How to define the light “color” from a given spectral distribution?

The following question may be naive and incomplete in some way I don't know. I'm not a specialist on spectroscopy, colours and light curves, color spaces, etc.

Suppose you have a simple power-law function ; $f(\omega, \alpha) = \omega^{\alpha}$, which describes the spectral distribution of light angular frequencies as this : $$\tag{1} I = \kappa \int_0^{\omega_{\text{max}}} f(\omega, \alpha) \, d\omega,$$ where the exponent $\alpha$ is a given constant (a characteristic of the spectral distribution) and $\omega_{\text{max}}$ is another constant (the maximal value of the angular frequency allowed). $I$ is the total bolometric intensity of light at the detector's location, in watt/m^2 (the detector is a theoretical ideal device). $\kappa$ is just another arbitrary constant.

Then the question is this :

Assuming that $\omega_{\text{max}}$ is an angular frequency (rad/sec) which is in the visible spectrum or above it (i.e. ultra-violet), how can we define the color of the light described by the $\alpha$ index and the maximal value $\omega_{\text{max}}$ ?

By color, I mean something that could be compared in some way with the perception that we would have of that "$\alpha$-light", in the visible spectrum only.

For example, if $\alpha = 0$, the spectral distribution would be "flat" (i.e. uniform). What would be the color of light if $\omega_{\text{max}}$ corresponds to pure violet light, and $0 \le \omega \le \omega_{\text{max}}$ ? I guess white light !

If $\alpha = 2$, then the distribution would favour the violet and blue frequencies over the orange and red frequencies, so the light would look like blueish in some way, isn't ?

I hope the question is clear enough and doesn't bring me to the all messy/complicated problems of human/eye/brain/psychology perception ! I'm looking for something simple and "physical" only, if it exists ! In other words : is there a simple approximate "trick" to define a "color" from $\alpha$ and $\omega_{\text{max}}$ alone ? I'm just looking for some kind of approximation, to give an idea of what color the light might have.

Unfortunately, since color is a matter of perception, this is almost purely a biology question. The problem is that, while we perceive sound rather faithfully (our ears basically just do a Fourier transform), our perception of light is extremely unfaithful. After all, the spectrum $f(\omega, \alpha)$ is an entire function, yet given any spectrum, we only perceive a single color.

More specifically, the human eye has three kinds of color receptors, with spectral sensitivities $g_i(\omega)$. The functions $g_i$ are vaguely peaked at the wavelengths of pure red, blue, and green light, but they each respond to a large range of wavelengths. The response of each color receptor will be something like $$R_i \propto \int f(\omega, \alpha) g_i(\omega) d\omega.$$ You'll need to consult biology papers for empirical measurements of the $g_i$. Then you need to take these responses, scale them appropriately, and use them to pick out a point in the CIE color space. This will identify approximately which color is perceived.

• Hmm, that's what I was afraid. So there is no simple approximate "trick" to define a "color" from $\alpha$ and $\omega_{\text{max}}$ alone ? Actually, I'm just looking for some kind of approximation, to give an idea of what color the light might have. – Cham Oct 25 '16 at 23:15
• @Cham Maybe the people on Biology.SE know. I'm not sure what the right way to approximate this is -- after all, color is not a physical quantity. – knzhou Oct 25 '16 at 23:21
• As a very cheap approximation you could take $f(\omega_{\text{red}})$, $f(\omega_{\text{green}})$, and $f(\omega_{\text{blue}})$ as your RGB values and map those to color space. – knzhou Oct 25 '16 at 23:22
• I agree that color isn't exactly "physical". Yet, we can associate a precise color for each frequency (or wavelength) in the visible spectrum. I know that not all colors are spectral (cyan ? magenta ? just as examples). How do you define the RGB values from my power-law function ? Do I have to normalize the function ? How ? – Cham Oct 25 '16 at 23:25
• There are definitions for the standard person's color response which you would need to look up in a table (see en.wikipedia.org/wiki/CIE_1931_color_space for details). There's a nice paper that writes each cone's response function in terms of simple gaussian's here ppsloan.org/publications/XYZJCGT.pdf (Wyman, Chris, Peter-Pike Sloan, and Peter Shirley. "Simple analytic approximations to the CIE XYZ color matching functions." Journal of Computer Graphics Techniques (JCGT) 2, no. 2 (2013): 1-11.) – Punk_Physicist Oct 25 '16 at 23:39

What you probably want is a result in terms of color temperature. The temperature of a color is the temperature of a black body that glows with that color.

Start with the Planck equation for the intensity as a function of frequency and temperature: $$I(\nu, T) = \frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT}-1}$$ ($h$ is Planck's constant, $k$ is Boltzmann's constant, $c$ is the speed of light) and find the temperature $T$ that best matches a scaled version of your $f(\omega, \alpha)$ for $0 < \omega < \omega_{max}$ (note that the Planck equation above uses normal frequency $\nu$, not angular $\omega$). Once you know the temperature, you can relate it to a visual color with the diagram below. A positive $\alpha$ would match a blue end of the spectrum, negative would match the red end, and $\alpha = 0$ would match the white area, where the peak frequency intensity is right in the middle of the visible spectrum (approximately constant).

• Very interesting ! Could be it. Now, how do you match the function $f(\omega, \alpha) = \omega^{\alpha} \, \Theta(\omega_{\text{max}} - \omega)$ (where $\Theta(x)$ is the Heaviside step function) and find its "effective" temperature ? As an example, take $\alpha = 2$ (and $\omega_{\text{max}} =$ some value). Can you give a detailed computation in this case ? – Cham Oct 26 '16 at 0:48
• I might be able to work something out later, but the gist is a least-squares fit between $AI(\nu, T)\Theta(\nu_{max} - \nu)$ and $\nu^\alpha\Theta(\nu_{max} - \nu)$ where $T$ and $A$ are varied. The scaling factor $A$ is needed because the Planck formula gives an intensity per surface area while you have intensity. My guess is this will have to be worked out numerically. – Mark H Oct 26 '16 at 1:00
• Your answer still need a detailed description of how to match a given function $f(\nu, \alpha) = \nu^{\alpha} \, \Theta(\nu_{\text{max}} - \nu)$ with the Planck distribution. – Cham Oct 26 '16 at 13:17
• @Cham I may or may not have time to do any detailed work in the coming days. – Mark H Oct 26 '16 at 21:41