22 September Relativity Lecture 6
 Conservation of 4momentum

Total 4momentum before collision equals 4momentum after collision

Pb1μ+Pb2μ+Pb3μ+...+Pbnμ = Pa1μ+Pa2μ+Pamμ
where Pb is a "before collision" and Pa an "after collision" momentum 4vector asociated with a particle, photon, neutrino, ball, etc.

Note that the number of 4momenta before and after may differ and recall the 4momentum contains both energy (rest mass energy and kinetic energy) and momentum. The conservation of P0 implies that rest mass and kinetic energy need not be conserved separately but may be converted one into the other. This is the conservation of mass-energy.

Threshold energies

Suppose a particle of mass m and kinetic energy T (m=0 and E=hν for photons) collides with a particle of mass M at rest to produce N particles of mass mi. What is the minimum enrgy T?

If we work in the center of momentum frame the total momentum is zero and we can produce all N particles at rest. We have

Pmμ + PMμ = ∑Piμ     i=1,N
Thus
P²m + P²M + 2Pμm gμν PνM = ∑Piμ gμν Pjν     i, j=1 to N
remember to use the "trick", evaluate the P² terms in their rest frames, the third term in the lab (rest for M before collision) frame and the RHS sum in the center of momentum frame where, remember, the velocities are all zero since we are after the threshold T. Then (Pμm PμM = (E/c,pi)*(Mc,0) = EM)
m²c² + M²c² +2EM = ∑mimjc²      (the sum is i, j = 1 to N)
Therefore
T = E-mc² = [∑mimjc² - m²c² - M²c²]/2M - mc²
   = [∑mimjc² - (m+M)²c²]/2M.

Example

The Berkley Bevatron was designed to have sufficient energy to produce antiprotons p- by bombarding hydrogen gas with accelerated protons p, vis:

p + p → p + p + (p + p-).
What minimum energy is required?
If M is the proton (and antiproton) mass
T = [16M²c² - 4M²c²]/2M = 6Mc² (almost 6 GeV).