Very interesting answers, thanks!
So, as i understand this, the luminance is just the addition of some fraction of every one of the color channels, depending on the ccd sensitivity and the filters (for the sake of simplicity let's say that every color contributes with 1/3 of the luminance, so that L = (R + G + B) / 3). So, if you get a luminance capture of a white star with a full histogram, then you take a pic with any of the color filters, the brightness of the filtered star will only reach 1/3 rd of the histogram. So, only 1/3 rd of the luminance data is from the blue channel, so it's not as good as a fully exposed blue filtered frame, but as the luminance is added to enhance an underexposed RGB capture, i think that the question is: is it better to use the underexposed blue filtered frames, or dedicate that time to capture more luminance frames, and then get the blue by substracting red and green from the luminance?
Obviously, if you had the fully exposed R, G, and B channels, then it would make no sense to capture the luminance, as the luminance of the rgb image will be as good as it can be.
I proposed the blue channel, because i think that the sensitivity is probably lower for that channel. Not sure if that's the case.
Alpal: That's a very good test! I think that the LRGB has more blue, but that's probably because you used 33% for all the colors, while the actual sensitivity of the CCD to every color likely varies. It's very similar, however, and probably just by boosting the blue channel a little bit, it will be identical.
Peter: don't forget that you have the luminance, and the real red and green channels. The luminance is made by the addition of the 3 channels, so by substracting the red and green from the luminance, you should get the real blue.
For example, to make things simple let's suppose that the CCD saturates with a value of 100, and that the sensitivity is R=30%, G=50%, B=20%.
You capture LRGB, and a particular pixel has a value of L=100, R=30, G=50, B=20. So, let's say that you discard the blue channel, so that you don't know its value. But you have LRG, so you can do the math B = L - R - G = 100 - 30 - 50 = 20.
By the way, i found this link about LRGB capture, it looks interesting:
http://www.robgendlerastropics.com/LRGB.html
Regards