What exactly does USRBDX do?

Hi dear FLUKA experts,

I am trying to understand what exactly does USRBDX do and how to interpret the plots generated by it.

In the attached input file, I used 1 GeV proton beam hitting into vacuum while setting two boundary detectors at different depths to test how USRBDX works.

The thing is that I found the explanations in FLUKA manual regarding WHAT(6) is a bit ambiguous. And the plots I generated have the fluence exceeding 1 which does not really make sense to me. I also followed “http://www.fluka.org/content/course/NEA/lectures/Scoring.pdf” to set up the detectors’ areas in the input file. Of course the results from the two different detectors generated by FLUKA are different. However, shouldn’t them be the same in principle? I am sending beam in vacuum after all.

Thank you for any help in advance.

Martin

Testing.inp (1.2 KB)

Dear Martin,

the description of WHAT(6) in the lecture you linked is maybe a bit misleading, but the one in the FLUKA manual is quite clear:

WHAT(6) = area of the detector in cm**2
   Default = 1.0 (fluence or current gets integrated over the boundary area)

https://flukafiles.web.cern.ch/flukafiles/manual/USRBDX.html

One comment on your USRBDX setup. It is always the best to have planar surface between two regions (like between Vacuum_1 and Vacuum_2). But since your geometry is fully vacuum, nothing will come from the side.

To understand the results and units better, we need to look into the processed _sum.lis file.

In the header you will see the total response:

Tot. resp. (Part/cmq/pr)  1.1368064E-02  +/-   0.000000     %
( -->      (Part/pr)      0.9999871      +/-   0.000000     % )

The 1.0 part/primary result is what we expect, and you get part/cmq/pr if you divide this by the area. (This means if you don’t specify the area on the USRBDX card, the part/cmq/pr result will be the same as the part/pr. You can always normalize later in Flair.)

Now lets investigate the numeric results of each bin. You will find the unit in the line:

Flux (Part/GeV/cmq/pr)

In your case you have one bin, with the result of 11.36821 Part/GeV/cmq/pr. If we multiply with the width of the bin 0.001 GeV and the area 87.9645920 cmq, we get: 1.0 Part/pr, as we expect.

The reason for the Part/GeV/cmq/pr unit is, if you plot the bins as a histogram, the area below it will equal to the total fluence.

Cheers,
David

2 Likes

Thank you David, I fully understand it now.