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I Instantaneous Lumi and Integrated Lumi

  1. Jul 28, 2016 #1

    ChrisVer

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    I have a pretty basic question...
    How can you convert the Instantaneous Luminosity [itex]L_t[/itex] to the integrated Luminosity [itex]L[/itex]?
    I know that the relation is the following:
    [itex]L = \int L_t dt[/itex]
    but if the time is [itex]\sim 25~ns[/itex] and [itex]L_t = 1.7 \times 10^{34} ~cm^{-2} s^{-1}[/itex], then I get an integrated luminosity of:
    [itex]L=42.5 \times 10^{25} ~cm^{-2} = 4.25 \times 10^{-13} fb^{-1}[/itex]
    which doesn't make sense as a number... dividing with the time gives a more sensible result but right now I don't see why.
     
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  3. Jul 28, 2016 #2

    mfb

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    You calculated the integrated luminosity from a single LHC bunch-crossing.
    If you want the integrated luminosity collected within a day, you have to take a day as time.

    Dividing luminosity by time gives wrong units, how can that be more sensible?
     
  4. Jul 28, 2016 #3

    ChrisVer

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    So in general it's impossible without knowing the time-span of your data-collection to translate your integrated luminosity to the instantaneous one? (the reverse of the above)

    As for the units, true... I got to use sleepnessless as an excuse.
     
  5. Jul 28, 2016 #4

    mfb

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    Can you tell how fast I went in my car, traveling 30 km, if I don't tell you how long I was traveling?
     
  6. Jul 28, 2016 #5

    Vanadium 50

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    Officer, I can't have been going 80 miles per hour! I've only been driving ten minutes!
     
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