aluque
commented
5 months ago Sorry for my very late reply. It somehow slipped under my radar and I only saw your comments when you closed the issue. In case you still need the clarifications:
- Here the normalisation is that the output corresponds to a single photon: to get physical results you must multiply by the total number of photons in the source. That is the reason that we divide by N here: otherwise the results would be proportional to the number of sampled photons, which is inconvenient.
- Project the image to a plane at unit distance from the camera. There the photon size is δu^2. You get a factor (cos θ) from Lambert's law (see e.g. https://en.wikipedia.org/wiki/Lambert%27s_cosine_law) because the pixel is seen at a slanted angle. The additional (cos θ)^2 appears because the distance from the camera to the pixel is R=1/(cos θ) and the solid angle is (area perpendicular to line of sight) / R^2.
- Yes, I think those units are correct.
Hope this clarifies these issues. Please do not hesitate to contact me again: hopefully I'll be quicker next time.
dbz123123
In the ”f = p(s, μscat) exp(-τ) trans / (sobs^2 N)"，why the ‘N’ need to be devided？ It seems that the weight in “weight f ” is just the energy of just a photon?
The comment in program said "The solid angle subtended by the pixel is roughly δu^2 * (cos θ)^3, where θ is the angle with the vertical. Actually cos θ = μobs[3]. The units would be here photons / sr / source photon." I just wonder *why the solid angle subtended by the pixel is expressed as δu^2 (cos θ)^3？Are there any books or theories i can find?** I looked for a while but cannot find any till now.
3.The units of the ' o.img[px, py, tid] += weight f / (o.δu^2 μobs[3]^3)' should be photons / sr/ source photon/m^2 ,Is it correct?
Looking forward to hearing from you soon! Thank you for your precious time!