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Forum: Space Applications
Re: Question Normalization of an isotropic flux in space (valentina)
Date: 03 Mar, 2009
From: Paul Nicholas Colin Gloster <Paul Nicholas Colin Gloster>

Valentina Fioretti submitted:
|----------------------------------|
|"[..]                             |
|                                  |
|I have an isotropic proton flux F:|
|                                  |
|F = prot /cm^2 sec sr"            |
|----------------------------------|

Dear Valentina Fioretti,

As you called that a flux, was that supposed to be
F = #protons / (centimeter^2 second steradian)
instead of what you seemed to type which was equivalent to
#protons second steradian / centimeter^2?


|------------------------------------------------------------------------|
|"[..]                                                                   |
|                                                                        |
|At the end of the run, I detect a number of counts C on the focal plane.|
|If N is the number of emitted protons, we call the fraction of          |
|detected/emitted particles Eff:                                         |
|                                                                        |
|Eff = C/N                                                               |
|                                                                        |
|In order to trasform the number C to a rate P in counts/sec, I apply    |
|this formula:                                                           |
|                                                                        |
|P = Eff x F x (4 x PI) x PI x r^2 counts/sec"                           |
|------------------------------------------------------------------------|

Eff can be treated as a dimensionless ratio (well strictly, its unit is count per proton). The unit of F, assuming F = #protons / (centimeter^2 second steradian), is count / (centimeter^2 second steradian).
Therefore, the unit of Eff x F is also count / (centimeter^2 second steradian).
For the unit of the P to be count/second, the unit of (4 x pi) x pi x r^2 would need to be centimeter^2 steradian.
It is not clear how you obtained the expression (4 x pi) x pi x r^2.



|-------------------------------------------------------------------------|
|"Then I divide the count rate P by the detecting area Adet, and the final|
|flux B is given by:                                                      |
|                                                                         |
|B = P/Adet counts/cm^2 sec"                                              |
|-------------------------------------------------------------------------|

If the unit of P is count/second then the unit of B is count/(cm^2 second), which you seemed to realize but you typed something else.


|--------------------------------------------------------------------|
|"The same normalization can be achieved using the simulation time:  |
|                                                                    |
|D = F x A_sphere x omega prot/sec                                   |
|                                                                    |
|where:                                                              |
|                                                                    |
|- A_sphere = 4 x PI x R^2 (the source sphere)                       |
|                                                                    |
|- omega = 2 x PI x (1 - cos(q)) (solide angle of the emission cone)"|
|--------------------------------------------------------------------|

The solid angle subtended by the detector at the source of the protons depends at least on the detector surface that faces the source and the distance between the detector and the source. You do not seem to have informed us of the distance. Omega = 2 x pi x (1 - cos(q)) may or may not apply for your situation.
For the correct value of Omega, the unit of F x A_sphere x Omega is indeed protoncount/second.



|--------------------------------------|
|"Then we calculate the simulated time:|
|                                      |
|Time = N/D (sec)"                     |
|--------------------------------------|

Agreed.



|-------------------------------------------|
|"The flux B on the focal plane is given by:|
|                                           |
|B = C/(Time x Adet) counts/cm^2 sec"       |
|-------------------------------------------|

Yes (with the unit being count/(centimeter^2 second)).


|----------------------------------------------------------------|
|"My normalization way is based on what shown by Sullivan (1971).|
|                                                                |
|Now my question is: is this the right way?                      |
|                                                                |
|[..]"                                                           |
|----------------------------------------------------------------|

Bear in mind that the flux reaching the detector (as opposed to merely the spacecraft) would be affected by the corresponding geometric factor of the detector, which can be affected by the particles' trajectories through the instrument (therefore different energies may have different geometric factors) and the geometry of the instrument.

It seems that you referred to Sullivan, J. D., 1971, "Geometrical factor and directional response of single and multi-element particle telescopes". I do not have a copy of this and I do not know enough about what you are trying to simulate so I do not say much more. You may wish to read
Sullivan, J. D.,
1972,
"Erratum: Geometrical factor and directional response of single and multi-element particle telescopes, Nucl. Instr. and Meth. 95 (1971) 5-11"
,
HTTP://DX.DOI.org/10.1016/0029-554X(72)90450-8
, "Nuclear Instruments and Methods", Volume 98, Issue 1, Page 187.

Regards,
Colin Paul Gloster

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