Solution:
 Concepts:
Relativistic collisions, energy and momentum conservation,
frame transformations
 Reasoning:
In relativistic collisions between free particles energy
and momentum are always conserved. Often the physics is best visualized in
the center of momentum frame.
 Details of the calculation:
M takes on its maximum value if in the CM
frame the reaction products are at rest.
[E^{2}/c^{2}  P^{2}]_{lab
before coll.} = [E^{2}/c^{2}]_{CM before coll.}
= [E^{2}/c^{2}]_{CM after coll.}
invariance of norm of (P_{0},P) energy conservation
lab, before:
(E_{2m }+ E_{m})^{2}/c^{2}
 P_{2m}^{2} = (E_{2m}^{2}+ E_{m}^{2}
+ 2 E_{2m}E_{m})/c^{2}  P_{2m}^{2}
= 4 m^{2}c^{2} + P_{2m}^{2} + m^{2}c^{2}
+ 2 E_{2m}m  P_{2m}^{2} = 5 m^{2}c^{2}
+ 2 E_{2m}m = 25 m^{2}c^{2},
since E_{2m}
= 2 mc^{2} + 8 mc^{2} = 10 m^{2}c^{2}.
CM, after:
E^{2}/c^{2} = 25 m^{2}c^{2},
E = 5 mc^{2}, Mc^{2} = (5  3) mc^{2} = 2 mc^{2}.
2 m_{ }is the maximum possible value for M.
Problem:
In a protonproton collision a π^{+} meson can be created through the
reaction
p_{1} + p_{2} > p + n + π^{+}.
In the center of mass (CM) frame of reference each proton has an initial
energy γm_{p}c^{2}, where m_{p} is the mass of the
proton and γ = (1  β^{2})^{1/2}, with β = v/c.
Take the
mass of the proton and the mass of the neutron to be 1837 m_{e} (m_{e}
= electron mass) and the mass of the pion to be 273 m_{e} = 0.1486 m_{p}.
Determine the minimum value of the initial velocity v for which π^{+}
creation is possible.
Solution:
 Concepts:
Energy and momentum conservation
 Reasoning:
In relativistic collisions between free particles energy
and momentum are always conserved.
 Details of the calculation:
Momentum conservation:
In the CM frame
the total momentum is zero before and after the collision. For the minimum
value of the initial velocity v the three particles are at rest after the
collision.
Energy conservation:
2γm_{p}c^{2} = 2.1486
m_{p}c^{2}.
γ = 1.0743.
v = 0.365 c = 1.1*10^{8}
m/s.
Problem:
HEPA is an asymmetric electron proton collider located near the city of
Hamburg in Germany. The energy of the electron beam is 26 GeV and the energy of
the proton beam is 820 GeV. Ignore baryon and lepton number conservation and
calculate
(a) the maximum number of neutral pions (mass of π^{0} = 134.98 MeV)
that can be produced in one protonelectron collision.
(b) What momentum would a beam of electrons incident on protons at rest need to
have to produce the same number of pions as in part (a)?
Solution:
 Concepts:
Relativistic collisions, energy and momentum conservation, frame
transformations
 Reasoning:
In relativistic collisions between free particles energy and momentum are
always conserved. Often the physics is best visualized in the center of
momentum frame.
 Details of the calculation:
(a) If in the CM frame the reaction products are at rest, they have the
largest mass M possible.
What is M?
[E^{2}/c^{2}  P^{2}]_{lab before coll.} =
[E^{2}/c^{2}]_{CM before coll.} = [E^{2}/c^{2}]_{CM
after coll.}
invariance of norm of (P_{0},P) energy
conservation
lab, before:
(E_{p }+ E_{e})^{2}/c^{2} – (P_{p} –
P_{e})^{2} = (E_{p}^{2 }+ E_{e}^{2}
+ 2E_{p}E_{e})/c^{2} – P_{p}^{2} – P_{e}^{2} +
2P_{e}P_{p}
= 85280.85 GeV^{2 }/c^{2}_{,
}since^{
}(E_{p }+ E_{e})^{2}/c^{2} = 715716 GeV^{2
}/c^{2
}E_{p}^{2}/c^{2}  m_{p}^{2}c^{2}
= P_{p}^{2}, P_{p}^{2} = (820^{2} 
0.938^{2}) GeV^{2}/c^{2} = 672399.1 GeV^{2}/c^{2},
E_{e}^{2}/c^{2} – m_{e}^{2}c^{2}
= P_{e}^{2}, P_{e}^{2} = (26^{2} –
(5.11*10^{4})^{2}) GeV^{2}/c^{2} = 676 GeV^{2}/c^{2}.
