Coulomb cross section and PDFs

Asked by John Doe

Hi everyone, there is a computation I've done in MG5 I find rather puzzling. First, the non-puzzling one. If I compute the cross section for, say, ta- b > b ta- /z h QED=2, in sm-full, for a "free" b quark, and in the entire phase space (no cuts), I obtain a non-sensical result: ME gives an absurdly large cross section and it apparently refuses to compute all of the events. This is as it should be, since the cross section for Coulomb scattering does not exist. The same thing happens if I set the PDF for the b quark to 1 (proton) and choose, say, CTEQ6m.

On the other hand, if I replace the b quark for a c quark, without the PDF the cross section blows up, but when I set the PDF to 1, the result seems convergent: ME evaluates all of the events and gives a result of order 10^4 which is apparently meaningful. Is this apparently convergent result just a numerical artifact?

But the problem is this: if I put a cut off on the scattering angle or the momentum transfer, as I make the cut off smaller the cross section should grow without bounds. This is so in the case of the b quark. But for the c quark the total cross section, being finite, puts an upper bound on the growth of the cut-off cross section. As far as I understand, this is incorrect.

It seems I'm not understanding this properly... Any hing would be greatly appreciated.

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Johan Alwall
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Best Johan Alwall (johan-alwall) said :
#1

Dear John,

I assume you ran with dynamical factorization scale choice. In this case, for massless final-state 2->2 scattering, the factorization scale is set to the pT of the final state particle. Since the PDFs can't handle a factorization scale below 2 GeV, we have put a hard cutoff at this scale. So in fact, there is an implicit pT cut of 2 GeV when you use massless final state particles and PDFs. If you set the factorization scale to fixed scale in the run_card, you should get your divergence.

All the best,
Johan

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John Doe (kwcpsn) said :
#2

Dear Johan, thanks a lot for your reply. That's exactly what I needed to know. Cheers!

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John Doe (kwcpsn) said :
#3

Thanks Johan Alwall, that solved my question.