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Muon production and decay from relativistic annihilation
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[QUOTE="caitphys, post: 6028412, member: 648586"] Thanks for the response! Okay, so my thinking for ##(E_{\bar p}+m_ p)^2 - P_{\bar p}^2 = 4E^2## was that using ##A^2=(\sum\nolimits E)^2-(\sum\nolimits P)^2##, Choosing my frame as that of the proton, since this seemed to minimise the impact of my confusion associated with its trajectory, for ##A^2## the first term is the energy of the antiproton and the rest mass of the proton, and the second just the momentum of the antiproton, given that in its rest frame the momentum of the proton shall be zero. Then, since ##A^2## is invariant, working in the ZMF frame after collision, as the muons have equal energies, the value of ##A^2## is merely the total energy squared, as total momentum will be zero, so ##(2E)^2##. As ##P=\sqrt {E^2-m^2}=34987.4MeV##, conservation of momentum would suggest the two muons have momentums of half this, ##17493,71##, although I am ignoring any momentum the proton may have. From this I could calculate the velocity of the muons, as by conservation of energy, they have half of the antiproton's energy each, ##17.5MeV##, and so using ##v=\frac {pc^2} {E}##, ##v=0.99964c## Finding γ and then using this to Lorentz contracting the distance (to 536.6m), I get a new survival probability of 0.44311 I've tried to use conservation of momentum here, but I'm not sure if it is fully realized, certainly the accompanying script for the problem stated that, "the main problem was that the students, almost uniformly, did not notice that the two muons would have to be going at an angle to the normal and that the flight path would therefore be longer". [/QUOTE]
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Muon production and decay from relativistic annihilation
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