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From: YANG Tao <yangt_at_ihep.ac.cn>

Date: Wed, 24 Oct 2018 19:50:03 +0800 (GMT+08:00)

Dear Vasilis,

Thanks for your patient interpretation.As you say, when I want to get dN/dθ, does it mean the only thing I would do is to multiply the output results(the 3rd column of ***tab.lis,i.e., dN/dΩ) with 2πsinθ? Besides, is the relation dΩ=2πsinθdθ (or Ω=2π(1-cosθ) always true in any condition especially the nonuniform distribution condition? If it is not always true, how to obtain the relation of dΩ and dθ?

Best regards!

Yang

-----原始邮件-----

发件人:"Vasilis Vlachoudis" <Vasilis.Vlachoudis_at_cern.ch>

发送时间:2018-10-24 16:47:10 (星期三)

收件人: "YANG Tao" <yangt_at_ihep.ac.cn>

抄送: "fluka-discuss_at_fluka.org" <fluka-discuss_at_fluka.org>

主题: [SPAM] RE: [SPAM] RE: [fluka-discuss]: Angular distribution of usryield

Hi Yang,

1) The conversion from dΩ to dθ is just a mathematical conversion and does not imply anything on the shape of

the distribution, just a different way of presenting it. Of course in the scoring we are not dealing with differentials (dθ)

but rather discrete intervals (Δθ), which the width choice must be done under the assumption that Δθ is close to zero

where your conversion formula holds.

2) You are right, the scoring intervals are in angle however according to Note 5 your end result is per steradian

inside this angular interval.

Vasilis

From: YANG Tao [yangt_at_ihep.ac.cn]

Sent: Monday, October 22, 2018 17:56

To: Vasilis Vlachoudis

Cc: fluka-discuss_at_fluka.org

Subject: Re: [SPAM] RE: [fluka-discuss]: Angular distribution of usryield

Dear Vasilis,

Thanks for your reply. The second question in my last email now is clear to me under your interpretation. But to the 1st question,I am still puzzled.

(1)I multiply the factor 2πsinθ with dθ to get dΩ, but which is based on the assumption of a isotropic distribution, however, this is a transcendental method, how can we know in advance the distribution is isotropic? Or dΩ=2πsinθdθ is only a geometric factor, the actual distribution is already calculated in the dN/dΩ, so to obtain dN/dθ we only just multiply the factor 2πsinθ with the output of fluka usryield card with an angular scoring?

(2)Another puzzled thing is that the distribution is given in steradian, but when I look at the output of the "gem_32_tab.lis" , the 1st column is from 0 to π(shown in attachment), which is still likely expressed in θ as I set in the input file, but the yield is referred to solid angle. As we all know , Ω will vary from 0 to 4π if θ from 0 to π (Ω=2π(1-cosθ)), why the output presents a maximum of π? If the 1st column is still θ, why can we get a flat curve like a isotropic distribution and why can we say the variable is Ω ?

Best regards!

Yang

-----原始邮件-----

发件人:"Vasilis Vlachoudis" <Vasilis.Vlachoudis_at_cern.ch>

发送时间:2018-10-22 16:47:10 (星期一)

收件人: "YANG Tao" <yangt_at_ihep.ac.cn>, "fluka-discuss_at_fluka.org" <fluka-discuss_at_fluka.org>

抄送:

主题: [SPAM] RE: [fluka-discuss]: Angular distribution of usryield

Hi Yang,

1. indeed as note 6 of USRYIELD states the results are always per steradian if an angular scoring is used. To obtain the dN/dtheta you have to calculate

the dtheta/domega when you know that the omega = 2pi(1-cos(theta))

2. As the header of the sum.lis is stating the total response is integrated over x1 (and not x2) you need to multiply with the dx2 to get the similar result

as in the usrbdx 0.2155*2e-3 ~= 4.3e-4

Cheers

Vasilis

From: owner-fluka-discuss_at_mi.infn.it [owner-fluka-discuss_at_mi.infn.it] on behalf of YANG Tao [yangt_at_ihep.ac.cn]

Sent: Sunday, October 21, 2018 04:04

To: fluka-discuss_at_fluka.org

Subject: [fluka-discuss]: Angular distribution of usryield

Dear fluka users,

Recently I simulate the angular distribution of secondary ions from (n,B) reaction, I using USRYIELD card to obtain this. However, I cannot interpret the results.

