# Relativistic Particle Decay of Higgs Boson

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1. Sep 12, 2014

### PerfecrtAzian

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. Sep 12, 2014

### Staff: Mentor

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.

3. Sep 12, 2014

### PerfecrtAzian

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.

4. Sep 13, 2014

### Staff: Mentor

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).

5. Sep 13, 2014

### nrqed

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