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I am trying to understand the proof of the conservation of specific intensity. Here we are trying to prove that specific intensity at dσ1 and dσ2 along the ray emitted by a source is the same. I don't understand this line:

Let dΩ1≪1 rad be the solid angle subtended by dσ2 as seen from the center of the surface dσ1 and dΩ2≪1 rad be the solid angle subtended by dσ1 as seen from the center of the surface dσ2.

Why are we using the solid angle subtended by dσ2 for dΩ1 instead of the solid angle subtended by area dσ1 at the source? Using the solid angle subtended between the two surfaces instead of the surface and source is the confusion here.

From https://www.cv.nrao.edu/~sransom/web/Ch2.html enter image description here

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The rays that come from area 1 and pass through area 2 (and hence the power emitted from area 1 that passes through area 2) will be emitted into a solid angle given by the (projected) area of area 2 divided by the separation of the two areas squared.

This is how $d\Omega_1$ is defined in the first equation and how it is used in the third.

The wording before the third equation might be confusing. $dP_1$ is the power flowing through the first area that also passes through the second area. This is proportional to the specific intensity of 1, the area of 1, but the solid angle of 2 as seen from 1. If energy is conserved (no absorption) then this must be the same as the power flowing through area 2 that has originated from 1 - this is the fourth equation - and is proportional to the specific intensity at 2, the area of 2 and the solid angle of 1 seen from 2.

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  • $\begingroup$ I don't agree with this answer. I think the derivation is just wrong. Why is the power flowing through one solid angle equal to the power flowing through the other? These are a different set of rays. If you consult more theoretical textbooks (e.g., Thorne and Blandford), you will see that they derive this from the conservation of phase space density (Liouville theorem). I would love to understand it with a ray diagram, but I'm not seeing it. (I ended up here after a day of trying to make sense of this type of derivation.) $\endgroup$
    – Sam Gralla
    Sep 16, 2022 at 20:36

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