# Re: Isotropic Source

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

• Next message: Francesco Cerutti: "Re: Problem with source.f"

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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