E_{CM} = 292 GeV
# of pions = 292/0.13498 = 2163
(b) Assume we want to create the same number of pions. Then we want
(E_{p }+ E_{e})^{2}/c^{2} – P_{e}^{2}
= (E_{p}^{2}+ E_{e}^{2} + 2E_{p}E_{e})/c^{2}
– P_{e}^{2} = (m_{p}^{2}c^{4} + 2m_{p}c^{2}E_{e})/c^{2}
+ m_{e}^{2}c
= 85280.85 GeV^{2 }/c^{2}.^{
}E_{e} = (85280.85 GeV^{2 } m_{p}^{2}c^{4}
 m_{e}^{2}c^{4})/(2m_{p}c^{2}) =
45458 GeV
Problem:
A particle of rest mass 1 MeV/c^{2} and kinetic
energy 2 MeV collides with a stationary particle of rest mass 2 MeV/c^{2}.
After the collision, the two particles stick together.
(a) What are the
energy, velocity and momentum of the incoming particle?
(b) What is the rest
energy of the outgoing combined particle?
Solution:
 Concepts:
Relativistic collisions, energy and momentum conservation,
frame transformations
 Reasoning:
In relativistic collisions between free particles energy
and momentum are always conserved. Often the physics is best visualized in
the center of momentum frame.
 Details of the calculation:
(a) Let mc^{2} = 1 MeV.
E_{m}
= mc^{2} + 2mc^{2} = 3mc^{2}The energy of the
incoming particle is 3 MeV.
γ = 3, 1  v^{2}/c^{2} =
1/9, v^{2}/c^{2} = 8/9, v = 0.9428c.
The velocity of
the incoming particle is 0.9428c.
E^{2} = m^{2}c^{4
}+ p^{2}c^{2}, 9m^{2}c^{4} = m^{2}c^{4
}+ p^{2}c^{2}, p^{2} = 8m^{2}c^{2},
p = 2.828 MeV/c.
The momentum of the incoming particle is 2.828 MeV/c.
(b) [E^{2}/c^{2}  p^{2}]_{lab before coll.}
= [E^{2}/c^{2}]_{CM before coll.} = [E^{2}/c^{2}]_{CM
after coll.}
invariance of norm of (P_{0},P)
energy conservation
lab, before:
(E_{m }+ E_{2m})^{2}/c^{2}
 P_{m}^{2} = (E_{2m}^{2}+ E_{m}^{2}
+ 2E_{2m}E_{m})/c^{2}  P_{m}^{2}
= 4m^{2}c^{2} + P_{m}^{2} + m^{2}c^{2}
+ 4E_{m}m  P_{m}^{2} = 5m^{2}c^{2}
+ 4E_{m}m = 17m^{2}c^{2}
since E_{m} = mc^{2}
+ 2mc^{2} = 3mc^{2}
CM, after:
E^{2}/c^{2} = 17m^{2}c^{2},
E = 17^{½}mc^{2} = 4.123 MeV.
4.123 MeV_{ }is
the rest energy of the outgoing particle.
Problem:
Consider the reaction p + p > p + p + Λ(bar) + Λ. Assume
that the masses of proton and Lambda are 1 GeV/c^{2}.
(a) Consider a laboratory frame in which one of the two
initial state protons is at rest. What is the "threshold" energy, i.e. the
minimum energy that the incident proton must have for the reaction to be
possible?
(b) For the reaction at threshold as described in part (a), what
is the mean distance that the Lambda travels before it decays? (The mean
lifetime of the lambda in its rest frame is τ ~ 2.6*10^{10} s).
(c)
What is the probability that at least one of the two lambdas travels the
distance determined in (b)?