� I set the first quantity polar angle θ (polar angle in the c.m.s. frame, relative to the primary beam direction)and second E, I get a similar isotropic distribution over the polar θ,indeed, I know this distribution is isotropic, but it corresponds to the solid angle Ω not the polar angle θ(i.e.,dΩ=2πsinθdθ). Suppose dN/dΩ=A(some constant),thus,dN/dθ=2Aπsinθ,so at the angle near 0° , we should get a minimal value, and get a maximal value near 90° (I think it is some counterintuitive). Maybe I misunderstand the output of USRYIELD card, so could anyone interpret the meaning of USRYIELD setting. I read the manual, it states:

"In the case of polar angle quantities (| ie| or | ia| = 14, 15, 17, 18, 24, 25) the differential yield is always referred to solid angle in steradian, although input is specified in radian or degrees."

Does it mean the results shown in figure1 is only dN/dΩ? I'm very confused about this, if it is the result of dN/dΩ, then how to obtain dN/dθ (the strange thing is that I set the SDUM as polar angle in the usryield input, now that it cannot give the polar angle distribution, why the sdum says it were polar angle)? I try to get dN/dθ only by multiplying the USRYIELD output results by the factor 2πsinθ (shown in figure2, indeed showing a minimal near 0°), which I think maybe not the correct method since we will loss the statistical meaning if we priori assume the distribution is isotropic in solid angle Ω.

� Besides, I find the total response of 32BIN unit of the input file is 0.2155161(gem_32_sum.lis), which I think is the total yield of α particle leaving from the layer since it is integrated over two quantities, but the value doesn't match the total response of 25BIN((gem_25_sum.lis)) result which is 4.3065209E-04(gem_25_sum.lis,this value seems to be the true yield basing on the reaction cross section ), do the two responding values have different meaning?

Any help is appreciate!

Best regards!

Yang

__________________________________________________________________________

You can manage unsubscription from this mailing list at https://www.fluka.org/fluka.php?id=acc_info

Received on Wed Oct 24 2018 - 15:27:07 CEST

Date: Wed, 24 Oct 2018 19:50:03 +0800 (GMT+08:00)

Dear Vasilis,

Thanks for your patient interpretation.As you say, when I want to get dN/dθ, does it mean the only thing I would do is to multiply the output results(the 3rd column of ***tab.lis,i.e., dN/dΩ) with 2πsinθ? Besides, is the relation dΩ=2πsinθdθ (or Ω=2π(1-cosθ) always true in any condition especially the nonuniform distribution condition? If it is not always true, how to obtain the relation of dΩ and dθ?

Best regards!

Yang

-----原始邮件-----

发件人:"Vasilis Vlachoudis" <Vasilis.Vlachoudis_at_cern.ch>

发送时间:2018-10-24 16:47:10 (星期三)

收件人: "YANG Tao" <yangt_at_ihep.ac.cn>

抄送: "fluka-discuss_at_fluka.org" <fluka-discuss_at_fluka.org>

主题: [SPAM] RE: [SPAM] RE: [fluka-discuss]: Angular distribution of usryield

Hi Yang,

1) The conversion from dΩ to dθ is just a mathematical conversion and does not imply anything on the shape of

the distribution, just a different way of presenting it. Of course in the scoring we are not dealing with differentials (dθ)

but rather discrete intervals (Δθ), which the width choice must be done under the assumption that Δθ is close to zero

where your conversion formula holds.

2) You are right, the scoring intervals are in angle however according to Note 5 your end result is per steradian

inside this angular interval.

Vasilis

From: YANG Tao [yangt_at_ihep.ac.cn]

Sent: Monday, October 22, 2018 17:56

To: Vasilis Vlachoudis

Cc: fluka-discuss_at_fluka.org

Subject: Re: [SPAM] RE: [fluka-discuss]: Angular distribution of usryield

Dear Vasilis,

Thanks for your reply. The second question in my last email now is clear to me under your interpretation. But to the 1st question,I am still puzzled.

