Why is there no blue peak in the photopic luminosity function?

A luminosity function "describes the average spectral sensitivity of human visual perception of brightness". My understanding is that basically, for the same power, we perceive green light as brightest. Red and blue are less bright, infrared and ultraviolet are invisible. The SI unit candela is defined in terms of the luminosity of green light at a wavelength of 555nm.

Anyway, how come the photopic luminosity function (in black): only has a single peak at about 555nm? Photopic vision uses cones, and is the kind of vision that is active in bright light. Cones, as opposed to rods, are sensitive to different wavelengths and give us colour vision. Here's a graph of the spectral sensitivity of cones in human eyes:

So how come the curve for photopic luminosity doesn't look like the integral of the spectral sensitivity curve? Or is it an integral, but the blue cones are actually so much less sensitive that the blue curve gets scaled down to almost nothing?

[Both images from wikipedia with CC0 and CC BY-SA 3.0 licenses]

• In your second graph the three curves have all been scaled to make their maximum equal to $1.0$. The sensitivities of the three types of cones are not all the same. Jul 4, 2019 at 6:01
• @JohnRennie I figured, but are they that different? There's a tiny bump around 425nm on the dotted luminosity curves, but it's really tiny. At least subjectively I don't feel that my sensitivity to blue is an order of magnitude less than green. If it is, then I guess that's the answer.
– craq
Jul 4, 2019 at 6:08
• Hyperphysics says: By population, about 64% of the cones are red-sensitive, about 32% green sensitive, and about 2% are blue sensitive so there are far fewer blue cones. But I guess the brain must do some image processing to keep the colour balance even. Jul 4, 2019 at 6:47
• @JohnRennie. I thought about this a bit more. In low light, I can imagine that a low density of blue cones lowers the sensitivity because many photons will not be detected. But in bright light, I would expect the lower cone density to mostly affect resolution, not perceived brightness.
– craq
Jul 5, 2019 at 20:38
• Also, if I look at a rainbow, I should be able to test the relative perceived brightness of the colours in sunlight. The spectral irradiance at 555nm is already higher than blue, and if the perceived brightness scales with the number of cones then it should appear at least 15x brighter... But that is not what I observe. When I look at a rainbow, I actually perceive blue and red much more clearly than green.
– craq
Jul 5, 2019 at 20:40

Actually, S cones are thought to not contribute to the photopic luminosity1,3 nor to mesopic one2. This is true for dim background, with which the CIE 1931 standard observer was defined. With intense background adaptation this is not generally true4.

To see it in real world color appearance models based on the CIE 1931 color space, consider the transformation matrix from XYZ to LMS color space from the Hunt and RLAB models for equal-energy illuminants (taken from this paragraph of the above link):

$$M_{\text{HPE}}=\begin{pmatrix} \hphantom{+}0.38971 & 0.68898 & - 0.07868 \\ - 0.22981 & 1.18340 & \hphantom{+}0.04641 \\ \hphantom{+}0.00000 & 0.00000 & \hphantom{+}1.00000 \end{pmatrix},$$

so that

$$C_{\text{LMS}}=M_{\text{HPE}}\cdot C_{\text{XYZ}}.$$

Its inverse, transforming from LMS to XYZ color space, will look like

$$M_{\text{HPE}}^{-1}=\begin{pmatrix} 1.910\hphantom{0} & - 1.112 & \hphantom{+}0.2019 \\ 0.3710 & \hphantom{+}0.629 & - 8.06\times 10^{-6} \\ 0.000\hphantom{0} & \hphantom{+}0.000 & \hphantom{+}1.000\hphantom{0} \end{pmatrix}.$$

Notice that the entry in the second row corresponding to contribution of the S cones is 5 orders of magnitude smaller than that for L cones. This is negligible compared to uncertainty of other matrix elements of $$M_{\text{HPE}}^{-1}$$. This confirms that, in (at least some) color appearance models dealing with cone responses, luminosity function has negligible contribution from S cones response.