Re: Isotropic Source

From: Katherine Harine (artsietopology@yahoo.com)
Date: Mon Oct 30 2006 - 16:04:06 CET

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    Alfredo,

    It was to be an isotropic source over a slab which had a finite
    thickness but extended out to infinity (in reality a large distance)
    for the x and y directions.
    For example a plate on the surface of the moon and parallel
    to the Moon's surface and assuming the Moon's surface was flat.
    My idea was that if I had an extended source over a square meter or so
    it didn't matter whether I had the source from one point or
    extended points because the relative contributions from each angle
    would be the same. Then I would just multiply the results by the
    correct fluence, while remembering that the incident fluence would
    be normalized to the 2 Pi hemisphere.

    Where am I going wrong?

    - Katherine

    ----- Original Message ----
    From: Alfredo Ferrari <alfredo.ferrari@cern.ch>
    To: Katherine Harine <artsietopology@yahoo.com>
    Cc: fluka-discuss@fluka.org
    Sent: Monday, October 30, 2006 6:52:16 AM
    Subject: Re: Isotropic Source

    Katherine

    that can work for a point like source, not for an extended one.
    Which source are you trying to model (ie that would not be suitable for
    generating a isotropic flux at the top of the atmosphere, it would rather
    generate an infinite fluence at 90 deg).

                     Ciao
                    Alfredo
    On Wed, 25 Oct 2006, Katherine Harine wrote:

    > In going over some runs I have done,
    > I want to make sure I have done the inputs correctly.
    > If I have an isotropic source incident down on a plate in
    > the x-y plane and I have the source a very small
    > distance above the plane, will the following statements in the source.f
    > file give me the isotropic flux incident on one side of the plate:
    >
    >
    > TZFLK (NPFLKA) = - FLRNDM(XXX)
    > XKPHI = TWOTWO * PIPIPI * FLRNDM(XXX)
    > TXFLK (NPFLKA) = SIN(ACOS(TZFLK(NPFLKA))) * COS(XKPHI)
    > TYFLK (NPFLKA) = SIN(ACOS(TZFLK(NPFLKA))) * SIN(XKPHI)
    >
    >
    > Next, since particles go only in the minus Z direction and since detectors will normalize results to one incident particle, will my flux
    > be 1.0 particle/cm^2/s integrated over the 2 Pi hemisphere?
    > That is, isn't my 4 Pi integrated flux equal to 2.0 particles/cm^2/s?
    >
    > I hope this makes sense. I read the recent discussion on
    > fluence and current on fluka-discuss and I hope I understand it correctly.
    >
    > - Katherine
    >
    >
    >
    >
    >
    >
    >
    >

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