# Definition of rapidity in run_card.dat

Asked by Laura Moreno on 2018-11-01

Dear all,

I am working with the process p p > H t j, and I try to reproduce the results from Madgraph with a code that takes the matrix elements and then performs the necessary integrals and applies cuts to obtain the same result. Up to now everything worked well, but when I try to apply cuts on the rapidity (eta) problems arise.

In my program, the pseudorapidity eta is defined as:

>> eta = atanh(p_z/|p|)

but then I don't get the same result as with Madgraph. Therefore I took a look at the definition of this parameter in Madgraph cards.

Now I am trying to understand the definition of the rapidity cut on Madgraph. I went to the file ./Source/kin_functions.f, where it says that for CM frame (in my case) we have:

>> rap = 0.5*log((p0 + p3)/(p0 - p3)) + cm_rap

the first term corresponds to the expression I knew and that I am using, i.e. atanh(p_z/|p|) given that I am dealing with a massless particle. But I cannot find the definition of the second term.

If I modify the definition of this parameter for simply the first term, then the results from my code reproduce perfectly (within uncertainties) the results from Madgraph, so the problem comes with the second term.

Could someone tell me what is exactly this second term and its mathematical expression? Thank you!

Best,
L.

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2018-11-07
2018-11-08
 Olivier Mattelaer (olivier-mattelaer) said on 2018-11-01: #1

Dear Laura,

cm_rap is the rapidity of the center of mass-frame in the lab frame.
For pp collision those two frame are NOT the same due to the PDF.

Cheers,

Olivier

> On 1 Nov 2018, at 14:42, Laura Moreno <email address hidden> wrote:
>
> New question #675845 on MadGraph5_aMC@NLO:
>
> Dear all,
>
> I am working with the process p p > H t j, and I try to reproduce the results from Madgraph with a code that takes the matrix elements and then performs the necessary integrals and applies cuts to obtain the same result. Up to now everything worked well, but when I try to apply cuts on the rapidity (eta) problems arise.
>
> In my program, the pseudorapidity eta is defined as:
>
>>> eta = atanh(p_z/|p|)
>
> but then I don't get the same result as with Madgraph. Therefore I took a look at the definition of this parameter in Madgraph cards.
>
> Now I am trying to understand the definition of the rapidity cut on Madgraph. I went to the file ./Source/kin_functions.f, where it says that for CM frame (in my case) we have:
>
>>> rap = 0.5*log((p0 + p3)/(p0 - p3)) + cm_rap
>
> the first term corresponds to the expression I knew and that I am using, i.e. atanh(p_z/|p|) given that I am dealing with a massless particle. But I cannot find the definition of the second term.
>
> If I modify the definition of this parameter for simply the first term, then the results from my code reproduce perfectly (within uncertainties) the results from Madgraph, so the problem comes with the second term.
>
> Could someone tell me what is exactly this second term and its mathematical expression? Thank you!
>
> Best,
> L.
>
> --

 Laura Moreno (laumova) said on 2018-11-01: #2

Dear Olivier,

So, you mean the expression used for the cut is then:

rap = 0.5*log((p0 + p3)/(p0 - p3)) (using momentum in LAB frame)
+ 0.5*log((p0 + p3)/(p0 - p3)) (using momentum in CM frame)

Is it like this? Because I have been working the whole time with the CM frame (where the sum of the momenta for the three final state particles is 0) and I use this momentum to get the values of the matrix elements for the phase space integration. But now I am confused by your statement, so could you please explain it a bit more? It would be really helpful, thanks.

I checked again the definition of rap in the ./Source/kin_functions.f, but it says that the first term is also computed with p being the momentum in the CM frame, so then what I said above makes no sense at all. Could you please tell me exactly how are both terms computed? Because they are not the same, but if the first is the rapidity using the CM momentum, shouldn't it be the same as the second term? I am sorry if I am asking some really easy stuff, but I am really confused right now.

Sorry for possible inconvenience and thank you again,
L.

