I Instantaneous Lumi and Integrated Lumi

[ChrisVer]

I have a pretty basic question...
How can you convert the Instantaneous Luminosity L_t to the integrated Luminosity L?
I know that the relation is the following:
L = \int L_t dt
but if the time is \sim 25~ns and L_t = 1.7 \times 10^{34} ~cm^{-2} s^{-1}, then I get an integrated luminosity of:
L=42.5 \times 10^{25} ~cm^{-2} = 4.25 \times 10^{-13} fb^{-1}
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.

 

[mfb]

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?

 

[ChrisVer]

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.

 

[mfb]

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)

Can you tell how fast I went in my car, traveling 30 km, if I don't tell you how long I was traveling?

 

[Vanadium 50]

Can you tell how fast I went in my car, traveling 30 km, if I don't tell you how long I was traveling?

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