### 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 pμ of the 4-vector (p0, p1, p2, p3) we have

particles_in pμ = ∑particles_out pμ, or ∑i (pi)μ = ∑j (pj)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
(P0,P)∙(P0,P) = (P0',P')∙(P0',P').
Here  P0 = ∑particles p0  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 4-vector
P02 - P2 = P02  = (∑p0)2 = (4 mc)2 = 16 m2c2.
Lab frame:

The "length" of the momentum 4-vector is 16 m2c2 before and after the collision.
Before the collision we have
P02 - P2 = (p0a + p0b)2 - (pa + pb)2 = p0a2 + p0b2 - pa2 + 2p0ap0b.
For any free particle we have p02- p2 = m2c2.  We therefore have:
P02 - P2 = 16m2c2 = 2m2c2 + 2p0ap0b,  p0ap0b = 7 m2c2.  p0amc = 7 m2c2.
Ea/c = 7 mc.  mc2 = 938 MeV,  Ea = 6.6 GeV.
The proton must have Ea - mc2 = 6 mc2 of kinetic energy to make the reaction possible.

#### Problem:

A particle of mass 2 m and kinetic energy 8 mc2 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.
[E2/c2 - P2]lab before coll. = [E2/c2]CM before coll. = [E2/c2]CM after coll.
invariance of norm of (P0,P)          energy conservation

lab, before:
(E2m + Em)2/c2 - P2m2 = (E2m2+ Em2 + 2 E2mEm)/c2 - P2m2
= 4 m2c2 + P2m2 + m2c2 + 2 E2mm - P2m2 = 5 m2c2  + 2 E2mm = 25 m2c2,
since E2m = 2 mc2 + 8 mc2 = 10 m2c2.

CM, after:
E2/c2 = 25 m2c2,  E = 5 mc2,  Mc2 = (5 - 3) mc2 = 2 mc2.

2 m is the maximum possible value for M.

#### Problem:

In a proton-proton collision a π+ meson can be created through the reaction

p1 + p2 --> p + n + π+.

In the center of mass (CM) frame of reference each proton has an initial energy γmpc2, where mp is the mass of the proton and γ = (1 - β2)-1/2, with β = v/c.
Take the mass of the proton and the mass of the neutron to be 1837 me (me = electron mass) and the mass of the pion to be 273 me = 0.1486 mp.
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:
2γmpc2 = 2.1486 mpc2.
γ = 1.0743.
v = 0.365 c = 1.1*108 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 π0 = 134.98 MeV) that can be produced in one proton-electron 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?
[E2/c2 - P2]lab before coll. = [E2/c2]CM before coll. = [E2/c2]CM after coll.
invariance of norm of (P0,P)          energy conservation
lab, before:
(Ep + Ee)2/c2 – (Pp – Pe)2 = (Ep2 + Ee2 + 2EpEe)/c2 – Pp2 – Pe2 + 2PePp
= 85280.85 GeV2 /c2,
since
(Ep + Ee)2/c2 = 715716 GeV2 /c2
Ep2/c2 - mp2c2 = Pp2,  Pp2 = (8202 - 0.9382) GeV2/c2 = 672399.1 GeV2/c2,
Ee2/c2 – me2c2 = Pe2,  Pe2 = (262 – (5.11*10-4)2) GeV2/c2 = 676 GeV2/c2.
ECM =  292 GeV
# of pions = 292/0.13498 = 2163
(b)  Assume we want to create the same number of pions.  Then we want
(Ep + Ee)2/c2 – Pe2 = (Ep2+ Ee2 + 2EpEe)/c2 – Pe2 = (mp2c4 + 2mpc2Ee)/c2 + me2c
= 85280.85 GeV2 /c2.
Ee = (85280.85 GeV2 - mp2c4 - me2c4)/(2mpc2) = 45458 GeV