Solution:
 Concepts:
Relativistic collisions, energy and momentum conservation, frame transformations
For each component p_{m} of the 4vector (p_{0},p_{1},p_{2},p_{3})
we have
∑_{particles_in} p_{μ} = ∑_{particles_out} p_{μ},
or ∑_{i} (p_{i})_{μ} = ∑_{j} (p_{j})_{μ},
where i denotes the
particles going into the collision and j denotes the particles emerging from the
collision.
For transformations between reference frames we have
(P_{0},P)∙(P_{0},P)
= (P_{0}',P')∙(P_{0}',P').
Here
P_{0} = ∑_{particles} p_{0} and P
= ∑_{particles} p.
 Reasoning:
In relativistic collisions between free particles energy
and momentum are always conserved. The proton has minimum energy when the
reaction products are at rest in the CM frame.
 Details of the calculation:
After the collision we have in the CM
frame
P_{0}^{2}  P^{2} = P_{0}^{2}
= (∑p_{0})^{2 }= (4mc)^{2} = 16m^{2}c^{2}.
The square of the magnitude of the momentum 4vector is 16m^{2}c^{2}
before and after the collision in any frame.
Before the collision we have
in the lab frame.
P_{0}^{2}  P^{2} = (p_{0a}
+ p_{0b})^{2}  (p_{a} + p_{b})^{2}
= p_{0a}^{2} + p_{0b}^{2}  p_{a}^{2}
+ 2p_{0a}p_{0b}.
For any free particle we have p_{0}^{2}
p^{2} = m^{2}c^{2}. We therefore have:
P_{0}^{2}
 P^{2} = 16m^{2}c^{2} = 2m^{2}c^{2}
+ 2p_{0a}p_{0b}, p_{0a}p_{0b} = 7m^{2}c^{2}.
p_{0a}mc = 7m^{2}c^{2}.
E_{a}/c = 7mc.
mc^{2} = 1 GeV
The "threshold" energy, for the incident proton is
7mc^{2}.
(b) The total energy before and after the collision is 8
mc^{2}.
The momentum before and after the collision is p^{2}c^{2}
= E_{proton}^{2} – m^{2}c^{4} = 48 m^{2}c^{4}.
The speed of the reaction products is v = pc/E =
((48)^{½}/8) c =
0.839 c
The mean lifetime of the lambda in the lab frame is γ*2.6*10^{10}
s = 4.77*10^{10} s.
The mean distance traveled is 0.12 m.
(c) For each lambda particle the probability that it is still present at
t = 4.77*10^{10} s is exp(1) = 0.368.
Assuming that we have
assigned initial probabilities for the occurrence of two events, A and B,
then we calculate the probability that at least one of these events in the
following way:
P(A ∨ B) = P(A) + P(B)  P(A · B)
The probability that
one or the other or both of two events will occur is equal to the
probability that the first will occur, plus the probability that the second
will occur, minus the probability that they both occur. (The final term in
this formula provides a necessary correction because we have already counted
the joint occurrence twice, once in each of the other terms.)
Therefore:
The probability that at least one is still present at t = 4.77*10^{10}
s is
0.736  0.368^{2} = 0.6.
Problem:
The PEPII collider at Stanford Linear
Accelerator Center creates electronpositron headon collisions. In the
laboratory frame the combination of electron energy E_{} = 9 GeV and
positron energy E_{+} = 3.1 GeV is resonant for the production of a
single Y particle.
(a) What is the rest mass of the produced particle
ϒ?
(b) What is the speed of Y in units of c?
Solution:
 Concepts:
Relativistic collisions, energy and momentum conservation
 Reasoning:
In relativistic collisions between free particles energy
and momentum are always conserved.
 Details of the calculation:
(a) E = E_{+} + E_{}, p
= p_{+} + p_{}.
p_{+}^{2}c^{2} =
E_{+}^{2}  m_{e}^{2}c^{4} = (3.1
GeV)^{2}  (0.511 MeV)^{2} = 9.61 GeV^{2}, p_{+}c
= 3.1 GeV.
p_{}^{2}c^{2} = E_{}^{2}
 m_{e}^{2}c^{4} = (9 GeV)^{2}  (0.511 MeV)^{2}
= 81 GeV^{2}, p_{}c = 9 GeV.
