cross section almost independent on alpha_S value

Asked by Maria Giulia Ratti

Dear experts,

shouldn't I expect that if I generate the process within SM

generate p p > j j

the cross section of the process should visibly depend on the value of alpha strong ?
If I change the default value in the param_card to a very large value as below

###################################
## INFORMATION FOR SMINPUTS
###################################
Block sminputs
    1 1.325070e+02 # aEWM1
    2 1.166390e-05 # Gf
    3 1.180000e+06 # aS

the total cross section changes negligibly, namely:

7.009e+08 ± 1.8e+06 for default aS
7.011e+08 ± 1.6e+06 for aS as above

Does this mean that the value of alphaS cannot be changed from the user ?

Thanks,
Maria Giulia

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MadGraph5_aMC@NLO Edit question
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Solved by:
Olivier Mattelaer
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Best Olivier Mattelaer (olivier-mattelaer) said :
#1

Hi,

In presence of PDF, then the value of alpha_S is automatically change to the value of alpha_S used in the PDF fit.
If you do a generation that does not use PDF, then the value of alpha_S is taken from the param_card.

Cheers,

Olivier

> On Jul 7, 2016, at 19:47, Maria Giulia Ratti <email address hidden> wrote:
>
> New question #296117 on MadGraph5_aMC@NLO:
> https://answers.launchpad.net/mg5amcnlo/+question/296117
>
> Dear experts,
>
> shouldn't I expect that if I generate the process within SM
>
> generate p p > j j
>
> this should visibly depend on the value of alpha strong ?
> If I change the default value in the param_card to a very large value as below
>
> ###################################
> ## INFORMATION FOR SMINPUTS
> ###################################
> Block sminputs
> 1 1.325070e+02 # aEWM1
> 2 1.166390e-05 # Gf
> 3 1.180000e+06 # aS
>
>
> the total cross section changes negligibly, namely:
>
> 7.009e+08 ± 1.8e+06 for default aS
> 7.011e+08 ± 1.6e+06 for aS as above
>
>
> Does this mean that the value of alphaS cannot be changed from the user ?
>
> Thanks,
> Maria Giulia
>
>
>
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Revision history for this message
Maria Giulia Ratti (maria-giulia-ratti) said :
#2

Thanks Olivier Mattelaer, that solved my question.