Re: the energy

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding John's solution below:
> ...
> Here goes:
>
>  p = [E,ps] = energy-momentum four-vector for particle #1 in lab
>               frame.
>
>  [1,0] = energy-momentum four-vector for particle #2 in lab frame.
>
>  [4,0] = energy-momentum four-vector for RHS in CM frame.
>
>  E^2 - ps^2 = 1    (invariant mass of particle #1)    [eq 4]
>
>  (E+1)^2 - ps^2 = mass of LHS = mass of RHS = 16
>
>  E^2 + 2E + 1 - ps^2 = 16     (by expanding the square)
>
>  2E + 1 = 15                  (by subtracting eq 4)
>
>  E = 7
>
>  KE = 6
>
> Note:  no gammas involved.  No Lorentz transformations.
> Note:  repeated, total reliance on the four-vector approach:
>  -- We add two four-vectors in the lab frame;
>  -- We add four four-vectors in the CM frame;
>  -- The norm of a four-vector is frame-independent.
>     (We use this idea at least three times.)

John, 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.

In any event, the problem has a high cuteness coefficient anyway.

David Bowman