Luminace, RGB, and SMPT ST 2084 EOTF

Question

I need a point clarified. According to wikipedia, SMPT ST 2084 EOTF is this:

I've seen this equation implemented in game engines where for L they feed the channels RGB, each of which is 0 to 10,000 and normalize them. Am confused because luminance from what I understand is supposed to be one quantity instead of 3. I understood luminance to be kind'a like speed in physics, and the RGB is the "direction" vector. Is it really correct to plug the RGB linear channels in into L (the "linear optical luminance") in this equation, or am I misunderstanding something? The way I think of it, we need to compute the luminance from the RGB triplet, and then plug that value into L, then from V we compute colors.

Where is my thinking wrong?

Thanks.

Answer 1

I fully understand the confusion (I was also a bit confused when I saw it). So, I decided to look at the ACES implementation of the ST 2084 EOTF. https://github.com/ampas/aces-dev/blob/v1.0.3/transforms/ctl/lib/ACESlib.Utilities_Color.ctl#L402

From there it is very clear: You perform the transfer function on the R, G and B values separately and not by first calculating the luminance. The confusion most likely comes from the fact that we are talking about two different kinds of luminance.

We have relative luminance, which is the luminance we use when talking about colours and can be calculated with $Y=0.2126R+0.7152G+0.0722B$

We also have luminance for which the SI unit is cd/m2 and basically means how bright something is.

You, and I when I first read your post, were thinking about relative luminance, but the Wikipedia article was talking about just luminance. In this case, it talks about the luminance of red light, green light and blue light separately (since a camera captures it separately and a monitor produces it separately). I'm pretty sure the ST 2048 spec is more clear about this, but that's behind a paywall.