# I Instantaneous Lumi and Integrated Lumi

1. Jul 28, 2016

### 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.

2. Jul 28, 2016

### Staff: Mentor

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?

3. Jul 28, 2016

### 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.

4. Jul 28, 2016

### Staff: Mentor

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

5. Jul 28, 2016