 Olivier Mattelaer (olivier-mattelaer) said on 2018-11-01: #3

No,

we compute the rapidity in the lab frame
so this is equal to
> rap = 0.5*log((p0 + p3)/(p0 - p3)) (using momentum in LAB frame)

but since we actually have (available) the momenta in the CM frame we use the formula
> rap = 0.5*log((p0 + p3)/(p0 - p3)) (using momentum in CM frame)
+ rap_cm (rapidity of the CM frame in the lab frame)

the rap_cm is given by 0.5*log((p0 + p3)/(p0 - p3) *where those momenta are the one of the frame and not of the particle.

Cheers,

Olivier

> On 1 Nov 2018, at 21:52, Laura Moreno <email address hidden> wrote:
>
> Question #675845 on MadGraph5_aMC@NLO changed:
>
>
> Laura Moreno is still having a problem:
> Dear Olivier,
>
> thank you for your answer, but I still have some doubts.
>
> So, you mean the expression used for the cut is then:
>
> rap = 0.5*log((p0 + p3)/(p0 - p3)) (using momentum in LAB frame)
> + 0.5*log((p0 + p3)/(p0 - p3)) (using momentum in CM frame)
>
> Is it like this? Because I have been working the whole time with the CM
> frame (where the sum of the momenta for the three final state particles
> is 0) and I use this momentum to get the values of the matrix elements
> for the phase space integration. But now I am confused by your
> statement, so could you please explain it a bit more? It would be really
>
> I checked again the definition of rap in the ./Source/kin_functions.f,
> but it says that the first term is also computed with p being the
> momentum in the CM frame, so then what I said above makes no sense at
> all. Could you please tell me exactly how are both terms computed?
> Because they are not the same, but if the first is the rapidity using
> the CM momentum, shouldn't it be the same as the second term? I am sorry
> if I am asking some really easy stuff, but I am really confused right
> now.
>
> Sorry for possible inconvenience and thank you again,
> L.
>
> --

 Laura Moreno (laumova) said on 2018-11-02: #4

Dear Olivier,

thank you for your answer. Still I have some doubts. I found in the file ./SubProcesses/genps.f the definition of the rap_cm, and it turns out to be:

rap_cm = 0.5D0*log((xbk(1)*ebeam(1))/(xbk(2)*ebeam(2))

I understand the parameters ebeam(i) that in the case of two colliding protons each of 6500 GeV (set in run_card.dat) both ebeam(1) and ebeam(2) take the same value. But I am confused about the meaning of the variable xbk(i). I guess it is the fraction of momenta that the parton takes from the initial proton.

If I check the cross-section for some given conditions using only the rapidity of the jet in the CM frame, I get the same result as with Madgraph. If I only consider the term rap_cm, I also get the same result as in Madgraph (modifying the corresponding line in the code). The problem comes when considering both of them simultaneosly, I do:

>> rap = 0.5D0*LOG((P(0,5)+P(3,5))/(P(0,5)-P(3,5))) +
* 0.5D0*LOG(x1/x2)

and I impose the cut using ABS(rap). But then I don't get the result that Madgraph yields. Is there any other thing I am missing and I don't see? Thank you and sorry for possible inconvenience.

Best,
L.

 Laura Moreno (laumova) said on 2018-11-07: #5

update:

I have also tried considering simply a constant addition to the rapidity in each case, so for example when I compute the value for the rapidity in both MadGraph and my code as:
>> rap = 0.5*LOG((P(0,5)+P(3,5))/(P(0,5)-P(3,5))) + 1.0D0

again there is the mismatch between Madgraph and the results for my code. I would like to know which if the expression for the rapidity is indeed:
>> rap = 0.5D0*LOG((P(0,5)+P(3,5))/(P(0,5)-P(3,5))) +
* 0.5D0*LOG(x1/x2)

and in this case, does it consider the absolute value in some case? Thank you in advance!

Best,
L.

 Olivier Mattelaer (olivier-mattelaer) said on 2018-11-08: #6

Hi,

I confirm that we use the following formula:
rap = 0.5D0*LOG((P(0,5)+P(3,5))/(P(0,5)-P(3,5))) +
* 0.5D0*LOG(x1/x2)
when both beam have the same energy.
In that case the momentum of P are taken in the rest-frame of the collision (not the one of the beam/lab frame)

Cheers,

Olivier