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 4-vector (p₀, p₁, p₂, p₃) 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
  (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 = (p_0a + p_0b)² - (p_a + p_b)² = p_0a² + p_0b² - p_a² + 2p_0ap_0b.
  For any free particle we have p₀²- p² = m²c².  We therefore have:
  P₀² - P² = 16m²c² = 2m²c² + 2p_0ap_0b,  p_0ap_0b = 7 m²c².  p_0amc = 7 m²c².
  E_a/c = 7 mc.  mc² = 938 MeV,  E_a = 6.6 GeV.
  The proton must have E_a - 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: 
  (E_2m + E_m)²/c² - P_2m² = (E_2m²+ E_m² + 2 E_2mE_m)/c² - P_2m² 
  = 4 m²c² + P_2m² + m²c² + 2 E_2mm - P_2m² = 5 m²c²  + 2 E_2mm = 25 m²c²,
  since E_2m = 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 γm_pc², where m_p 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 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_pc² = 2.1486 m_pc².
  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: 
  (E_p + E_e)²/c² - (P_p - P_e)² = (E_p² + E_e² + 2E_pE_e)/c² - P_p² - P_e² + 2P_eP_p 
  = 85280.85 GeV² /c²_,
  since
  (E_p + E_e)²/c² = 715716 GeV² /c²
  E_p²/c² - m_p²c² = P_p²,  P_p² = (820² - 0.938²) GeV²/c² = 672399.1 GeV²/c²,
  E_e²/c² - m_e²c² = P_e²,  P_e² = (26² - (5.11*10^-4)²) GeV²/c² = 676 GeV²/c².
  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/c² - P_e² = (E_p²+ E_e² + 2E_pE_e)/c² - P_e² = (m_p²c⁴ + 2m_pc²E_e)/c² + m_e²c
  = 85280.85 GeV² /c².
  E_e = (85280.85 GeV² - m_p²c⁴ - m_e²c⁴)/(2m_pc²) = 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.
  E_m = 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: 
  (E_m + E_2m)²/c² - P_m² = (E_2m²+ E_m² + 2E_2mE_m)/c² - P_m² 
  = 4m²c² + P_m² + m²c² + 4E_mm - P_m² = 5m²c²  + 4E_mm = 17m²c²
  since E_m = mc² + 2mc² = 3mc²

  CM, after:
  E²/c² = 17m²c²,  E = 17^½mc² = 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 p_m of the 4-vector (p₀,p₁,p₂,p₃) 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
  (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 = (p_0a + p_0b)² - (p_a + p_b)² = p_0a² + p_0b² - p_a² + 2p_0ap_0b.
  For any free particle we have p₀²- p² = m²c².  We therefore have:
  P₀² - P² = 16m²c² = 2m²c² + 2p_0ap_0b,  p_0ap_0b = 7m²c².  p_0amc = 7m²c².
  E_a/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 = E_proton² - 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 - m_e²c⁴ = (3.1 GeV)² - (0.511 MeV)² = 9.61 GeV², p₊c = -3.1 GeV.
  p_-²c² = E_-² - m_e²c⁴ = (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(45^o) = γ'm'v'
  Energy conservation:              2 γmc² = γ'm'c².
  Dividing we obtain  v cos(45^o) = 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 = (E_B- - E_B+)/( E_B- + E_B+) = 0.39 gives the speed of the ϒ (4S) in the lab.
  γ = (1 - β²)^-½ = 1.087.
  Lorentz transformation of 4-vector 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.