your formula assumes that broadband R intensity is evenly distributed across the entire range, correct?
Yes, it's the only thing you can do. You must assume that you will have some residuals. But hey, as the images above, it seems to work.
Once you have estimated the actual Ha signal you're making the image more red in those areas that have Ha.
Yes, but think on it as it is: I'm multiplying the H-alpha emission intensity. I'm not doing a curve adjustment putting the cleaned H-alpha as mask, or something like. I simply add the clean H-alpha image to the R one, when they are linear.
It sounds like Chris is trying to do the same thing but not doing a very good job explaining it. As far as I can tell there is no normalization for the R subtraction based on pass bands of R and Ha filters. Perhaps it was in the talk but this is so essential that it should have been in the slides as well.
Not at all. My formula is not
Ha_clean = Ha - R*k, because the R image has
all the H-alpha emission embedded. Thus a simple subtraction doesn't solve the problem completely.
OTOH, I really suspect that Chris is doing all the process with non-linear images. After photographing M82, I can say you that the images are not linear. I think he's doing a different histogram transformation for the R and H-alpha images. In other words, he is giving the same brightness to the galaxy body in both images, and then subtracting the R to the H-alpha image (at least, that's what my eyes are telling me after looking the slides). This is *very* risky, because you can create or modify inadequately emission structures when you subtract the R image. This non-linear processing would disagree you equation:
Ha - (Ha + other red data) = - other red data
because the images are not linear. So you can make:
Ha - (Ha + other red data) = Ha_cleaned_image
Best regards,
Vicent.
