Dear Paola,
Thank you very much for your patient replies!
However, I have still a question about the naming convention of the polar angle.
It's clear about the meaning of zero polar angle and the 180 degrees polar angle, namely the forward direction and the backward direction, respectively.
However, for the perpendicular direction to the beam, there should be two directions, 90 degrees and 270 degrees. So where is the 270 degrees?
I guess FLUKA makes a symmetry treatment for the polar angle. The particles emitted in the perpendicular direction to the beam, whatever 90 degrees and 270 degrees, are classified to a same polar angle. Is it true?
Many thanks again and hope get your replies!
Zhiyuan
> -----原始邮件-----
> 发件人: "Paola Sala" <paola.sala_at_mi.infn.it>
> 发送时间: 2020-10-13 14:49:38 (星期二)
> 收件人: "\"梅志远\"" <meizhiyuan_at_hust.edu.cn>
> 抄送: "fluka discuss" <fluka-discuss_at_fluka.org>
> 主题: Re: [fluka-discuss]: Puzzles about the polar angle in USRYIELD card
>
> Hello
> maybe there is a problem of naming convention.
> the polar angle is the angle between the direction of the particle and the
> direction of the beam. If the particle is emitted in the forward direction
> along the beam axis, it has zero polar angle. If emitted in a direction
> perprndicular to the beam, it has 90 degrees (pi) angle. Uf totally
> backward--> 180 degrees or 2pi.
> The 4pi is the total solid angle, where the additional 2pi comes from the
> integration on the azimuthal angle, the one measured in the plane
> perpendicular to the beam.
> Hope this helps
> Paola
> > Dear FLUKA experts,I want to obtain the photon yield around a Cu target
> > shot by an electron beam. I'd like to plot the yield with an angular
> > distribution. So I use the USRYIELD card.
> >
> >
> >
> >
> > I indeed get a plot as shown below. But I'm puzzled about the definition
> > of the polar angle, namely the X axis of the plot.
> >
> >
> >
> >
> > (1) Yes, I know the Upper limit of the polar angle is forced to π, why?
> > why not the natural upper limit of polar angle, 2π? In my opinion, the
> > maximal polar angle should be 2π to cover the full angular distribution.
> > For a polar angle of π, only half the angular distribution is obtained.
> > Maybe it works for usual Symmetric simulations. BUT for an asymmetrical
> > yield angular distribution, it cannot work. Did I get a wrong
> > understanding?
> >
> >
> >
> >
> > (2) How can I associate the X asis (namely the polar angle) to the solid
> > angle in sr?
> >
> >
> >
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> > Any help is appreciate!
> >
> > Best regards!
> >
> > Zhiyuan
> >
> >
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> >
> >
> >
> >
> > --
> >
> >
> > Zhiyuan Mei
> > meizhiyuan_at_hust.edu.cn
> >
> >
> > /---------------------------------------------------
> > Institute of Applied Electromagnetic Engineering(IAEE)
> > Huazhong University of Science and Technology(HUST)
> > Address: Luoyu Road 1037, Wuhan, China
> > Tel: +86-155-2701-7168
> > /---------------------------------------------------
> >
> >
> >
> >
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> >
>
>
> Paola Sala
> INFN Milano
> tel. Milano +39-0250317374
> tel. CERN +41-227679148
--
Zhiyuan Mei
meizhiyuan_at_hust.edu.cn
/---------------------------------------------------
Institute of Applied Electromagnetic Engineering(IAEE)
Huazhong University of Science and Technology(HUST)
Address: Luoyu Road 1037, Wuhan, China
Tel: +86-155-2701-7168
/---------------------------------------------------
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Received on Tue Oct 13 2020 - 11:56:01 CEST