How Do You Convert Proton Flux from 1/(cm^2 s sr) to 1/(cm^2 s)?

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Discussion Overview

The discussion revolves around converting proton flux measurements from units of 1/(cm2 s sr) to 1/(cm2 s). Participants explore the concept of solid angles and how to calculate the appropriate factor for conversion based on a given distance and effective area of a detector.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant asks how to convert proton flux from 1/(cm2 s sr) to 1/(cm2 s).
  • Another participant suggests multiplying by the number of steradians (sr) to achieve the conversion.
  • The original poster provides a detailed calculation of the solid angle based on a 1m distance from the source to the detector with an effective area of 1 cm2, leading to a calculated solid angle of approximately 9.92 x 10-5 sr.
  • A later reply expresses agreement with the original poster's calculations.

Areas of Agreement / Disagreement

Participants generally agree on the method of conversion involving the solid angle, but the exact value of the solid angle and its implications for the conversion remain open to further discussion.

Contextual Notes

The discussion does not resolve potential uncertainties regarding the assumptions made in the calculations, such as the uniformity of the proton source or the specific geometry of the setup.

nykon
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Dear forum users,

I have a quite simple question. I have got a flux of protons measured in 1/(cm^2 s sr), I would like to have it in 1/(cm^2 s).


nykon
 
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Then you should multiple by the number of sr
 
zhermes said:
Then you should multiple by the number of sr

Yes, I have got problem with sr concept:p Let's say that the distance between source and the detector is 1m. My detector has effective area of 1 cm^2. So by how many sr I should multiply my flux?

My answer would be:

The area of the sphere with radius of 1m^2 is:

S = 40000 \pi [cm^2]

the effectiv area of the detector is D = 1cm^2, so:

\frac{D}{S} = \frac{1}{40 000 \pi} = \frac{1}{125600} = 7.9 \times 10^{-6}

So the soild angle of my detector should be:

θ_{D} = 4\pi \times 7.9 \times 10^{-6}[sr]

So finally:

θ_{D} = 9.92 \times 10^{-5} [sr]

Am I right?


nykon
 
Last edited:
Yeah, I think that looks great!
 

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