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I Cross sections

  1. Nov 8, 2016 #1
    If given the cross section for ZZ* to two leptons, is it possible to work out the cross section for ZZ* to just electrons or muons from it? So excluding the tau decay mode, unless ##ZZ^{*} \rightarrow \tau + \tau## can't happen anyway? I'm guessing it wouldn't just be two thirds of the ##ZZ \rightarrow ll## cross section, because of the different masses and probably various other factors. Do you just have to go and experimentally measure the cross section?

    Apologies for the poor phrasing. I'm having trouble expressing what my question actually is, but mainly I don't understand how cross sections work. Especially if the cross section for Z to electron or muon actually is two thirds the cross section for Z to leptons. And actually ##\frac{2}{3}## wouldn't include the neutrinos so maybe I mean ##\frac{2}{6}##.
     
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  3. Nov 8, 2016 #2

    Vanadium 50

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    Those aren't cross-sections. Perhaps you mean branching fractions?
     
  4. Nov 8, 2016 #3

    mfb

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    You mean branching fractions, and I guess the ZZ* is in a Higgs decay. You can calculate the cross-section for "pp collisions produce Higgs which then decays to X", but if you just want to compare how frequent different decay modes are, it is easier to just consider branchning fractions.

    Each Z in the decay decays to two leptons (or other stuff), so in total you get 4 leptons. Lepton universality leads to (nearly) identical branching fractions for electrons, muons and taus. "Nearly" because the masses are not identical, but all three leptons are very light compared to both Higgs and Z, so this effect does not matter.

    A comment on notation: "Leptons" in experimental particle physics usually means "charged leptons", and informally often just electron and muon (taus are messy to measure).
     
  5. Nov 17, 2016 #4

    ChrisVer

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    Why would [itex]ZZ \rightarrow ll[/itex] happen but not [itex]ZZ \rightarrow \tau \tau[/itex]?

    not more messy than neutrinos (Missing ET)
     
  6. Nov 17, 2016 #5
    Why did I think the tau mode might not occur? I misunderstood the meaning of branching fractions, and thought ##Z \rightarrow ll## was the probability of ##Z \rightarrow \mu^+ \mu^- + Z \rightarrow e^+ e^- + Z \rightarrow \tau^+ \tau^-##. So I thought I had to find a way to extract from that just the probability of ## Z\rightarrow \mu^+ \mu^- + Z \rightarrow e^+ e^-## that I needed, and the only way I thought that was doable was if the ##\tau## decay wasn't part of that branching fraction for some reason. A misunderstanding.
     
  7. Nov 17, 2016 #6

    ChrisVer

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    well the probabilities are the same for all [itex](Z \rightarrow e e) = (Z \rightarrow \mu\mu) = (Z\rightarrow \tau\tau)[/itex] by lepton universality (as mentioned by mfb)... Now it all goes down to what you mean by [itex]Z \rightarrow ll[/itex] because there are three ways to read it:
    1. take just an instance and for "saving" typing you mean any of the three ( [itex]Z \rightarrow ee/\mu\mu/\tau\tau[/itex] ), so in that case the above equality would be written [itex](Z \rightarrow e e) = (Z \rightarrow \mu\mu) = (Z\rightarrow \tau\tau) = Z\rightarrow ll[/itex]
    2. take [itex]l[/itex] to denote "light leptons" and so you excluded the taus from the very beginning, in that case since the two probabilities (to 2 e or to 2 muons) are complementary you just need to add them up:
    [itex]Z \rightarrow ll = (Z \rightarrow ee )+ (Z \rightarrow \mu\mu) = 2 ( Z \rightarrow ee) = 2 (Z \rightarrow \mu \mu )[/itex]
    3. you meant all 3 leptons, and in the same way:
    [itex]Z \rightarrow ll = (Z \rightarrow ee )+ (Z \rightarrow \mu\mu) + (Z \rightarrow \tau\tau)= 3 ( Z \rightarrow ee) = 3 (Z \rightarrow \mu \mu ) + 3 (Z \rightarrow \tau\tau)[/itex]

    generally if you want you can also consider neutrinos, but we generally avoid writing neutrinos with [itex]l[/itex] (although they are in the lepton families).
     
  8. Nov 17, 2016 #7

    mfb

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    Two neutrinos are messy as well. Taus have their lifetime to make things even more messy (vertex association).
     
  9. Nov 17, 2016 #8

    ChrisVer

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    For 2 neutrinos I agree... in fact I am not sure how a leptonically decaying tau from a W could be distinguished from an electron or muon... For ditaus that's possible since Z (or H) other leptonic decays don't give you neutrinos.

    Well tau displacement is actually used for their reconstruction (for the tau decays to 3 charged mesons), so I think we trust our vertex association. In fact the efficiency of tau vertex association is almost independent to the pile-up conditions (https://atlas.web.cern.ch/Atlas/GROUPS/PHYSICS/PUBNOTES/ATL-PHYS-PUB-2015-045/ Fig 1b) .
     
  10. Nov 17, 2016 #9

    mfb

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    17% tau -> mu, with a reconstruction and identification efficiency worse than for muons, and with more background from pileup
    17% tau -> e, with a reconstruction and identification efficiency worse than for electrons, and with more background from pileup
    15% tau -> 3 charged hadrons, with a reconstruction efficiency that is even worse than for the leptonic decays
    The remaining decays are typically one charged hadron, for the worst efficiency.
    And Z->tau tau means you have to identify both taus.

    Yes it is possible, but the efficiency is worse, you have more background and you quickly get 10+ subchannels which all need their own selection, background estimates, efficiency evaluations and so on: very messy.
     
  11. Nov 17, 2016 #10

    ChrisVer

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    I am not trying to say that taus are as good to be reconstructed as the basic leptons... the higher level objects in the sequence of reconstruction are always worse than the basic ones because they inherit from them... as Missing Transverse Momentum (MET) is bad because you rely a lot on measuring the hard objects of the events (e, muons, photons, taus, jets)...
    hmm... depends on the tau decaying channel...
    and yes it's not so easy to work with all the channels. There are different groups that work on each channel which shows the complexity...
     
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