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Question Re: Normalization of an isotropic flux in space 

Forum: Space Applications
Re: Question Normalization of an isotropic flux in space (valentina)
Re: Feedback Re: Normalization of an isotropic flux in space (Giovanni Santin)
Date: 03 Mar, 2009
From: valentina <valentina>

Dear Giovanni,

thank you for your quick reply!

The fact that you agree with my approach calm me down a lot.

>I see a problem in your macro, as you use "iso" for the angular distribution. In >fact, if you want to generate an isotropic source in a volume starting from a >surface, you have to bias the angular distribution with a cosine-law function, >that means using "cos" in GPS.

Yes, you'r right. The reason is that I am running a cross-checking simulation, and my colleagues use "iso"... I will change it.

>Also, it seems to me that the 2nd formula you propose is a factor 2 different >from the 1st one (am I right?). Indeed, that might come from the fact that you >should integrate the cosine-biased emission, not an isotropic one.
>In my calculations, also, if I limit the max emission angle theta to a certain q, >I normally get a sin2(q) instead of (1 - cos q), as a result of the integration >of the cos-biased distribution.

Yes, the 2nd formula is a more intuitive but approximated approach. Since q is this case a very small angle, the approximation does not affect the result.

From the 1st formula, the simulation time is given by:

Time_1 = N / [F(prot/(cm2 s sr)) x (4 x PI) x (PI x (R x tan(q))^2)]

From the 2nd formula, the time is given by:

Time_2 = N / [F x (4. x PI x R^2) x (2 x PI x (1 - cos(q)))]

For q = 0.01 rad, Time_1 is about Time_2.

If a take q = 0.9, Time_1 is equal to Time_2 for the following formulas:

Time_1 = N / [F(prot/(cm2 s sr)) x (4 x PI) x (PI x (R x sin(q))^2)]

Time_2 = N / [F x (4. x PI x R^2) x (2 x PI x (1/2 - (cos(q)^2)/2))]

(I take the cosin-law function)

Apart from the differences with these two formulas, my problem is that another normalization approach has been presented to me.

Since the intersection of infinite cones originating from the source sphere (radius R) would give the surface of the small sphere of radius r, the simulation time is given by:

Time_B = N/[F(prot/(cm2 s sr)) x (4 x PI) x (4 x PI x (R x sin(q))^2)]

Note the factor 4, that gives the area of the sphere surface and not of the plane of radius r.

The simulation time is 4 times lower than "my" simulation time.

Consider of having an isotropic flux, integrated over 4PI, of:

F = 2.31 #protons / (cm^2 sec)

If we run a simulation with a simple geometry made of a total absorbing cube placed within the small sphere (of radius r), what is the proton count rate that you expect on the cube surface?

If you use the last normalization, you obtain 2.31 counts/(cm^2 sec). My objection is that the proton count rate on the cube surface is not the flux integrated over 4PI, but (using the cosin law) over PI, because every cube side sees only PI sr.

What do you think of this different normalization? And, the most important, how one could prove the accuracy of one normalization approach respect to the other?

Thank you Giovanni!

Best regards, Valentina

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1 Note: Re: Normalization of an isotropic flux in space   (Juan - 04 Mar, 2009)
(_ Ok: Re: Normalization of an isotropic flux in space   (valentina - 04 Mar, 2009)
(_ None: Re: Normalization of an isotropic flux in space   (Juan - 04 Mar, 2009)
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