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I believe you left out the effects of a very important
fact here. Protons are *not* elementary particles.
When a proton interacts deeply with another proton asymptotic
freedom becomes relevant. In the collision what may appear to us
to be a proton interacting with another proton is in fact a single
quark from one proton interacting with a single quark of the other
proton. The remaining 4 quarks are oblivious bystanders to what
is going on with the 2 colliding/interacting quarks. This means
that all the KE in these by-standing quarks is essentially
*unavailable* for creating particle-antiparticle pairs.
OK, I'll bite. Why then does the proton-smacking-proton
calculation give the empirically-correct answer?!
AM-history01-b.htmlI'm told this calculation was used in the early 1950s as the
basis for the design of the Bevatron ... which worked!
http://livefromcern.web.cern.ch/livefromcern/antimatter/history/
This was a decade or more before quarks were conceived of.
In any event, the problem has a high cuteness coefficient anyway.
Oh yeah.
structure in those days. (I arrived at Berkeley in the fall of
'56).
The kinematics of the Bevatron had nothing to do with quarks,
gluons or QCD. The kinematics is identical to that for car-car
collision where one of the cars was parked before the collision. A
lot of the energy in the final state just comes from center-of-mass
motion.
To see this (this is where John's gamma's can be helpful),
first do the calculation for a collider, where the net spatial
momentum is zero before and after the collision. The final energy
(3 protons and a p-bar, all at rest in the center of momentum
system) must equal the the original energy. The square of the
total energy in the center-of-momentum system is an invariant,
S=(E^{2}-p^{{2})=16, since p=0. Now calculate the same quantity in
the lab system where one of the two initial protons is at rest, to
get the canonical result.