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Relativistic Particle Decay of Higgs Boson

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  1. Sep 12, 2014 #1
    1. The problem statement, all variables and given/known data

    A Higgs boson has mass 125 GeV/c2, decaying into a pair of Z bosons, mass 91 GeV/c2. In the lab frame, one of the Z bosons is at rest. Determine the kinetic energy for the other Z boson in this laboratory frame.

    2. Relevant equations

    E = γmc2
    γ=1/√(1-β2)

    3. The attempt at a solution

    My basic issue surrounding the problem is the derived value for gamma.
    First, treating the problem in the center mass frame, I stated:

    Mhc2=2γmzc2 thus γ=125Gev/(2*91GeV)= 0.6868

    this is where my dilemma lies. Seeing that the Z boson is at rest in the lab frame I assumed γ= 0.6868; however as Kh= (γ-1)E

    (0.6868-1)(125 GeV)= -39.14 GeV

    What confuses me is that the Kinetic Energy would be negative, thus I assumed my approach to the problem was erroneous. Thus I alternately attempted to derive velocity from gamma in order to perform a lorentz transformation, but then as

    γ= 0.6868
    v=c√(1-(1/γ2)

    there would be a negative within the square root.
     
  2. jcsd
  3. Sep 12, 2014 #2

    mfb

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    If this is the full, exact problem statement, then the problem is very misleading - a Higgs boson with a mass of 125 GeV does not have enough energy to produce two Z bosons with a mass of 91 GeV each, just by energy conservation.

    The decay actually happens in nature, but at least one of the Z bosons has to be virtual for that (appear like a less massive Z boson) - and the problem statement doesn't seem to be appropriate to consider this case.
     
  4. Sep 12, 2014 #3
    mfb,

    Thanks for the feedback. That is what I thought too when reading over this problem myself. The problem is identical to the question posed by my professor, so I wanted to confirm whether it was a personal miscalculation or a general issue with the problem itself.
     
  5. Sep 13, 2014 #4

    mfb

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    It does not matter if it is at rest or not. The availability of decay channels is independent of the reference frame you use to describe it.
    There is always a frame where the Higgs is at rest, where it is easier to see that the decay is not possible (... in the way required here).
     
  6. Sep 13, 2014 #5

    nrqed

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    You are right, of course. I stand corrected. I should not post at 1 in the morning :-)
     
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