Particle Decay III
Consider a particle M at rest that decays into two particles, m1=m/3 with b=4/5 g=5/3, and m2=2m1=2m/3 with b,g to be determined.
First we apply conservation of momentum to evaluate the speed of m2:

m1g1b1c + m2g2b2c = 0

(m/3)(5/3)(4/5)c + (2m/3)gbc = 0

(gb)² = b²/(1-b²) = (-2/3)²

b = -2/13½ = -0.5547,    g = 13½/3 = 1.20185

How fast does particle 2 fly away from particle 1's viewpoint? Use the addition law of velocities to get

b' = (0.5547+0.8)/(1+0.5547*0.8) = 0.938314,    g' = 2.891973

What is the penalty for looking at the decay in the (m/3)'s b = 0.8 frame?

The momentum before decay is Mgv = M(5/3)(4/5)c = 1.33333Mc

After the decay the momentum is (2m/3)g'b'c = (2/3)*2.891973*0.938314*mc = 1.809052mc

which implies m = (1.333/1.809)M = 0.737034M.

Ha! The "lost mass" is the "binding energy", the mass equivalent of the kinetic energies of the particles.

Conservation of energy states

"before" = Mc² = "after" = m1g1c² + m2g2c²

and we have M = (2/3)m*1.20185 + (1/3)m*1.6667

so m = 0.737034M like we implied above.