E = 12.1 GeV, pc = 5.9 GeV
m^{2}c^{4} = E^{2}  p^{2}c^{2}, mc^{2}
= 10.56 GeV.
(b) E = γmc^{2}. pc = γmv.
pc/E = v/c. v = 0.49c.
Problem:
Two identical particles are each moving with a speed 0.8 c
and are on a collision course. The velocity vectors of the two particles are at
90 degrees to each other. After the collision the two particles create one new
particle. What is a ratio of the rest mass of the new particle to the sum of
the rest masses of the initial particles?
Solution:

Concepts:
Energy and momentum conservation

Reasoning:
In relativistic collisions between free
particles energy and momentum are always conserved.

Details of the calculation:
Set up the coordinate
system so that the xcomponents of the particles are equal and the
ycomponents have equal magnitude but opposite sign.
Momentum
conservation: 2 γmv cos(45^{o})
= γ'm'v'
Energy conservation:
2 γmc^{2} = γ'm'c^{2}.
Dividing we obtain v cos(45^{o})
= v'.
m'/(2m) = γ/ γ' = [(1 – v^{2}/(2c^{2}))/(1 – v^{2}/c^{2})]^{½}
= [(1 – 0.32)/(1 – 0.64)]^{½} = 1.37
Problem:
The Japanese Bfactory
collided electron beams and positron beams with adjustable and unequal beam
energies to produce an excited state of the ϒ meson, called the ϒ (4S). The
rest mass of the ϒ (4S) is 10.58 GeV/c^{2}. Subsequently, the ϒ (4S)
meson decays into a pair of Bmesons: B^{+} and B^{}. The rest
masses of the oppositely charged B^{+} and B^{} mesons each are
5.28 GeV/c^{2}.
(a) If the beam energy of the
electron beam is chosen as E^{} = 8 GeV and the center of mass energy of the
colliding beams equals the mass of the ϒ (4S) meson, calculate the energy of the
positron beam, E^{+}, in the laboratory frame. Momenta perpendicular to
the beam direction are zero.
(b) Calculate the magnitudes
of the momenta for the B^{+} and B^{} mesons in the rest frame
of the ϒ (4S).
(c) Assume the B^{+}
is emitted in the direction of the electron beam. What are the magnitudes of
themomenta for the B^{+} and B^{} mesons in the laboratory
frame?
Solution:
 Concepts:
Relativistic collisions
 Reasoning:
Energy and momentum are conserved in every frame.
 Details of the calculation:
(a) Let M demote the mass of the ϒ (4S).
energy conservation: E = E^{} + E^{+} =
γMc^{2},
momentum conservation: pc = E^{}  E^{+} = γMvc.
pc/E = β = v/c = (E^{}  E^{+})/( E^{}
+ E^{+}).
1 
β^{2} = 4E^{+}E^{}/(E^{}
+ E^{+})^{2}. γ^{2}
= (E^{} + E^{+})^{2}/(4E^{+}E^{}).
γ^{2}M^{2}c^{4}
= (E^{} + E^{+})^{2}. M^{2}c^{4}/(4E^{+}E^{})
= 1, E^{+} = M^{2}c^{4}/(4E^{}).
E^{+}
= 3.5 GeV .
(b) For each B meson:
p^{2}c^{2} = E^{2} – m^{2}c^{4}. p^{2}c^{2}
= (5.29 GeV)^{2} – (5.28 GeV )^{2}.
p = 0.325 GeV/c.
(c) β = v/c = (E_{B}  E_{B+})/( E_{B} + E_{B+}) = 0.39 gives the speed of the ϒ (4S) in the
lab.
γ = (1 – β^{2})^{½} = 1.087.
Lorentz transformation of
4vector momentum:
p_{B+} = γβ
*5.28 GeV/c + γ*0.325 GeV/c = 2.59 GeV/c.
p_{B}^{ }= γβ *5.28 GeV/c  γ*0.325 GeV/c
= 1.81 GeV/c.