AFB: Cos(theta) values.

Asked by daniel camargo on 2018-08-13

Hey guys,

I´m trying to calculate the forward-backward asymmetry in some models with Z-primes. It is supposed to be smooth, but I did not found it after generate the events with MG5 for two regions assuming |cos(theta)| <0 (backward) and |cos(theta)| > 0 (forward) and subtract them.... My guess is that the integrals in the FWD/BWD regions do not use the same |cos(theta)| value. It is possible to check by dumping the value of cos(theta) in two separate files and see whether they only differ by a sign? Or any hint in order to have a smooth AFB?

Thanks!

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Last query:
2018-08-13
Last reply:
2018-08-14

Hi,

For LHC this value is exactly zero right?
For Tevatron, it is not. Do you know how small is the assymetry compare to the two contributions?
And how many events did you generate in each contribution before doing the difference?
(It should be large enough to have significant digit in such difference)

Cheers,

Olivier

> On 13 Aug 2018, at 21:13, daniel camargo <email address hidden> wrote:
>
> New question #672164 on MadGraph5_aMC@NLO:
> https://answers.launchpad.net/mg5amcnlo/+question/672164
>
> Hey guys,
>
> I´m trying to calculate the forward-backward asymmetry in some models with Z-primes. It is supposed to be smooth, but I did not found it after generate the events with MG5 for two regions assuming |cos(theta)| <0 (backward) and |cos(theta)| > 0 (forward) and subtract them.... My guess is that the integrals in the FWD/BWD regions do not use the same |cos(theta)| value. It is possible to check by dumping the value of cos(theta) in two separate files and see whether they only differ by a sign? Or any hint in order to have a smooth AFB?
>
> Thanks!
>
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

daniel camargo (dacamargov) said : #2

Hey Oliver,

For LHC this value is exactly zero right? --> for Z yes, but for the Z´ its not. I´m trying to get something smooth like in the figures 8 and 9 of https://arxiv.org/abs/1503.02672

I guess the issue is not about the amount of events, I generated 1 million, but about the random way to pick up the cos(theta) values for each set, forward and backward.

Cheers,

Hi,

> For LHC this value is exactly zero right? --> for Z yes, but for the Z´
> its not. I´m trying to get something smooth like in the figures 8 and 9
> of https://arxiv.org/abs/1503.02672

Ok this is NOT a "simple" cos \theta (at least not the one I was guessing from your first email).
The definition of the forward/backward assymetry is quite complex in that paper.
(as it should be to be non zero at LHC)

We do not have any random number associated to that observable (that i would not call cos\theta).
So the only way to plot it write the events and do the plots from those events.
You can add that print in cuts.f if you want (note that some symmetry might not be fully correct at that stage --in particular everything which is related to forward/backward assymetry. Since the events might be flipped at a later stage of the computation.)

Cheers,

Olivier

> On 13 Aug 2018, at 21:43, daniel camargo <email address hidden> wrote:
>
> Question #672164 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/672164
>
> daniel camargo posted a new comment:
> Hey Oliver,
>
> For LHC this value is exactly zero right? --> for Z yes, but for the Z´
> its not. I´m trying to get something smooth like in the figures 8 and 9
> of https://arxiv.org/abs/1503.02672
>
> I guess the issue is not about the amount of events, I generated 1
> million, but about the random way to pick up the cos(theta) values for
> each set, forward and backward.
>
> Cheers,
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

daniel camargo (dacamargov) said : #4

Hey Oliver,

Well, the way the AFB is calculated there is quite tricky and requires the implementation of the dilepton rapidity that I already did in MadAnalysis. I made an script that takes both forward and backward cross-sections and calculate from them the AFB, the thing with the dilepton rapidity is just to know the orientation of the hardest lepton with the parton, but anyway I managed to did that....

To recap, for the computation of the forward cross-section I demand MG5 to take cos(theta) > 0 (costhmin 3 1 0d0) and save the events, after it I calculate the backward cross-section and demand MG5 to take cos(theta) < 0 (costhmax 3 1 0d0) and again save the events. Then, subtract them and normalize by the sum, it will be the approx AFB which for a dilepton rapidity of |yll| > 0.8 coincides with the original AFB up to ~10% of error. The issue here is that I can not ensure that I´m subtracting events with the same |cos(theta)|, to do it correctly I have to be sure they just differ for a sing... Any hint in how I can control the cos(theta) values to do it?

Cheers,

Hi Daniel,

What is this syntax?
> costhmin 3 1 0d0

This is not a MG5aMC command that I know.
costhmin is not define in the run_card.

This sounds more a syntax of MA5.
In that case your statement:
"I demand MG5
> to take cos(theta) > 0 (costhmin 3 1 0d0) and save the events, after it
> I calculate the backward cross-section and demand MG5 to take
> cos(theta) < 0 (costhmax 3 1 0d0) and again save the events."
should rather be
> I demand MG5 to generate a sample and MA5 to split it in two, one with
> cos(theta) > 0 (costhmin 3 1 0d0) and one with
> cos(theta) < 0 (costhmax 3 1 0d0)

If my understanding is correct, then I will assign this question to MA5 author.
(Note that i think that they are all in holliday for the moment)

Cheers,

Olivier

> On 14 Aug 2018, at 14:42, daniel camargo <email address hidden> wrote:
>
> Question #672164 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/672164
>
> daniel camargo posted a new comment:
> Hey Oliver,
>
> Well, the way the AFB is calculated there is quite tricky and requires
> the implementation of the dilepton rapidity that I already did in
> MadAnalysis. I made an script that takes both forward and backward
> cross-sections and calculate from them the AFB, the thing with the
> dilepton rapidity is just to know the orientation of the hardest lepton
> with the parton, but anyway I managed to did that....
>
> To recap, for the computation of the forward cross-section I demand MG5
> to take cos(theta) > 0 (costhmin 3 1 0d0) and save the events, after it
> I calculate the backward cross-section and demand MG5 to take
> cos(theta) < 0 (costhmax 3 1 0d0) and again save the events. Then,
> subtract them and normalize by the sum, it will be the approx AFB which
> for a dilepton rapidity of |yll| > 0.8 coincides with the original AFB
> up to ~10% of error. The issue here is that I can not ensure that I´m
> subtracting events with the same |cos(theta)|, to do it correctly I have
> to be sure they just differ for a sing... Any hint in how I can control
> the cos(theta) values to do it?
>
> Cheers,
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

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