Cross sections of bins and combination

[ChrisVer]

suppose you have 2 bins each with cross section $\sigma_1, \sigma_2$...
if you combine those bins, is it a logical assumption to say that the cross section will also be added?
I suppose from the equality of the luminosity one can get:
$\frac{1}{2}[ N_1 / \sigma_1 + N_2 /\sigma_2] = N_{1+2}/ \sigma_{1+2}$
Obviously $N_{1+2} = N_1 + N_2$ (the entries of the 2 bins is equal to the sum of the entries of each bin)
However by that I obtain:
$\frac{N_1}{\sigma_1} + \frac{N_2}{\sigma_2} =2 \frac{N_1 + N_2}{\sigma_{1+2}}$
$\sigma_{1+2} = \frac{2 \sigma_1 \sigma_2 (N_1 + N_2)}{N_1 \sigma_2 + N_2 \sigma_1}$

Isn't this result irrational?

 

[mfb]

The two ratios N/σ should be the same, so you can simplify your expression.

If the cross sections in the bins are measured with different datasets, adding the numbers doesn't make sense, but adding cross sections still works. Just make sure they are not normalized to bin width.

 

[ChrisVer]

Just make sure they are not normalized to bin width.

Had some doubts with that too [in case the bin widths vary]... I wasn't going to take into account the widths...but generally speaking:
I don't understand how the cross section would come to get normalized to bin widths...You obtain it by integrating a histogram; so how would the bin width matter?

 

[mfb]

If your bin is from 20 to 30 GeV, and your cross section is 10 pb, then your cross-section is 1pb/GeV.
If your bin is from 20 to 30 GeV, and your cross section is given as 1pb/GeV, then your cross section is 10 pb
As easy as that. Cross sections can be added, cross sections per GeV cannot (directly).

 

[ChrisVer]

obviously the one is the $d \sigma /dx$ (in your example $x=E$) while the other is the $\sigma$. But OK I think I got what you wanted to pass.