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Special H boson with very high cross section?

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    (a) Explain the results.
    (b) Why is the cross section much higher? Suggest the dominant decay product pair.


    2013_B4_Q2.png

    2. Relevant equations


    3. The attempt at a solution

    Part(a)
    For a centre of mass energy of ##170GeV##, you produce pairs of oppositely charged ##W^{\pm},W^{\mp}## bosons. It is kinematically forbidden to produce ##Z,Z## pair and forbidden by charge to produce ##Z,W^{\pm}## pairs.
    First group of 4 hadronic jets can be explained by production of 2 pairs of quark/anti-quark stream. Number of combinations = 3x4 = 12 because of 3 colours possible. ~45%
    2013_B4_Q2_2.png
    Second group can be explained by a quark/anti-quark stream and a lepton/neutrino stream. Number of possibilities = 3x2 + 2x3 = 12. (3 types of leptons, 2 charge types) ~45%
    2013_B4_Q2_3.png
    Third group can be explained by 2 pairs of lepton/neutrino stream. Number of possibilities = 3 (leptons must be oppositely charged) ~10%
    2013_B4_Q2_4.png

    Part (b)

    This question stumbled me. Is there something special happening at 125 GeV? I suppose the heaviest quark-anti-quark pair it can decay to is the bottom/anti-bottom quark? Why is the cross section so much higher? Why is it that in part (a) only {u,d,c,s} quarks/anti-quarks are produced and not bottom (b) quarks??
     
  2. jcsd
  3. Apr 20, 2015 #2

    mfb

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    The Z has a large width - you will get ZZ events with one Z a bit off-shell, but I guess you can ignore that here.
    Why 4? And what about the other W?
    I don't understand what you are adding here.
    What about the other W?

    The LHC found a particle at this energy.
    Right.
    What does the Higgs couple to?
    That is a mistake you have to fix in (a).
     
  4. Apr 20, 2015 #3
    Ok, now thinking about it - each W boson can produce 5 possible types of quarks (u,d,s,c,b) so possibilities are 5x5 = 25? But 3 types of colour so 25x3 = 75.

    For the second picture, 5 quark possibilities, 3 types of colour, 3 types of leptons, 2 charges of leptons so 5x3x3x2 = 90? Doesn't match the first one.

    For the third picture, 3 types of leptons but the W bosons must be oppositely charged, so 3 types for the other W boson. So 3x3 = 9

    The numbers are all wrong...

    Right, should have noticed they were talking about the Higgs! The Higgs boson couples to mass, so it naturally has an extremely large cross section with the heaviest possible quark, bottom quark. Why is it 40,000 times as large though?
     
  5. Apr 21, 2015 #4

    mfb

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    Sorry, I was thinking of Z decays with my comment about five quarks. No b in W decays as the top is too heavy.

    You don't have so many options for the W. Let's consider the W+ boson: it either decays to up+antidown or charm+antistrange. Just two options, multiplied by three for colors => 6.
    The W- has 6 options as well.

    That is the dominant decay mode for the 125 GeV Higgs, right.
    What is the electron to muon mass ratio? Do you see how those numbers could be related?
     
  6. Apr 21, 2015 #5
    Why can't the W+ boson decay to charm/antidown? It is calibbo suppressed by ##sin 13^o ## but it is possible? Also why can't W+ boson decay to ##u\bar u, d\bar d, c\bar c## pairs?

    First picture
    6 options for each W.
    Total = 6x6=36 options

    Second picture:
    Quark producing W can either be W+ or w- = 2x6 = 12 options
    For the lepton decay, 3 options.
    Total = 12x3=36 options

    Third Picture
    3 options for each.
    Total = 3x3 = 9 options


    Mass of muon = 200 times Mass of electrons

    I suppose ##\sigma \propto m^2##?
     
  7. Apr 22, 2015 #6

    mfb

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    It can, but the probability is small. Small enough to neglect it.
    Quark + antiquark of the same type would violate charge conservation.

    The new numbers look right.
    Right.
     
  8. Apr 22, 2015 #7
    Why is ##\sigma \propto m^2##? I know that ##\Gamma \propto m^5##. Cross section is also (Rate of reactions/Incident Flux) ##\sigma = \frac{\Gamma}{J}##.
     
  9. Apr 22, 2015 #8

    mfb

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    You have two massive particles in the interaction.
    For a better description, see the Lagrangian and QFT.
    You are mixing unrelated concepts here.
     
  10. Apr 22, 2015 #9
    I thought lifetime ##\tau \propto \frac{1}{\Gamma}## and rate of reactions is decay width ##\Gamma##?
     
  11. Apr 22, 2015 #10

    mfb

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    The decay width is not the reaction rate. They are related if you take the partial width for the right process (here: electron/positron or muon/antimuon).
    Also, the total width is not just the mass (the Higgs mass! Not the decay products) to the fifth power, it is a more complicated function as you have to take multiple processes with different thresholds into account.
     
  12. Apr 25, 2015 #11
    True. Upon doing some reading I realized that there is indeed a relation between cross section and rates, using Fermi's golden rule:

    [tex]\frac{d\sigma}{d\Omega} = \frac{\Gamma}{\nu} = \frac{1}{\nu}\frac{2\pi}{\hbar} |M_{fi}|^2\frac{dN}{dE}[/tex]

    where ##\nu## is incoming flux.

    And for EM interactions like ##e^+e^-\rightarrow \mu^+\mu^-## or ##e^+e^-\rightarrow \tau^+\tau^-##, it is ##\sigma = \frac{4 \pi}{3}\left(\frac{\alpha \hbar c}{W} \right)^2## where ##\alpha = \frac{g_{EM}^2}{4\pi}## and ##W## is the centre of mass energy.
     
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