Angle pd decay between two muons between Z boson?

[philip041]

Question leads on from others where s^1/2 = 200GeV mH = 70GeV EH = 91.5GeV and p = 59GeV, (three momentum).

The reaction was e+e- -> HZ0

where H is higgs boson and Z0 is the Z0 boson with mass 91GeV

It says the Z0 decays to two muon and antimuon. They are observed to be equal energy in the lab frame, what is the angle between them in the detector? They are massless.

It says use the the three-momentum already worked out. I haven't a clue how to describe the particles after the decay, I didn't think you can predict where they go, isn't it random?

 

[Meir Achuz]

The energy of each mu is 92/2, which equals each mu momentum.
The total momentum is 59.
cos(theta/2)=45/59, not quite random.

 

[philip041]

The energy of each mu is 92/2, which equals each mu momentum.
The total momentum is 59.
cos(theta/2)=45/59, not quite random.

Is the total momentum because the Higgs and Z boson have equal momentum? If the muons decay off the Z why is their energy equals to the energy of the Higgs?

 

[Meir Achuz]

e+e are colliding beams, so the Z and H do have the same momentum.
I misread the problem, and made a silly mistake. You need the energy of the Z, which is
200-92, and not the 92 I wrote.