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Re: the energy



Ken Caviness wrote:
would you be
willing to show us your steps? I can't get it to come out simpler than the
above, in which the algebra is easy but the 4-vector-ness of the approach is
not obvious.

OK, you've put in enough work to earn it.
The statement of the problem can be found at
http://lists.nau.edu/cgi-bin/wa?A2=ind0410&L=phys-l&F=&S=&P=24553

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.)