RE: 1/N*dN/d(pt) and 1/sigma* dsigma/d(pt)

Hello Wajid,

Assuming luminosity is independent of pt (which seems likely!) then it would seem that the two expressions are the same and the integral is 1 (if you integrate over the full range of pt).

If you want plot it you need to chose what to use for bin contents; a reasonable choice (but may be not what you want) would be the fraction in each bin, i.e.,

Yi = 1/N Integral_i dN/d(pt)

where Integral_i mean in integration over the pt boundaries of the i'th bin.

-Art S.

-----Original Message-----

From: owner-roottalk_at_lxbuild091.cern.ch on behalf of Wajid Ali Khan Sent: Sun 1/3/2010 9:32 PM
To: roottalk_at_lxbuild091.cern.ch
Subject: [ROOT] 1/N*dN/d(pt) and 1/sigma* dsigma/d(pt)  

Dear Root Experts,

I need your help. Any body can explain me what is the main difference between the following

1/N*dN/d(pt) and 1/sigma* dsigma/d(pt)

Is the same N=luminosity*Sigma.
If I want to normalize my histogram per entry per unite of Xaxis with say pt. What would be the quantity plotted on Y axis will it be 1/N*dN/d(pt) or some other quantity.

One thing more I want to get the area under the fitted histogram how can I get that area.

I would be indeed thank full to you.

Best Regards

Wajid Ali
National Center for Physics, Islamabad
Phone: 051-2077336 Received on Mon Jan 04 2010 - 09:14:12 CET