(1)I multiply the factor 2πsinθ with dθ to get dΩ, but which is based on the assumption of a isotropic distribution, however, this is a transcendental method, how can we know in advance the distribution is isotropic? Or dΩ=2πsinθdθ is only a geometric factor, the actual distribution is already calculated in the dN/dΩ, so to obtain dN/dθ we only just multiply the factor 2πsinθ with the output of fluka usryield card with an angular scoring?

(2)Another puzzled thing is that the distribution is given in steradian, but when I look at the output of the "gem_32_tab.lis" , the 1st column is from 0 to π(shown in attachment), which is still likely expressed in θ as I set in the input file, but the yield is referred to solid angle. As we all know , Ω will vary from 0 to 4π if θ from 0 to π (Ω=2π(1-cosθ)), why the output presents a maximum of π? If the 1st column is still θ, why can we get a flat curve like a isotropic distribution and why can we say the variable is Ω ?

Best regards!

Yang

-----原始邮件-----

发件人:"Vasilis Vlachoudis" <Vasilis.Vlachoudis_at_cern.ch>

发送时间:2018-10-22 16:47:10 (星期一)

收件人: "YANG Tao" <yangt_at_ihep.ac.cn>, "fluka-discuss_at_fluka.org" <fluka-discuss_at_fluka.org>

抄送:

主题: [SPAM] RE: [fluka-discuss]: Angular distribution of usryield

Hi Yang,

1. indeed as note 6 of USRYIELD states the results are always per steradian if an angular scoring is used. To obtain the dN/dtheta you have to calculate

the dtheta/domega when you know that the omega = 2pi(1-cos(theta))

2. As the header of the sum.lis is stating the total response is integrated over x1 (and not x2) you need to multiply with the dx2 to get the similar result

as in the usrbdx 0.2155*2e-3 ~= 4.3e-4

Cheers

Vasilis

From: owner-fluka-discuss_at_mi.infn.it [owner-fluka-discuss_at_mi.infn.it] on behalf of YANG Tao [yangt_at_ihep.ac.cn]

Sent: Sunday, October 21, 2018 04:04

To: fluka-discuss_at_fluka.org

Subject: [fluka-discuss]: Angular distribution of usryield

Dear fluka users,

Recently I simulate the angular distribution of secondary ions from (n,B) reaction, I using USRYIELD card to obtain this. However, I cannot interpret the results.

� I set the first quantity polar angle θ (polar angle in the c.m.s. frame, relative to the primary beam direction)and second E, I get a similar isotropic distribution over the polar θ,indeed, I know this distribution is isotropic, but it corresponds to the solid angle Ω not the polar angle θ(i.e.,dΩ=2πsinθdθ). Suppose dN/dΩ=A(some constant),thus,dN/dθ=2Aπsinθ,so at the angle near 0° , we should get a minimal value, and get a maximal value near 90° (I think it is some counterintuitive). Maybe I misunderstand the output of USRYIELD card, so could anyone interpret the meaning of USRYIELD setting. I read the manual, it states:

"In the case of polar angle quantities (| ie| or | ia| = 14, 15, 17, 18, 24, 25) the differential yield is always referred to solid angle in steradian, although input is specified in radian or degrees."

Does it mean the results shown in figure1 is only dN/dΩ? I'm very confused about this, if it is the result of dN/dΩ, then how to obtain dN/dθ (the strange thing is that I set the SDUM as polar angle in the usryield input, now that it cannot give the polar angle distribution, why the sdum says it were polar angle)? I try to get dN/dθ only by multiplying the USRYIELD output results by the factor 2πsinθ (shown in figure2, indeed showing a minimal near 0°), which I think maybe not the correct method since we will loss the statistical meaning if we priori assume the distribution is isotropic in solid angle Ω.

� Besides, I find the total response of 32BIN unit of the input file is 0.2155161(gem_32_sum.lis), which I think is the total yield of α particle leaving from the layer since it is integrated over two quantities, but the value doesn't match the total response of 25BIN((gem_25_sum.lis)) result which is 4.3065209E-04(gem_25_sum.lis,this value seems to be the true yield basing on the reaction cross section ), do the two responding values have different meaning?

Any help is appreciate!

Best regards!

Yang

__________________________________________________________________________

You can manage unsubscription from this mailing list at https://www.fluka.org/fluka.php?id=acc_info

Received on Wed Oct 24 2018 - 15:27:07 CEST

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