#### Problem:

A particle of rest mass 1 MeV/c2 and kinetic energy 2 MeV collides with a stationary particle of rest mass 2 MeV/c2.  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 mc2 = 1 MeV.
Em = mc2 + 2mc2 = 3mc2
The energy of the incoming particle is 3 MeV.
γ = 3,  1 - v2/c2 = 1/9,  v2/c2 = 8/9,  v = 0.9428c.
The velocity of the incoming particle is 0.9428c.
E2 = m2c4 + p2c2,  9m2c4 = m2c4 + p2c2, p2 = 8m2c2,  p = 2.828 MeV/c.
The momentum of the incoming particle is 2.828 MeV/c.
(b)  [E2/c2 - p2]lab before coll. = [E2/c2]CM before coll. = [E2/c2]CM after coll.
invariance of norm of (P0,P)          energy conservation

lab, before:
(Em + E2m)2/c2 - Pm2 = (E2m2+ Em2 + 2E2mEm)/c2 - Pm2
= 4m2c2 + Pm2 + m2c2 + 4Emm - Pm2 = 5m2c2  + 4Emm = 17m2c2
since Em = mc2 + 2mc2 = 3mc2

CM, after:
E2/c2 = 17m2c2,  E = 17½mc2 = 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/c2.

(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 pm of the 4-vector (p0,p1,p2,p3) we have
particles_in pμ = ∑particles_out pμ, or ∑i (pi)μ = ∑j (pj)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
(P0,P)∙(P0,P) = (P0',P')∙(P0',P').
Here  P0 = ∑particles p0  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
P02 - P2 = P02  = (∑p0)2 = (4mc)2 = 16m2c2.
The square of the magnitude of the momentum 4-vector is 16m2c2 before and after the collision in any frame.
Before the collision we have in the lab frame.
P02 - P2 = (p0a + p0b)2 - (pa + pb)2 = p0a2 + p0b2 - pa2 + 2p0ap0b.
For any free particle we have p02- p2 = m2c2.  We therefore have:
P02 - P2 = 16m2c2 = 2m2c2 + 2p0ap0b,  p0ap0b = 7m2c2.  p0amc = 7m2c2.
Ea/c = 7mc.  mc2 = 1 GeV
The "threshold" energy, for the incident proton is 7mc2.
(b) The total energy before and after the collision is 8 mc2.
The momentum before and after the collision is  p2c2 = Eproton2 – m2c4 = 48 m2c4.
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 - 0.3682 = 0.6.

#### Problem:

The PEP-II collider at Stanford Linear Accelerator Center creates electron-positron head-on 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+2c2 = E+2 - me2c4 = (3.1 GeV)2 - (0.511 MeV)2 = 9.61 GeV2, p+c = -3.1 GeV.
p-2c2 = E-2 - me2c4 = (9 GeV)2 - (0.511 MeV)2 = 81 GeV2, p-c = 9 GeV.
E = 12.1 GeV, pc = 5.9 GeV
m2c4 = E2 - p2c2,  mc2 = 10.56 GeV.
(b)   E = γmc2.  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 x-components of the particles are equal and the y-components have equal magnitude but opposite sign.
Momentum conservation:       2 γmv cos(45o) = γ'm'v'
Energy conservation:              2 γmc2 = γ'm'c2.
Dividing we obtain  v cos(45o) = v'.
m'/(2m) = γ/ γ' = [(1 – v2/(2c2))/(1 – v2/c2)]½ = [(1 – 0.32)/(1 – 0.64)]½ = 1.37

#### Problem:

The Japanese B-factory 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/c2.  Subsequently, the ϒ (4S) meson decays into a pair of B-mesons: B+ and B-. The rest masses of the oppositely charged B+ and B- mesons each are 5.28 GeV/c2

(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 the-momenta 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+ = γMc2
momentum conservation: pc =  E- - E+ = γMvc.
pc/E = β = v/c = (E- - E+)/( E- + E+).
1 - β2 = 4E+E-/(E- + E+)2.  γ2 = (E- + E+)2/(4E+E-).
γ2M2c4 = (E- + E+)2.  M2c4/(4E+E-) = 1,  E+ = M2c4/(4E-).
E+ = 3.5 GeV  .
(b)  For each B meson p2c2 = E2 – m2c4.  p2c2 = (5.29 GeV)2 – (5.28 GeV )2.
p = 0.325 GeV/c.
(c)  β = v/c = (EB- - EB+)/( EB- + EB+) = 0.39 gives the speed of the ϒ (4S) in the lab.
γ = (1 – β2) = 1.087.
Lorentz transformation of 4-vector momentum:
pB+ = γβ *5.28 GeV/c + γ*0.325 GeV/c = 2.59 GeV/c.
pB- = γβ *5.28 GeV/c -  γ*0.325 GeV/c = 1.81 GeV/c.