WHIZARD

how to generate beam with transverse momentum

Asked by Andrea Paolo Puppin

hi, sorry to bother you again,
this is my code:
process eeprovamultifasci = A, e1 => e1,e1,E1

beams = A, e1 => gaussian

beams_momentum = 5 keV, 5 GeV

gaussian_spread1= 5%

gaussian_spread2= 10%

plot E_Theta_Px_Py_E1

plot E_Theta_Px_Py_e1

plot E_Theta_Px_Py_e2

sample_format = lhef

  scan real y=(0=>20/+1){

     real x= y*0.0001-0.001
      integrate (eeprovamultifasci)

     beams_theta = 0,x

     real z=(x*x)/(0.0003)^2

     n_events=10000*exp(-z)

    #?rebuild_events = true

 $sample = sprintf "eeff_%1.3f_%1.3f" (real (x), real (y))

analysis= record E_Theta_Px_Py_E1 (eval E["E1"], eval Theta["E1"], eval Px["E1"],eval Py["E1"]) and

          record E_Theta_Px_Py_e1 (eval E[extract index 1 [e1]], eval Theta[extract index 1 [e1]], eval Px[extract index 1 [e1]],eval Py[extract index 1 [e1]])

          and

          record E_Theta_Px_Py_e2 (eval E[extract index 2 [e1]], eval Theta[extract index 2 [e1]], eval Px[extract index 2 [e1]],eval Py[extract index 2 [e1]])

  simulate (eeprovamultifasci)
}
as you can see this file generates a lot of simulation(and i have for some beams theta angle a number of events and so different luminosity for every simulation).
What i wanna try to study instead are two colliding beams with beams theta= 0, 0 the problem in this way is that my incoming electron and photon beams have no transverse momentum(if the beams theta is 0,0 infact in the mokka files i have for input particles only longitudinal momentum) .
but i wanna run a simulation where i fixe only one luminosity, and it creates two beams incoming (with beam theta=0,0) but not with only longitunal momentum but even transverse momentum(infact in this way it represents better a real beam), is possible to do what i asked?thank really again in advance,
kind regards

Question information

Language:
English Edit question
Status:
Solved
For:
WHIZARD Edit question
Assignee:
Juergen Reuter Edit question
Last query:
Last reply:
Revision history for this message
Juergen Reuter (j.r.reuter) said : 		#1

Sorry, I do not understand the question. There is a transverse momentum, as you asked in the last question.

Revision history for this message
Juergen Reuter (j.r.reuter) said : 		#2

Hi Andrea,
your input file is again faulty. The beams_theta statement has to come _before_ the integrate statement, otherwise it will
be ignored. You could have seen this by looking into the logfile of your process and/or the corresponding Whizard run. Please do this before you open up a new issue.
Cheers,
    JRR

Revision history for this message
Andrea Paolo Puppin (2andre3) said : 		#3

thank you very much.
I have one doubt, if i set the code in this way
process eeprovamultifasci = A, e1 => e1,e1,E1

beams = A, e1 => gaussian

beams_momentum = 5 keV, 5 GeV
 beams_theta = 0, 1 rad ......
1)the first question is if WHIZARD can read the command rad for angles
2) the second question is this: when i set my electron angle to 1 rad WHIZARD generates an incoming electron beam with py=0 and px,pz different from zero so that px/p =1 rad?.
so the question is essentialy how whizard calculates the angle theta and how it creates the incoming beam with the condition theta=1 rad

Revision history for this message
Juergen Reuter (j.r.reuter) said : 		#4

Yes, rad for radian is accepted. For asymmetric beams, you might have to specify two angles, for each beam separately. The polar angle is the angle w.r.t. the z axis.

Revision history for this message
Andrea Paolo Puppin (2andre3) said : 		#5

Thank you very much!!,
and for the second answer can you tell me if it’s correct what i wrote?
Kind regards

Inviato da Posta<https://go.microsoft.com/fwlink/?LinkId=550986> per Windows

Da: Juergen Reuter<mailto:<email address hidden>>
Inviato: domenica 21 maggio 2023 16:10
A: <email address hidden><mailto:<email address hidden>>
Oggetto: Re: [Question #706714]: how to generate beam with transverse momentum

Your question #706714 on WHIZARD changed:
https://answers.launchpad.net/whizard/+question/706714

    Status: Open => Answered

Juergen Reuter proposed the following answer:
Yes, rad for radian is accepted. For asymmetric beams, you might have to
specify two angles, for each beam separately. The polar angle is the
angle w.r.t. the z axis.

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Revision history for this message
Juergen Reuter (j.r.reuter) said : 		#6

I answered your second question: more explicitly, it is not correct. The relation is given by the cosine or sine of the polar angle (in radians or degrees). You can easily check from simple setups in your event files. I will mark this as solved now, because there is nothing more to say.

Revision history for this message
Juergen Reuter (j.r.reuter) said : 		#7

Closing this one for good.