Question 6.2: Antiprotons can be produced in a proton-proton collision acc......
Antiprotons can be produced in a proton-proton collision according to the reaction
p + p → p + p + p + \overline{p} , (6.124)
where p denotes a proton and \overline{p} an antiproton. If the target proton is at rest in the laboratory frame, what is the threshold energy (minimum energy) of the incoming proton in order that the reaction be possible?
According to (6.110) a free particle has the lowest energy possible when at rest. Therefore, the minimum energy of the incoming proton is the one that allows the formation of the reaction products at rest. However, the reaction products cannot be formed at rest in the laboratory frame because this would violate the conservation of momentum. As a consequence, the reaction products can only be formed at rest in the centre-of-mass frame, for it is in this frame that the total momentum is zero. Denoting by P^{(F)} the total four-momentum of the reaction products, its components in the centre of- mass frame are
E=\frac{mc^{2}}{\sqrt{{1-v^{2}}/{c^{2}}} } (6.110)
P^{(F)}= (4mc, 0, 0, 0) (in the centre-of-mass frame) (6.125)
since each of the reaction products has four-momentum with components (mc, 0, 0, 0), where m is mass of the proton (equal to that of the antiproton). The conservation of four-momentum is written
P + Q =P^{(F)}, (6.126)
where P and Q are, respectively, the four-momenta of the incoming proton and of the target proton. Squaring (6.126) there results
P² + 2P · Q + Q² =P^{(F)^{2}} \Longrightarrow m²c² + 2P · Q + m²c² =P^{(F)^{2}}, (6.127)
where we have used (6.94). The other scalar quantities involved in this last equation may be calculated in the inertial frame that is found the most convenient. Thus, calculating the right-hand side of (6.127) in the centre-of-mass frame with the help of (6.125), we get
P^{\mu }P_{\mu } = m²c². (6.94)
P^{(F)^{2}}= 16m²c², (6.128)
where (6.35) has been used. Finally, in the laboratory frame we have
V² ≡ V · V = (V^{0})² − (V^{1})² − (V²)² − (V³)² . (6.35)
P = \left(\frac{E}{c}, p\right), Q = (mc, 0, 0, 0) (in the laboratory frame) , (6.129)
whence, by (6.34),
V ·W = V_{\mu}W^{\mu} = V_{0}W^{0} + V_{1}W^{1} + V_{2}W^{2} + V_{3}W^{3} , (6.34)
P · Q = \frac{E}{c} mc = mE . (6.130)
Substituting (6.128) and (6.130) into (6.127) we immediately obtain
E = 7mc² (6.131)
as the threshold energy for the reaction (6.124). In order to have this energy, the incoming proton must have a speed of 0.99c.