MadGraph5_aMC@NLO

Turn off grouping of final leptons only

Asked by Johann Brehmer

Hi,

the automatic grouping of subprocesses in MadGraph is generally a great feature. However, the grouping of final-state e and mu leads to problems when cutting on same-flavour lepton pairs, i.e. when mmll > 0.

Is there any way to turn off the grouping of final electrons and muons, while initial light quarks are still grouped?

In case this depends on the model, I'm working with the HEL model from the FeynRules database, http://feynrules.irmp.ucl.ac.be/wiki/HEL .

Thanks a lot in advance,
Johann Brehmer

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Olivier Mattelaer
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Best Olivier Mattelaer (olivier-mattelaer) said : 		#1

Hi,

If the lepton doesn’t have the same mass parameter,
then the grouping will not be performed.

So you need to modify your model to allow a non zero muon mass.

Cheers,

Olivier

On Nov 18, 2014, at 2:06 PM, johann.brehmer <email address hidden> wrote:

> New question #257807 on MadGraph5_aMC@NLO:
> https://answers.launchpad.net/mg5amcnlo/+question/257807
>
> Hi,
>
> the automatic grouping of subprocesses in MadGraph is generally a great feature. However, the grouping of final-state e and mu leads to problems when cutting on same-flavour lepton pairs, i.e. when mmll > 0.
>
> Is there any way to turn off the grouping of final electrons and muons, while initial light quarks are still grouped?
>
> In case this depends on the model, I'm working with the HEL model from the FeynRules database, http://feynrules.irmp.ucl.ac.be/wiki/HEL .
>
> Thanks a lot in advance,
> Johann Brehmer
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Revision history for this message
Johann Brehmer (johann.brehmer) said : 		#2

Thanks a lot, Olivier, for the fast help.

I thought I had tried this already, but it turns out that m_mu = 0 was hard-coded into the model, which I had not noticed.

Cheers,
Johann

Revision history for this message
Johann Brehmer (johann.brehmer) said : 		#3

Thanks Olivier Mattelaer, that solved my question.